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whenever bMta and that (X E rank a. We must show that (X = rank x for some x with xMta. By Theorem 7V(b), (X E (rank br for some b E a, so (X E rank b. If (X = rank b, we are done, so suppose not. By the inductive hypothesis (X E rank b S {rank x I xMtb} so that we have (X = rank x & xMtb & b E a for some x. Since xMta, the proof of (1:r) is complete. From (1:r) and the equation E(a) = {E(x) I xMta} we can conclude (from the uniqueness clause in recursion) that E(a) = rank a for all a in TC S. By applying ({r) to a set containing S (in place of S itself) we obtain rank S = {rank a I a E... E S} = ran E as claimed. Natural Models 249 Theorem 9E gives one of the (many) possible ways of introducing ordinal numbers. It produces the rank operation directly (without reference to ordinals). Thus one could then define an ordinal number to be something that is the rank of some set. It then follows, for example, that rank ('/. = ('/. for any ordinal ('/.. And all this can be carried through without any mention of well-ordered structures! One can then proceed to develop the usual properties of ordinal numbers (e.g., those in Theorem 7M) by using the rank operation. The end result of such an approach would be equivalent to the more traditional approach followed in Chapter 7. Exercises 1. Prove Theorem 9C without using the axiom of choice. 2. Give an intrinsic characterization of these relations R with the property that Rt is a partial ordering. (Konig) Assume that R is a well-founded relation such that {x I xRy} 3. is finite for each y. Prove that {x I xRty} is finite for each y. 4. Show that for any set S, TC S = S u U{TC x I XES}. NATURAL MODELS ) In this section we want to consider the question of whether there can be a set M that is in a certain sense a miniature model of the class V of all sets. Consider
a formula (J of the language of set theory; for example (J might be one of our axioms. (Recall from Chapter 2 the ways of constructing formulas.) We can convert (J to a new formula (JM by replacing expressions "Ix and 3x by "Ix E M and 3x E M. Then the proposition (true or false) that (J asserts of V is asserted by (JM of just M. Example Let (J be the pairing axiom, VuVv3BVx[XEB <'> x=uorx=v]. Then (JM is (Vu E M)(Vv E M)(3B E M)(Vx E M)[x E B <'> X = U or x = v]. In something closer to English, this says that you can take any u and v inside the set M, and be assured of the existence of a set B inside the set M whose members belonging to the set M are exactly u and v. If somebody has the delusion that M is the class of all sets, then (JM asserts what he believes the pairing axiom to assert. 250 9. Special Topics The formula (JM is called the relativization of (J to M. We say that (J is true in M (or that M is a model of (J) iff (JM is true. Now it is conceivable that there might be some set M that was a model for all of our axioms. It is also conceivable that no such M exists. In this section we will be particularly concerned with the question of whether ~, for some lucky oc, might be a model of all of our axioms. Logical Notes (1) What we here call a model might also be called an "E-model." There is a more general concept of a model (M, E) involving not only reinterpretation of "land 3 (to refer to M), but also a reinterpretation of E (by means of E). We will not delve into this more general concept; we only caution you that it exists. (2) There is a fascinating theorem of mathematical logic, called "G6del's second incompleteness theorem." It implies, among other things, that if our axioms are consistent (as we sincerely believe them to be), then we can never use them to prove the existence of a model of the axioms. Such a model may exist (as a set), but
a proof (from our axioms) that it is there doe~ not exist. So do not expect such a proof in this section! We will now go through the axioms one by one, checking to see whether its relativization is true in ~ for suitable oc. 1. Extensionality. We claim that the relativization of extensionality to any transitive set (and ~ is always a transitive set) is true. That is, consider any transitive set M and any sets A and B in M. Our claim is tha t if ("Ix E M)(x E A <'> X E B), then A = B. Well, if x E A, then x E A E M, and by the transitivity of M we have x E M. Hence we can apply (r:r) to conclude that x E B. Thus A s;; B, and similarly B s;; A. So A = B. 2. Empty set. We claim that the relativization of the empty set axiom to ~ is true provided that oc i= O. We must show that (for oc i= 0), (3B E ~)(Vx E ~) x rt B. What should we take for B? The empty set, of course. Since 1 E oc, we have 0 E VI s;; ~, so the empty set is in ~. And certainly nothing in ~ (or elsewhere) belongs to the empty set. 3. Pairing. We claim that the relativization of the pairing axiom to V). is true for any limit ordinal A.. To prove this, consider any u and v in V).. Since V). = UaE).~' we have u E ~ and v E Vp for some oc and P Natural Models 251 less than A.. Either ('I. E /3 or /3 E ('I.; by the symmetry we may suppose that ('I. §. /3. Then {u, v} ~ Vp and- {u, v} E J.P + s;; V),. Thus in V), there is a set B (namely {u, v}) such that (V'x E V;,)(x E B <'> X = U or x = v). And this is what we want. 4. Union. The union axiom is true in ~ for any ('I.. For suppose that A E ~. Then
A s;; J.P for some /3 less than ('I.. Since Vp is a transitive set, X E UA => x E b E A for some b => XEJ.P. SO UA s;; Vp, whence UA E ~. Thus we have a set B in ~ (namely, UA) such that for any x in ~ (or elsewhere), x E B <'> (3b)(x E b E A) <'> (3b E ~)(x E b E A) since ~ is transitive. 5. Power set. As with the pairing axiom, the power set axiom is true in V), whenever A. is a limit ordinal. To prove this, consider any set a in V),. Then rank a < A., so that rank f!J>a = (rank ar < A. and f!J>a E V),. Thus we have a set B in V), (namely f!J>a) such that for any x in V)" XEB <'> xs;;a <'> V't(tEx => tEa) <'> (V'tEV),)(tEX => tEa) since V), is transitive. 6. Subset. All subset axioms are true in ~ for any ('I.. For consider any set c in ~ and any formula rp not containing B. We seek a set B in ~ such that for any x in ~, x E B <'> X E C & rpv,. This tells us exactly what to take for B, namely {x Eel rpv,}. Since B s;; c E ~, we have B E ~. 252 9. Special Topics 7. Infinity. The infinity axiom is true in ~ for any ('/. greater than w. We have w ~ V",. whence w E ~ for any larger ('/.. As with the other axioms. this implies that the infinity axiom is true in ~. 8. Choice. The axiom of choice (version I) is true in ~ for any ('/.. If R is a relation in ~. then any subset of R is in ~. In particular. the subfunction F of R provided by the axiom of choice,{version I) is in ~. (For some other versions of the axiom of choice. we need to assume that ('/. is a limit ordinal. Although Theorem 6M proves
