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b contains the set BN of homogeneous polynomials of degree N. Because mB and b are graded ideals, BN mB C b H) BN D mBN C BN \ b: In detail: the first inclusion says that an f 2 BN can be written f D g C h with g 2 mB and h 2 b. On equating homogeneous components, we find that fN D gN C hN. Moreover: fN D f ; if g D P mi bi, mi 2 m, bi 2 B, then gN D P mi biN ; and hN 2 b because b is homogeneous. Together these show f 2 mBN C BN \ b. Let M D BN =BN \ b, regarded as an A-module. The displayed equation says that M D mM. The argument in the proof of Nakayama’s lemma (1.3) shows that.1 C m/M D 0 for some m 2 m. Because A! B=b is injective, the image of 1 C m in B=b is nonzero. But d. Elimination theory 161 M D BN =BN \ b B=b, which is an integral domain, and so the equation.1 C m/M D 0 2 b for all i, which contradicts the implies that M D 0. Hence BN b, and so X N i assumption that Z D V.b/ is nonempty. Remarks 7.23. Every complete curve is projective. 7.24. Every nonsingular complete surface is projective (Zariski), but there exist singular complete surfaces that are not projective (Nagata). 7.25. There exist nonsingular complete three-dimensional varieties that are not projective (Nagata, Hironaka). 7.26. A nonsingular complete irreducible variety V is projective if and only if every finite set of points of V is contained in an open affine subset of V (Conjecture of Chevalley; proved by Kleiman1; see 6.22 for the necessity). d. Elimination theory When given a system of polynomial equations to solve, we first use some of the equations to eliminate some of the variables; we then find the solutions of the reduced system, and go back to find the solutions of the original system. Elimination theory does this more systematically. Note that the
fact that P system of polynomial equations n is complete has the following explicit restatement: for each./ 8 ˆ< ˆ: P1.X1; : : : ; XmI Y0; : : : ; Yn/ D 0 ::: Pr.X1; : : : ; XmI Y0; : : : ; Yn/ D 0 such that each Pi is homogeneous in the Yj, there exists a system of polynomial equations./ 8 ˆ< ˆ: R1.X1; : : : ; Xm/ D 0 ::: Rs.X1; : : : ; Xm/ D 0 with the following property; an m-tuple.a1; : : : ; am/ is a solution of (*) if and only if there exists a nonzero n-tuple.b0; : : : ; bn/ such that.a1; : : : ; am; b0; : : : ; bn/ is a solution of (). In other words, the polynomials Pi.a1; : : : ; amI Y0; : : : ; Yn/ have a common zero if and only if Rj.a1; : : : ; am/ D 0 for all j. The polynomials Rj are said to have been obtained from the polynomials Pi by elimination of the variables Yi. Unfortunately, the proof we gave of the completeness of P n, while short and elegant, gives no indication of how to construct (*) from (). The purpose of elimination theory is to provide an algorithm for doing this. 1Kleiman, Steven L., Toward a numerical theory of ampleness. Ann. of Math. (2) 84 1966 293–344 (Theorem 3, p. 327, et seq.). See also, Hartshorne, Robin, Ample subvarieties of algebraic varieties. Lecture Notes in Mathematics, Vol. 156 Springer, 1970, I 9 p45. 162 7. COMPLETE VARIETIES Elimination theory: special case Let P D s0X m C s1X m1 C C sm and Q D t0X n C t1X n1 C C tn be polynomials. The resultant of P and Q is defined to be the determinant s0 t0 s1 s0 t1 t0
: : : : : : : : : : : : : : : : : : sm tn sm tn rows m rows There are n rows with s0 : : : sm and m rows with t0 : : : tn, so that the matrix is.m C n/.m C n/; all blank spaces are to be filled with zeros. The resultant is a polynomial in the coefficients of P and Q. PROPOSITION 7.27. The resultant Res.P; Q/ D 0 if and only if (a) both s0 and t0 are zero; or (b) the two polynomials have a common root. PROOF. If (a) holds, then Res.P; Q/ D 0 because the first column is zero. Suppose that ˛ is a common root of P and Q, so that there exist polynomials P1 and Q1 of degrees m 1 and n 1 respectively such that P.X/ D.X ˛/P1.X/; Q.X/ D.X ˛/Q1.X/: Using these equalities, we find that P.X/Q1.X/ Q.X/P1.X/ D 0: (33) On equating the coefficients of X mCn1; : : : ; X; 1 in (33) to zero, we find that the coefficients of P1 and Q1 are the solutions of a system of m C n linear equations in m C n unknowns. The matrix of coefficients of the system is the transpose of the matrix s0 t0 0 B B B B B B @ s1 s0 t1 t0 sm tn sm tn : : : : : : : : : : : : : : : : : : 1 C C C C C C A : : : : : : The existence of the solution shows that this matrix has determinant zero, which implies that Res.P; Q/ D 0. Conversely, suppose that Res.P; Q/ D 0 but neither s0 nor t0 is zero. Because the above matrix has determinant zero, we can solve the linear equations to find polynomials P1 and Q1 satisfying (33). A root ˛ of P must be also be a root of P1 or of Q. If the former, cancel X ˛ from the left hand side of
(33), and consider a root ˇ of P1=.X ˛/. As deg P1 < deg P, this argument eventually leads to a root of P that is not a root of P1, and so must be a root of Q. d. Elimination theory 163 The proposition can be restated in projective terms. We define the resultant of two homogeneous polynomials P.X; Y / D s0X m C s1X m1Y C C smY m; Q.X; Y / D t0X n C C tnY n; exactly as in the nonhomogeneous case. PROPOSITION 7.28. The resultant Res.P; Q/ D 0 if and only if P and Q have a common zero in P 1. PROOF. The zeros of P.X; Y / in P (a).1 W 0/ in the case that s0 D 0; (b).a W 1/ with a a root of P.X; 1/. 1 are of the form: Since a similar statement is true for Q.X; Y /, 7.28 is a restatement of 7.27. Now regard the coefficients of P and Q as indeterminates. The pairs of polynomials mCnC2. Consider the closed subset.P; Q/ are parametrized by the space A mCnC2 is the set mCnC2 P V.P; Q/ in A defined by Res.P; Q/ D 0. Thus, not only have we shown that the projection of V.P; Q/ is closed, but we have given an algorithm for passing from the polynomials defining the closed set to those defining its projection. 1. The proposition shows that its projection on A nC1 D A mC1 A Elimination theory does this in general. Given a family of polynomials Pi.T1; : : : ; TmI X0; : : : ; Xn/; homogeneous in the Xi, elimination theory gives an algorithm for finding polynomials Rj.T1; : : : ; Tm/ such that the Pi.a1; : : : ; amI X0; : : : ; Xn/ have a common zero if and only if Rj.a1; : : : ; am/ D 0 for all j
. (Theorem 7.22 shows only that the Rj exist.) Maple can find the resultant of two polynomials in one variable: for example, entering “resultant..x C a/5;.x C b/5; x/” gives the answer.a C b/25. Explanation: the polynomials have a common root if and only if a D b, and this can happen in 25 ways. Macaulay doesn’t seem to know how to do more. Elimination theory: general case In this subsection, we give a proof of Theorem 7.22, following Cartier and Tate 1978,2 which is a more explicit proof than that given above. Throughout, k is a field (not necessarily algebraically closed) and K is an algebraically closed field containing k. THEOREM 7.29. For any graded ideal a in kŒX0; : : : ; Xn, exactly one of the following statements is true: (a) there exists an integer d0 0 such that a contains every homogeneous polynomial of degree d d0; (b) the ideal a has a nontrivial zero in KnC1. PROOF. Statement (a) says that the radical of a contains.X0; : : : ; Xn/, and so the theorem is a restatement of 6.2(a), which we deduced from the strong Nullstellensatz. For a direct proof of it, see the article of Cartier and Tate. 2Cartier, P., Tate, J., A simple proof of the main theorem of elimination theory in algebraic geometry. Enseign. Math. (2) 24 (1978), no. 3-4, 311–317. 164 7. COMPLETE VARIETIES THEOREM 7.30. Let R D L Rd be a graded k-algebra such that R0 D k, R is generated as a k-algebra by R1, and Rd is finite-dimensional for all d. Then exactly one of the following statements is true: d 2N (a) there exists an integer d0 0 such that Rd D 0 for all d d0; (b) no Rd D 0, and there exists a k-algebra homomorphism R! K whose kernel is not equal to RC defD L d 1 Rd. PROOF. The hypotheses on R say that it is a
quotient of kŒX0; : : : ; Xn by a graded ideal. Therefore 7.30 is a restatement of 7.29. Let P1; : : : ; Pr be polynomials in kŒT1; : : : ; TmI X0; : : : ; Xn with Pj homogeneous of degree dj in the variables X0; : : : ; Xn. Let J be the ideal.P1; : : : ; Pr / in kŒT1; : : : ; TmI X0; : : : ; Xn, and let A be the ideal of polynomials f in kŒT1; : : : ; Tm with the following property: there exists an integer N 1 such that all lie in J. n.K/ P n.K/. The projection of V into THEOREM 7.31. Let V be the zero set of J in A A n.K/ is the zero set of A. Consider the ring B D kŒT1; : : : ; TmI X0; : : : ; Xn and its subring B0 D kŒT1; : : : ; Tm. Then B is a graded B0-algebra with Bd the B0-submodule generated by the monomials of degree d in X0; : : : ; Xn, and J is a homogeneous (graded) ideal in B. Let A D L Ad be the quotient graded ring B=J D L Bd =.Bd \ J /. Let S be the ideal of elements a d 2N of A0 such that aAd D 0 for all sufficiently large d. d 2N THEOREM 7.32. A ring homomorphism 'W A0! K extends to a ring homomorphism W A! K not annihilating the ideal AC defD L d 1 Ad if and only if '.S/ D 0. Following Cartier and Tate, we leave it to reader to check that 7.32 is equivalent to 7.31. Proof of Theorem 7.32 We shall prove 7.32 for any graded ring A D L tions: d 0 Ad satisfying the following two condi- (a) as an A0-algebra, A is generated by A1; (b) for every d 0, Ad is finitely generated as
an A0-module. In the statement of the theorem, K is any algebraically closed field. The proof proceeds by replacing A with other graded rings with the properties (a) and (b) and also having the property that no Ad is zero. Let 'W A0! K be a homomorphism such that '.S/ D 0, and let P D Ker.'/. Then P is a prime ideal of A0 containing S. Step 1. Let J be the ideal of elements a of A for which there exists an s 2 A0 X P such that sa D 0. For every d 0, the annihilator of the A0-module Ad is contained in S, hence in P, and so J \ Ad ¤ Ad. The ideal J is graded, and the quotient ring A0 D A=J has the required properties. Step 2. Let A00 be the ring of fractions of A0 whose denominators are in ˙ defD A0 0 Let A00 L d be the set of fractions with numerator in A0 d 0 A00 d is a graded ring with the required properties, and A00 X P. d and denominator in ˙. Then A00 D 0 is a local ring with maximal ideal P00 defD P0 A0 0. e. The rigidity theorem; abelian varieties 165 Step 3. Let R be the quotient of A00 by the graded ideal P00 A00. As A00 finitely generated module over the local ring A00 Therefore R is graded ring with the required properties, and k D R0 0, Nakayama’s lemma shows that A00 defD A00 d is a nonzero ¤ P00A00 d. 0=P00 is a field. d Step 4. At this point R satisfies the hypotheses of Theorem 7.30. Let " be the composite of the natural maps A! A0! A00! R. In degree 0, this is nothing but the natural map from A0 to k with kernel P. As'has the same kernel, it factors through "0, making K into an algebraically closed extension of k. Now, by Theorem 7.30, there exists a k-algebra homomorphism f W R! K such that f.RC/ ¤ 0. The composite map D f ı " has the required properties. For more on elimination theory, see Chapter 8, Section 5,
� Demazure 2012 quotes Dieudonn´e as saying: “Il faut ´eliminer la th´eorie de l’´elimination.” 166 7. COMPLETE VARIETIES As '.V; w0/ D '.v0; w0/, we see that w0 2 W X C. Therefore W X C is nonempty, and so it is dense in W. As V fwg is complete and U is affine, '.V fwg/ must be a point whenever w 2 W X C (see 7.10); in fact '.V fwg/ D '.v0; w/ D g.w/: We have shown that'and g ı q agree on the dense subset V.W X C / of V W, and therefore on the whole of V W. COROLLARY 7.35. Let 'W V W! T be a regular map, and assume that V is complete, that V and W are irreducible, and that T is separated. If there exist points v0 2 V, w0 2 W, t0 2 T such that '.V fw0g/ D ft0g D '.fv0g W /, then '.V W / D ft0g. PROOF. With g as in the proof of the theorem, '.v; w/ D g.w/ D '.v0; w/ D t0: In more colloquial terms, the corollary says that if'collapses a vertical and a horizontal slice to a point, then it collapses the whole of V W to a point, which must therefore be “rigid”. DEFINITION 7.36. An abelian variety is a complete connected group variety. THEOREM 7.37. Every regular map ˛W A! B of abelian varieties is the composite of a homomorphism with a translation; in particular, a regular map ˛W A! B such that ˛.0/ D 0 is a homomorphism. PROOF. After composing ˛ with a translation, we may suppose that ˛.0/ D 0. Consider the map 'W A A! B; '.a; a0/ D ˛.a C a0/ ˛.a/ ˛.a0/: Then '.A 0/
D 0 D '.0 A/ and so'D 0. This means that ˛ is a homomorphism. COROLLARY 7.38. The group law on an abelian variety is commutative. PROOF. Commutative groups are distinguished among all groups by the fact that the map taking an element to its inverse is a homomorphism: if.gh/1 D g1h1, then, on taking inverses, we find that gh D hg. Since the negative map, a 7! aW A! A, takes the identity element to itself, the preceding corollary shows that it is a homomorphism. f. Chow’s Lemma The next theorem is a useful tool in extending results from projective varieties to complete varieties. It shows that a complete variety is not far from a projective variety. THEOREM 7.39 (CHOW’S LEMMA). For every complete irreducible variety V, there exists a surjective regular map f W V 0! V from a projective algebraic variety V 0 to V such that, for some dense open subset U of V, f induces an isomorphism f 1.U /! U (in particular, f is birational). f. Chow’s Lemma 167 Write V as a finite union of nonempty open affines, V D U1 [: : :[Un, and let U D T Ui. Because V is irreducible, U is a dense in V. Realize each Ui as a dense open subset of a projective variety Pi. Then P defD Q i Pi is a projective variety (6.26). We shall construct an algebraic variety V 0 and regular maps f W V 0! V and gW V 0! P such that (a) f is surjective and induces an isomorphism f 1.U /! U ; (b) g is a closed immersion (hence V 0 is projective). Let '0 (resp. 'i ) denote the given inclusion of U into V (resp. into Pi ), and let'D.'0; '1; : : : ; 'n/W U! V P1 Pn; be the diagonal map. We set U 0 D '.U / and V 0 equal to the closure of U 0 in V P1 Pn. The projection maps pW V P!
V and qW V P! P restrict to regular maps f W V 0! V and gW V 0! P. Thus, we have a commutative diagram ': (34) PROOF OF (a) In the upper-left triangle of the diagram (34), the maps'and '0 are isomorphisms from U onto its images U 0 and U. Therefore f restricts to an isomorphism U 0! U. Note that U 0 D f.u; '1.u/; : : : ; 'n.u// j u 2 U g; which is the graph of the map.'1; : : : ; 'n/W U! P. Therefore, U 0 is closed in U P (5.28), and so U 0 D V 0 \.U P / D f 1.U /: The map f is dominant, and f.V 0/ D p.V /, which is closed because P is complete. Hence f is surjective. PROOF OF (b) We first show that g is an immersion. As this is a local condition, it suffices to find g! Vi is an open subsets Vi P such that S q1.Vi / V 0 and each map V 0 \ q1.Vi / immersion. We set Vi D p1 i.Ui / D P1 Ui Pn where pi is the projection map P! Pi. We first show that the sets q1.Vi / cover V 0. The sets Ui cover V, hence the sets f 1.Ui / cover V 0, and so it suffices to show that q1.Vi / f 1.Ui / 168 7. COMPLETE VARIETIES for all i. Consider the diagrams q1.Vi / V P pi ıq Ui 'i Pi f 1.Ui / f V P pi ıq Ui 'i Pi '0 U'V P pi ıq Ui 'i Pi : The diagram at left is cartesian, i.e., it realizes q1.Vi / as the fibred product q1.Vi / D.V P / Pi Ui ; and so it suffices to show that the middle diagram commutes. But U 0 is dense in V 0, hence in f 1.Ui /, and so it suffices to prove that the middle diagram commutes with f 1.Ui / replaced
by U 0. But then it becomes the diagram at right, which obviously commutes. We next show that V 0 \ q1.Vi / g! Vi is an immersion for each i. Recall that and so Vi D Ui P i ; where P i D Y j ¤i Pj : q1.Vi / D V Ui P i V P: Let i denote the graph of the map Ui P i pi! Ui,! V : Being a graph, i is closed in V Ui P i and the projection map V Ui P i! Ui P i restricts to an isomorphism i! Ui P i. In other words, i is closed in q1.Vi /, and the projection map q1.Vi /! Vi restricts to an isomorphism i! Vi. As i is closed in q1.Vi / and contains U 0, it contains V 0 \ q1.Vi /, and so the projection map q1.Vi /! Vi restricts to an immersion V 0 \ q1.Vi /! Vi. Finally, V P is complete because V and P are, and so V 0 is complete (7.3). Hence g.V / is closed (7.7), and so g is a closed immersion. Notes 7.40. Let V be a complete variety, and let V1; : : : ; Vs be the irreducible components of V. Each Vi is complete (7.4), and so there exists a surjective birational regular map V 0! Vi i with V 0 i projective (7.39). Now F V 0 i is projective 6.26, and the composite G! G V 0 i Vi! V is surjective and birational. 7.41. Chow (1956, Lemma 1)4 proved essentially the statement 7.42 by essentially the above argument. He used the lemma to prove that all homogeneous spaces are quasiprojective. See also EGA II, 5.6.1. 4Chow, Wei-Liang. On the projective embedding of homogeneous varieties. Algebraic geometry and topology. A symposium in honor of S. Lefschetz, pp. 122–128. Princeton University Press, Princeton, N. J., 1957. g. Analytic spaces; Chow’s theorem 169 g. Analytic spaces; Chow’s theorem
We summarize a little of Serre, Jean-Pierre. G´eom´etrie alg´ebrique et g´eom´etrie analytique. Ann. Inst. Fourier, Grenoble 6 (1955–1956), 1–42, commonly referred to as GAGA. 7.42. The following is more general than Theorem 7.39: for every algebraic variety V, there exists a projective algebraic variety V 0 and a birational regular map'from an open dense subset U of V 0 onto V whose graph is closed in V 0 V ; the subset U equals V 0 if and only if V is complete. Ibid. p. 12. A subset V of C n is analytic if every v 2 V admits an open neighbourhood U in C n such that V \ U is the zero set of a finite collection of holomorphic functions on U. An analytic subset is locally closed. Let V 0 be an open subset of an analytic set V. A function f W V 0! C is holomorphic if, for every v 2 V 0, there exists an open neighbourhood U of v in C n and a holomorphic function h on U such that f D h on V 0 \ U. The holomorphic functions on open subsets of V define on V the structure of a C-ringed space. DEFINITION 7.43. An analytic space is a C-ringed space.V; two conditions: O V / satisfying the following (a) there exists an open covering V D S Vi of V such that, for each i, the C-ringed space V jVi / is isomorphic to an analytic set equipped with its sheaf of holomorphic.Vi ; O functions; (b) the topological space V is Hausdorff. PROPOSITION 7.44. An algebraic variety V is complete if and only if V.C/ is compact in the complex topology. PROOF. The proof uses Chow’s lemma (ibid. Proposition 6, p. 12). There is a natural functor V V an from algebraic varieties over C to complex analytic spaces (ibid. 2). We omit the definition of a coherent sheaf of V -modules. O THEOREM 7.45. Let V be a projective variety over C. Then the functor equivalence from the category of coherent modules, under which locally free modules correspond. In particular,.
