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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... |
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 ... |
: : : : : : : : : : : : : : : : : : 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 an... |
(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 m1... |
. (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 t... |
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 idea... |
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 o... |
� 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); i... |
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 t... |
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/; : : ... |
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... |
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 bir... |
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 ... |
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 varie... |
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 Norma... |
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, a... |
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). PR... |
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. ... |
�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) ... |
/, 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; : : : ; ... |
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 factor... |
. 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... |
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. PROO... |
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, ... |
/ ˝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... |
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 ... |
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 se... |
/, 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... |
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, an... |
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... |
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. ... |
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... |
.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 p... |
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 ... |
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,.... |
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 : Bec... |
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 containe... |
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 w... |
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 t... |
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 i... |
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 '... |
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... |
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).... |
.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... |
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 cu... |
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 va... |
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 th... |
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... |
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 /; ... |
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 integr... |
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... |
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 \ Z... |
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... |
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 su... |
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 al... |
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, ... |
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 com... |
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 ... |
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/ — b... |
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, a... |
. 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 i... |
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 ... |
). 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... |
(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... |
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... |
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. Ther... |
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-M... |
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 THE... |
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... |
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 k... |
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... |
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 clos... |
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/ ... |
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 ... |
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 ... |
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 ... |
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 fo... |
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/... |
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 polynomi... |
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 m... |
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 ... |
=.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 ... |
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... |
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... |
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 subs... |
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 central... |
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 homeomorp... |
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 speci... |
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, sinc... |
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 ... |
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.... |
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... |
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(... |
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(... |
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 ... |
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 fundame... |
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. Firstl... |
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 loc... |
˜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 ... |
[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) → (... |
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