that the several versions are equivalent. the proof to 6M uses some of the above-listed axioms. And those axioms may fail in ~ if ('/. is not a limit ordinal.) 9. The regularity axiom is true in ~ for any ('/.. Consider any nonempty set A in ~. Let m be a member of A having least possible rank. Then m E ~ and m n A = 0. (Note that we do not need --to use the regularity axiom in this proof. in contrast to the situation with all other axioms.) We have now verified the following result. Theorem 9F If A. is any limit ordinal greater than w. then V). is a model of all of our axioms except the replacement axioms. Our axioms (including replacement) are called the Zermelo-Fraenkel axioms. whereas the Zermelo axioms are obtained by omitting replacement. (Recall the historical notes of Chapter 1.) The above theorem asserts that V). is a model of the Zermelo axioms whenever A. is a limit ordinal greater than w. The smallest limit ordinal greater than w is w. 2. Informally. we can describe the ordinal w. 2 by the equation w·2 = {O. 1.2 •...• W. w+. w+ + •••• } listing the smaller ordinals. It is the ordinal of the set Z of integers under the peciIliar ordering 0<1<2<",< -1< -2< -3<···. V", contains the sets of finite rank. Any set in V", is finite. its members are all finite. and so forth. We can informally describe V",.2 as V",.2 = V", u [!),V", u [!),[!),V",.... In V",.2 all the sets needed in elementary mathematics can be found. The real numbers are all there. as are all functions from reals to reals Natural Models 253 (compare Exercise 27 of Chapter 7). This reflects the fact that we needed only the Zermelo axioms to construct the reals. What sets are not in V",. 2? Well, the only ordinals in V"'. 2 are those less than w. 2, by Exercise 26 of Chapter 7. Since w. 2 is obviously a countable ordinal, V
""2 contains only ordinals that are countable, and not even all of those. Lemma 9G There is a well-ordered structure in V""2 whose ordinal number is not in V",.2' Proof Start with an uncountable set S in V"'. 2' such as IR or f!J>w. By the well-ordering theorem, there exists some well ordering < on S. Now < ~ S x S ~ f!J>f!J>S by Lemma 3B, so < is also in V",,2' (The rank of < is only two steps above the rank of S.) Hence by going up two more steps, we have (S, <) in V""2' But the ordinal number of (S, <), being uncountable, cannot be in V""2' -l Corollary 9H Not all of the replacement axioms are true in V",.2' Sketch of Proof Let (J be the formula of set theory: "For any well-ordered structure (S, <) there exists an ordinal ('I. such that (S, <) is isomorphic to «('I., Ea)'" Then (J can be proved from the Zermelo-Fraenkel axioms-we proved in Chapter 7 (see Theorem 70). But we claim that (J is not true in V",.2' This claim follows from Lemma 9G, together with some argument to the effect that if it appears from inside V",.2 that ('I. is the ordinal number of (S, <), then it really is. it back But if V""2 were a model for the Zermelo-Fraenkel axioms, it would have to be a model for any theorem that follows from those axioms. (This was the meaning of "theorem" back in Chapter 1.) So V""2 cannot be a model of the Zermelo-Fraenkel axioms. Since it is a model of the Zermelo axioms, it must be replacement axioms that fail in V""2' -l Corollary 91 Not all of the replacement axioms are theorems of the Zermelo axioms. Proof Any theorem of the Zermelo axioms must be true in any model of those axioms, such as V""2' But by the preceding result, not all
replacement axioms are true in V",.2' -l The standard abbreviation for the Zermelo-Fraenkel axioms is "ZF" (or" ZFC" if we want to emphasize that the axiom of choice is included). The above corollary shows that there is a sense in which ZF is strictly stronger than the Zermelo axioms. It is our only excursion into the metamathematics of set theory. 254 9. Special Topics Definition A cardinal number K is said to be inaccessible iff it meets the following three conditions. (a) K is greater than ~o. (b) For any cardinal A. less than K. we have 2). < K. (Here cardjpal exponentiation is used.) (c) It is not possible to represent K as the supremum of fewer than K smaller ordinals. That is. if S is a set of ordinals less than K and if card S < K. then the ordinal sup S is less than K. Cardinals meeting clause (c) are called regular; we will have more to say about such cardinals in the section on cofinality. Examples Conditions (b) and (c) are true when K = ~o (by Corollary 6K and Exercise 13 of Chapter 6). But of course condition (a) fails. Conditions (a) and (c) hold when K = ~ 1 (since a countable union of countable ordinals is countable). But condition (b) fails for ~1' Conditions (a) and (b) hold when K = :lea. but condition (c) fails since :lea = Uneea:ln' What is an example of a cardinal meeting all three conditions? Are there any inaccessible cardinals at all? We will return to these questions after the next theorem. First we need the following lemma. which relates the beth numbers to the ~ sets. Lemma 9J For any ordinal number 0(. Proof We use transfinite induction on 0(. card Vea +a = :l... Case I 0( = O. We must show that card Vea = ~o. But this is clear. since Vea is the union of ~o finite sets of increasing size. Case I I 0( is a successor ordinal f3 +. Then and its cardinality is 2cardV..,+p - 2::lp - :l p+
.... by the inductive hypothesis. Case III 0( is a limit ordinal. Then w + 0( is also a limit ordinal and so w + 0( = sup{w + bib E O(}. Consequently (Exercise 9). Vea+a = U{VeaH I b E O(}. Since card VeaH = :l6' the cardinality of the union is at least :la' On the other hand. it is at most (card O(). :la' Since card 0( § 0( § :la (by the mono tonicity of the beth operation). this product is just :l... -I Natural Models 255 Lemma 9K Assume that I<: is inaccessible. (a) (b) If ('I. is an ordinal less than 1<:, then :1.. is less than 1<:. If A E V"' then card A < 1<:. Proof We prove part (a) by transfinite induction on ('I.. So suppose that the condition holds for all ordinals less than ('I., where ('I. is less than 1<:. Case I Case II ('I. = O. Then:1o = ~o < I<: since I<: is uncountable. ('I. = 13+ for some p. Then:1p < I<: by the inductive hypothesis, and :1.. = 2::J p < I<: by inaccessibility.,~ase III g" is a limit ordin~l. Then by the inductive hypothesis, is not the :1y < I<: for all y E ('I.. Then :1.. = sup{:1y lYE ('I.} < 1<:, since I<: supremum of fewer than I<: (and card ('I. < 1<:) smaller ordinals. This proves part (a). For part (b), suppose that A E V"' Then A ~ ~ for some ('I. less than 1<:. Hence we have card A ~ card ~ ~ :1.. < I<: by the preceding lemma and part (a). Theorem 9L If I<: is an inaccessible cardinal number, then all of the ZFaxioms (including the replacement axioms) are true in V"' Proof All of the Zermelo axioms are true in v..: by Theorem 9F, since I<
: is an uncountable limit ordinal. The only axioms left to worry about are the replacement axioms. Consider a set A in V" and any formula tp(x, y) such that ('Ix E A) 'IYl 'IY2[tp(X, Yl) & tp(x, Y2) => Yl = Y2] is true in V"' Define the function F by F = {(x, y) E A x V" I tp(x, y) is true in VJ. Notice that F is a function, by what we have said about tp. The domain of F is some subset of A. Let B = ran F; thus for any Y in V" Y E B <'> (3x E A) tp(x, y) is true in V". What we must prove is that BE V"' because then it will follow that 3B 'Iy[y E B <'> (3x E A) tp(x, y)] is true in V". To show that B E V"' consider the set S = {rank F(x) I x E dom F}. 256 9. Special Topics in S are all (because F(x) E V.J, and The ordinals card S ~ card dom F ::;; card A < I<: by part (b) of the above lemma. So by the inaccessibility, the ordinal than less I<: rt. = sup S = sup{rank F(x) I x E dom F} is less than 1<:. But any member F(x) of B has rank no more than rt., so F(x) E ~+. Hence B ~ ~+ and so rank B § rt.+ E 1<:. Thus BE V"' as desired. -l Now back to the question: Are there any inaccessible cardinals at all? If so, then by the above theorem, there are models of the ZFaxioms. Because we cannot hope to prove the existence of such a model (due to G6del's second incompleteness theorem, mentioned earlier in this section), it follows that we cannot hope to prove from our axioms that inaccessible cardinals exist. On the other hand, we intuitively want the cardinal and ordinal numbers to go on forever. To deny the existence of inaccessible numbers would appear to impose an