V an;.V; V -modules to the category of coherent O O F F V /. 7! an is an V anO V an/'O PROOF. This summarizes the main results of GAGA (ibid. Th´eor´eme 2,3, p. 19, p. 20). THEOREM 7.46 (CHOW’S THEOREM). Every closed analytic subset of a projective variety is algebraic. PROOF. Let V be a projective space, and let Z be a closed analytic subset of V an. A theorem Zan is a coherent analytic sheaf on V an, and so there exists a of Henri Cartan states that coherent algebraic sheaf is Zariski closed, Zan. The support of F and equals Z (ibid. p. 29). O on V such that an D O F F THEOREM 7.47. Every compact analytic subset of an algebraic variety is algebraic. 170 7. COMPLETE VARIETIES PROOF. Let V be an algebraic variety, and let Z be a compact analytic subset. By Chow’s lemma (7.42), there exists a projective variety V 0, a dense open subset U of V 0, and a surjective regular map 'W U! V whose graph is closed in V V 0. Let 0 D \.Z V 0/. As Z and V 0 are compact and is closed, 0 is compact, and so its projection V 00 on V 0 is also compact. On the other hand, V 00 D f 1.Z/, which shows that it is an analytic subset of U, and therefore also of V 0. According to Chow’s theorem, it is a Zariski closed subset of V 0 (hence an algebraic variety). Now Z D f.V 00/ is constructible (Zariski sense; see 9.7 below), and therefore its Zariski closure coincides with its closure for the complex topology, but (by assumption) it is closed. COROLLARY 7.48. Let V and W be algebraic varieties over C. If V is complete, then every holomorphic map f W V an! W an is algebraic. PROOF. Apply the preceding theorem to the graph of f. EXAMPLE 7.49. The graph of z 7! ezW C! C C is closed in C C but it is
not Zariski closed. h. Nagata’s Embedding Theorem A necessary condition for a prevariety to be an open subvariety of a complete variety is that it be separated. An important theorem of Nagata says that this condition is also sufficient. THEOREM 7.50. Every variety V admits an open immersion V,! W into a complete variety W. If V is affine, then one can embed V,! A n. The proof in the general case is quite difficult. See: n,! P in P n, and take W to be the closure of V Nagata, Masayoshi. Imbedding of an abstract variety in a complete variety. J. Math. Kyoto Univ. 2 1962 1–10; A generalization of the imbedding problem of an abstract variety in a complete variety. J. Math. Kyoto Univ. 3 1963 89–102. For a modern exposition, see: L¨utkebohmert, W. On compactification of schemes. Manuscripta Math. 80 (1993), no. 1, 95–111. In the 1970s, Deligne translated Nagata’s work into the language of schemes. His personal notes are available in three versions. Deligne, P., Le th´eor`eme de plongement de Nagata, Kyoto J. Math. 50, Number 4 (2010), 661-670. Conrad, B., Deligne’s notes on Nagata compactifications. J. Ramanujan Math. Soc. 22 (2007), no. 3, 205–257. Vojta, P., Nagata’s embedding theorem, 19pp., 2007, arXiv:0706.1907. See also: Temkin, Michael. Relative Riemann-Zariski spaces. Israel J. Math. 185 (2011), 1–42. A little history When he defined abstract algebraic varieties, Weil introduced the term “complete variety” to denote the algebraic geometer’s analogue of a compact manifold. h. Nagata’s Embedding Theorem 171 Exercises 7-1. Identify the set of homogeneous polynomials F.X; Y / D P aij X i Y j, 0 i; j m, with an affine space. Show that the subset of reducible polynomials is closed. 7-2. Let V and
W be complete irreducible varieties, and let A be an abelian variety. Let P and Q be points of V and W. Show that any regular map hW V W! A such that h.P; Q/ D 0 can be written h D f ı p C g ı q where f W V! A and gW W! A are regular maps carrying P and Q to 0 and p and q are the projections V W! V; W. CHAPTER 8 Normal Varieties; (Quasi-)finite maps; Zariski’s Main Theorem We begin by studying normal varieties. These varieties have some of the good properties of nonsingular varieties, and it is easy to show that every variety is birationally equivalent to a normal variety. After studying finite and quasi-finite maps, we discuss the celebrated Zariski’s Main Theorem (ZMT), which says that every quasi-finite map of algebraic varieties can be obtained from a finite map by removing a closed subset from the source variety. In its original form, the theorem says that a birational regular map to a normal algebraic variety fails to be a local isomorphism only at points where the fibre has dimension > 0. a. Normal varieties Recall (1.42) that an integrally closed domain is an integral domain that is integrally closed in its field of fractions. Moreover, that an integral domain A is normal if and only if Am is normal for every maximal ideal m in A (see 1.49). DEFINITION 8.1. A point P on an algebraic variety V is normal if closed domain. An algebraic variety is said to be normal if all of its points are normal. V;P is an integrally O Since the local ring at a point lying on two irreducible components can’t be an integral domain (see 3.14), a normal variety is a disjoint union of its irreducible components, which are therefore its connected components. PROPOSITION 8.2. The following conditions on an irreducible variety V are equivalent. (a) The variety V is normal. (b) For all open affine subsets U of V, the ring V.U / is an integrally closed domain. O (c) For all open subsets U of V, a rational function on V that satisfies a monic polynomial equation on U
whose coefficients are regular on U is itself regular on U. PROOF. The equivalence of (a) and (b) follows from 1.49. (a) H) (c). Let U be an open subset of V, and let f 2 k.V / satisfy f n C a1f n1 C C an D 0; ai 2 V.U /; O (equality in k.V /). Then ai 2 This implies that f 2 O V.U / (5.11). V.U / O P for all P 2 U, and so f 2 O P for all P 2 U. O 173 174 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM (c) H) (b). The condition applied to an open affine subset U of V implies that integrally closed in k.V /. V.U / is O A regular local noetherian ring is normal — this is a difficult result that we don’t prove here (see CA 22.5 for references). Conversely, a normal local domain of dimension one is regular. Thus nonsingular varieties are normal, and normal curves are nonsingular. However, a normal surface need not be nonsingular: the cone X 2 C Y 2 Z2 D 0 is normal, but it is singular at the origin — the tangent space at the origin is k3. The singular locus of a normal variety V must have dimension dim V 2 (see 8.12 below). For example, a normal surface can only have isolated singularities — the singular locus can’t contain a curve. In particular, the surface Z3 D X 2Y (see 4.42) is not normal. The normalization of an algebraic variety Let E F be a finite extension of fields. The extension E=F is said to be normal if the minimal polynomial of every element of E splits in E. Let F al be an algebraic closure of F containing E. The composite in F al of the fields E, 2 Aut.E=F /, is normal over F (and is called the normal closure of F in F al). If E is normal over F, then E is Galois over EAut.E=F / (FT 3.10), and EAut.E=F / is purely inseparable over F (because
HomF.EAut.E=F /; F al/ consists of a single element). PROPOSITION 8.3. Let A be a finitely generated k-algebra. Assume that A is an integral domain, and let E be a finite field extension of its field of fractions F. Then the integral closure A0 of A in E is a finite A-algebra (hence a finitely generated k-algebra). PROOF. According to the Noether normalization theorem (2.45), A contains a polynomial subalgebra A0 and is finite over A0. Now E is a finite extension of F.A0/ and A0 is the integral closure of A0 in E, and so we only need to consider the case that A is a polynomial ring kŒX1; : : : ; Xd. Let QE denote the normal closure of E in some algebraic closure of F containing E, and let QA denote the integral closure of A in QE. If QA is finitely generated as an A-module, then so is its submodule A0 (because A is noetherian). Therefore we only need to consider the case that E is normal over F. According to the above discussion, E E1 F with E Galois over E1 and E1 purely inseparable over F. Let A1 denote the integral closure of A in E1. Then A0 is a finite A1-algebra (1.51), and so it suffices to show that A1 is a finite A-algebra. Therefore we only need to consider the case that E is purely inseparable over F. In this case, k has characteristic p ¤ 0, and, for each x 2 E, there is a power q.x/ of p such that xq.x/ 2 F. As E is finitely generated over F, there is a single power q of p such that xq 2 F for all x 2 E. Let F al denote an algebraic closure of F containing E. For each i, there is a unique Yi 2 F al such that Y q i D Xi. Now F D k.X1; : : : ; Xd / E k.Y1; : : : ; Yd / and A D kŒX1; : : : ; Xd A0 kŒY1; : : : ; Yd because k
ŒY1; : : : ; Yd contains A and is integrally closed (1.32, 1.43). Obviously kŒY1; : : : ; Yd is a finite A-algebra, and this implies, as before, that A0 is a finite A-algebra. a. Normal varieties 175 COROLLARY 8.4. Let A be as in 8.3. If Am is normal for some maximal ideal m in A, then Ah is normal for some h 2 A X m. PROOF. Let A0 be the integral closure of A in its field of fractions. Then A0 D AŒf1; : : : ; fm 0 D Am, and so there exists an h 2 A X m such that, for some fi 2 A0. Now.A0/m for all i, hfi 2 A. Now A0 h D Ah, and so Ah is normal. 1.47D.Am/ The proposition shows that if A is an integral domain finitely generated over k, then the integral closure A0 of A in a finite extension E of F.A/ has the same properties. Therefore, Spm.A0/ is an irreducible algebraic variety, called the normalization of Spm.A/ in E. This construction extends without difficulty to nonaffine varieties. PROPOSITION 8.5. Let V be an irreducible algebraic variety, and let K be a finite field extension of k.V /. Then there exists an irreducible algebraic variety W with k.W / D K and a regular map 'W W! V such that, for all open affines U in V, '1.U / is affine and kŒ'1.U / is the integral closure of kŒU in K. The map'(or just W ) is called the normalization of V in K. PROOF. For each v 2 V, let W.v/ be the set of maximal ideals in the integral closure of in K. Let W D F For an open affine subset U of V, v O v2V W.v/, and let 'W W! V be the map sending the points of W.v/ to v. '1.U /'spm.kŒU 0/; where kŒU 0 is the integral closure of k�
�U in K. We endow W with the k-ringed space structure for which.'1.U /; W j'1.U //'Spm.kŒU 0/. O W / is an algebraic variety with the required properties. A routine argument shows that.W; O EXAMPLE 8.6. (a) The normalization of the cuspidal cubic V W Y 2 D X 3 in k.V / is the map A 1! V, t 7!.t 2; t 3/ (see 3.29). (b) The normalization of the nodal cubic 4.10) in k.V / is the map 1! V, t 7!.t 2 1; t 3 t/. A PROPOSITION 8.7. The normal points in an irreducible algebraic variety form a dense open subset. PROOF. Corollary 8.4 shows that the set of normal points is open, and it remains to show that it is nonempty. Let V be an irreducible algebraic variety. According to (3.37, 3.38), V is birationally equivalent to a hypersurface H in A d C1, d D dim V, H W a0X m C a1X m1 C C am; ai 2 kŒT1; : : : ; Td ; a0 ¤ 0; m 2 NI moreover, T1; : : : ; Td can be chosen to be a separating transcendence basis for k.V / over k. Therefore the discriminant D of the polynomial a0X m C C am is nonzero (it is an element of kŒT1; : : : ; Td ). Let A D kŒT1; : : : ; Td ; then kŒH D AŒX=.a0X m C C am/ D AŒx. Let y D c0 C C cm1xm1; ci 2 k.T1; : : : ; Td /; (35) be an element of k.H / integral over A. For each j 2 N, Trk.H /=F.A/.yxj / is a sum of conjugates of yxj, and hence is integral over A (cf. the proof of 1.44). As it lies in F.A
/, it 176 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM is an element of A. On multiplying (35) with xj and taking traces, we get a system of linear equations c0 Tr.xj / C c1 Tr.x1Cj / C C cm1 Tr.xm1Cj / D Tr.yxj /; j D 0; : : : ; m 1: By Cramer’s rule (p. 26), det.Tr.xi Cj // cl 2 A; l D 0; : : : ; m 1: But det.Tr.xi Cj // D D,1 and so cl 2 AŒD1. Hence kŒH becomes normal once we invert the nonzero element D. We have shown that H contains a dense open normal subvariety, which implies that V does also. PROPOSITION 8.8. For every irreducible algebraic variety V, there exists a surjective regular map 'W V 0! V from a normal algebraic variety V 0 to V such that, for some dense open subset U of V;'induces an isomorphism '1.U /! U (in particular'is birational). PROOF. Proposition 8.7 shows that the normalization of V in k.V / has this property. 8.9. More generally, for a dominant map 'W W! V of irreducible algebraic varieties, there exists a normalization of V in W. For each open affine U in V we have kŒU.'1.U /; W / k.W /: O The integral closure kŒU 0 of.U; V / in.'1.U /; W / is a finite kŒU -algebra (because O it is a kŒU -submodule of the integral closure of kŒU in k.W /). The normalization of V in W is a regular map '0W V 0! V such that, for every open affine U in V, O.'01.U /; O V 0/ D Spm.kŒU 0/: In particular, '0 is an affine map. For example, if W and V are affine, then V 0 D Spm.k
ŒV 0/, where kŒV 0 is the integral closure of kŒV in kŒW. There is a commutative triangle W'V 0 '0 j V: b. Regular functions on normal varieties DEFINITION 8.10. An algebraic variety V is factorial at a point P if domain. The variety V is factorial if it is factorial at all points P. P is a factorial O When V is factorial, it does not follow that V.U / is factorial for all open affines U in V. O A prime divisor Z on a variety V is a closed irreducible subvariety of codimension 1. Let Z be a prime divisor on V, and let P 2 V ; we say that Z is locally principal at P if there exists an open affine neighbourhood U of P and an f 2 kŒU such that I.Z \ U / D.f /; the regular function f is then called a local equation for Z at P. If P … Z, then Z is locally principal at P because then we can choose U so that Z \ U D ;, and I.Z \ U / D.1/. 1See, for example, 2.34 of my notes Algebraic Number Theory. b. Regular functions on normal varieties 177 PROPOSITION 8.11. An irreducible variety V is factorial at a point P if and only if every prime divisor on V is locally principal at P. PROOF. Recall that an integral domain is factorial if and only if every prime ideal of height 1 is principal (1.24, 3.52). PROPOSITION 8.12. The codimension of the singular locus in a normal variety is at least 2. PROOF. Let V be a normal algebraic variety of dimension d, and suppose that its singular locus has an irreducible component W of codimension 1. After replacing V with an open subvariety, we may suppose that it is affine and that W is principal, say, W D.f / (see 8.11). There exists a nonsingular point P on W (4.37). Let.U; f1/; : : : ;.U; fd 1/ be germs of functions at P (on V ) whose restrictions to W generate the maximal ideal in W
. More intrinsically we can define discrete valuation ring (4.20), which is defined to be Z to be the set of rational functions on V that are defined an open subset U of V meeting O O O Z. Let ordZ be the valuation k.V / onto! Z with valuation ring element of Z, then O a D unit ordZ.a/: The divisor of a nonzero element f of k.V / is defined to be div.f / D X ordZ.f / Z: Z; thus, if is a prime O 178 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM The sum is over all the prime divisors of V, but in fact ordZ.f / D 0 for all but finitely many Z. In proving this, we can assume that V is affine (because it is a finite union of affines), say V D Spm.R/. Then k.V / is the field of fractions of R, and so we can write f D g= h with g; h 2 R, and div.f / D div.g/ div.h/. Therefore, we can assume f 2 R. The zero set of f, V.f / either is empty or is a finite union of prime divisors, V D S Zi (see 3.42) and ordZ.f / D 0 unless Z is one of the Zi. The map f 7! div.f /W k.V /! Div.V / is a homomorphism. A divisor of the form div.f / is said to be principal, and two divisors are said to be linearly equivalent, denoted D D0, if they differ by a principal divisor. When V is nonsingular, the Picard group Pic.V / of V is defined to be the group of divisors on V modulo principal divisors. (The definition of the Picard group of a general algebraic variety agrees with this definition only for nonsingular varieties; it may differ for normal varieties.) THEOREM 8.14. Let V be a normal variety, and let f be rational function on V. If f has no zeros or poles on an open subset U of V, then f is regular on U. PROOF. We may assume that V is connected, hence irred
ucible. Now apply the following statement (proof omitted): a noetherian domain is normal if and only if Ap is a discrete valuation ring for all prime ideals p of height 1 and A D T ht.p/D1 Ap. COROLLARY 8.15. A rational function on a normal variety, regular outside a subset of codimension 2, is regular everywhere. PROOF. This is a restatement of the theorem. COROLLARY 8.16. Let V and W be affine varieties with V normal, and let 'W V X Z! W be a regular map defined on the complement of a closed subset Z of V. If codim.Z/ 2, then'extends to a regular map on the whole of V. PROOF. We may suppose that W is affine, and embed it as a closed subvariety of A map V X Z! W,! A V. Therefore V X Z! A n. The n is given by n regular functions on V X Z, each of which extends to n, and its image is contained in W. n extends to A c. Finite and quasi-finite maps Finite maps DEFINITION 8.17. A regular map 'W W! V of algebraic varieties is finite if there exists a finite covering V D S i Ui of V by open affines such that, for each i, the set '1.Ui / is affine and kŒ'1.Ui / is a finite kŒUi -algebra. EXAMPLE 8.18. Let V be an irreducible algebraic variety, and let 'W W! V be the normalization of V in a finite extension of k.V /. Then'is finite. This follows from the definition 8.5 and Proposition 8.3. The next lemma shows that, for maps of affine algebraic varieties, the above definition agrees with Definition 2.39. c. Finite and quasi-finite maps 179 LEMMA 8.19. A regular map 'W W! V of affine algebraic varieties is finite if and only if kŒW is a finite kŒV -algebra. PROOF. The necessity being obvious, we prove the sufficiency. For simplicity, we shall assume in the proof that V and W are irreducible. Let.Ui /i be a finite family
of open affines covering V and such that, for each i, the set '1.Ui / is affine and kŒ'1.Ui / is a finite kŒUi -algebra. Each Ui is a finite union of basic open subsets of V. These are also basic open subsets of Ui, because D.f / \ Ui D D.f jUi /, and so we may assume that the original Ui are basic open subsets of V, say, Ui D D.fi / with fi 2 A. Let A D kŒV and B D kŒW. We are given that.f1; : : : ; fn/ D A and that Bfi is a finite Afi -algebra for each i. We have to show that B is a finite A-algebra. Let fbi1; : : : ; bimi g generate Bfi as an Afi -module. After multiplying through by a power of fi, we may assume that the bij lie in B. We shall show that the family of all bij generate B as an A-module. Let b 2 B. Then b=1 2 Bfi, and so b D ai1 f ri i n / D A because any maximal ideal containing.f r1 1 ; : : : ; f rn bimi, some aij 2 A and ri 2 N: bi1 C C aimi f ri i The ideal.f r1 have to contain.f1; : : : ; fn/ D A. Therefore, 1 ; : : : ; f rn n / would 1 D h1f r1 1 C C hnf rn n, some hi 2 A: Now b D b 1 D h1 bf r1 1 C C hn bf rn n D h1.a11b11 C C a1m1b1m1/ C C hn.an1bn1 C C anmnbnmn/, as required. LEMMA 8.20. Let 'W W! V be a regular map with V affine, and let U be an open affine in V. There is a canonical isomorphism of k-algebras PROOF. Let U 0 D '1.U /. The map is defined by the kŒV -bilinear pairing.W; W
/ ˝kŒV kŒU!.'1.U /; O W /: O.f; g/ 7!.f jU 0; g ı 'jU 0/W.W; W / kŒU!.U 0; O W /: O When W is also affine, it is an isomorphism (see 5.31, 5.32). Let W D S Wi be a finite open affine covering of W, and consider the commutative diagram: 0 0.W; O W / ˝kŒV kŒU Q.Wi ; i O W / ˝kŒV kŒU Q.Wij ; i;j W / ˝kŒV kŒU O.U 0; W / O Q.U 0 \ Wi ; i W / O.U \ Wij ; W /. O O Here Wij D Wi \ Wj. The bottom row is exact because W is a sheaf, and the top row is W is a sheaf and kŒU is flat over kŒV.2 The varieties Wi and Wi \ Wj are exact because all affine, and so the two vertical arrows at right are products of isomorphisms. This implies that the first is also an isomorphism. 2A sequence 0! M 0! M! M 00 is exact if and only if 0! Am ˝A M 0! Am ˝A M! Am ˝A M 00 is'OU;P'OV;P'kŒV mP exact for all maximal ideals m of A (1.21). This implies the claim because kŒU mP for all P 2 U. O 180 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM PROPOSITION 8.21. Let 'W W! V be a regular map of algebraic varieties. If'is finite, then, for every open affine U in V, '1.U / is affine and kŒ'1.U / is a finite kŒU -algebra. PROOF. Let Vi be an open affine covering of V (which we may suppose to be finite) such defD '1.Vi / is an aff
ine subvariety of W for all i and kŒWi is a finite kŒVi -algebra. that Wi Let U be an open affine in V, and let U 0 D '1.U /. Then.U 0; W / is a subalgebra of Q W /, and so it is an affine k-algebra finite over kŒU.3 We have a morphism O i.U 0 \ Wi ; O of varieties over V U 0 canonical Spm..U 0; W // O (36) V which we shall show to be an isomorphism. We know that each of the maps U 0 \ Wi! Spm..U 0 \ Wi ; W // O is an isomorphism. But Spm..U 0 \Wi ; Therefore the canonical morphism is an isomorphism over each Vi, and so it is an isomorphism. W // is the inverse image of Vi in Spm..U 0; O O W //. SUMMARY 8.22. Let 'W W! V be a regular map, and consider the following condition on an open affine subset U of V : () '1.U / is affine and kŒ'1.U / is a finite over kŒU. The map'is finite if () holds for the open affines in some covering of V, in which case (*) holds for all open affines of V. PROPOSITION 8.23. (a) Closed immersions are finite. (b) The composite of two finite morphisms is finite. (c) The product of two finite morphisms is finite. PROOF. (a) Let Z be a closed subvariety of a variety V, and let U be an open affine subvariety of V. Then Z \ U is a closed subvariety of U. It is therefore affine, and the map Z \ U! U corresponds to a map A! A=a of rings, which is obviously finite. This proves (a). As to be finite is a local condition, it suffices to prove (a) and (b) for maps of affine varieties. Then the statements become statements in commutative algebra. (b) If B is a finite A-algebra and C is a finite B-algebra, then C is a finite A-algebra. To see this,
note that if fbi g is a set of generators for B as an A-module, and fcj g is a set of generators for C as a B-module, then fbi cj g is a set of generators for C as an A-module. (c) If B and B 0 are respectively finite A and A0-algebras, then B ˝k B 0 is a finite A ˝k A0-algebra. To see this, note that if fbi g is a set of generators for B as an A-module, and fb0 g is a set of generators j for B ˝A B 0 as an A ˝ A0-module. g is a set of generators for B 0 as an A0-module, then fbi ˝ b0 j 3Recall that a module over a noetherian ring is noetherian if and only if it is finitely generated, and that a submodule of a noetherian module is noetherian. Therefore, a submodule of a finitely generated module over a noetherian ring is finitely generated. c. Finite and quasi-finite maps 181 1 X f0g,! A By way of contrast, open immersions are rarely finite. For example, the inclusion 1 is not finite because the ring kŒT; T 1 is not finitely generated as a kŒT A module (any finitely generated kŒT -submodule of kŒT; T 1 is contained in T nkŒT for some n). THEOREM 8.24. Finite maps of algebraic varieties are closed. PROOF. It suffices to prove this for affine varieties. Let 'W W! V be a finite map of affine varieties, and let Z be a closed subset of W. The restriction of'to Z is finite (by 8.23a and b), and so we can replace W with Z; we then have to show that Im.'/ is closed. The map corresponds to a finite map of rings A! B. This will factors as A! A=a,! B, from which we obtain maps Spm.B/! Spm.A=a/,! Spm.A/: The second map identifies Spm.A=a/ with the closed subvariety V.a/ of Spm.A
/, and so it remains to show that the first map is surjective. This is a consequence of the going-up theorem (1.53). The base change of a finite map Recall that the base change of a regular map 'W V! S is the map '0 in the diagram: V S W '0 W 0 V'S: PROPOSITION 8.25. The base change of a finite map is finite. PROOF. We may assume that all the varieties concerned are affine. Then the statement becomes: if A is a finite R-algebra, then A ˝R B=N is a finite B-algebra, which is obvious. PROPOSITION 8.26. Finite maps of algebraic varieties are proper. PROOF. The base change of a finite map is finite, and hence closed. COROLLARY 8.27. Let 'W V! S be finite; if S is complete, then so also is V. PROOF. Combine 7.19 and 8.26. Quasi-finite maps Recall that the fibres of a regular map 'W W! V are the closed subvarieties '1.P / of W for P 2 V. As for affine varieties (2.39), we say that a regular map of algebraic varieties is quasi-finite if all of its fibres are finite. PROPOSITION 8.28. A finite map 'W W! V is quasi-finite. PROOF. Let P 2 V ; we wish to show '1.P / is finite. After replacing V with an affine neighbourhood of P, we can suppose that it is affine, and then W will be affine also. The map'then corresponds to a map ˛W A! B of affine k-algebras, and a point Q of W maps to P if and only ˛1.mQ/ D mP. But this holds if and only if mQ ˛.mP /, and so the points of W mapping to P are in one-to-one correspondence with the maximal ideals of 182 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM B=˛.mP /B. Clearly B=˛.mP /B is generated as a k-vector space
by the image of any generating set for B as an A-module, and so it is a finite k-algebra. The next lemma shows that it has only finitely many maximal ideals. LEMMA 8.29. A finite k-algebra A has only finitely many maximal ideals. PROOF. Let m1; : : : ; mn be maximal ideals in A. They are obviously coprime in pairs, and so the Chinese Remainder Theorem (1.1) shows that the map A! A=m1 A=mn; a 7!.: : : ; ai mod mi ; : : :/; is surjective. It follows that dimk A X dimk.A=mi / n — here dimk means dimension as a k-vector space. Finite and quasi-finite maps of prevarieties are defined as for varieties. Examples 8.30. The projection from the curve XY D 1 onto the X axis (see p. 71) is quasi-finite but not finite — its image is not closed in A 1, and kŒX; X 1 is not finite over kŒX. 8.31. The map t 7!.t 2; t 3/W A 1! V.Y 2 X 3/ A 2 from the line to the cuspidal cubic is finite because the image of kŒX; Y in kŒT is kŒT 2; T 3, and f1; T g is a set of generators for kŒT as a kŒT 2; T 3-module (see 3.29). 8.32. The map A 1! A 1, a 7! am is finite. 8.33. The obvious map.A 1 with the origin doubled /! A 1 is quasi-finite but not finite (the inverse image of A 1 is not affine). 8.34. The map A of A 2 is not affine (see 3.33). The map 2 X foriging,! A 2 is quasi-finite but not finite, because the inverse image 2 X f.0; 0/g t fOg! A 2 A sending O to.0; 0/ is bijective but not finite (here fOg D Spm.k/ D A 0). 8.35. The map in 8.31, and the Frobenius map.
t1; : : : ; tn/ 7!.W A n! A n in characteristic p ¤ 0; are examples of finite bijective regular maps that are not isomorphisms. c. Finite and quasi-finite maps 183 8.36. Let V D A 2 D Spm.kŒX; Y / and let f be the map defined on the ring level by X 7! X D A Y 7! XY 2 C Y C 1 D B: Then f is (obviously) quasi-finite, but it is not finite. For this we have to show that kŒX; Y is not integral over its subring kŒA; B. The minimal polynomial of Y over kŒA; B is AY 2 C Y C 1 B D 0; which shows that it is not integral over kŒA; B (see 1.44). Alternatively, one can show directly that Y can never satisfy an equation Y s C g1.A; B/Y s1 C C gs.A; B/ D 0; gi.A; B/ 2 kŒA; B; by multiplying the equation by A. 8.37. Let V be the hyperplane X n C T1X n1 C C Tn D 0 in A nC1, and consider the projection map.a1; : : : ; an; x/ 7!.a1; : : : ; an/W V! A n: The fibre over a point.a1; : : : ; an/ 2 A n is the set of solutions of X n C a1X n1 C C an D 0; and so it has exactly n points, counted with multiplicities. The map is certainly quasi-finite; it is also finite because it corresponds to the finite map of k-algebras, kŒT1; : : : ; Tn! kŒT1; : : : ; Tn; X=.X n C T1X n1 C C Tn/: See also the more general example p. 51. 8.38. Let V be the hyperplane T0X n C T1X n1 C C Tn D 0 in A nC2. The projection map.a0; : : : ; an; x/ 7!.a0; : : : ; an/
W V '! A nC1 has finite fibres except for the fibre above o D.0; : : : ; 0/, which is A 1. Its restriction to V X '1.o/ is quasi-finite, but not finite. Above points of the form.0; : : : ; 0; ; : : : ; / some of the roots “vanish off to 1”. (Example 8.30 is a special case of this.) See also the more general example p. 51. 8.39. Let P.X; Y / D T0X n C T1X n1Y C C TnY n; and let V be its zero set in P A nC1 X fog is finite. 1.A nC1 X fog/. In this case, the projection map V! 184 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM d. The fibres of finite maps Let 'W W! V be a finite dominant morphism of irreducible varieties. Then dim.W / D dim.V /, and so k.W / is a finite field extension of k.V /. Its degree is called the degree of the map '. The map'is said to be separable if the field k.W / is separable over k.V /. Recall that jS j denotes the number of elements in a finite set S. THEOREM 8.40. Let 'W W! V be a finite surjective regular map of irreducible varieties, and assume that V is normal. ˇ'1.P /ˇ ˇ'1.P /ˇ ˇ D deg.'/ is an open subset of V, and it (a) For all P 2 V, ˇ ˇ deg.'/. (b) The set of points P of V such that ˇ is nonempty if'is separable. Before proving the theorem, we give examples to show that we need W to be separated and V to be normal in (a), and that we need k.W / to be separable over k.V / for the second part of (b). EXAMPLE 8.41. (a) The map fA 1 with origin doubled g! A 1 has degree one and is one-to-one except over
the origin where it is two-to-one. (b) Let C be the curve Y 2 D X 3 C X 2, and consider the map t 7!.t 2 1; t.t 2 1//W A 1! C. It is one-to-one except that the points t D ˙1 both map to 0. On coordinate rings, it corresponds to the inclusion kŒx; y,! kŒT, x 7! T 2 1 y 7! T.T 2 1/, and so is of degree one. The ring kŒx; y is not integrally closed — in fact kŒT is the integral closure of kŒx; y in its field of fractions k.x; y/ D k.T /. (c) The Frobenius map.a1; : : : ; an/ 7!.ap 1 ; : : : ; ap n /W A n! A n in characteristic p ¤ 0 is bijective on points, but has degree pn. The field extension corresponding to the map is k.X1; : : : ; Xn/ k. / which is purely inseparable. LEMMA 8.42. Let Q1; : : : ; Qr be distinct points on an affine variety V. Then there is a regular function f on V taking distinct values at the Qi. PROOF. We can embed V as closed subvariety of A statement with V D A n — almost any linear form will do. n, and then it suffices to prove the d. The fibres of finite maps 185 PROOF (OF 8.40). In proving (a) of the theorem, we may assume that V and W are affine, and so the map corresponds to a finite map of k-algebras, kŒV! kŒW. Let '1.P / D fQ1; : : : ; Qr g. According to the lemma, there exists an f 2 kŒW taking distinct values at the Qi. Let F.T / D T m C a1T m1 C C am be the minimal polynomial of f over k.V /. It has degree m Œk.W / W k.V / D deg ', and it has coefficients in kŒV because V is normal (see 1.44). Now F
.f / D 0 implies F.f.Qi // D 0, i.e., f.Qi /m C a1.P / f.Qi /m1 C C am.P / D 0: Therefore the f.Qi / are all roots of a single polynomial of degree m, and so r m deg.'/. In order to prove the first part of (b), we show that, if there is a point P 2 V such that '1.P / has deg.'/ elements, then the same is true for all points in an open neighbourhood of P. Choose f as in the last paragraph corresponding to such a P. Then the polynomial T m C a1.P / T m1 C C am.P / D 0 () has r D deg'distinct roots, and so m D r. Consider the discriminant disc F of F. Because () has distinct roots, disc.F /.P / ¤ 0, and so disc.F / is nonzero on an open neighbourhood U of P. The factorization kŒV! kŒV ŒT =.F / T 7!f! kŒW gives a factorization W! Spm.kŒV ŒT =.F //! V: Each point P 0 2 U has exactly m inverse images under the second map, and the first map is finite and dominant, and therefore surjective (recall that a finite map is closed). This proves that '1.P 0/ has at least deg.'/ points for P 0 2 U, and part (a) of the theorem then implies that it has exactly deg.'/ points. We now show that if the field extension is separable, then there exists a point such that '1.P / has deg'elements. Because k.W / is separable over k.V /, there exists an f 2 kŒW such that k.V /Œf D k.W /. Its minimal polynomial F has degree deg.'/ and its discriminant is a nonzero element of kŒV. The diagram W! Spm.kŒV ŒT =.F //! V shows that j'1.P /j deg.'/ for P a point such that disc.f /.P / ¤ 0. Let E F be a finite extension of fields. The elements of
E separable over F form a subfield F sep of E, and the separable degree of E over F is defined to be the degree of F sep over F. The separable degree of a finite surjective map 'W W! V of irreducible varieties is the separable degree of k.W / over k.V /. THEOREM 8.43. Let 'W W! V be a finite surjective regular map of irreducible varieties, and assume that V is normal. ˇ'1.P /ˇ ˇ sepdeg.'/, with equality holding on a dense open subset. (a) For all P 2 V, ˇ (b) For all i, is closed in V. Vi D fP 2 V j ˇ ˇ'1.P /ˇ ˇ i g 186 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM PROOF. If'is separable, this was proved in 8.40. If'is purely inseparable, then'is one-to-one because, for some q, the Frobenius map V.q1/ F! V factors through '. To prove the general case, factor'as the composite of a purely inseparable map with a separable map. ASIDE 8.44. A finite map from a variety onto a normal variety is open (hence both open and closed). For an elementary proof, see Theorem 63.12 of Musili, C., Algebraic geometry for beginners. Texts and Readings in Mathematics, 20. Hindustan Book Agency, New Delhi, 2001. e. Zariski’s main theorem In this section, we explain a fundamental theorem of Zariski. Statement and proof One obvious way to construct a nonfinite quasi-finite map is to take a finite map W! V and remove a closed subset of W. Zariski’s Main Theorem (ZMT) shows that, for algebraic varieties, every quasi-finite map arises in this way. THEOREM 8.45 (ZARISKI’S MAIN THEOREM). Every quasi-finite map of algebraic varieties 'W W! V factors into W! V with '0 finite and j an open immersion:,! V 0 '0 j
W open immersion V 0 quasi-finite finite V: When'is a dominant map of irreducible varieties, the statement is true with '0W V 0! V equal to the normalization of V in W (in the sense of 8.9). The key result needed to prove 8.45 is the following statement from commutative algebra. For a ring A and a prime ideal p in A,.p/ denotes the field of fractions of A=p. THEOREM 8.46 (LOCAL VERSION OF ZMT). Let A be a commutative ring, and let i W A! B be a finitely generated A-algebra. Let q be a prime ideal of B, and let p D i 1.q/. Finally, let A0 denote the integral closure of A in B. If Bq=pBq is a finite.p/-algebra, then there exists an f 2 A0 not in q such that the map A0 f! Bf is an isomorphism. PROOF. The proof is quite elementary, but intricate — see 17 of my notes CA. Recall that a point v in a topological space V is isolated if fvg is an open subset of V. The isolated points v of an algebraic variety V are those such that fvg is both open and closed. Thus they are the irreducible components of V of dimension 0. Let 'W W! V be a continuous map of topological spaces. We say that w 2 W is isolated in its fibre if it is isolated in the subspace '1.'.w// of W. Let 'W A! B be a homomorphism of finitely generated k-algebras, and consider spm.'/W spm.B/! spm.A/; then n 2 spm.B/ is isolated in its fibre if and only if Bn=mBn is a finite k-algebra; here m D '1.n/. PROPOSITION 8.47. Let 'W W! V be a regular map of algebraic varieties. The set W 0 of points of W isolated in their fibres is open in W. e. Zariski’s main theorem 187 PROOF. Let w 2 W 0. Let Ww and Vv be open affine neighbourhoods of w and v D '.w/ such that '.Ww / Vv, and let A
D kŒVv and B D kŒWw. Let n D ff 2 B j f.w/ D 0g — it is the maximal ideal in B corresponding to w. Let A0 be the integral closure of A in B. Theorem 8.46 shows that there exists an f 2 A0'Bf. Write A0 as the union of the finitely generated A-subalgebras not in m such that A0 f Ai of A0 containing f : A0 D [ i Ai : Because A0 is integral over A, each Ai is finite over A (see 1.35). We have Bf'A0 f D [ Aif : i Because Bf is a finitely generated A-algebra, Bf D Aif for all sufficiently large Ai. As the Ai are finite over A, Bf is quasi-finite over A, and spm.Bf / is an open neighbourhood of w consisting of quasi-finite points. PROPOSITION 8.48. Every quasi-finite map of affine algebraic varieties 'W W! V factors into W j! V 0 '0! V with j a dominant open immersion and '0 finite. PROOF. Let A D kŒV and B D kŒW. Because'is quasi-finite, Theorem 8.46 shows that there exist fi 2 A0 such that the sets spm.Bfi / form an open covering of W and A0'Bfi for all i. As W quasicompact, finitely many sets spm.Bfi / suffice to cover fi W. The argument in the proof of (8.47) shows that there exists an A-subalgebra A00 of A0, finite over A, which contains f1; : : : ; fn and is such that Bfi for all i. Now the map W D Spm.B/! Spm.A00/ is an open immersion because it is when restricted to Spm.Bfi / for each i. As Spm.A00/! Spm.A/ D V is finite, we can take V 0 D Spm.A00/.'A00 fi Recall (Exercise 8-3) that a regular map 'W W! V is affine if '1.U / is affine whenever U is an open affine subset of V. PROPOSITION
8.49. Let 'W W! V be an affine map of irreducible algebraic varieties. Then the map j W W! V 0 from W into the normalization V 0 of V in W (8.9) is an open immersion. PROOF. Let U be an open affine in V. Let A D kŒU and B D kŒ'1.U /. In this case, the normalization A0 of A in B is finite over A (because it is contained in the normalization of A in k.W /, which is finite over A (8.3)). Thus, in the proof of 8.48 we can take A00 D A0, and then '1.U /! Spm.A0/ is an open immersion. As Spm.A0/ is an open subvariety of V 0 and the sets '1.U / cover W, this implies that j W W! V 0 is an open immersion. As V 0! V is finite, this proves Theorem 8.45 in the case that'is an affine map of irreducible varieties. To deduce the general case of Theorem 8.45 from 8.44 requires an additional argument. See Theorem 12.83 of G¨ortz, U. and Wedhorn, T., Algebraic Geometry I., Springer Spektrum, Wiesbaden, 2020. NOTES 8.50. Let 'W W! V be a quasi-finite map of algebraic varieties. In 8.45, we may replace V 0 with the closure of the image of j. Thus, there is a factorization'D '0 ı j with '0 finite and j a dominant open immersion. 188 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM 8.51. Theorem 8.45 is false for prevarieties (see 8.33). However, it is true for separated maps of prevarieties. (A regular map 'W V! S of algebraic prevarieties is separated if the image V =S of the map v 7!.v; v/W V! V S V is closed; the map'is separated if V is separated.) 8.52. Assume that V is normal in 8.45. Then '0 is open (
8.44), and so'is open. Thus, every quasi-finite map from an algebraic variety to a normal algebraic variety is open. Applications to finite maps Zariski’s main theorem allows us to give a geometric criteria for a regular map to be finite. PROPOSITION 8.53. Every quasi-finite regular map 'W W! V of algebraic varieties with W complete is finite. PROOF. The map j W W,! V 0 in 8.45 is an isomorphism of W onto its image j.W / in V 0. If W is complete, then j.W / is closed (7.7), and so the restriction of '0 to j.W / is finite. PROPOSITION 8.54. Every proper quasi-finite map 'W W! V of algebraic varieties is finite. PROOF. Factor'into W j,! W 0 ˛! V with ˛ finite and j an open immersion. Factor j into W w7!.w;jw/! W V W 0.w;w 0/7!w 0 The image of the first map is j, which is closed because W 0 is a variety (see 5.28; W 0 is separated because it is finite over a variety — exercise). Because'is proper, the second map is closed. Hence j is an open immersion with closed image. It follows that its image is a connected component of W 0, and that W is isomorphic to that connected component.! W 0: NOTES 8.55. When W and V are curves, every surjective map W! V is closed. Thus it is easy to give examples of closed surjective quasi-finite, but nonfinite, maps. Consider, for example, the map A 1 X f0g t A 0! A 1; sending each a 2 A because the map is only closed, not universally closed. 1 X f0g to a and O 2 A 0 to 0. This doesn’t violate the Proposition 8.54, Applications to birational maps Recall (p. 116) that a regular map 'W W! V of irreducible varieties is said to be birational if it induces an isomorphism k.V /! k.W / on the fields of rational functions. 8.56. One may ask how a birational regular map 'W W! V can
fail to be an isomorphism. Here are three examples. (a) The inclusion of an open subset into a variety is birational. e. Zariski’s main theorem 189 (b) The map (8.31) from A 1 to the cuspidal cubic, 1! C; A t 7!.t 2; t 3/; is birational. Here C is the cubic Y 2 D X 3, and the map kŒC! kŒA 1 D kŒT identifies kŒC with the subring kŒT 2; T 3 of kŒT. Both rings have k.T / as their fields of fractions. (c) For any smooth variety V and point P 2 V, there is a regular birational map 'W V 0! V such that the restriction of'to V 0 X '1.P / is an isomorphism onto V X P, but '1.P / is the projective space attached to the vector space TP.V /. See the section on blow-ups below. The next result says that, if we require the target variety to be normal (thereby excluding example (b)), and we require the map to be quasi-finite (thereby excluding example (c)), then we are left with (a). PROPOSITION 8.57. Let 'W W! V be a birational regular map of irreducible varieties. If V is normal and the map'is quasi-finite, then'is an isomorphism from W onto an open subvariety of V. PROOF. Factor'as in the Theorem 8.45 (so, in particular, '0W V 0! V is the normalization of V in W ). For each open affine subset U of V, kŒ'01.U / is the integral closure of kŒU in k.W /. Because'is birational, the inclusion k.V / k.V 0/ D k.W / is an equality. Now kŒU is integrally closed in k.V / (because V is normal), and so U D '01.U / (as varieties). We have shown that '0W V 0! V is an isomorphism locally on the base V, and hence an isomorphism. 8.58. In topology, a continuous bijective map 'W
W! V need not be a homeomorphism, but it is if W is compact and V is Hausdorff. Similarly, a bijective regular map of algebraic varieties need not be an isomorphism. Here are three examples: (a) In characteristic p, the Frobenius map.x1; : : : ; xn/ 7!.xp 1 ; : : : ; xp n /W A n! A n is bijective and regular, but it is not an isomorphism even though A n is normal. (b) The map t 7!.t 2; t 3/ from A 1 to the cuspidal cubic (see 8.56b) is bijective, but not an isomorphism. (c) Consider the regular map A 1 sending x to 1=x for x ¤ 0 and 0 to 0. Its graph is the union of.0; 0/ and the hyperbola xy D 1, which is a closed subvariety of 1 is a bijective, regular, birational map, A but it is not an isomorphism even though A 1. The projection.x; y/ 7! xW! A 1 is normal. 1! A 1 A If we require the map to be birational (thereby excluding example (a)), V to be normal (thereby excluding example (b)), and the varieties to be irreducible (thereby excluding example (c)), then the map is an isomorphism. PROPOSITION 8.59. Let 'W W! V be a bijective regular map of irreducible algebraic varieties. If the map'is birational and V is normal, then'is an isomorphism. PROOF. The hypotheses imply that'is an isomorphism of W onto an open subset of V (8.57). Because'is bijective, the open subset must be the whole of V. 190 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM In fact, example (a) can be excluded by requiring that'be generically separable (instead of birational). PROPOSITION 8.60. Let 'W W! V be a bijective regular map of irreducible varieties. If V is normal and k.W / is separably generated over k.