unnatural ceiling on "forever." With these thoughts in mind, Alfred Tarski in 1938 proposed for consideration as an additional set-theoretic axiom the statement: For any cardinal number there is a larger inaccessible cardinal number. This is an example of a "large cardinal axiom." Despite the fact that it literally concerns huge sets, one can prove from it new facts about natural numbers! The interest in various large cardinal axioms and their consequences motivates an important part of current research in set theory. Exercises 5. Show that a set S belongs to Vea iff TC S is finite. (Thus S belongs to Vea iff S is finite and all the members of.,. of members of S are finite. Because of this fact, Vea is often called HF, the collection of hereditarily finite sets.) 6. Are the replacement axioms true in Vea? Which axioms are not true in Vea? 7. Let:F be the collection of finite subsets of w. Let g be a one-to-one correspondence between wand :F with the property that whenever mE g(n), then m is less than n. (For example, we can use g(n) = {m E w I the coefficient of 2m in the binary representation of n is I}.) Define the binary relation E on w by Show that (w, E) is isomorphic to (Vea, EO). mEn <'> mE g(n). Cofinality 257 8. The proof to Lemma 9G used the axiom of choice (in the form of the well-ordering theorem). Give a proof that does not use the axiom of choice. 9. Assume that A. is a limit ordinal. Show that ~+). = U{~H I bE A.}. 10. Define the set S = w u YJw u YJYJw u.... Prove that S is a model of the Zermelo axioms. 11. Assume that I<: is inaccessible. Show that :1" = I<: and card v..: = 1<:. COFINALITY Any limit ordinal is the supremum of the set of all smaller ordinals; this merely says that A. = UA. for any limit ordinal A.. But we do not need to take all smaller ordinals. We can find a proper subset S of
A. such that A. is the supremum Us of S. How small can we take S to be? That will depend on what A. is. Definition The co finality of a limit ordinal A., denoted cf A., is the smallest cardinal I<: such that A. is the supremum of I<: smaller ordinals. The cofinality of nonlimit ordinals is defined by setting cf 0 = 0 and cf (X + = 1. Thus to find the cofinality of A., we seek a set S of smaller ordinals (i.e., S ~ A.) for which A. = sup S. Such a set S is said to be cofinal in A.. There will be many sets that are cofinal in A.; for example, A. itself is such a set. There will not exist any smallest (with respect to inclusion) set S cofinal in A.. But there will be a smallest possible cardinality for S, and this smallest cardinality is by definition cf A.. Example Obviously cf A. ::;; card A., since A. is cofinal in itself. Sometimes equality holds; for example, cf w = ~o. (We could not have cf w < ~o' since a finite union of natural numbers would be finite.) On the other hand, sometimes cf A. i= card A.. For example, Theorem 9N will show that limit ordinal) has cofinality ~o' being the supremum of ~w (as a {~o' ~1' ~2'... }. Definition A cardinal I<: is said to be regular iff cf I<: = 1<:, and is said to be singular if cf I<: < 1<:. The above example shows that ~o is regular whereas ~w is singular. Recall that an inaccessible cardinal is required to be regular. Note on Style The definition of cofinality has an awkward three-part format. Can we not give a simple definition covering zero, the successor 258 9. Special Topics ordinals, and the limit ordinals in one sweep? Yes, we can do so by using the idea of the "strict supremum." The cofinality of any ordinal rt. is the least cardinal number" such that there exists a subset S of rt. having cardinality" and rt. is the least ordinal strictly greater than every member of S. (Exercise 12 asks you to verify that this is correct
.) Theorem 9M ~a+1 is a regular cardinal number for every rt.. Proof Assume that ~a + 1 = sup S, where S is a set of ordinals, each of which is less than ~a+1' Then each member of S has cardinality at most ~... Therefore (compare Exercise 26 of Chapter 6) we have ~a+1 = card US:::;; (card S)· ~a' But this implies that card S ~ ~a+1' -l Since ~o is also a regular cardinal, there remain only the cardinals ~). where A. is a limit ordinal. Theorem 9N For any limit ordinal A., cf ~). = cf A.. Proof First of all, we claim that cf~). :::;; cf A.. We know that A. is the supremum of some set S ~ A. with card S = cf A.. It suffices to show that ~). = sUP{~a I rt. E S}. But this is Theorem 8E, applied to the aleph operation. Second, we claim that cf A. :::;; cf ~).. Suppose that ~). is the supremum of some set A of smaller ordinals. Let B = {y E A. I ~y is the cardinality of some ordinal in A}. Then card B :::;; card A. To complete the proof it suffices to show that sup B = A.. Any rt. in A has cardinality at most ~suP S' so rt. E ~(suP 8)+ l' Hence ~). = sup A :::;; ~(SUP 8)+ 1 and so A. §. (sup B) + 1. Since A. is a limit ordinal, A. §. sup B, whence equality holds. -l For example, cf~w = cfw = ~o. And cf~(} = cfn = ~l (where n is the first uncountable ordinal). In particular, both ~w and ~n are singular. For any limit ordinal A., cf~). = cf A. §. A. §.~).. If ~). is regular, then equality holds throughout. But does this ever happen? If it happens, A. is said to be weakly inaccessible. We cannot prove (from our axioms, if they are consistent) that any weakly inaccessible cardinals exist. If
any inaccessible cardinals exist, then they are also weakly inaccessible (Exercise 15). There is another way of characterizing cofinality that uses increasing sequences of ordinals instead of sets of ordinals. For an ordinal number rt., define an rt.-sequence to be simply a function with domain rt.. For example, an ordinary infinite sequence is an w-sequence, and a finite sequence is an n-sequence for some natural number n. For an rt.-sequencef, it is customary Cofinality 259 to write f~ in place of the Eulerian f (0· An oc-sequence f into the ordinals is called increasing iff it preserves order: eE'1 => f~Ef~· An increasing oc-sequence of ordinals is said to converge to 13 iff 13 = sup ranf, i.e., 13 = sup{!~ leE oc}. From a nonincreasing sequence we can extract an increasing subsequence (possibly with a smaller domain): Lemma 9P Assume thatf is an oc-sequence (not necessarily increasing) into the ordinal numbers. There is an increasing f3-sequence g into the ordinal numbers for some 13 §. oc with sup ranf = sup ran g. Proof We will construct g so that it is a subsequence of f. That is, we will have g~ = fh~ for a certain increasing sequence h. First define the sequence h by transfinite recursion over (oc, Ea)' Suppose that hd is known for every bEe. Then define h~ = the least y such thatfh6 Efy for all bEe, if any such y exists; if no such y exists, then the sequence halts. Let 13 = dom h, and define Then by construction, bEe => fh6Efh~ => g6 E g~, so g IS Increasing. Also h is increasing, because h is one-to-one and h~ is the least ordinal meeting certain conditions that become more stringent as e increases. We have ran g ~ ran f, so certainly sup ran g E sup ranf. On the other hand we have - for all b §. h~ by the leastness of h~. In particular, e §. h~, so f~ §. g~. Hence if 13 = oc, then clearly sup ranf §. sup ran g.