V /, then'is an isomorphism. PROOF. Because'is bijective, dim.W / D dim.V / (see Theorem 9.9 below) and the separable degree of k.W / over k.V / is 1 (apply 8.40 to the variety V 0 in 8.45). Hence'is birational, and we may apply 8.59. 8.61. In functional analysis, the closed graph theorem states that, if a linear map 'W W! V between two Banach spaces has a closed graph defD f.w; 'w/ j w 2 W g, then'is continuous (q.v. Wikipedia). One can ask (cf. mo113858) whether a similar statement is true in algebraic geometry. Specifically, if 'W W! V is a map (in the set-theoretic sense) of algebraic varieties V; W whose graph is closed (for the Zariski topology), then is'a regular map? The answer is no in general. For example, even in characteristic zero, the map.t 2; t 3/! tW C! A 1 inverse to that in 8.56(b) has closed graph but is not regular. In characteristic p, the inverse of the Frobenius map x 7! xp provides another counterexample. For a third counterexample, see 8.58(c). The projection from to W is a bijective regular map, and so'will be regular if is an isomorphism. According to 8.60, is an isomorphism if the varieties are irreducible, W is normal, and is generically separable. In particular, a map between irreducible normal algebraic varieties in characteristic zero is regular if its graph is closed. A condition for an algebraic monoid to be a group A monoid variety is an algebraic variety G together with the structure of a monoid defined by regular maps mW G G! G; eW A 0! G: LEMMA 8.62. Let.G; m; e/ be an algebraic monoid. The map TeG ˚ TeG'T.e;e/.G G/.d m/.e;e/! Te.G/ is addition. PROOF. The first isomorphism is.X; Y / 7!.d˛/
e.X/ C.dˇ/e.Y /, where ˛ is the map x 7!.x; e/W G! G G and ˇ is x 7!.e; x/. To compute.d m/.e;e/..dˇ/e.X/C.d˛/e.Y //; note that m ı ˛ D idG D m ı ˇ. PROPOSITION 8.63. Let.G; m; e/ be an algebraic monoid over k. If.G.k/; m.k// is a group with identity element e, then.G; m/ is an algebraic group, that is, the map a 7! a1 is regular. PROOF. Let a 2 G.k/. The translation map LaW x 7! ax is an isomorphism G! G because it has an inverse La1. Therefore G is homogeneous as an algebraic variety: for any two points in jGj, there is an isomorphism G! G mapping one to the other. It follows that G is nonsingular, in particular, normal. The map.x; y/ 7!.x; xy/W G G! G G f. Stein factorization 191 is regular, a bijection on k-points, and induces an isomorphism on the tangent spaces at.e; e/ (apply the lemma). It is therefore an isomorphism of algebraic varieties over k. Therefore, its inverse.x; y/ 7!.x; x1y/ is regular, and so.x; y/ 7! x1yW G G! G is regular. This implies that.G; m/ is an algebraic group. Note that it is necessary in the proposition that G be reduced: consider G D Spec kŒT =.T n/, n > 1, with the trivial monoid structure G G! e! G. Variants of Zariski’s main theorem Mumford, 1966,4 III, 9, lists the following variants of ZMT. Original form (8.57) Let 'W W! V be a birational regular map of irreducible varieties. If V is normal and'is quasi-finite, then'is an isomorphism of W onto an open subvariety of V. Topological
form Let V be a normal variety over C, and let v 2 V. Let S be the singular locus of V. Then the complex neighbourhoods U of v such that U X U \ S is connected form a base for the system of complex neighbourhoods of v. Power series form Let V be a normal variety, and let an irreducible closed subset of V (cf. p. 177). If domain, then so also is its completion. O V;Z be the local ring attached to O V;Z is an integrally closed integral Grothendieck’s form (8.45) Every quasi-finite map of algebraic varieties factors as the composite of an open immersion with a finite map. Connectedness theorem Let 'W W! V be a proper birational map, and let v be a (closed) normal point of V. The '1.v/ is a connected set (in the Zariski topology). The original form of the theorem was proved by Zariski using a fairly direct argument whose method doesn’t seem to generalize.5 The power series form was also proved by Zariski, who showed that it implied the original form. The last two forms are much deeper and were proved by Grothendieck. See the discussion in Mumford 1966. NOTES. The original form of the theorem (8.57) is the “Main theorem” of Zariski, O., Foundations of a general theory of birational correspondences. Trans. Amer. Math. Soc. 53, (1943). 490–542. f. Stein factorization The following important theorem shows that the fibres of a proper map are disconnected only because the fibres of finite maps are disconnected. THEOREM 8.64 (STEIN FACTORIZATION). Every proper map 'W W! V of algebraic varieties factors into W '1! W 0 '2! V with '1 proper with connected fibres and '2 finite. 4Introduction to Algebraic Geometry, Harvard notes. Reprinted as “The Red Book of Varieties and Schemes” (with the introduction of misprints) by Springer 1999. 5See Lang, S., Introduction to Algebraic Geometry, 1958, V 2, for Zariski’s original statement and proof of this theorem. See Springer, T.A., Linear Algebraic Groups, 1998, 5.2
.8, for a direct proof of (8.59). 192 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM When V is affine, this is the factorization W! Spm. W.W //! V: O The first major step in the proof of the theorem is to show that'on V. Here'O W is the sheaf of V -algebras on V, O U W.'1.U //: O W is a coherent sheaf O O To say that'of regular map '2W Spm.' O attached by Spm to the map of k-algebras kŒU! W is coherent means that, on every open affine subset U of V, it is the sheaf U -algebras defined by a finite kŒU -algebra. This, in turn, means that there exists a W /! V that, over every open affine subset U of V, is the map W.'1.U //: O The Stein factorization is then O '1! W 0 defD Spm.' W '2! V: W / O By construction, '2 is finite and '1W W! W 0 has the property that W is an isomorphism. That its fibres are connected is a consequence of the following extension of Zariski’s connectedness theorem to non birational maps. W 0! '1 O O THEOREM 8.65. Let 'W W! V be a proper map such that the map isomorphism. Then the fibres of'are connected. O V!'W is an O See Hartshorne 1977, III, 11. NOTES. The Stein factorization was originally proved by Stein for complex spaces (q.v. Wikipedia). g. Blow-ups Under construction. Let P be a nonsingular point on an algebraic variety V, and let Tp.V / be the tangent space at P. The blow-up of V at P is a regular map QV! V that replaces P with the projective space P.TP.V //. More generally, the blow-up at P replaces P with P.CP.V //, where CP.V / is the geometric tangent cone at P. Blowing up the origin in An Let O be the origin in
A.a1W : : : W an/. Let be the graph of, and let fA The map W fA n defined by the projection map A at O. n, and let W A n! A n X fOg! P n be the closure of in A n1 be the map.a1; : : : ; an/ 7! n1. n n is the blow-up of A n1! A n P n P Blowing up a point on a variety Examples 8.66. The nodal cubic 8.67. The cuspidal cubic h. Resolution of singularities 193 h. Resolution of singularities Let V be an algebraic variety. A desingularization of V is birational regular map W W! V such that W is nonsingular and is proper; if V is projective, then W should also be projective, and should induce an isomorphism W X 1.Sing.V //! V X Sing.V /: In other words, the nonsingular variety W is the same as V except over the singular locus of V. When a variety admits a desingularization, then we say that resolution of singularities holds for V. Note that with “nonsingular” replaced by “normalization”, the normalization of V (see 8.5) provides such a map (resolution of abnormalities). Nagata’s embedding theorem 7.50 shows that it suffices to prove resolution of singularities for complete varieties, and Chow’s lemma 7.39 then shows that it suffices to prove resolution of singularities for projective varieties. From now on, we shall consider only projective varieties. Resolution of singularities for curves was first obtained using blow-ups (see Chapter 7 of Fulton’s book, Algebraic Curves). Zariski introduced the notion of the normalization of a variety, and observed that the normalization W QV! V of a curve V in k.V / is a desingularization of V. There were several proofs of resolution of singularities for surfaces over C, but the first to be accepted as rigorous is that of Walker (patching Jung’s local arguments; 1935). For a surface V, normalization gives a surface with only point singularities (8.12), which can then be blown up. Zariski showed that the desingularization of a surface in
characteristic zero can be obtained by alternating normalizations and blow-ups. The resolution of singularities for three-folds in characteristic zero is much more difficult, and was first achieved by Zariski (Ann. of Math. 1944). His result was extended to nonzero characteristic by his student Abhyankar and to all varieties in characteristic zero by his student Hironaka. The resolution of singularities for higher dimensional varieties in nonzero characteristic is one of the most important outstanding problems in algebraic geometry. In 1996, de Jong proved a weaker result in which, instead of the map being birational, k.W / is allowed to be a finite extension of k.V /. A little history Normal varieties were introduced by Zariski in a paper, Amer. J. Math. 61, 1939, p. 249–194. There he noted that the singular locus of a normal variety has codimension at least 2 and that the system of hyperplane sections of a normal variety relative to a projective embedding is complete (i.e., is a complete rational equivalence class). Zariski’s introduction of the notion of a normal variety and of the normalization of a variety was an important insertion of commutative algebra into algebraic geometry. It is not easy to give a geometric intuition for “normal”. One criterion is that a variety is normal if and only if every surjective finite birational map onto it is an isomorphism (8.57). See mo109395 for a discussion of this question. 194 8. NORMAL VARIETIES; (QUASI-)FINITE MAPS; ZARISKI’S MAIN THEOREM Exercises 8-1. Prove that a finite map is an isomorphism if and only if it is bijective and ´etale. (Cf. Harris 1992, 14.9.) 8-2. Give an example of a surjective quasi-finite regular map that is not finite (different from any in the notes). 8-3. Let 'W W! V be a regular map with the property that '1.U / is an open affine subset of W whenever U is an open affine subset of V (such a map is said to be affine). Show that if V is separated, then so also is W. 8-4. For every n 1, find a finite map 'W W! V with the following
property: for all 1 i n, defD fP 2 V j '1.P / has i pointsg Vi is a nonempty closed subvariety of dimension i. CHAPTER 9 Regular Maps and Their Fibres Consider again the regular map 'W A 2,.x; y/ 7!.x; xy/ (Exercise 3-3). The line Y D c maps to the line Y D cX. As c runs over the elements of k, this line sweeps out the whole x; y-plane except for the y-axis, and so the image of'is 2! A which is neither open nor closed, and, in fact, is not even locally closed. The fibre C D.A 2 X fy-axisg/ [ f.0; 0/g; '1.a; b/ D 8 < : point.a; b=a/ Y -axis ; if a ¤ 0 if.a; b/ D.0; 0/ if a D 0, b ¤ 0: From this unpromising example, it would appear that it is not possible to say anything about the image of a regular map or its fibres. However, it turns out that almost everything that can go wrong already goes wrong in this example. We shall show: (a) the image of a regular map is a finite union of locally closed sets; (b) the dimensions of the fibres can jump only over closed subsets; (c) the number of elements (if finite) in the fibres can drop only on closed subsets, provided the map is finite, the target variety is normal, and k has characteristic zero. a. The constructibility theorem THEOREM 9.1. Let 'W W! V be a dominant regular map of irreducible affine algebraic varieties. Then '.W / contains a dense open subset of V. PROOF. Because'is dominant, the map f 7! f ı 'W kŒV! kŒW is injective (3.34). According to Lemma 9.4 below, there exists a nonzero a 2 kŒV such that every homomorphism ˛W kŒV! k such that ˛.a/ ¤ 0 extends to a homomorphism ˇW kŒW! k with ˇ.1/ ¤ 0. In particular, for P 2 D.
a/, the homomorphism g 7! g.P /W kŒV! k extends to a nonzero homomorphism ˇW kŒW! k. The kernel of ˇ is a maximal ideal of kŒW whose zero set is a point Q of W such that '.Q/ D P. Before beginning the proof of Lemma 9.4, we should look at an example. EXAMPLE 9.2. Let A be an affine k-algebra, and let B D AŒT =.f / with f D amT m C C a0. When does a homomorphism ˛W A! k extend to B? The extensions of ˛ correspond to roots of the polynomial ˛.am/T m C C˛.a0/ in k, and so there exists an extension unless this is a nonzero constant polynomial. In particular, ˛ extends if ˛.am/ ¤ 0. 195 196 9. REGULAR MAPS AND THEIR FIBRES LEMMA 9.3. Let A B be finitely generated k-algebras. Assume that A and B are integral domains, and that B is generated by a single element, say, B D AŒt'AŒT =a. Let c A be the set of leading coefficients of the polynomials in a. Then every homomorphism ˛W A! k such that ˛.c/ ¤ 0 extends to a homomorphism B! k. PROOF. Note that c is an ideal in A. If a D 0, then every homomorphism ˛ extends. Thus we may assume that a ¤ 0. Let f D amT m C C a0 be a nonzero polynomial of minimum degree in a such that ˛.am/ ¤ 0. Because B ¤ 0, we have that m 1. Extend ˛ to a homomorphism Q˛W AŒT! kŒT by sending T to T. The k-submodule of kŒT generated by Q˛.a/ is an ideal (because T P ci Q˛.gi / D P ci Q˛.gi T //. Unless Q˛.a/ contains a nonzero constant, it generates a proper ideal
in kŒT, which will have a zero c in k (2.11). The homomorphism AŒT Q˛! kŒT h7!h.c/! k; T 7! T 7! c then factors through AŒT =a D B and extends ˛. In the contrary case, a contains a polynomial g.T / D bnT n C C b0; ˛.bi / D 0.i > 0/; ˛.b0/ ¤ 0: On dividing f.T / into g.T /, we find that ad mg.T / D q.T /f.T / C r.T /; d 2 N; q; r 2 AŒT ; deg r < m: On applying Q˛ to this equation, we obtain ˛.am/d ˛.b0/ D Q˛.q/ Q˛.f / C Q˛.r/: Because Q˛.f / has degree m > 0, we must have Q˛.q/ D 0, and so Q˛.r/ is a nonzero constant. After replacing g.T / with r.T /, we may assume n < m. If m D 1, such a g.T / can’t exist, and so we may suppose m > 1 and (by induction) that the lemma holds for smaller values of m. For h.T / D cr T r C cr1T r1 C C c0, let h0.T / D cr C C c0T r. Then the Amodule generated by the polynomials T sh0.T /, s 0, h 2 a, is an ideal a0 in AŒT. Moreover, a0 contains a nonzero constant if and only if a contains a nonzero polynomial cT r, which implies t D 0 and A D B (since B is an integral domain). If a0 does not contain nonzero constants, then set B 0 D AŒT =a0 D AŒt 0. Then a0 contains the polynomial g0 D bn C C b0T n, and ˛.b0/¤ 0. Because deg g0 < m, the induction hypothesis implies that ˛ extends to a hom
omorphism B 0! k. Therefore, there is a c 2 k such that, for all h.T / D cr T r C cr1T r1 C C c0 2 a, h0.c/ D ˛.cr / C ˛.cr1/c C C c0cr D 0: On taking h D g, we see that c D 0, and on taking h D f, we obtain the contradiction ˛.am/ D 0. LEMMA 9.4. Let A B be finitely generated k-algebras. Assume that A and B are integral domains, and let b be a nonzero element of B. Then there exists a nonzero a 2 A with the following property: every homomorphism ˛W A! k from A into k such that ˛.a/ ¤ 0 extends to a homomorphism ˇW B! k such that ˇ.b/ ¤ 0. a. The constructibility theorem 197 PROOF Suppose that we know the proposition in the case that B is generated by a single element, and write B D AŒx1; : : : ; xn. Then there exists an element bn1 2 AŒx1; : : : ; xn1 with the following property: every homomorphism ˛W AŒx1; : : : ; xn1! k such that ˛.bn1/ ¤ 0 extends to a homomorphism ˇW B! k such that ˇ.b/ ¤ 0. Then there exists a bn2 2 AŒx1; : : : ; xn2 etc. Continuing in this fashion, we obtain an element a 2 A with the required property. Thus we may assume B D AŒx. Let a be the kernel of the homomorphism T 7! x, AŒT! AŒx. Case (i). The ideal a D.0/. Write b D f.x/ D a0xn C a1xn1 C C an; ai 2 A; and take a D a0. If ˛W A! k is such that ˛.a0/ ¤ 0, then there exists a c 2 k such that f.c/ ¤ 0, and we can take ˇ to be
the homomorphism P di xi 7! P ˛.di /ci. Case (ii). The ideal a ¤.0/. Let f.T / D amT m C C a0; am ¤ 0; be an element of a of minimum degree. Let h.T / 2 AŒT represent b. As b is nonzero, h … a. Because f is irreducible over the field of fractions of A, it and h are coprime over that field. Hence there exist u; v 2 AŒT and c 2 A X f0g such that uh C vf D c: It follows now that cam satisfies our requirements, for if ˛.cam/ ¤ 0, then ˛ can be extended to ˇW B! k by the preceding lemma, and ˇ.u.x/ b/ D ˇ.c/ ¤ 0, and so ˇ.b/ ¤ 0. ASIDE 9.5. It is also possible to deduce Theorem 9.1 from the generic freeness theorem (CA 21.11). In order to generalize 9.1 to arbitrary maps of arbitrary varieties, we need the notion of a constructible set. Let W be a topological space. A subset C of W is said to constructible if it is a finite union of sets of the form U \ Z with U open and Z closed. Obviously, if C is constructible in W and V W, then C \ V is constructible in V, and it is constructible in W if V is open or closed. A constructible subset of A n is one that is definable by a finite number of polynomials. More precisely, it is defined by a finite number of statements of the form f.X1; : : : ; Xn/ D 0; g.X1; : : : ; Xn/ ¤ 0 1 that omits an infinite set. combined using only “and” and “or” (or, better, statements of the form f D 0 combined using “and”, “or”, and “not”). The next proposition shows that a constructible set C that is dense in an irreducible variety V must contain a nonempty open subset of V. Contrast Q, which is dense in R (real topology),
but does not contain an open subset of R, or an infinite subset of A PROPOSITION 9.6. Let C be a constructible set whose closure NC is irreducible. Then C contains a nonempty open subset of its closure NC. PROOF. We are given that C D S.Ui \ Zi / with each Ui open and each Zi closed. We may assume that each set Ui \ Zi in this decomposition is nonempty. Clearly NC S Zi, and as NC is irreducible, it must be contained in one of the Zi. For this i C Ui \ Zi Ui \ NC Ui \ C Ui \.Ui \ Zi / D Ui \ Zi : Thus Ui \ Zi D Ui \ NC is a nonempty open subset of NC contained in C. 198 9. REGULAR MAPS AND THEIR FIBRES THEOREM 9.7. Every regular map 'W W! V sends constructible sets to constructible sets. PROOF We first show that it suffices to prove the theorem with W and V affine. Write V as a finite union of open affines, and then write the inverse image of each of the affines as a finite union of open affines. In this way, we get W D S i 2I Wi with each Wi open affine and '.Wi / contained in an open affine of V. If C is a constructible subset of W, then '.C / D S i 2I '.C \Wi /, and so '.C / is constructible if each set '.C \Wi / is constructible. Now assume that W and V are affine, and let C be a constructible subset of W. Let Wi be the irreducible components of W. They are closed in W, and so C \ Wi is constructible in W. As '.W / D S '.C \ Wi /, it is constructible if the '.C \ Wi / are. Hence we may suppose that W is irreducible. Moreover, C is a finite union of its irreducible components. As these are closed in C, they are constructible in W. We may therefore assume that C is also irreducible; NC is then an irreducible closed subvariety of W. We prove the theorem by induction on the dimension of W. If dim.W / D 0, then the statement is obvious because
W is a point. If NC ¤ W, then dim. NC / < dim.W /, and '.C / is '! V. We may therefore assume constructible by the induction hypothesis applied to NC that NC D W. Replace V with '.C /. According to Proposition 9.6, C contains a dense open subset U 0 of W, and Theorem 9.1 applied to U 0 '! V shows that '.C / contains a dense open subset U of V. Write '.C / D U [ '.C \ '1.V U //: Then '1.V U / is a proper closed subset of W (the complement of V U is dense in V and'is dominant). As C \ '1.V U / is constructible in '1.V U /, the set '.C \ '1.V U // is constructible in V by induction, which completes the proof. ASIDE 9.8. Let X be a subset of C closure of X for the Zariski topology is equal to its closure for the complex topology. n. If X is constructible for the Zariski topology on C n, then the b. The fibres of morphisms We wish to examine the fibres of a regular map 'W W! V. We can replace V by the closure of '.W / in V and so assume that'is dominant. THEOREM 9.9. Let 'W W! V be a dominant regular map of irreducible varieties. Then (a) dim.W / dim.V /; (b) if P 2 '.W /, then dim.'1.P // dim.W / dim.V / for every P 2 V, with equality holding exactly on a nonempty open subset U of V. (c) The sets are closed in '.W /. Vi D fP 2 V j dim.'1.P // i g In other words, for P on a dense open subset U of V, the fibre '1.P / has the expected dimension dim.W / dim.V /. On the closed complement of U (possibly empty), the dimension of the fibre is > dim.W / dim.V /, and it may jump further on closed subsets. Before proving the theorem, we should look at an example. b. The fibres of morphisms 199 EXAMPLE 9.10. Consider the subvariety W V A m defined by
r linear equations m X j D1 aij Xj D 0; aij 2 kŒV ; i D 1; : : : ; r; and let'be the projection W! V. For P 2 V, '1.P / is the set of solutions of system of equations m X aij.P /Xj D 0; aij.P / 2 k; i D 1; : : : ; r; j D1 and so its dimension is m rank.aij.P //. Since the rank of the matrix.aij.P // drops on closed subsets, the dimension of the fibre jumps on closed subsets. More precisely, for each r 2 N, fP 2 V j rank.aij.P // rg is a closed subset of V (see Exercise 2-2); hence, for each r 0 2 N, fP 2 V j dim '1.P / r 0g is closed in V. PROOF. (a) Because the map is dominant, there is a homomorphism k.V /,! k.W /, and obviously tr degkk.V / tr degkk.W / (an algebraically independent subset of k.V / remains algebraically independent in k.W /). (b) In proving the first part of (b), we may replace V by any open neighbourhood of P. In particular, we can assume V to be affine. Let m be the dimension of V. From (3.47) we know that there exist regular functions f1; : : : ; fm such that P is an irreducible component of V.f1; : : : ; fm/. After replacing V by a smaller neighbourhood of P, we can suppose that P D V.f1; : : : ; fm/. Then '1.P / is the zero set of the regular functions f1 ı '; : : : ; fm ı ', and so (if nonempty) has codimension m in W (see 3.45). Hence dim '1.P / dim W m D dim.W / dim.V /: In proving the second part of (b), we can replace both W and V with open affine subsets. Since'is dominant, kŒV! kŒW is injective, and we may regard it as an inclusion (we identify a function x on V with x
ı'on W /. Then k.V / k.W /. Write kŒV D kŒx1; : : : ; xM and kŒW D kŒy1; : : : ; yN, and suppose V and W have dimensions m and n respectively. Then k.W / has transcendence degree n m over k.V /, and we may suppose that y1; : : : ; ynm are algebraically independent over kŒx1; : : : ; xm, and that the remaining yi are algebraic over kŒx1; : : : ; xm; y1; : : : ; ynm. There are therefore relations Fi.x1; : : : ; xm; y1; : : : ; ynm; yi / D 0; i D n m C 1; : : : ; N; (37) with Fi.X1; : : : ; Xm; Y1; : : : ; Ynm; Yi / a nonzero polynomial. We write Nyi for the restriction of yi to '1.P /. Then kŒ'1.P / D kŒ Ny1; : : : ; NyN : The equations (37) give an algebraic relation among the functions x1; : : : ; yi on W. When we restrict them to '1.P /, they become equations: Fi.x1.P /; : : : ; xm.P /; Ny1; : : : ; Nynm; Nyi / D 0; i D n m C 1; : : : ; N: 200 9. REGULAR MAPS AND THEIR FIBRES If these are nontrivial algebraic relations, i.e., if none of the polynomials Fi.x1.P /; : : : ; xm.P /; Y1; : : : ; Ynm; Yi / is identically zero, then the transcendence degree of k. Ny1; : : : ; NyN / over k will be n m. Thus, regard Fi.x1; : : : ; xm; Y1; : : : ; Ynm; Yi / as a polynomial in the Y ’s with coefficients polynomials in the x’s. Let Vi be the closed subvariety of V defined
by the simultaneous vanishing of the coefficients of this polynomial — it is a proper closed subset of V. Let U D V X S Vi — it is a nonempty open subset of V. If P 2 U, then none of the polynomials Fi.x1.P /; : : : ; xm.P /; Y1; : : : ; Ynm; Yi / is identically zero, and so for P 2 U, the dimension of '1.P / is n m, and hence D n m by (a). Finally, if for a particular point P, dim '1.P / D n m, then we can modify the above argument to show that the same is true for all points in an open neighbourhood of P. (c) We prove this by induction on the dimension of V — it is obviously true if dim V D 0. We know from (b) that there is an open subset U of V such that dim '1.P / D n m ” P 2 U: Let Z be the complement of U in V ; thus Z D VnmC1. Let Z1; : : : ; Zr be the irreducible components of Z. On applying the induction to the restriction of'to the map '1.Zj /! Zj for each j, we obtain the result. Recall that a regular map 'W W! V of algebraic varieties is closed if, for example, W is complete (7.7). PROPOSITION 9.11. Let 'W W! V be a regular surjective closed map of varieties, and let n 2 N. If V is irreducible and all fibres '1.P / of'are irreducible of dimension n, then W is irreducible of dimension dim.V / C n. PROOF. Let Z be an irreducible closed subset of W, and consider the map 'jZW Z! V ; it has fibres.'jZ/1.P / D '1.P / \ Z. There are three possibilities. (a) '.Z/ ¤ V. Then '.Z/ is a proper closed subset of V. (b) '.Z/ D V, dim.Z/ < n C dim.V /. Then (b) of (9.9) shows that there is a nonempty open subset U of V such that for P
2 U, dim.'1.P / \ Z/ D dim.Z/ dim.V / < n: Thus, for P 2 U, the fibre '1.P / is not contained in Z. (c) '.Z/ D V, dim.Z/ n C dim.V /. Then 9.9(b) shows that dim.'1.P / \ Z/ dim.Z/ dim.V / n for all P ; thus '1.P / Z for all P 2 V, and so Z D W ; moreover dim Z D dim V C n. Now let Z1; : : : ; Zr be the irreducible components of W. I claim that (c) holds for at least one of the Zi. Otherwise, there will be an open subset U of V such that for P in U, '1.P / is contained in none of the Zi ; but '1.P / is irreducible and '1.P / D S.'1.P / \ Zi /, and so this is impossible. CAUTION. It is possible for all the fibres of regular map W! V to be reducible without 2y2 D 0 is irreducible, W being reducible. The variety in A but the fibres of the projection to the first factor (obtained by fixing the values of y1 and y2) are all reducible. Pass to the projective closure to extend this to P 2 with equation x2 1y1 x2 2 A 2 P 2. c. Flat maps and their fibres 201 c. Flat maps and their fibres Flat maps Let A be a ring, and let B be an A-algebra. If the sequence of A-modules 0! N 0 ˛! N ˇ! N 00! 0 is exact, then the sequence of B-modules B ˝A N 0 1˝˛! B ˝A N 1˝ˇ! B ˝A N 00! 0 is exact,1 but B˝A N 0! B˝A N need not be injective. For example, when we tensor the exact sequence of kŒX-modules 0! kŒX f 7!Xf! kŒX f 7!f mod.X/! kŒX=.X /! 0 with k, we get the sequence 0! k
id! k! 0: k DEFINITION 9.12. An A-algebra B is flat if M! N injective H) B ˝A M! B ˝A N injective. It is faithfully flat if, in addition, B ˝A M D 0 H) M D 0: Therefore, an A-algebra B is flat if and only if the functor M B ˝A M from A- modules to B-modules is exact. EXAMPLE 9.13. (a) Let S be a multiplicative subset of A. Then S 1A is a flat A-algebra (1.18). (b) Every open immersion is flat (obvious). (c) The composite of two flat maps is flat (obvious). PROPOSITION 9.14. Let A! A0 be a homomorphism of rings. If A! B is flat, then so also is A0! B ˝A A0. PROOF. For any A0-module M,.B ˝A A0/ ˝A0 M'B ˝A.A0 ˝A0 M /'B ˝A M: In other words, tensoring an A0-module M with B ˝A A0 is the same as tensoring M (regarded as an A-module) with B. Therefore it preserves exact sequences. 1The surjectivity of 1 ˝ ˇ is obvious. Let B ˝A N! Q be the cokernel of 1 ˝ ˛. Because.1 ˝ ˇ/ ı.1 ˝ ˛/ D 1 ˝.ˇ ı ˛/ D 0; there is a unique A-linear map f W Q! B ˝A N 00 such that f ı D 1 ˝ ˇ. We shall construct an inverse g to f. Let b 2 B, and let n 2 N. If ˇ.n/ D 0, then n D ˛.n0/ for some n0 2 N 0; hence b ˝ n D b ˝ ˛.n0/, and so.b ˝ n/ D 0. It follows by linearity that.b ˝ n1/ D.
b ˝ n2/ if ˇ.n1/ D ˇ.n2/, and so the A-bilinear map B N! Q;.b; n/ 7!.b ˝ n/ factors through B N 00. It therefore defines an A-linear map gW B ˝A N 00! Q. To show that f and g are inverse, it suffices to check that g ı f D idQ on elements of the form.b ˝ n/ and that f ı g D idB˝AN 00 on elements of the form b ˝ ˇ.n/ — both are obvious. 202 9. REGULAR MAPS AND THEIR FIBRES PROPOSITION 9.15. A homomorphism ˛W A! B of rings is flat if and only if, for all maximal ideals n in B, the map A˛1.n/! Bn is flat. PROOF. Let n be a prime ideal of B, and let m D ˛1.n/ — it is a prime ideal in A. If A! B is flat, then so is Am! Am ˝A B'S 1 m B (9.14). The map Bn is flat (9.13a), and so the composite Am! Bn is flat (9.13c). For the converse, let N 0! N be an injective homomorphism of A-modules, and let n be a maximal ideal of B. Then Am ˝A.N 0! N / is injective (9.13). Therefore, the map Bn ˝A.N 0! N /'Bn ˝Am.Am ˝A.N 0! N // is injective, and so the kernel M of B ˝A.N 0! N / has the property that Mn D 0. Let x 2 M, and let a D fb 2 B j bx D 0g. For each maximal ideal n of B, x maps to zero in Mn, and so a contains an element not in n. Hence a D B, and so x D 0. PROPOSITION 9.16. A flat homomorphism 'W A! B is faithfully flat if and only if every maximal ideal m of A is of the form '1.n/ for some maximal
ideal n of B. PROOF. ): Let m be a maximal ideal of A, and let M D A=m; then B ˝A M'B='.m/B: As B ˝A M ¤ 0, we see that '.m/B ¤ B. Therefore '.m/ is contained in a maximal ideal n of B. Now '1.n/ is a proper ideal in A containing m, and hence equals m. (: Let M be a nonzero A-module. Let x be a nonzero element of M, and let a D ann.x/ defD fa 2 A j ax D 0g. Then a is an ideal in A, and M 0 defD Ax'A=a. Moreover, B ˝A M 0'B='.a/ B and, because A! B is flat, B ˝A M 0 is a submodule of B ˝A M. Because a is proper, it is contained in a maximal ideal m of A, and therefore '.a/ '.m/ n for some maximal ideal n of A. Hence '.a/ B n ¤ B, and so B ˝A M B ˝A M 0 ¤ 0. COROLLARY 9.17. A flat local homomorphism A! B of local rings is faithfully flat. PROOF. Let m and n be the (unique) maximal ideals of A and B. By hypothesis, nc D m, and so the statement follows from the proposition. Properties of flat maps LEMMA 9.18. Let B be an A-algebra, and let p be a prime ideal of A. The prime ideals of B contracting to p are in natural one-to-one correspondence with the prime ideals of B ˝A.p/. PROOF. Let S D A X p. Then.p/ D S 1.A=p/. Therefore we obtain B ˝A.p/ from B by first passing to B=pB and then making the elements of A not in p act invertibly. After the first step, we are left with the prime ideals q of B such that qc p, and after the second step only with those such that qc \ S D ;, i.e., such that qc D p. PROPOSITION 9.19. Let B be a faithfully flat A-algebra
. Every prime ideal p of A is of the form qc for some prime ideal q of B. c. Flat maps and their fibres 203 PROOF. The ring B ˝A.p/ is nonzero, because.p/ ¤ 0 and A! B is faithfully flat, and so it has a prime (even maximal) ideal q. For this ideal, qc D p. SUMMARY 9.20. A flat homomorphism 'W A! B is faithfully flat if the image of spec.'/W spec.B/! spec.A/ includes all maximal ideals of A, in which case it includes all prime ideals of A. PROPOSITION 9.21 (GOING-DOWN THEOREM FOR FLAT MAPS). Let A! B be a flat homomorphism. Let p p0 be prime ideals in A, and let q be a prime ideal in B such that qc D p. Then q contains a prime ideal q0 such that q0c D p0: B A q q0 p p0: PROOF. Because A! B is flat, the homomorphism Ap! Bq is flat, and because pAp D.qBq/c, it is faithfully flat (9.16). The ideal p0Ap is prime (1.14), and so there exists a prime ideal of Bq lying over p0Ap (by 9.19). The contraction of this ideal to B is contained in q and contracts to p0 in A. DEFINITION 9.22. A regular map 'W W! V of algebraic varieties is flat if, for all P 2 W, the map W;P is flat, and it is faithfully flat if it is flat and surjective. OV;'.P /! O PROPOSITION 9.23. A regular map 'W W! V of affine algebraic varieties is flat (resp. faithfully flat) if and only if the map f 7! f ı 'W kŒV! kŒW is flat (resp. faithfully flat). PROOF. Apply (9.15) and (9.16). PROPOSITION 9.24. Let 'W W! V be a flat map of affine algebraic varieties. Let S S 0 be closed irreducible subsets of V, and let T be a closed irreducible subset of W such that '.