But if 13 is less than oc, it is only because there exists no y for which fy exceeds every gd for b E 13. Again we conclude that sup ran f §. sup ran g. -l This lemma is used in the proof of the following theorem. Theorem 9Q Assume that A is a limit ordinal. Then there is an increasing (cf A)-sequence into A that converges to A. 260 9. Special Topics Proof By definition of cofinality, A is the supremum of cf A smaller ordinals. That is, there is a function f from cf A into A such that A = sup ranf. Apply the lemma to obtain an increasing f3-sequence converging to A for some 13 § cf A. But it is impossible that 13 E cf A, lest we have A represented as the supremum of card 13 smaller ordinals, contradicting the leastness of cf A. -l Corollary 9R For a limit ordinal A, we can characterize cf A as the least ordinal (X such that some increasing (X-sequence into A converges to A. We know that cf A :=; A. What about cf cf A? Or cf cf cf A? These are all equal to cf A, by the next theorem. Theorem 9S For any ordinal A, cf A is a regular cardinal. Proof We must show that cf cf,1= cf A. We may assume that A is a limit ordinal, since 0 and 1 are regular. By the preceding theorem, there is an increasing (cf A)-sequencef that converges to A. Suppose that cf A is the supremum of some set S of smaller ordinals; we must show that cf A :=; card S. (This will prove that cf A :=; cf cf A and will be done.) To do this, consider the set f[S] = {fa I (X E S}. This is a subset of A having the same cardinality as S. It suffices to show thatf[S] is cofinal in A, since this implies that cf A :=; cardf[S] = card S. Consider then any (X E A. Since f converges to A, we have (X E f(f3) E A for some 13 E cf A. Since cf A: = sup S, we have 13 EyE S for some y. Hence (X E f(f3)
E f(y) E f[S]. So f[S] is indeed cofinal in A. Now we want to study the particular case where A is an infinite cardinal number. In this case, we can give an "all cardinal" characterization of cf A that makes no direct mention of ordinal numbers. Theorem 9T Assume that A is an infinite cardinal. Then cf A is the least cardinal number" such that A can be decomposed into the union of" sets, each having cardinality less than A. Proof We know that A is the union of cf A smaller ordinals, and any smaller ordinal (X satisfies card (X E (X E A = card A. Co finality 261 Hence A. can be represented as the union of cf A. sets of cardinality less than A.. It remains to show that we cannot make do with fewer than cf A. such sets. So consider an arbitrary decomposition A. = Ud where each member of d has cardinality less than A.. Let I<: = card d; we seek to prove that cf A. :=; 1<:. We can express d as the range of a I<:-sequence of sets: d = {A~ leE I<:}. Thus A. = U{A~ leE I<:} and card A~ < A.. Define the cardinal Jl = sup{card A~ leE I<:}. Then each card A~ :=; Jl and so A. = card U{A~ leE I<:} :=; Jl. I<: (compare Exercise 26 of Chapter 6). Case I A.:=; K. Then cf A. :=; A. :=; 1<:, so we are done. Case I I I<: < A.. Then since A. :=; Jl. 1<:, we can conclude that A. = Jl. Thus we have A. = sup{card A~ leE I<:}, so A. is the supremum of I<: smaller ordinals. Hence cf A. :=; 1<:. -l Cantor's theorem tells us that 2" is always greater than 1<:. The following theorem tells us that even cf 2" is greater than 1<:. Konig's Theorem Assume that I<: is an infinite cardinal number. Then I<: < cf 2". Proof Suppose to the contrary that cf 2" :=; 1<:
. Then 2", and consequently any set of size 2", is representable as the union of I<: sets each of size less than 2". The particular set to consider is "S, where S is some set of size 2". (Thus card "S = (2")" = 2".) We have, then, a representation of the form where card A~ < 2". For anyone e in 1<:, {g(e) I g E A~} £; S. Furthermore we have proper inclusion, because card{g(e) I g E A~} :=; card A~ < 2", whereas card S = 2". So we can choose some point This construction yields a I<:-sequence s, i.e., s E "S. But s rt A~ for any eEl<:, by the construction. -l Corollary 9U 2 No i= ~ w • Proof By Konig's theorem, cf 2 No is uncountable. But cf ~w = ~o· -j 262 Exercises 9. Special Topics 12. For any set S of ordinals, define the strict supremum of S, ssup S, to be the least ordinal strictly greater than every member of S. Show that the cofinality of any ordinal ('/. (zero, successor, or limit) is the least cardinal I<: such that ('/. is the strict supremum of I<: smaller ordinals. 13. Show that cf :J). = cf A. for any limit ordinal A.. 14. Assume that R is a well-founded relation, I<: is a regular infinite cardinal, and card{x I xRy} < I<: for each y. Prove that card{x I xRty} < I<: for each y. 15. Prove that any inaccessible cardinal is also weakly inaccessible. 16. Assume that A. is weakly inaccessible, i.e., it is a limit ordinal for which ~). is regular. Further assume that the generalized continuum hypothesis holds. Show that A. is an inaccessible cardinal. 17. (Konig) Assume that for each i in a set I we are given sets Ai and Bi with card Ai < card Bi · Show that card U i E I Ai < card Xi E I Bi · [Suggestion: Iff maps U i E I Ai onto Xi E I Bp then select a point bi in Bi not
in the projection off [AJ.] 18. Consider the operation assigning to each ordinal number ('/. its cofinality cf ('/.. Is this operation monotone? Continuous? Normal? (Give proofs or counterexamples. ) 19. Assume that I<: is a regular cardinal and that S is a subset of V" with card S < 1<:. Show that S E V". 20. Consider a normal operation assigning ta to each ordinal number ('/., and assume that A. is a limit ordinal. Show that cf t). = cf A.. NOTATION, LOGIC, AND PROOFS APPENDIX In Chapters 1 and 2 we described how formulas of the language of set theory could be built up from component parts. We start with the indivisible "atomic" formulas, such as 'x E S' and 'x = y'. Once we know what objects are named by the letters 'x', 'y', and'S', we can consider the truth or falsity of such formulas. 1 Given any formulas rp and t/J, we can combine them in various ways to obtain new ones. For example, r(rp & t/J)' will be a new formula. The intended meaning of the ampersand can be captured by giving a truth table (Table 1). And similarly we can specify by this table that 'or' is to mean "one or the other or both." A simpler case is ','; the truth value of r (, rp)' is determined by the last column of Table 1. 1 In this appendix we utilize the convention of forming the name of an expression by the use of single quotation marks. For example, x might be a set, but 'x' is a letter used to name that set. We also use utilize corners, e.g., if cp is the formula 'x = x', then r(cp & cpr is the formula '(x = x & x = x)'. 263 264 Appendix. Notation, Logic, and Proofs TABLE 1 q> T T F F 1/1 T F T F (q> & 1/1) (q> or 1/1) (---'q Example We can show that the truth value, T or F, of "(I (tp & t/I))' is always the same as the truth value of "((Itp) or (It/I))' by making a table (Table 2) illustrating