T / is a dense subset of S. Then there exists a closed irreducible subset T 0 of W containing T and such that '.T 0/ is a dense subset of S 0. PROOF. Let p D I.S/, p0 D I.S 0/, and q D I.T /. Then p p0 because S S 0. Moreover '! S is dominant and so the map kŒS D kŒV =p! kŒT =q is injective. qc D p because T According to (9.21), there exists a prime ideal q0 in kŒW contained in q and such that q0c D p0. Now V.q0/ has the required properties. THEOREM 9.25 (GENERIC FLATNESS). For every regular map 'W W! V of irreducible '! U is algebraic varieties, there exists a nonempty open subset U of V such that '1.U / faithfully flat. PROOF. We may assume that W and V are affine, say, V D Spm.A/ and W D Spm.B/. Let F be the field of fractions of A. We regard B as a subring of F ˝A B. As F ˝A B is a finitely generated F -algebra, the Noether normalization theorem (2.45) shows that there exist elements x1; : : : ; xm of F ˝A B such that F Œx1; : : : ; xm is a polynomial ring over F and F ˝A B is a finite F Œx1; : : : ; xm-algebra. After multiplying each xi by an element of A, we may suppose that it lies in B. Let b1; : : : ; bn generate B as an Aalgebra. Each bi satisfies a monic polynomial equation with coefficients in F Œx1; : : : ; xm. 204 9. REGULAR MAPS AND THEIR FIBRES Let a 2 A be a common denominator for the coefficients of these polynomials. Then each bi is integral over Aa. As the bi generate Ba as an Aa-algebra, this shows that Ba is a finite AaŒx1; : : : ; xm-algebra (1.36
). Therefore, after replacing A with Aa and B with Ba, we may suppose that B is a finite AŒx1; : : : ; xm-algebra. injective B finite F ˝A B finite E ˝AŒx1;:::;xm B finite AŒx1; : : : ; xm F Œx1; : : : ; xm E defD F.x1; : : : ; xm/ A F: Let E D F.x1; : : : ; xm/ be the field of fractions of AŒx1; : : : ; xm, and let b1; : : : ; br be elements of B that form a basis for E ˝AŒx1;:::;xm B as an E-vector space. Each element of B can be expressed as a linear combination of the bi with coefficients in E. Let q be a common denominator for the coefficients arising from a set of generators for B as an AŒx1; : : : ; xm-module. Then b1; : : : ; br generate Bq as an AŒx1; : : : ; xmq-module. In other words, the map.c1; : : : ; cr / 7! P ci bi W AŒx1; : : : ; xmr q! Bq () is surjective. This map becomes an isomorphism when tensored with E over AŒx1; : : : ; xmq, which implies that each element of its kernel is killed by a nonzero element of AŒx1; : : : ; xmq and so is zero (because AŒx1; : : : ; xnq is an integral domain). Hence the map () is an isomorphism, and so Bq is free of finite rank over AŒx1; : : : ; xmq. Let a be some nonzero coefficient of the polynomial q, and consider the maps Aa! AaŒx1; : : : ; xm! AaŒx1; : : : ; xmq! Baq: The first and third arrows realize their targets as nonzero free modules over their sources, and so are faithfully flat. The middle arrow is flat by
(9.13). Let m be a maximal ideal in Aa. Then mAaŒx1; : : : ; xm does not contain the polynomial q because the coefficient a of q is invertible in Aa. Hence mAaŒx1; : : : ; xmq is a proper ideal of AaŒx1; : : : ; xmq, and so the map Aa! AaŒx1; : : : ; xmq is faithfully flat (apply 9.16). This completes the proof. LEMMA 9.26. Let V be an algebraic variety. A constructible subset C of V is closed if it has the following property: let Z be a closed irreducible subset of V ; if Z \ C contains a dense open subset of Z, then Z C. PROOF. Let Z be an irreducible component of NC. Then Z \ C is constructible and it is dense in Z, and so it contains a nonempty open subset U of Z (9.6). Hence Z C. THEOREM 9.27. A flat map 'W W! V of algebraic varieties is open. PROOF. Let U be an open subset of W. Then '.U / is constructible (9.7) and the goingdown theorem (9.21) implies that V X '.U / satisfies the hypotheses of the lemma. Therefore V X '.U / is closed. COROLLARY 9.28. Let 'W W! V be a regular map of irreducible algebraic varieties. Then there exists a dense open subset U of W such that '.U / is open, U D '1.'U /, and U '! '.U / is flat. c. Flat maps and their fibres 205 '! U PROOF. According to 9.25, there exists a dense open subset U of V such that '1.U / is flat. In particular, '.'1.U // is open in V (9.27). Note that '1.'.'1.U // D '1.U /. Let U 0 D '1.U /. Then U 0 is a dense open subset of W, '.U 0/ is open, U 0 D '1.'U 0/, and U 0 '! '.U 0/ is flat. Fibres and flatness The notion of
flatness allows us to sharpen our earlier results. PROPOSITION 9.29. Let 'W W! V be a dominant map of irreducible algebraic varieties. Let P 2 '.W /. Then dim '1.P / dim.W / dim.V /; (38) and equality holds if'is flat. PROOF. The inequality was proved in 9.9. If'is flat, then we shall prove (more precisely) that, if Z is an irreducible component of '1.P /, then dim.Z/ D dim.W / dim.V /: After replacing V with an open neighbourhood of P and W with an open subset intersecting Z, we may suppose that both V and W are affine. Let V V1 Vm D fP g be a maximal chain of distinct irreducible closed subsets of V (so m D dim.V /). Now '.Z/ D fP g, and so (see 9.24) there exists a chain of irreducible closed subsets W W1 Wm D Z such that '.Wi / is a dense subset of Vi. Let Z Z1 Zn be a maximal chain of distinct irreducible closed subsets of V (so n D dim.Z/). The existence of the chain W W1 Wm Z1 Zn shows that dim.W / m C n D dim.V / C dim.Z/: Together with (38), this implies that we have equality. PROPOSITION 9.30. Let 'W W! V be a dominant map of irreducible algebraic varieties. Let P 2 '.W /. Then dim '1.P / dim.W / dim.V /: There exists a dense open subset U of W such that '.U / is open in V, U D '1.'.U //, and equality holds for all P 2 '.U /. PROOF. Let U be an open subset of W as in 9.28. 206 9. REGULAR MAPS AND THEIR FIBRES PROPOSITION 9.31. Let 'W W! V be a dominant map of irreducible varieties. Let S be a closed irreducible subset of V, and let T be an irreducible component of '1.S/ such that '.T / is dense in S. Then dim.T / dim.
S / C dim.W / dim.V /; and equality holds if'is flat. PROOF. The inequality can be proved by a similar argument to that in 9.9 — see, for example, Hochschild 1981, X, Theorem 2.1.2 The equality can be deduced by the same argument as in 9.29. PROPOSITION 9.32. Let 'W W! V be a dominant map of irreducible varieties. There exists '! '.U / a nonempty open subset U of W such that '.U / is open, U D '1.'U /, and U is flat. If S is a closed irreducible subset of V meeting '.U /, and T is an irreducible component of '1.S/ meeting U, then dim.T / D dim.S / C dim.W / dim.V /: PROOF. Let U be an open subset of W as in 9.28. FINITE MAPS PROPOSITION 9.33. Let V be an irreducible algebraic variety. A finite map 'W W! V is flat if and only if is independent of P 2 V. Q7!P X dimk O Q=mP Q O PROOF. It suffices to prove this with V affine, in which case it follows from CA 12.6 (equivalence of (d) and (e)). The integer dimk O Q is the multiplicity of Q in its fibre. The theorem says that a finite map is flat if and only if the number of points in each fibre (counting multiplicities) is constant. Q=mP O For example, let V be the subvariety of A nC1 defined by an equation X m C a1X m1 C C am D 0; ai 2 kŒT1; : : : ; Tn and let 'W V! A set of points.P; c/ with c a root of the polynomial n be the projection map (see p. 51). The fibre over a point P of A n is the X m C a1.P /X m1 C C am.P / D 0: The multiplicity of.P; c/ in its fibre is the multiplicity of c as a root of the polynomial. Therefore P Q D m for every P, and so the map '
is flat. Q=mP Q7!P dimk O O 2Hochschild, Gerhard P., Basic theory of algebraic groups and Lie algebras. Springer, 1981. c. Flat maps and their fibres 207 Criteria for flatness THEOREM 9.34. Let 'W A! B be a local homomorphism of noetherian local rings, and let m be the maximal ideal of A. If A is regular, B is Cohen-Macaulay, and dim.B/ D dim.A/ C dim.B=mB/; then'is flat. PROOF. See Matsumura 1986, 23.1.3 9.35. We don’t define the notion of being Cohen-Macaulay here (see ibid. p. 134), but merely list some of its properties. (a) A noetherian ring A is Cohen-Macaulay if and only if Am is Cohen-Macaulay for every maximal ideal m of A (this is part of the definition). (b) Zero-dimensional and reduced one-dimensional noetherian rings are Cohen-Macaulay (ibid. p. 139). (c) Regular noetherian rings are Cohen-Macaulay (ibid. p. 137). (d) Let 'W A! B be a flat local homomorphism of noetherian local rings, and let m be the maximal ideal of A. Then B is Cohen-Macaulay if and only if both A and B=mB are Cohen-Macaulay (ibid. p. 181). PROPOSITION 9.36. Let 'W A! B be a finite homomorphism noetherian rings with A regular. Then'is flat if and only if B is Cohen-Macaulay. PROOF. Note that B=mB/ is zero-dimensional,4 hence Cohen-Macaulay, for every maximal ideal m of A (9.35b), and that ht.n/ D ht.nc/ for every maximal ideal n of B. If'is flat, then B is Cohen-Macaulay by (9.35d). Conversely, if B is Cohen-Macaulay, then'is flat by (9.34). EXAMPLE 9.37. Let A be a finite kŒX1; : : : ; Xn-algebra (cf.
2.45). The map kŒX1; : : : ; Xn! A is flat if and only if A is Cohen-Macaulay. An algebraic variety V is said to be Cohen-Macaulay if V;P is Cohen-Macaulay for all P 2 V. An affine algebraic variety V is Cohen-Macaulay if and only if kŒV is Cohen-Macaulay (9.35a). A nonsingular variety is Cohen-Macaulay (9.35c). O THEOREM 9.38. Let V and W be algebraic varieties with V nonsingular and W CohenMacaulay. A regular map 'W W! V is flat if and only if dim '1.P / D dim W dim V for all P 2 V. PROOF. Immediate consequence of (9.34). (39) 3Matsumura, Hideyuki, Commutative ring theory. Cambridge University Press, Cambridge, 1986. 4Note that C defD B=mB D B ˝A A=m is a finite k-algebra. Therefore it has only finitely many maximal ideals. Every prime ideal in C is an intersection of maximal ideals (2.18), but a prime ideal can equal a finite intersection of ideals only if it equals one of the ideals. 208 9. REGULAR MAPS AND THEIR FIBRES ASIDE 9.39. The theorem fails with “nonsingular” weakened to “normal”. Let Z=2Z act on W defD A 2 by.x; y/ 7!.x; y/. The quotient of W by this action is the quadric cone V A 3 defined by T V D U 2. The quotient map 'W W! V is.x; y/ 7!.t; u; v/ D.x2; xy; y2/. The variety W is nonsingular, and V is normal because kŒV D kŒX; Y G (cf. CA 23.12). Moreover'is finite, and so its fibres have constant dimension 0, but it is not flat because X Q7!P dimk O Q=mP Q D O 3 2 if P D.0; 0; 0/ otherwise (see 9.33). See mo117043. d. Lines on surfaces As an application of
some of the above results, we consider the problem of describing the set 3. To avoid possible problems, we assume for the rest of lines on a surface of degree m in P of this chapter that k has characteristic zero. We first need a way of describing lines in P 3 as being one-dimensional subspaces in k4, and lines in P 3. Recall that we can associate with each n an affine cone over QV in knC1. This allows us to think of points projective variety V P 3 as being two-dimensional in P subspaces in k4. To such a subspace W k4, we can attach a one-dimensional subspace V2 W in V2 k4 k6, that is, to each line L in P 5. Not 3 should form a every point in P 3 corresponds to choosing a four-dimensional set. (Fix two planes in P point on each of the planes.) We shall show that there is natural one-to-one correspondence 5. Rather between the set of lines in P than using exterior algebras, I shall usually give the old-fashioned proofs. 5 should be of the form p.L/ — heuristically, the lines in P 3 and the set of points on a certain hyperspace ˘ P 3, we can attach point p.L/ in P 3; giving a line in P Let L be a line in P 3 and let x D.x0 W x1 W x2 W x3/ and y D.y0 W y1 W y2 W y3/ be distinct points on L. Then p.L/ D.p01 W p02 W p03 W p12 W p13 W p23/ 2 P 5; pij defD ˇ ˇ ˇ ˇ xi xj yi yj ˇ ˇ ˇ ˇ ; depends only on L. The pij are called the Pl¨ucker coordinates of L, after Pl¨ucker (18011868). In terms of exterior algebras, write e0, e1, e2, e3 for the canonical basis for k4, so that x, regarded as a point of k4 is P xi ei, and y D P yi ei ; then V2 k4 is a 6-dimensional vector space with basis ei ^ej, 0 i < j 3, and
x^y D P pij ei ^ej with pij given by the above formula. We define pij for all i; j, 0 i; j 3 by the same formula — thus pij D pj i. LEMMA 9.40. The line L can be recovered from p.L/ as follows: L D f.P j aj p1j W P j aj p0j W P j aj p2j W P PROOF. Let QL be the cone over L in k4 — it is a two-dimensional subspace of k4 — and let x D.x0; x1; x2; x3/ and y D.y0; y1; y2; y3/ be two linearly independent vectors in QL. Then QL D ff.y/x f.x/y j f W k4! k linearg: j aj p3j / j.a0 W a1 W a2 W a3/ 2 P 3g: Write f D P aj Xj ; then f.y/x f.x/y D.P aj p0j ; P aj p1j ; P aj p2j ; P aj p3j /: d. Lines on surfaces 209 LEMMA 9.41. The point p.L/ lies on the quadric ˘ P 5 defined by the equation X01X23 X02X13 C X03X12 D 0: PROOF. This can be verified by direct calculation, or by using that x0 x1 x2 x3 y0 y1 y2 y3 x0 x1 x2 x3 y0 y1 y2 y3.p01p23 p02p13 C p03p12/ (expansion in terms of 2 2 minors). LEMMA 9.42. Every point of ˘ is of the form p.L/ for a unique line L. PROOF. Assume p03 ¤ 0; then the line through the points.0 W p01 W p02 W p03/ and.p03 W p13 W p23 W 0/ has Pl¨ucker coordinates.p01p03 W p02p03 W p2 03 W p01p23 p02p13 „ … ƒ‚ p03p12 W p03p13 W p03
p23/ D.p01 W p02 W p03 W p12 W p13 W p23/: A similar construction works when one of the other coordinates is nonzero, and this way we get inverse maps. Thus we have a canonical one-to-one correspondence flines in P 3g $ fpoints on ˘ gI 3 with the points of an algebraic variety. We that is, we have identified the set of lines in P may now use the methods of algebraic geometry to study the set. (This is a special case of the Grassmannians discussed in 6.) We next consider the set of homogeneous polynomials of degree m in 4 variables, F.X0; X1; X2; X3/ D X ai0i1i2i3X i0 0 : : : X i3 3 : i0Ci1Ci2Ci3Dm LEMMA 9.43. The set of homogeneous polynomials of degree m in 4 variables is a vector space of dimension 3Cm m PROOF. See the footnote p. 141. Let D 3Cm m 1 D.mC1/.mC2/.mC3/ as the projective space attached to the vector space of homogeneous polynomials of degree m in 4 variables (p. 145). Then we have a surjective map 1, and regard P 6! fsurfaces of degree m in P 3g; P.: : : W ai0i1i2i3 W : : :/ 7! V.F /; F D X ai0i1i2i3X i0 0 X i1 1 X i2 2 X i3 3 : The map is not quite injective — for example, X 2Y and XY 2 define the same surface — as being (possibly but nevertheless, we can (somewhat loosely) think of the points of P degenerate) surfaces of degree m in P 3. Let m ˘ P P 5 P be the set of pairs.L; F / consisting of a line L in P 3 lying on the surface F.X0; X1; X2; X3/ D 0. 210 9. REGULAR MAPS AND THEIR FIBRES THEOREM 9.44. The set m is an irreducible closed subset of ˘ P projective variety. The dimension
of m is m.mC1/.mC5/ C 3. 6 ; it is therefore a EXAMPLE 9.45. For m D 1; m is the set of pairs consisting of a plane in P 3 and a line on the plane. The theorem says that the dimension of 1 is 5. Since there are 13 planes in P 3, and each has 12 lines on it, this seems to be correct. PROOF. We first show that m is closed. Let p.L/ D.p01 W p02 W : : :/ F D X ai0i1i2i3X i0 0 X i3 3 : From 9.40 we see that L lies on the surface F.X0; X1; X2; X3/ D 0 if and only if F.P bj p0j W P bj p1j W P bj p2j W P bj p3j / D 0, all.b0; : : : ; b3/ 2 k4: Expand this out as a polynomial in the bj with coefficients polynomials in the ai0i1i2i3 and pij. Then F.:::/ D 0 for all b 2 k4 if and only if the coefficients of the polynomial are all zero. But each coefficient is of the form P.: : : ; ai0i1i2i3; : : : I p01; p02 W : : :/ with P homogeneous separately in the a’s and p’s, and so the set is closed in ˘ P the discussion in 6.51). (cf. It remains to compute the dimension of m. We shall apply Proposition 9.11 to the projection map.L; F / m ˘ P L'˘: For L 2 ˘, '1.L/ consists of the homogeneous polynomials of degree m such that L V.F / (taken up to nonzero scalars). After a change of coordinates, we can assume that L is the line X0 D 0 X1 D 0; i.e., L D f.0; 0; ; /g. Then L lies on F.X0; X1; X2; X3/ D 0 if and only if X0 or X1 occurs in each nonzero monomial term in F,
i.e., F 2 '1.L/ ” ai0i1i2i3 D 0 whenever i0 D 0 D i1: Thus '1.L/ is a linear subspace of P ; in particular, it is irreducible. We now compute its dimension. Recall that F has C 1 coefficients altogether; the number with i0 D 0 D i1 is m C 1, and so '1.L/ has dimension.m C 1/.m C 2/.m C 3/ 6 1.m C 1/ D m.m C 1/.m C 5/ 6 1: We can now deduce from 9.11 that m is irreducible and that dim.m/ D dim.˘ / C dim.'1.L// D m.m C 1/.m C 5/ 6 C 3; as claimed. d. Lines on surfaces 211 Now consider the other projection. By definition 1.F / D fL j L lies on V.F /g: EXAMPLE 9.46. Let m D 1. Then D 3 and dim 1 D 5. The projection W 1! P 3 is surjective (every plane contains at least one line), and (9.9) tells us that dim 1.F / 2. In fact of course, the lines on any plane form a 2-dimensional family, and so 1.F / D 2 for all F. THEOREM 9.47. When m > 3, the surfaces of degree m containing no line correspond to an open subset of P. PROOF. We have dim m dim P D m.m C 1/.m C 5/ 6 C 3.m C 1/.m C 2/.m C 3/ 6 C 1 D 4.m C 1/: Therefore, if m > 3, then dim m < dim P P. This proves the claim., and so.m/ is a proper closed subvariety of We now look at the case m D 2. Here dim m D 10, and D 9, which suggests that should be surjective and that its fibres should all have dimension 1. We shall see that this is correct. A quadric is said to be nondegenerate if it is defined by an irreducible polynomial of degree 2. After a change of variables, any nondegenerate quadric will be defined by an equation This is just the image of the
Segre mapping (see 6.26) XW D Y Z:.a0 W a1/,.b0 W b1/ 7!.a0b0 W a0b1 W a1b0 W a1b1/ W P 1 P 1! P 3: There are two obvious families of lines on P vertical family; each is parametrized by P two families of lines on the quadric: 1 P 1, namely, the horizontal family and the 1, and so is called a pencil of lines. They map to t0X D t1Z t0Y D t1W and t0X D t1Y t0Z D t1W: Since a degenerate quadric is a surface or a union of two surfaces, we see that every quadric surface contains a line, that is, that W 2! P 9 is surjective. Thus (9.9) tells us that all the fibres have dimension 1, and the set where the dimension is > 1 is a proper closed subset. In fact the dimension of the fibre is > 1 exactly on the set of reducible F ’s, which we know to be closed (this was a homework problem in the original course). It follows from the above discussion that if F is nondegenerate, then 1.F / is isomorphic to the disjoint union of two lines, 1.F / P 1. Classically, one defines a regulus to be a nondegenerate quadric surface together with a choice of a pencil of lines. One can show that the set of reguli is, in a natural way, an algebraic variety R, and that, over the set of nondegenerate quadrics, factors into the composite of two regular maps: 2 1.S / D pairs,.F; L/ with L on set of reguli; D set of nondegenerate quadrics. 212 9. REGULAR MAPS AND THEIR FIBRES 1/, The fibres of the top map are connected, and of dimension 1 (they are all isomorphic to P and the second map is finite and two-to-one. Factorizations of this type occur quite generally (see the Stein factorization theorem, 8.64). We now look at the case m D 3. Here dim 3 D 19; D 19 W we have a map W 3! P THEOREM 9.48. The set of cubic surfaces containing exactly