every possibility. TABLE Example The truth value of" (tp & (t/I or 0))' is always the same as the truth value of "((tp & t/I) or (tp & 0))'. The table for this must have eight lines to cover all possibilities. You are invited to contemplate the relationship of this example to Fig. 5 in Chapter 2. The formula "(tp => t/I)' is to be read as "if tp, then t/I." The exact meaning is determined by Table 3. Note in particular that the formula is "vacuously" true whenever tp is false. The formula is sometimes read as "tp implies t/I." This usage of the word" implies" is a conventional part of mathematical jargon, but it is not exactly the way the rest of the world used the word. TABLE Appendix. Notation, Logic, and Proofs 265 Example The truth value of C (tp => t/J)' is always the same as the truth value of its contrapositive C((,t/J) => ('tp)),. (You should check this.) Sometimes when it is desired to prove a formula of the form C (tp => t/J)', it is actually more convenient to prove C ( (, t/J) => (, tp))'. And this suffices to establish the truth of C(tp => t/J)'. Example If () is true, then the truth value of C ( (, tp) => (, ()))' is the same as the truth value of tp. Sometimes to prove tp, we proceed in an indirect way: We show that C(,tp), would lead to a result contradicting what is already known to be true, i.e., we use a "proof by contradiction." Example The truth value of C (, (tp => t/J))' is the same as the truth value of C (tp & (, t/J))'. This is reflected in that to give a counterexample to show that C(tp => t/J)' is false, we give an example in which tp is true and t/J is false. the fact To construct any interesting formulas, we must use (in addition to features already mentioned) the phrases "for all x" and "for some x." These phrases can be symbolized by '''Ix' and '3x', respectively. If tp is a formula (in which '
x' occurs but '''Ix' and '3x' do not), then the condition for cVx tp' to be true is that tp should be true no matter what, x' names (in the universe of all sets). And the condition for C 3x tp' to be true is that there is at least one thing (in the universe of all sets) such that tp is true when 'x' names that thing. These criteria do not reduce to simple tables. There is no mechanical procedure that can be substituted for clear thinking. Example The truth value of C (, "Ix tp)' is the same as the truth value of C 3x(, tp)'. To deny that tp is true of everything is to assert that there is at least one thing of which tp is false, and of which C (, tp)' is consequently true. Similarly the truth value of C (, 3x tp)' is the same as the truth value of CVx('tp)'. Example Suppose that C 3x Vy tp' is true. Then, reading from left to right, we can say that there is some set x for which CVy tp' is true. That is, there is at least one set that, when we bestow the name' x' upon it, we then find that no matter what set' y' names, tp is true. This guarantees that the weaker statement CVy 3x tp' is true. This weaker sta.tement demands only that for any set (upon which we momentarily bestow the name 'y ') there exists some set (which we call' x') such that tp is true. The difference here is that the choice of x may depend on y, whereas C 3x Vy tp' demands that there be an x fixed in advance that is successful with every y. For example, it is true of the real numbers that for every number y there is some number x (e.g., x = y + 1) with y < x. But the stronger statement, that there 266 Appendix. Notation, Logic, and Proofs is some number x such that for every number y we have y < x, is false. Or to give a different sort of example, it may well be true that every boy has some girl whom he admires, and yet there is no one girl so lucky as to be admired by every boy. Next
we want to give an example of a proof. Actually, the book is full of proofs. But what we want to do now is to illustrate how one constructs a proof, without knowing in advance how to do it. After one knows how the proof goes, it can be written out in the conventional form followed by contemporary mathematics books. But before one knows how the proof goes, it is a matter of looking at the assumptions and the conclusions and trying to gain insight into the connection between them. Suppose, for example, one wants to prove the assertion: a E B => [lila E [lII[lIIU B. We will present the process of constructing the proof as a discussion among three mental states: Prover, Referee, and Commentator. P: We assume that a E B. C: We might as well, because the assertion to be proved is vacuously true if a rt= B.. It is to be proved that [lila E [lII[lIIUB. R: C: Always keep separate the facts you have available and the facts that you want to establish. I have a E B, but it is not obvious how to utilize that fact. P: R: Look at the goal. It suffices to show that [lila ~ [lIIUB, by the definition of [lII. P: Well, in that case I will assume that c E [lila. R: Then the goal is to show that C E [lIIUB. C: The point is to consider an arbitrary member (call it c) of [lila, and show that it belongs to [lIIUB. Then we can conclude that [lila ~ [lIIUB. P: c~ a. R: The goal is to get c ~ UB C: This pair of sentences is a recasting of their previous pair, using definitions. P: Since I am expected to prove that c ~ UB, I will consider any X EC. R: The goal now is to get x E UB. P: Looking back, I see that xEc~aEB. R: But you want x E UB. P: x E a E B, so I have x E UB. Appendix. Notation, Logic, and Proofs 267 C: The job is done. The information Prover has is the same as the information Referee wants. Now that the mental discussion is over, the proof can be written out
in the conventional style. Here it is: Proof Assume that a E B. To show that f!l>a E f!l>f!l>UB, it suffices to show that f!l>a S f!l>UB. SO consider any C E f!l>a. Then we have XEC => xEcSaEB => xEaEB => xEUB. Thus c S UB and c E f!l>UB. Since c was arbitrary in f!l>a, we have shown that f!l>a S f!l>UB, as desired. -l We conclude this appendix with some comments on a task mentioned in Chapter 2. In that chapter there is a fairly restrictive definition of what aformula is. In particular, to obtain a legal formula, it is necessary to get rid of the defined symbols, such as '0', 'U', 'f!l>', etc. We will indicate a mechanical procedure for carrying out this elimination process. Suppose then, we have a statement C _ f!l>t _'in which' f!l>' occurs. We can rewrite it first as 'v'a(a = f!l>t => _ a _) and then rewrite' a = f!l>t' as 'v'X(X E a <'> x E f!l>t). This reduces the problem to eliminating' f!l>' from statements of the form 'x E f!l>t'. And this is easy; 'x E f!l>t' can be replaced by 'x S t' or by ''v'y(y EX=> YEt)'. This process is similar for other symbols. For the union symbol, we reduce the problem to eliminating 'U' from 'x E Ut'. And the definition of 'U' tells us how to do this; we replace 'x E U t' by '3y(x EyE t)'. SELECTED REFERENCES FOR FURTHER STUDY Georg Cantor. Contributions to the Founding of the Theory of Transfinite Numbers, translated by P. Jourdain. Dover, New York, 1955. This is a translation of two papers that Cantor originally published in 1895 and 1897. The papers treat cardinal and ordinal numbers and their arithmetic. Jourdain's introduction, first published in 1915, discusses Cantor's work in historical perspective. Paul J. Cohen.
Set Theory and the Continuum Hypothesis. Benjamin, New York, 1966. This gives a highly condensed account of the main results in the metamathematics of set theory. Frank R. Drake. Set Theory, An Introduction North-Holland Pub!., Amsterdam, 1974. to Large Cardinals. This book treats the implications that large cardinals have for the metamathematics of set theory, along with other related topics. Abraham A. Fraenkel, Yehoshua Bar-Hillel, and Azriel Levy. Foundations of Set Theory, 2nd rev. ed. North-Holland Pub!., Amsterdam, 1973. This book considers the p~ilosophical background for set theory 269 270 Selected References for Further Study and the foundations of mathematics. Different approaches to set theory are developed and compared. Among the topics discussed are the limitations ofaxiomatizations, the role of classes in set theory, and the history of the axiom of choice. Thomas J. Jech. The Axiom of Choice. North-Holland Pub!., Amsterdam, 1973. Centered on the topic of its title, this book contains much material on the metamathematics of set theory. K. Kuratowski and A. Mostowski. Set Theory. PWN-Polish Scientific Pub!., Warsaw, and North-Holland Pub!., Amsterdam, 1968. In addition to the standard topics of set theory, this book has considerable material on the applications of set theory, particularly to topology. The footnotes provide a guide to the history of set-theoretic ideas. Waof'aw Sierpinski. Cardinal and Ordinal Numbers, 2nd ed. PWN-Polish Scientific Pub!., Warsaw, 1965. This is a classic book on cardinal and ordinal arithmetic, the axiom of choice, and related topics. The approach is nonaxiomatic. Jean van Heijenoort (editor). From Frege to Godel, A Source Book in Mathematical Logic, 1879-1931. Harvard Univ. Press, Cambridge, Massachusetts, 1967. This is a collection of forty-six fundamental papers in logic and set theory, translated into English and supplied with commentaries. There is an 1899 letter from Cantor to Dedekind and the 1902 correspondence between Russell and Frege. Zermelo's 1908 paper giving the first axiomatization of set theory is included, as are papers by Fraenkel,
Skolem, and von Neumann. LIST OF AXIOMS Extensionality axiom 'v'A'v'B['v'x(XEA <'> XEB) => A=B] Empty set axiom Pairing axiom Union axiom Power set axiom 3B 'v'x x rt B v.u'v'v 3B 'v'x(x E B <'> X = U or x = v) 'v'A 3B 'v'x[x E B <'> (3b E A)x E b] 'v'a 3B 'v'x(x E B <'> X ~ a) Subset axioms For each formula qJ not containing B, the following is an axiom: 271 272 Infinity axiom Choice axiom List of Axioms 3A[0 E A & (Va E A) a+ E A] (V relation R)(3 function F)(F ~ R & dom F = dom R) Replacement axioms For any formula tp(x, y) not containing the letter B, the following is an axiom: Vt 1... Vtk VA[(Vx E A)Vy1VY2(tp(X, Yl) & tp(x, Y2) => 3B Vy(y E B <'> (3x E A)tp(x, y))] => Yl = Y2) Regularity axiom (VA =1= 0)(3m E A) m n A = 0 INDEX A Abelian group, 95, 122 Absolute value, 109, 118 Absorption law, 164 Abstraction, 4-6, 20-21, 30, 37 Addition, see Arithmetic Additive inverse, 95, 104, 117 Aleph, 137,212 Algebra of sets, 27-31 Algebraic numbers, 161 Antinomies, see Paradoxes Archimedean, 120 Arithmetic of cardinal numbers, 138-143, 149, 164 of integers, 92-100 of natural numbers, 79-82, 85 of order types, 222-226 of ordinal numbers, 227-239 of rational numbers, 103-110 of real numbers, 114-119 Associative laws, 28, see also Arithmetic Atomic formula, 263 Atoms, 7-9, 13 Aussonderung axioms 15,21, see also Subset axioms
Axiom of choice, see Choice axiom Axiomatic method, 10-11,66, 125 Axioms of set theory, 15, 166,271-272, see also individual axioms B Bar-Hillel, Yehoshua, 269 Bernays, Paul, 15 Bernstein, Felix, 148 Bernstein's theorem, 147 Berry, G., 6 Berry paradox, 5 Beth, 214, 254 Binary operation, 79 Binary relation, 42 Borel, Emile, 148 Bound, 114, 171 Bounded, 114, 212 273 274 Burali-Forti, Cesare, 15, 194 Burali-Forti theorem, 194 Burrill, Claude, 112 C Cancellation laws for cardinal numbers, 141 for integers, 100 for natural numbers, 86 for ordinal numbers, 234 for rational numbers, 110 for real numbers, 118 Canonical map, 58 Cantor, Georg, 14-15, 112, 148, 166. 194, 269-270 Cantor-Bernstein theorem, 148 Cantor normal form, 238 Cantor's theorem, 132, 261 Cardinal comparability, 151 Cardinal numbers. 136-137, 197. 199,221 arithmetic of. 138-143, 149, 164 ordering of, 145 Cardinality. 137 Cartesian product. 37 infinite. 54 Cauchy sequence, 112 Chain. 151 descending. 173 Characteristic function, 131 Choice axiom, 11. 49. 55.151-155.198 Choice function. 151 Class. 6. 10. 15.20 Closed. 68, 70, 107. 216 Closure. 78 transitive. 178. 244 Cofinal. 257 Cofinality. 257. 260 Cohen. Paul. 166. 269 Commutative diagram. 59 Commutative laws, 28. see also Arithmetic Commutative ring. 122 Comparability. 151 Compatible. 60 Complement, 27. see also Relative complement Complete ordered field. 119. 123 Composition. 44. 47 Comprehension. see Abstraction Connected. 63 Index Constructed, 175-177, 180,211 Constructive method, 66, 125 Constructivity, 154 Continuous, 216 Continuous functions. 165 Continuum hypothesis, 165,215 generalized, 166,215 Contradiction, proof by, 265 Contrapositive, 265 Converge, 259 Coordinate, 37, 54 Coordinate plane, 40 Corners, 263 Correspondence, 129 Countable, 159 Counterexample, 87, 265 Counting, 136, 197 Cut, 112-113 D