27 lines corresponds to an 19; the remaining surfaces either contain an infinite number of lines or a open subset of P nonzero finite number 27. 19: EXAMPLE 9.49. (a) Consider the Fermat surface C X 3 2 Let be a primitive cube root of one. There are the following lines on the surface, 0 i; j 2: X 3 0 X0 C i X1 D 0 X2 C j X3 D 0 X0 C i X2 D 0 X1 C j X3 D 0 X0 C i X3 D 0 X1 C j X2 D 0: There are three sets, each with nine lines, for a total of 27 lines. (b) Consider the surface X1X2X3 D X 3 0 : In this case, there are exactly three lines. To see this, look first in the affine space where X0 ¤ 0 — here we can take the equation to be X1X2X3 D 1. A line in A 3 can be written in parametric form Xi D ai t C bi, but a direct inspection shows that no such line lies on the surface. Now look where X0 D 0, that is, in the plane at infinity. The intersection of the surface with this plane is given by X1X2X3 D 0 (homogeneous coordinates), which is the union of three lines, namely, X1 D 0; X2 D 0; X3 D 0: Therefore, the surface contains exactly three lines. (c) Consider the surface X 3 1 C X 3 2 D 0: Here there is a pencil of lines: t0X1 D t1X0 t0X2 D t1X0: (In the affine space where X0 ¤ 0, the equation is X 3 C Y 3 D 0, which contains the line X D t, Y D t, all t:/ We now discuss the proof of Theorem 9.48. If W 3! P 19 were not surjective, then 19, and the nonempty fibres would all have.3/ would be a proper closed subvariety of P dimension 1 (by 9.9), which contradicts two of the above examples. Therefore the map is 19 where the fibres have dimension 0; outside surjective, and there is an open subset U of P U, the fibres have dimension > 0. Given that every cubic surface has at least one line, it is
closed subset has been removed from each (3.36). The main problem of birational algebraic geometry is to classify algebraic varieties up to birational equivalence by finding a particularly good representative in each equivalence class. For curves this is easy: in each birational equivalence class there is exactly one nonsingular projective curve (up to isomorphism). More precisely, the functor V k.V / is a contravariant equivalence from the category of nonsingular projective algebraic curves over k and dominant maps to the category of fields finitely generated and of transcendence degree 1 over k. For surfaces, the problem is already much more difficult because many surfaces, even projective and nonsingular, will have the same function field. For example, every blow-up of a point on a surface produces a birationally equivalent surface. A nonsingular projective surface is said to be minimal if it cannot be obtained from another such surface by blowing up. The main theorem for surfaces (Enriques 1914, Kodaira 1966) says that a birational equivalence class contains either (a) a unique minimal surface, or (b) a surface of the form C P 1 for a unique nonsingular projective curve C. In higher dimensions, the problem becomes very involved, although much progress has been made — see Wikipedia: MINIMAL MODEL PROGRAM. Exercises 9-1. Let G be a connected group variety, and consider an action of G on a variety V, i.e., a regular map G V! V such that.gg0/v D g.g0v/ for all g; g0 2 G and v 2 V. Show that 214 9. REGULAR MAPS AND THEIR FIBRES each orbit O D Gv of G is open in its closure NO, and that NO X O is a union of orbits of strictly lower dimension. Deduce that each orbit is a nonsingular subvariety of V, and that there exists at least one closed orbit. 9-2. Let G D GL2 D V, and let G act on V by conjugation. According to the theory of Jordan canonical forms, the orbits are of three types: (a) Characteristic polynomial X 2 C aX C b; distinct roots. (b) Characteristic polynomial X 2 C aX C b; minimal polynomial the same; repeated roots. (c) Characteristic polyn
omial X 2 C aX C b D.X ˛/2; minimal polynomial X ˛. For each type, find the dimension of the orbit, the equations defining it (as a subvariety of V ), the closure of the orbit, and which other orbits are contained in the closure. (You may assume, if you wish, that the characteristic is zero. Also, you may assume the following (fairly difficult) result: for any closed subgroup H of an group variety G, G=H has a natural structure of an algebraic variety with the following properties: G! G=H is regular, and a map G=H! V is regular if the composite G! G=H! V is regular; dim G=H D dim G dim H.) [The enthusiasts may wish to carry out the analysis for GLn.] 9-3. Find 3d 2 lines on the Fermat projective surface ; d 3;.p; d / D 1; p the characteristic. 9-4. (a) Let 'W W! V be a quasi-finite dominant regular map of irreducible varieties. Show that there are open subsets U 0 and U of W and V such that '.U 0/ U and 'W U 0! U is finite. (b) Let G be a group variety acting transitively on irreducible varieties W and V, and let 'W W! V be G-equivariant regular map satisfying the hypotheses in (a). Then'is finite, and hence proper. Solutions to the exercises 1-1 Use induction on n. For n D 1, use that a nonzero polynomial in one variable has only finitely many roots (which follows from unique factorization, for example). Now suppose n > 1 and write f D P gi X i n with each gi 2 kŒX1; : : : ; Xn1. If f is not the zero polynomial, then some gi is not the zero polynomial. Therefore, by induction, there exist.a1; : : : ; an1/ 2 kn1 such that f.a1; : : : ; an1; Xn/ is not the zero polynomial. Now, by the degree-one case, there exists a b such that f.a1; : : : ; an1; b/ ¤ 0. 1-2.X
C 2Y; Z/; Gaussian elimination (to reduce the matrix of coefficients to row echelon form);.1/, unless the characteristic of k is 2, in which case the ideal is.X C 1; Z C 1/. 2-1 W D Y -axis, and so I.W / D.X/. Clearly,.X 2; XY 2/.X/ rad.X 2; XY 2/ and rad..X// D.X/. On taking radicals, we find that.X/ D rad.X 2; XY 2/. 2-2 The d d minors of a matrix are polynomials in the entries of the matrix, and the set of matrices with rank r is the set where all.r C 1/.r C 1/ minors are zero. 2-3 Clearly V D V.Xn X n 1 ; : : : ; X2 X 2 1 /. The map Xi 7! T i W kŒX1; : : : ; Xn! kŒT 1! V.] induces an isomorphism kŒV! kŒT. [Hence t 7!.t; : : : ; t n/ is an isomorphism of affine varieties A 2-4 We use that the prime ideals are in one-to-one correspondence with the irreducible closed subsets Z of A 2. For such a set, 0 dim Z 2. Case dim Z D 2. Then Z D A Case dim Z D 1. Then Z ¤ A 2, and the corresponding ideal is.0/. 2, and so I.Z/ contains a nonzero polynomial f.X; Y /. If I.Z/ ¤.f /, then dim Z D 0 by (2.64, 2.62). Hence I.Z/ D.f /. Case dim Z D 0. Then Z is a point.a; b/ (see 2.63), and so I.Z/ D.X a; Y b/. 2-6 The statement Homkalgebras.A ˝ Q k/ ¤ ; can be interpreted as saying that a certain set of polynomials has a zero in k.6 If the polynomials have a common zero in C, then the ideal they generate in CŒX1; : : : does not contain 1. A fortiori, the
ideal they generate in QŒX1; : : : does not contain 1, and so the Nullstellensatz (2.11) implies that the polynomials have a common zero in k. Q k; B ˝ 2-7 Regard HomA.M; N / as an affine space over k; the elements not isomorphisms are the zeros of a polynomial; because M and N become isomorphic over kal, the polynomial is not identically zero; therefore it has a nonzero in k (Exercise 1-1). 6Choose bases for A and B as Q-vector spaces. Now a linear map from A to B is given by a matrix M. The condition on the coefficients of the matix for the map to be a homomorphism of algebras is polynomial. 215 216 SOLUTIONS TO THE EXERCISES 1! A 3-1 A map ˛W A 1 is continuous for the Zariski topology if the inverse images of finite sets are finite, whereas it is regular only if it is given by a polynomial P 2 kŒT, so it is easy to give examples, e.g., any map ˛ such that ˛1.point/ is finite but arbitrarily large. 3-3 The image omits the points on the Y -axis except for the origin. The complement of the image is not dense, and so it is not open, but any polynomial zero on it is also zero at.0; 0/, and so it not closed. 3-4 Let i be an element of k with square 1. The map.x; y/ 7!.x C iy; x iy/ from the circle to the hyperbola has inverse.x; y/ 7!..x C y/=2;.x y/=2i/. The k-algebra kŒX; Y =.XY 1/'kŒX; X 1, which is not isomorphic to kŒX (too many units). 3-5 No, because both C1 and 1 map to.0; 0/. The map on rings is kŒx; y! kŒT ; x 7! T 2 1; y 7! T.T 2 1/; which is not surjective (T is not in the image). 1. Then f
jU0 D P.X/ 2 kŒX, where X is the regular function 5-1 Let f be regular on P.a0W a1/ 7! a1=a0W U0! k, and f jU1 D Q.Y / 2 kŒY, where Y is.a0W a1/ 7! a0=a1. On U0 \ U1, X and Y are reciprocal functions. Thus P.X / and Q.1=X/ define the same function on U0 \ U1 D A 1 X f0g. This implies that they are equal in k.X/, and must both be constant. V / D Q.Vi ; Vi / — to give a regular function on FVi is the same 5-2 Note that.V; as to give a regular function on each Vi (this is the “obvious” ringed space structure). Thus, if V is affine, it must equal Specm.Q Ai /, where Ai D.Vi ; Vi /, and so V D FSpecm.Ai / (use the description of the ideals in A B on in Section 1a). Etc.. O O O 5-5 Let H be an algebraic subgroup of G. By definition, H is locally closed, i.e., open in its Zariski closure NH. Assume first that H is connected. Then NH is a connected algebraic group, and it is a disjoint union of the cosets of H. It follows that H D NH. In the general case, H is a finite disjoint union of its connected components; as one component is closed, they all are. 4-1 (b) The singular points are the common solutions to 8 < : 4X 3 2XY 2 D 0 4Y 3 2X 2Y D 0 X 4 C Y 4 X 2Y 2 D 0: H) X D 0 or Y 2 D 2X 2 H) Y D 0 or X 2 D 2Y 2 Thus, only.0; 0/ is singular, and the variety is its own tangent cone. 4-2 Directly from the definition of the tangent space, we have that Ta.V \ H / Ta.V / \ Ta.H /. As dim Ta.V \ H / dim V \ H D dim V 1 D dim Ta.V / \ Ta.H /;
we must have equalities everywhere, which proves that a is nonsingular on V \ H. (In particular, it can’t lie on more than one irreducible component.) The surface Y 2 D X 2 C Z is smooth, but its intersection with the X-Y plane is singular. No, P needn’t be singular on V \ H if H TP.V / — for example, we could have H V or H could be the tangent line to a curve. SOLUTIONS TO THE EXERCISES 217 4-4 We can assume V and W to affine, say I.V / D a kŒX1; : : : ; Xm I.W / D b kŒXmC1; : : : ; XmCn: If a D.f1; : : : ; fr / and b D.g1; : : : ; gs/, then I.V W / D.f1; : : : ; fr ; g1; : : : ; gs/. Thus, T.a;b/.V W / is defined by the equations.df1/a D 0; : : : ;.dfr /a D 0;.dg1/b D 0; : : : ;.dgs/b D 0; which can obviously be identified with Ta.V / Tb.W /. 4-5 Take C to be the union of the coordinate axes in A irreducible, then this is more difficult... ) n. (Of course, if you want C to be 4-6 A matrix A satisfies the equations if and only if.I C "A/tr J.I C "A/ D I Atr J C J A D 0: Such an A is of the form M N P Q with M; N; P; Q n n-matrices satisfying N tr D N; P tr D P; M tr D Q. The dimension of the space of A’s is therefore n.n C 1/ 2 (for N ) C n.n C 1/ 2 (for P ) C n2 (for M; Q) D 2n2 C n: 4-7 Let C be the curve Y 2 D X 3, and consider the map A 1! C, t 7!.t 2; t 3/. The corresponding map on rings kŒX; Y
=.Y 2/! kŒT is not an isomorphism, but the map on the geometric tangent cones is an isomorphism. 4-8 The singular locus Vsing has codimension 2 in V, and this implies that V is normal. [Idea of the proof: let f 2 k.V / be integral over kŒV, f … kŒV, f D g= h, g; h 2 kŒV ; for any P 2 V.h/ X V.g/, 4-9 No! Let a D.X 2Y /. Then V.a/ is the union of the X and Y axes, and I V.a/ D.XY /. For a D.a; b/, P is not integrally closed, and so P is singular.] O.dX 2Y /a D 2ab.X a/ C a2.Y b/.dXY /a D b.X a/ C a.Y b/. If a ¤ 0 and b D 0, then the equations.dX 2Y /a D a2Y D 0.dXY /a D aY D 0 have the same solutions. 6-1 Let P D.a W b W c/, and assume c ¤ 0. Then the tangent line at / is @F @X P X C @F @Y P Y @F @X P C a c @F @Y P b c Z D 0: 218 SOLUTIONS TO THE EXERCISES Now use that, because F is homogeneous, F.a; b; c/ D 0 H) @F @X P a C @F @Y P C @F @Z P c D 0. (This just says that the tangent plane at.a; b; c/ to the affine cone F.X; Y; Z/ D 0 passes through the origin.) The point at 1 is.0 W 1 W 0/, and the tangent line is Z D 0, the line at 1. [The line at 1 meets the cubic curve at only one point instead of the expected 3, and so the line at 1 “touches” the curve, and the point at 1 is a point of inflexion.] 6-2 The equation defining the conic must be irreducible (otherwise the conic is singular). After a linear change of
variables, the equation will be of the form X 2 C Y 2 D Z2 (this is proved in calculus courses). The equation of the line in aX C bY D cZ, and the rest is easy. [Note that this is a special case of Bezout’s theorem (6.37) because the multiplicity is 2 in case (b).] 6-3 (a) The ring kŒX; Y; Z=.Y X 2; Z X 3/ D kŒx; y; z D kŒx'kŒX; which is an integral domain. Therefore,.Y X 2; Z X 3/ is a radical ideal. (b) The polynomial F D Z XY D.Z X 3/ X.Y X 2/ 2 I.V / and F D ZW XY. If ZW XY D.Y W X 2/f C.ZW 2 X 3/g; then, on equating terms of degree 2, we would find ZW XY D a.Y W X 2/; which is false. 6-4 Let P D.a0W : : : W an/ and Q D.b0W : : : W bn/ be two points of P that the hyperplane LcW Pci Xi D 0 pass through P and not through Q is that n, n 2. The condition Pai ci D 0; Pbi ci ¤ 0: The.n C 1/-tuples.c0; : : : ; cn/ satisfying these conditions form a nonempty open subset of the hyperplane H W P ai Xi D 0 in A nC1. On applying this remark to the pairs.P0; Pi /, we find that the.n C 1/-tuples c D.c0; : : : ; cn/ such that P0 lies on the hyperplane Lc but not P1; : : : ; Pr form a nonempty open subset of H. 6-5 The subset C D f.a W b W c/ j a ¤ 0; b ¤ 0g [ f.1 W 0 W 0/g 2 is not locally closed. Let P D.1 W 0 W 0/. If the set C were locally closed, then P 2 such that U \ C is closed. When we look in of P would have an open neighbourhood U in P
U0, P becomes the origin, and C \ U0 D.A 2 X fX-axisg/ [ foriging. 2 a finite number of The open neighbourhoods U of P are obtained by removing from A curves not passing through P. It is not possible to do this in such a way that U \ C is closed in U (U \ C has dimension 2, and so it can’t be a proper closed subset of U ; we can’t have U \ C D U because any curve containing all nonzero points on X-axis also contains the origin). SOLUTIONS TO THE EXERCISES 219 6-6 Let Pcij Xij D 0 be a hyperplane containing the image of the Segre map. We then have Pcij ai bj D 0 for all a D.a0; : : : ; am/ 2 kmC1 and b D.b0; : : : ; bn/ 2 knC1. In other words, aC bt D 0 for all a 2 kmC1 and b 2 knC1, where C is the matrix.cij /. This equation shows that aC D 0 for all a, and this implies that C D 0. 7-2 Define f.v/ D h.v; Q/ and g.w/ D h.P; w/, and let'D h.f ı p C g ı q/. Then '.v; Q/ D 0 D '.P; w/, and so the rigidity theorem (7.35) implies that'is identically zero. 8-2 For example, consider.A 1 X f1g/! A 1 x7!xn! A 1 for n > 1 an integer prime to the characteristic. The map is obviously quasi-finite, but it is not finite because it corresponds to the map of k-algebras X 7! X nW kŒX! kŒX;.X 1/1 which is not finite (the elements 1=.X 1/i, i 1, are linearly independent over kŒX, and so also over kŒX n). 8-3 Assume that V is separated, and consider two regular maps f; gW Z W. We have to show that the set on which f and g agree is closed in Z. The set where'ı f and'ı g
agree is closed in Z, and it contains the set where f and g agree. Replace Z with the set where'ı f and'ı g agree. Let U be an open affine subset of V, and let Z0 D.' ı f /1.U / D.' ı g/1.U /. Then f.Z0/ and g.Z0/ are contained in '1.U /, which is an open affine subset of W, and is therefore separated. Hence, the subset of Z0 on which f and g agree is closed. This proves the result. [Note that the problem implies the following statement: if 'W W! V is a finite regular map and V is separated, then W is separated.] 8-4 Let V D A n, and let W be the subvariety of A n A 1 defined by the polynomial Qn i D1.X Ti / D 0: The fibre over.t1; : : : ; tn/ 2 A union of the linear subspaces defined by the equations n is the set of roots of Q.X ti /. Thus, Vn D A n; Vn1 is the Ti D Tj ; 1 i; j n; i ¤ j I Vn2 is the union of the linear subspaces defined by the equations Ti D Tj D Tk; 1 i; j; k n; i; j; k distinct, and so on. 9-1 Consider an orbit O D Gv. The map g 7! gvW G! O is regular, and so O contains an open subset U of NO (9.7). If u 2 U, then gu 2 gU, and gU is also a subset of O which is open in NO (because P 7! gP W V! V is an isomorphism). Thus O, regarded as a topological subspace of NO, contains an open neighbourhood of each of its points, and so must be open in NO. We have shown that O is locally closed in V, and so has the structure of a subvariety. From (4.37), we know that it contains at least one nonsingular point P. But then gP is nonsingular, and every point of O is of this form. From set theory, it is clear that NO X O is a union of orbits. Since NO X O is a proper closed
subset of NO, all of its subvarieties must have dimension < dim NO D dim O. Let O be an orbit of lowest dimension. The last statement implies that O D NO. 9-2 An orbit of type (a) is closed, because it is defined by the equations Tr.A/ D a; det.A/ D b; (as a subvariety of V ). It is of dimension 2, because the centralizer of 0 0 An orbit of type (b) is of dimension 2, but is not closed: it is defined by the equations, which has dimension 2. ˛ 0 0 ˇ, ˛ ¤ ˇ, is Tr.A/ D a; det.A/ D b root of X 2 C aX C b. An orbit of type (c) is closed of dimension 0: it is defined by the equation A D ˛ 0 0 ˛. An orbit of type (b) contains an orbit of type (c) in its closure. 9-3 Let be a primitive d th root of 1. Then, for each i; j, 1 i; j d, the following equations define lines on the surface X0 C i X1 D 0 X2 C j X3 D 0 X0 C i X2 D 0 X1 C j X3 D 0 X0 C i X3 D 0 X1 C j X2 D 0: There are three sets of lines, each with d 2 lines, for a total of 3d 2 lines. 9-4 (a) Compare the proof of Theorem 9.9. (b) Use the transitivity, and apply Proposition 8.26. 220 algebra affine, 65 finite, 13 finitely generated, 13 algebraically dependent, 35 algebraically independent, 35 A analytic space, 169 axiom n, 37 separation, 101 base change, 112 basis transcendence, 35 birationally equivalent, 73, 116 boundary, 55 codimension, 55 complete intersection ideal-theoretic, 79 local, 79 set-theoretic, 79 component of a function, 50 cone, 131 affine over a set, 132 content of a polynomial, 24 convergent, 60 Cramer’s rule, 26 curve, 55 elliptic, 38, 130, 135 degree of a hypersurface, 150 of a map, 184 of a projective variety, 152 derivation, 91 desingularization, 193 differential, 87
y) = ˜f2(y). Then there is an evenly covered open neighbourhood U ⊆ X of f (y). Let ˜U be such that ˜f1(y) ∈ ˜U, p( ˜U ) = U and p\| ˜U : ˜U → U is a homeomorphism. Let V = ˜f −1 2 ( ˜U ). We will show that ˜f1 = ˜f2 on V. 1 ( ˜U ) ∩ ˜f −1 Indeed, by construction p\| ˜U ◦ ˜f1\|V = p\| ˜U ◦ ˜f2\|V. Since p\| ˜U is a homeomorphism, it follows that ˜f1\|V = ˜f2\|V. Now we show S is closed. Suppose not. Then there is some y ∈ ¯S \ S. So ˜f1(y) = ˜f2(y). Let U be an evenly covered neighbourhood of f (y). Let p−1(U ) = Uα. Let ˜f1(y) ∈ Uβ and ˜f2(y) ∈ Uγ, where β = γ. Then V = ˜f −1 2 (Uγ) is an open neighbourhood of y, and hence intersects S by deﬁnition of closure. So there is some x ∈ V such that ˜f1(x) = ˜f2(x). But ˜f1(x) ∈ Uβ and ˜f2(x) ∈ Uγ, and hence Uβ and Uγ have a non-trivial intersection. This is a contradiction. So S is closed. 1 (Uβ) ∩ ˜f −1 We just had a uniqueness statement. How about existence? Given a map, is there guarantee that we can lift it to something? Moreover, if I have ﬁxed a “copy” of X I like, can I also lift my map to that copy? We will later come up with a general criterion for when lifts exist. However, it turns out homotopies can always be lifted. Lemma (Homotopy lifting lemma). Let p : ˜X → X be a covering space, H : Y × I → X be a hom
otopy from f0 to f1. Let ˜f0 be a lift of f0. Then there exists a unique homotopy ˜H : Y × I → ˜X such that (i) ˜H( ·, 0) = ˜f0; and (ii) ˜H is a lift of H, i.e. p ◦ ˜H = H. This lemma might be diﬃcult to comprehend at ﬁrst. We can look at the special case where Y = ∗. Then a homotopy is just a path. So the lemma specializes to Lemma (Path lifting lemma). Let p : ˜X → X be a covering space, γ : I → X a path, and ˜x0 ∈ ˜X such that p(˜x0) = x0 = γ(0). Then there exists a unique path ˜γ : I → ˜X such that (i) ˜γ(0) = ˜x0; and (ii) ˜γ is a lift of γ, i.e. p ◦ ˜γ = γ. This is exactly the picture we were drawing before. We just have to start at a point ˜x0, and then everything is determined because locally, everything upstairs in ˜X is just like X. Note that we have already proved uniqueness. So we just need to prove existence. 24 3 Covering spaces II Algebraic Topology In theory, it makes sense to prove homotopy lifting, and path lifting comes immediately as a corollary. However, the proof of homotopy lifting is big and scary. So instead, we will prove path lifting, which is something we can more easily visualize and understand, and then use that to prove homotopy lifting. Proof. Let Observe that (i) 0 ∈ S. S = {s ∈ I : ˜γ exists on [0, s] ⊆ I}. (ii) S is open. If s ∈ S and ˜γ(s) ∈ Vβ ⊆ p−1(U ), we can deﬁne ˜γ on some small neighbourhood of s by ˜γ(t) = (p\|Vβ )−1 ◦ γ(t) (iii) S is closed. If s ∈ S, then pick