, 224 Hereditarily finite, 256 Hilbert, David, 166 Historical notes, 6,11,14-16,36, 43n, 67, 70, 111,112,124,131,148,153,154,166, 194, 206, 256 Honesty, 124 Identity element, 94-95, 97. 104, 106, 116, 119, 226, 229 Identity function, 40, 47 Identity principle, 2 Iff, 2 Image, 44, 50 Imply, 264 Inaccessible, 254-256 weakly, 258 Inclusion, 3 proper, 85, 167 Increasing, 259, see also Monotone Index set, 51, 54 Induction, 69 strong, 87 transfinite. 174, 200, 242 Inductive, 68, 174 Infimum, 171 Infinite 15, 133, 157 cardinal, 137 Infinity axiom, 68 Initial ordinal. 199 Initial segment, 173 Injection. 43 276 Integers, 75,92, 121 arithmetic of, 92-100 ordering of, 98 Integral domain, 97, 122 Intensional, 125 Intersection, 3, 21, 24-25 indexed,51 Interval, 44, 130 Into, 43 Inverse, 44, 46, see also Additive inverse, Multiplicative inverse left, 48 right, 48 Inverse image, 45, 51 Irreflexive, 63 Isomorphic, 77, 184 Isomorphic embedding, 100, 110, 119 Isomorphism, 184, 186 Isomorphism type, 207, 221 Jech, Thomas, 270 Jourdain, Philip, 269 J K Konig, Denes, 249 Konig, Julius, 262 Konig's theorem, 261 Kronecker, Leopold, 14 Kuratowski, Kazimierz, 36, 153,270 L Landin, Joseph, 113 Language of set theory, see Formula Large cardinal axiom, 256 Largest, 171 Least, 87, 171 Least common multiple, 172 Least upper bound, 114, 171, see also Supremum Length,160 Less than, 83 Levy, Azriel, 269 Lexicographic ordering, 64,185 Hebrew, 224 Index Limit ordinal, 203 Linear algebra, 58, 153 Linear ordering, 62, 170 Logarithm theorem, 237 Logic, 12, 186,263-267 Logical consequence, 12 Logicism, 15 Loset, 170 Lower bound, 171 M Map, 43 Mathematics, 126 Maximal, 151, 171 Maximum, 171 Measurement, 136, 187
, 220 Member. I, II Mendelson, Elliott, 119 Metamathematics, 16, 166,253 Minimal, 170, 241 Minimum, 171 Mirimanoff, Dmitry, 15, 206 Model,250 Monotone, 216, see also Increasing Monotonicity, 30 Mostowski, Andrzej, 270 Multiplication, see Arithmetic Multiplicative axiom, 151 Multiplicative inverse, 102, 107, 119 N n-ary relation, 42 n-tuples,41 Naive set theory, II Natural map, 58 Natural numbers, 21, 66, 68 arithmetic of, 79-82, 85 ordering of, 83 Neumann, John von, see von Neumann, John Normal,216 Notation, 13, see also Formula Number, concept of, 123-127 Numbers, see Cardinal numbers, Integers, Natural numbers, Ordinal numbers, Rational numbers, Real numbers Numeration theorem, 197 Numerology, 5 277 Index o Odd,83 One-to-one, 43 One-to-one correspondence, 129 Onto, 43 Operation binary, 79 on ordinals, 215-216 Order-preservation in arithmetic for cardinal numbers, 149 for integers, 99 for natural numbers, 85 for ordinal numbers, 234 for rational numbers, 109 for real numbers, 118-119 Order type, 189,222, see also Isomorphism type arithmetic of, 222-226 Ordered field, 109, 119, 123 Ordered Il-tuples, 41-42 Ordered pair, 35-36 Ordering of cardinal numbers, 145 of integers, 98 of natural numbers, 83 of ordinal numbers, 192 of rational numbers, 108 of real numbers, 113 Orderings, 39-40 lexicographic, 64, 185, 224 linear, 62, 170 partial, 168 total,62 well,l72 Ordinal numbers, 8, 15, 182, 189, 191-194, 249 arithmetic of, 227-239 initial,199 limit, 203 ordering of, 192 successor, 203 P Pair set, 2, 19 ordered, 35-36 Pairing axiom, 18 Paradoxes, 5, 6, 11, 15, 21, 194 Partial ordering, 168 Partial well ordering, 245 Partition, 55, 57 Peano, Giuseppe, 16, 70 Peano induction postulate, 71 Peano system, 70, 76-77 Peano's postulates, 70 Perfect number, 5 Permutation, 144 Pigeonhole principle,
134 Pioneer, 221 Poset, 170 Positive, 99, 109 Power set, 4, 19, 141 Power set axiom, 18 Pre, 52 Primitive notions, 11 Proof, 12, 266 Proper class, 6, 15 Proper subset, 85, 134 Pure sets. 9 Pythagoreans, 111 Quadruples, 42 Quantifiers, 265 Quotient, 58 Quotient field, 107 Q R Range, 40 Rank, 9, 204, 206, 247-248 Rational numbers, 102, 121 arithmetic of, 103-110 ordering of, 108 Real numbers, 113, 121 arithmetic of, 114-119 ordering of, 113 Recursion, 73 transfinite, 175, 177,210,245 Rel1exive, 56 Regular, 254, 257 Regularity, 8, 205-206 Regularity axiom, 15, 84, 206 Relation, 40 binary, 42 equivalence, 56 n-ary, 42 ordering, 62 278 Relative complement, 21, 27 Relativization, 250 Replacement axioms, 15, 179,253 Restriction, 44, 189 Richard, Jules, 6 Ring, 122, 154 Russell, Bertrand, 6, 15, 155, 206, 270 Russell's paradox, 6, II, 15,21 s Schema, 33, 177 Schroder, Ernst, 148 Schroder-Bernstein theorem, 147 Segment, 173 Sequence, 52, 54, 160,258 Set, I, 6--8, II SierpiIiski, Waclaw, 270 Single-rooted, 43 Single-valued, 42 Singleton, 19 Singular, 257 Size, 128 Skolem, Thoralf. 15,270 Smallest, 171 Space-filling curves, 149 Strong induction, 87 Structure, 170, 186 Subsequence, 259 Subset, 3 proper, 85, 134 Subset axioms, 21 Subtraction, see Arithmetic Subtraction theorem, 235 Successor. 68 Successor ordinal, 203 Supremum, 171, 193, 216 strict, 258, 262 Symmetric, 56 Symmetric difference. 32 T Tarski, Alfred. 256 Teichmiiller-Tukey lemma. 158 Ternary, 42 Theorem. 12 Thread,54 Total ordering, 62 Index Transcendental numbers, 161, 164 Transfinite induction, 174, 242 schema, 200 Transfinite numbers, 14, see also Cardinal numbers, Ordinal numbers Transfinite recursion, 175, 177
on ordinals, 210 on well-founded relations, 245 Transitive closure, 178, 244 Transitive extension, 244 Transitive relation, 56, 243 Transitive set, 71-73 Trichotomy, 62-63, 99, see also Ordering Triples, 41 True, 250 Truth table, 29, 264 Truth value, 264 Tuples, 41, 54 Type, see Order type Types (Russell), 206 u Unary, 42 Undefined notions, see Primitive notions Union, 3, 19,23 indexed,51 Union axiom, 18, 24 Universal set, 10, 22 Upper bound, 114, 171 Use and mention, 263 v Vacuous truth, 3, 264 Value. 43 van Heijenoort, Jean, 270 Veblen theorem, 218 Vector. 58, 153 Venn diagram, 29 von Neumann, John, 15, 67. 206. 270 von Neumann-Bernays set theory. 10, 15 W Well defined. 18. 59-60. 92. 98 Wen founded. 241-242 Index Well ordering, 172 by epsilon, 191 of natural,lUmbers, 86 partial, 245 Well-ordering theorem, 196 Wiener, Norbert, 36 Word,160 Woset,l72 279 Z Zermelo, Ernst, 6, 15,67,270 Zermelo-Fraenkel set theory, 10, 15,252 Zermelo set theory, 252 Zero, 203 Zero divisor, 97, 107 ZF, see Z~rmelo-Fraenkel set theory Zorn's lemma, 151, 153-154, 195I. We first have to check that I ̸= R. Suppose not. So there are b(f1,ℓ1), · · ·, b(fr,ℓr) with g1, · · ·, gr ∈ R such that g1b(f1,ℓ1) + · · · + grb(fr,ℓr) = 1. (∗) We will attempt to reach a contradiction by constructing a homomorphism ϕ that sends each b(fi,ℓi) to 0. Let E be a splitting field of f1f2 · · · fr. So in E[t], for each i, we can write fi = deg fi j=1 (t − αi,j). Then we define a homomorphism ϕ : R → E