an evenly covered neighbourhood U of γ(s). 2, 1] ∩ S = ∅. So S is 2 ∈ S. So (s − ε Suppose γ((s − ε, s)) ⊆ U. So s − ε closed. Since S is both open and closed, and is non-empty, we have S = I. So ˜γ exists. How can we promote this to a proof of the homotopy lifting lemma? At every point y ∈ Y, we know what to do, since we have path lifting. So ˜H(y, · ) is deﬁned. So the thing we have to do is to show that this is continuous. Steps of the proof are (i) Use compactness of I to argue that the proof of path lifting works on small neighbourhoods in Y. (ii) For each y, we pick an open neighbourhood U of y, and deﬁne a good path lifting on U × I. (iii) By uniqueness of lifts, these path liftings agree when they overlap. So we have one big continuous lifting. With the homotopy lifting lemma in our toolkit, we can start to use it to do stuﬀ. So far, we have covering spaces and fundamental groups. We are now going to build a bridge between these two, and show how covering spaces can be used to reﬂect some structures of the fundamental group. At least one payoﬀ of this work is that we are going to exhibit some non-trivial fundamental groups. We have just showed that we are allowed to lift homotopies. However, what we are really interested in is homotopy as paths. The homotopy lifting lemma does not tell us that the lifted homotopy preserves basepoints. This is what we are going to show. Corollary. Suppose γ, γ : I → X are paths x0 x1 and ˜γ, ˜γ : I → ˜X are lifts of γ and γ respectively, both starting at ˜x0 ∈ p−1(x0). If γ γ as paths, then ˜γ and ˜γ are homotopic as paths. In particular, ˜γ(1) = ˜γ(1). 25 3 Covering spaces II Algebraic Topology Note that if we
cover the words “as paths” and just talk about homotopies, then this is just the homotopy lifting lemma. So we can view this as a stronger form of the homotopy lifting lemma. Proof. The homotopy lifting lemma gives us an ˜H, a lift of H with ˜H( ·, 0) = ˜γ. cx0 γ H γ cx1 lift c˜x0 ˜γ ˜H ˜γ c˜x1 In this diagram, we by assumption know the bottom of the ˜H square is ˜γ. To show that this is a path homotopy from ˜γ to ˜γ, we need to show that the other edges are c˜x0, c˜x1 and ˜γ respectively. Now ˜H( ·, 1) is a lift of H( ·, 1) = γ, starting at ˜x0. Since lifts are unique, we must have ˜H( ·, 1) = ˜γ. So this is indeed a homotopy between ˜γ and ˜γ. Now we need to check that this is a homotopy of paths. We know that ˜H(0, · ) is a lift of H(0, · ) = cx0. We are aware of one lift of cx0, namely c˜x0. By uniqueness of lifts, we must have ˜H(0, · ) = c˜x0. Similarly, ˜H(1, · ) = c˜x1. So this is a homotopy of paths. So far, our picture of covering spaces is like this: x0 x1 Except... is it? Is it possible that we have four copies of x0 but just three copies of x1? This is obviously possible if X is not path connected — the component containing x0 and the one containing x1 are completely unrelated. But what if X is path connected? Corollary. If X is a path connected space, x0, x1 ∈ X, then there is a bijection p−1(x0) → p−1(x1). Proof. Let γ : x0 x1 be a path. We want to use this to construct a bijection between each preimage of x0 and each preimage of x1. The obvious thing to do is to use lifts
of the path γ. 26 3 Covering spaces II Algebraic Topology γ x0 x1 Deﬁne a map fγ : p−1(x0) → p−1(x1) that sends ˜x0 to the end point of the unique lift of γ at ˜x0. The inverse map is obtained by replacing γ with γ−1, i.e. fγ−1. To show this is an inverse, suppose we have some lift ˜γ : ˜x0 ˜x1, so that fγ(˜x0) = ˜x1. Now notice that ˜γ−1 is a lift of γ−1 starting at ˜x1 and ending at ˜x0. So fγ−1 (˜x1) = ˜x0. So fγ−1 is an inverse to fγ, and hence fγ is bijective. Deﬁnition (n-sheeted). A covering space p : ˜X → X of a path-connected space X is n-sheeted if \|p−1(x)\| = n for any (and hence all) x ∈ X. Each covering space has a number associated to it, namely the number of sheets. Is there any number we can assign to fundamental groups? Well, the index of a subgroup might be a good candidate. We’ll later see if this is the case. One important property of covering spaces is the following: Lemma. If p : ˜X → X is a covering map and ˜x0 ∈ ˜X, then p∗ : π1( ˜X, ˜x0) → π1(X, x0) is injective. Proof. To show that a group homomorphism p∗ is injective, we have to show that if p∗(x) is trivial, then x must be trivial. Consider a based loop ˜γ in ˜X. We let γ = p ◦ ˜γ. If γ is trivial, i.e. γ cx0 as paths, the homotopy lifting lemma then gives us a homotopy upstairs between ˜γ and c˜x0. So ˜γ is trivial. As we have originally said, our objective is to make our fundamental group act on something. We are almost
there already. Let’s look again at the proof that there is a bijection between p−1(x0) and p−1(x1). What happens if γ is a loop? For any ˜x0 ∈ p−1(x0), we can look at the end point of the lift. This end point may or may not be our original ˜x0. So each loop γ “moves” our ˜x0 to another ˜x 0. However, we are not really interested in paths themselves. We are interested in equivalence classes of paths under homotopy of paths. However, this is ﬁne. If γ is homotopic to γ, then this homotopy can be lifted to get a homotopy between ˜γ and ˜γ. In particular, these have the same end points. So each (based) homotopy class gives a well-deﬁned endpoint. 27 3 Covering spaces II Algebraic Topology ˜x0 ˜x 0 x0 ˜γ γ ˜X p X Now this gives an action of π1(X, x0) on p−1(x0)! Note, however, that this will not be the sort of actions we are familiar with. We usually work with left-actions, where the group acts on the left, but now we will have right-actions, which may confuse you a lot. To see this, we have to consider what happens when we perform two operations one after another, which you shall check yourself. We write this action as ˜x0 · [γ]. When we have an action, we are interested in two things — the orbits, and the stabilizers. This is what the next lemma tells us about. Lemma. Suppose X is path connected and x0 ∈ X. (i) The action of π1(X, x0) on p−1(x0) is transitive if and only if ˜X is path connected. Alternatively, we can say that the orbits of the action correspond to the path components. (ii) The stabilizer of ˜x0 ∈ p−1(x0) is p∗(π1( ˜X, ˜x0)) ⊆ π1(X, x0). (iii) If ˜X is path connected, then
there is a bijection p∗(π1( ˜X, ˜x0))\π1(X, x0) → p−1(x0). Note that p∗(π1( ˜X, ˜x0))\π1(X, x0) is not a quotient, but simply the set of cosets. We write it the “wrong way round” because we have right cosets instead of left cosets. Note that this is great! If we can ﬁnd a covering space p and a point x0 such that p−1(x0) is non-trivial, then we immediately know that π1(X, x0) is non-trivial! Proof. (i) If ˜x0, ˜x 0 ∈ p−1(x0), then since ˜X is path connected, we know that there is some ˜γ : ˜x0 ˜x 0. Then we can project this to γ = p ◦ ˜γ. Then γ is a path from x0 x0, i.e. a loop. Then by the deﬁnition of the action, ˜x0 · [γ] = ˜γ(1) = ˜x 0. (ii) Suppose [γ] ∈ stab(˜x0). Then ˜γ is a loop based at ˜x0. So ˜γ deﬁnes [˜γ] ∈ π1( ˜X, ˜x0) and γ = p ◦ ˜γ. (iii) This follows directly from the orbit-stabilizer theorem. 28 3 Covering spaces II Algebraic Topology We now want to use this to determine that the fundamental group of a space is non-trivial. We can be more ambitious, and try to actually ﬁnd π1(X, x0). In the best possible scenario, we would have π1( ˜X, ˜x0) trivial. Then we have a bijection between π1(X, x0) and p−1(x0). In other words, we want our covering space ˜X to be simply connected. Deﬁnition (Universal cover). A covering map p : ˜X → X is a universal cover
if ˜X is simply connected. We will look into universal covers in depth later and see what they really are. Corollary. If p : ˜X → X is a universal cover, then there is a bijection : π1(X, x0) → p−1(x0). Note that the orbit-stabilizer theorem does not provide a canonical bijection between p−1(x0) and p∗π1( ˜X, ˜x0)\π1(X, x0). To obtain a bijection, we need to pick a starting point ˜x0 ∈ p−1(x0). So the above bijection depends on a choice of ˜x0. 3.2 The fundamental group of the circle and its applica- tions Finally, we can exhibit a non-trivial fundamental group. We are going to consider the space S1 and a universal covering R. R p−1(1) p 1 S1 Then our previous corollary gives Corollary. There is a bijection π1(S1, 1) → p−1(1) = Z. What’s next? We just know that π1(S1, 1) is countably inﬁnite, but can we work out the group structure? We can, in fact, prove a stronger statement: Theorem. The map : π1(S1, 1) → p−1(1) = Z is a group isomorphism. Proof. We know it is a bijection. So we need to check it is a group homomorphism. The idea is to write down representatives for what we think the elements should be. 29 3 Covering spaces II Algebraic Topology ˜u2 u2 2 1 0 −1 −2 R p S1 Let ˜un : I → R be deﬁned by t → nt, and let un = p ◦ ˜un. Since R is simply connected, there is a unique homotopy class between any two points. So for any [γ] ∈ π1(S1, 1), if ˜γ is the lift to R at 0 and ˜γ(1) = n, then ˜γ ˜un as paths. So [γ] = [un]. To show that this has the right group operation, we can easily
see that um · un = ˜um+n, since we are just moving by n + m in both cases. Therefore ([um][un]) = ([um · um]) = m + n = ([um+n]). So is a group isomorphism. What have we done? In general, we might be given a horrible, crazy loop in S1. It would be rather diﬃcult to work with it directly in S1. So we pull it up to the universal covering R. Since R is nice and simply connected, we can easily produce a homotopy that “straightens out” the path. We then project this homotopy down to S1, to get a homotopy from γ to un. It is indeed possible to produce a homotopy directly inside S1 from each loop to some un, but that would be tedious work that involves messing with a lot of algebra and weird, convoluted formulas. With the fundamental group of the circle, we do many things. An immediate application is that we can properly deﬁne the “winding number” of a closed curve. Since C \ {0} is homotopy equivalent to S1, its fundamental group is Z as well. Any closed curve S1 → C \ {0} thus induces a group homomorphism Z → Z. Any such group homomorphism must be of the form t → nt, and the winding number is given by n. If we stare at it long enough, it is clear that this is exactly the number of times the curve winds around the origin. Also, we have the following classic application: Theorem (Brouwer’s ﬁxed point theorem). Let D2 = {(x, y) ∈ R2 : x2 +y2 ≤ 1} be the unit disk. If f : D2 → D2 is continuous, then there is some x ∈ D2 such that f (x) = x. Proof. Suppose not. So x = f (x) for all x ∈ D2. 30 3 Covering spaces II Algebraic Topology f (x) x g(x) We deﬁne g : D2 → S1 as in the picture above. Then we know that g is continuous and g is a retraction from D2 onto S1. In other words, the following composition is the identity: S1 ι D2
idS1 g S1 Then this induces a homomorphism of groups whose composition is the identity: Z ι∗ g∗ Z {0} idZ But this is clearly nonsense! So we must have had a ﬁxed point. But we have a problem. What about D3? Can we prove a similar theorem? Here the fundamental group is of little use, since we can show that the fundamental group of Sn for n ≥ 2 is trivial. Later in the course, we will be able to prove this theorem for higher dimensions, when we have developed more tools to deal with stuﬀ. 3.3 Universal covers We have deﬁned universal covers mysteriously as covers that are simply connected. We have just shown that p : R → S1 is a universal cover. In general, what do universal covers look like? Let’s consider a slightly more complicated example. What would be a universal cover of the torus S1 × S1? An obvious guess would be p × p : R × R → S1 × S1. How can we visualize this? First of all, how can we visualize a torus? Often, we just picture it as the surface of a doughnut. Alternatively, we can see it as a quotient of the square, where we identify the following edges: 31 3 Covering spaces II Algebraic Topology Then what does it feel like to live in the torus? If you live in a torus and look around, you don’t see a boundary. The space just extends indeﬁnitely for ever, somewhat like R2. The diﬀerence is that in the torus, you aren’t actually seeing free space out there, but just seeing copies of the same space over and over again. If you live inside the square, the universe actually looks like this: As we said, this looks somewhat likes R2, but we know that this is not R2, since we can see some symmetry in this space. Whenever we move one unit horizontally or vertically, we get back to “the same place”. In fact, we can move horizontally by n units and vertically by m units, for any n, m ∈ Z, and still get back to the same place. This space has a huge translation symmetry. What is this symmetry? It is exactly Z × Z. We see that if we live inside the torus S1 ×
S1, it feels like we are actually living in the universal covering space R × R, except that we have an additional symmetry given by the fundamental group Z × Z. Hopefully, you are convinced that universal covers are nice. We would like to say that universal covers always exist. However, this is not always true. Firstly, we should think — what would having a universal cover imply? Suppose X has a universal cover ˜X. Pick any point x0 ∈ X, and pick an evenly covered neighbourhood U in X. This lifts to some ˜U ⊆ ˜X. If we draw a teeny-tiny loop γ around x0 inside U, we can lift this γ to ˜γ in ˜U. But we know that ˜X is simply connected. So ˜γ is homotopic to the constant path. Hence γ is also homotopic to the constant path. So all loops (contained in U ) at x0 are homotopic to the constant path. It seems like for every x0 ∈ X, there is some neighbourhood of x0 that is simply connected. Except that’s not what we just showed above. The homotopy from ˜γ to the constant path is a homotopy in ˜X, and can pass through anything 32 3 Covering spaces II Algebraic Topology in ˜X, not just ˜U. Hence the homotopy induced in X is also a homotopy in X, not a homotopy in U. So U itself need not be simply connected. What we have is a slightly weaker notion. Deﬁnition (Locally simply connected). X is locally simply connected if for all x0 ∈ X, there is some neighbourhood U of x0 such that U is simply connected. As we mentioned, what we actually want is a weaker condition. Deﬁnition (Semi-locally simply connected). X is semi-locally simply connected if for all x0 ∈ X, there is some neighbourhood U of x0 such that any loop γ based at x0 is homotopic to cx0 as paths in X. We have just argued that if a universal cover p : ˜X → X exists, then X is semi-locally simply connected. This is really not interesting, since we don’t care if a space is semi-locally simply connected. What is important is that
this is the other direction. We still need one more additional condition: Deﬁnition (Locally path connected). A space X is locally path connected if for any point x and any neighbourhood V of x, there is some open path connected U ⊆ V such that x ∈ U. It is important to note that a path connected space need not be locally path connected. It is an exercise for the reader to come up with a counterexample. Theorem. If X is path connected, locally path connected and semi-locally simply connected, then X has a universal covering. Note that we can alternatively deﬁne a universal covering as a covering space of X that is also a covering space of all other covers of X. If we use this deﬁnition, then we can prove this result easily using Zorn’s lemma. However, that proof is not too helpful since it does not tell us where the universal covering comes from. Instead, we will provide a constructive proof (sketch) that will hopefully be more indicative of what universal coverings are like. Proof. (idea) We pick a basepoint x0 ∈ X for ourselves. Suppose we have a universal covering ˜X. Then this lifts to some ˜x0 in ˜X. If we have any other point ˜x ∈ ˜X, since ˜X should be path connected, there is a path ˜α : ˜x0 ˜x. If we have another path, then since ˜X is simply connected, the paths are homotopic. Hence, we can identify each point in ˜X with a path from ˜x0, i.e. {points of ˜X} ←→ {paths ˜α from ˜x0 ∈ ˜X}/. This is not too helpful though, since we are deﬁning ˜X in terms of things in ˜X. However, by path lifting, we know that paths ˜α from ˜x0 in ˜X biject with paths α from x0 in X. Also, by homotopy lifting, homotopies of paths in X can be lifted to homotopies of paths in ˜X. So we have {points of ˜X} ←→ {paths α from x0 ∈ X}/. So we can produce our
˜X by picking a basepoint x0 ∈ X, and deﬁning ˜X = {paths α : I → X such that α(0) = x0}/. The covering map p : ˜X → X is given by [α] → α(1). One then has to work hard to deﬁne the topology, and then show this is simply connected. 33 3 Covering spaces II Algebraic Topology 3.4 The Galois correspondence Recall that at the beginning, we wanted to establish a correspondence between covering spaces and fundamental groups. We have already established the result that covering maps are injective on π1. Therefore, given a (based) covering space p : ( ˜X, ˜x0) → (X, x0), we can give a subgroup p∗π1( ˜X, ˜x0) ≤ π1(X, x0). It turns out that as long as we deﬁne carefully what we mean for based covering spaces to be “the same”, this is a one-to-one correspondence — each subgroup corresponds to a covering space. We can have the following table of correspondences: Covering spaces Fundamental group (Based) covering spaces ←→ Subgroups of π1 Number of sheets ←→ Index Universal covers ←→ Trivial subgroup We now want to look at some of these correspondences. Recall that we have shown that π1(X, x0) acts on p−1(x0). However, this is not too interesting an action, since p−1(x0) is a discrete group with no structure. Having groups acting on a cube is fun because the cube has some structure. So we want something more “rich” for π1(X, x0) to act on. We note that we can make π1(X, x0) “act on” the universal cover. How? Recall that in the torus example, each item in the fundamental group corresponds to translating the whole universal covering by some amount. In general, a point on ˜X can be thought of as a path α on X starting from x0. Then it is easy to make a loop γ : x0 x0 act on this: use the concatenation γ · α: [γ] · [α] =
[γ · α]. α ˜x0 ˜γ ˜x 0 γ α x0 ˜X p X We will use this idea and return to the initial issue of making subgroups correspond to covering spaces. We want to show that this is surjective — every subgroup arises from some cover. We want to say “For any subgroup H ≤ π1(X, x0), there is a based covering map p : ( ˜X, ˜x0) → (X, x0) such that p∗π1( ˜X, ˜x0) = H”. Except, this cannot possibly be true, since by taking the trivial subgroup, this would imply that there is a universal covering for every space. So we need some additional assumptions. Proposition. Let X be a path connected, locally path connected and semilocally simply connected space. For any subgroup H ≤ π1(X, x0), there is a based covering map p : ( ˜X, ˜x0) → (X, x0) such that p∗π1( ˜X, ˜x0) = H. 34 3 Covering spaces II Algebraic Topology Proof. Since X is a path connected, locally path connected and semi-locally simply connected space, let ¯X be a universal covering. We have an intermediate group H such that π1( ˜X, ˜x0) = 1 ≤ H ≤ π1(X, x0). How can we obtain a corresponding covering space? Note that if we have ¯X and we want to recover X, we can quotient ¯X by the action of π1(X, x0). Since π1(X, x0) acts on ¯X, so does H ≤ π1(X, x0). Now we can deﬁne our covering space by taking quotients. We deﬁne ∼H on ¯X to be the orbit relation for the action of H, i.e. ˜x ∼H ˜y if there is some h ∈ H such that ˜y = h˜x. We then let ˜X be the quotient space ¯X/∼H. We can now do the messy algebra to show that this is the covering space we want. We have just showed that every subgroup comes from some covering

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