by ϕ(t(fi,j)) = αi,j ϕ(tλ) = 0 otherwise This induces a homomorphism ˜ϕ : R[t] → E[t]. Now apply ˜ϕ( ˜fi) = ˜ϕ(fi) − deg fi j=1 ˜ϕ(t − t(fi,j)) deg fi j=1 = fi − = 0 (t − αi,j) So ϕ(b(fi,ℓi)) = 0 as b(fi,ℓi) is a coefficient of fi. Now we apply ϕ to (∗) to obtain ϕ(g1b(f1,ℓ1) + · · · + grb(fr,ℓr)) = ϕ(1). But this is a contradiction since the left had side is 0 while the right is 1. Hence we must have I ̸= R. We would like to quotient by I, but we have to be a bit more careful, since the quotient need not be a field. Instead, pick a maximal ideal M containing I, and consider L = R/M. Then L is a field. Moreover, since we couldn’t 20 2 Field extensions II Galois Theory have quotiented out anything in K (any ideal containing anything in K would automatically contain all of R), this is a field extension L/K. We want to show that L is an algebraic closure. Now we show that L is algebraic over K. This should all work out smoothly, since that’s how we constructed L. First we pick α ∈ L. Since L = R/M and R is generated by the terms tλ, there is some (f1, j1), · · ·, (fr, jr) such that α ∈ K(¯t(fi,ji), · · ·, ¯t(fr,jr)). So α is algebraic over K if each ¯t(fi,ji) is algebraic over K. To show this, note that ˜fi = 0, since we’ve quotiented out each of its coefficients. So by definition, 0 = fi(t) − deg fi j=1 (t − ¯t(fi,j)). So fi(¯t(fi,ji)) = 0. So done. Finally, we
have to show that L is algebraically closed. Suppose L ⊆ E is a finite (and hence algebraic) extension. We want to show that L = E. Consider arbitrary β ∈ E. Then β is algebraic over L, say a root of f ∈ L[t]. Since every coefficient of f can be found in some finite extension K(¯t(fi,ji), · · ·, ¯t(fr,jr)), there is a finite extension F of K that contains all coefficients of f. Since F (β) is a finite extension of F, we know F (β) is a finite and hence algebraic extension of K. In particular, β is algebraic in K. Let Pβ be the minimal polynomial of β over K. Since all polynomials in K split over L by construction (f (t) = (t − ¯t(f,j))), its roots must in L. In particular, β ∈ L. So L = E. Theorem (Uniqueness of algebraic closure). Any field K has a unique algebraic closure up to K-isomorphism. This is the same proof as the proof that the splitting field is unique — given two algebraic closures, we take the largest subfield of the algebraic closures that biject with each other. However, since there could be infinitely many subfields, we have to apply Zorn’s lemma to obtain the maximal such subfield. Proof. (sketch) Suppose L, L′ are both algebraic closures of K. Let H = {(F, ψ) : K ⊆ F ⊆ L, ψ ∈ HomK(F, L′)}. We define a partial order on H by (F1, ψ1) ≤ (F2, ψ2) if F1 ≤ F2 and ψ1 = ψ2\|F1. We have to show that chains have upper bounds. Given a chain {(Fα, ψα)}, we define F = Fα, ψ(x) = ψα(x) for x ∈ Fα. Then (F, ψ) ∈ H. Then applying Zorn’s lemma, there is a maximal element of H, say (F, ψ). Finally, we have to prove that F = L, and that ψ(L) = L′
. Suppose F ̸= L. Then we attempt to produce a larger ˜F and a K-isomorphism ˜F → ˜F ′ ⊆ L′. Since F ̸= L, there is some α ∈ L \ F. Since L is an algebraic extension of K, there is some irreducible g ∈ K[t] such that deg g > 0 and g(α) = 0. Now there is an isomorphism F [t]/⟨g⟩ → F (α) defined by ¯t → α. The isomorphism ψ : F → F ′ then extends to an isomorphism µ : F [t] → F ′[t] 21 2 Field extensions II Galois Theory and thus to F[t]/⟨g⟩ → F ′[t]/⟨µ(g)⟩. Then if α′ is a root of µ(g), then we have F ′[t]//⟨µ(g)⟩ ∼= F ′(α′). So this gives an isomorphism F (α) → F (α′). This contradicts the maximality of ϕ. By doing the argument the other way round, we must have ψ(L) = L′. So done. 2.6 Separable extensions Here we will define what it means for an extension to be separable. This is done via defining separable polynomials, and then an extension is separable if all minimal polynomials are separable. At first, the definition of separability might seem absurd — surely every polynomial should be separable. Indeed, polynomials that are not separable tend to be weird, and our theories often break without separability. Hence it is important to figure out when polynomials are separable, and when they are not. Fortunately, we will end up with a result that tells us exactly when a polynomial is not separable, and this is just a very small, specific class. In particular, in fields of characteristic zero, all polynomials are separable. Definition (Separable polynomial). Let K be a field, f ∈ K[t] non-zero, and L a splitting field of f. For an irreducible f, we
say it is separable if f has no repeated roots, i.e. \| Rootf (L)\| = deg f. For a general polynomial f, we say it is separable if all its irreducible factors in K[t] are separable. It should be obvious from definition that if P is separable and Q \| P, then Q is also separable. Note that some people instead define a separable polynomial to be one with no repeated roots, so (x − 2)2 over Q would not be separable under this definition. Example. Any linear polynomial t − a (with a ∈ K) is separable. This is, however, not a very interesting example. To get to more interesting examples, we need even more preparation. Definition (Formal derivative). Let K be a field, f ∈ K[t]. (Formal) differentiation the K-linear map K[t] → K[t] defined by tn → ntn−1. The image of a polynomial f is the derivative of f, written f ′. This is similar to how we differentiate real or complex polynomials (in case that isn’t obvious). The following lemma summarizes the properties of the derivative we need. Lemma. Let K be a field, f, g ∈ K[t]. Then (i) (f + g)′ = f ′ + g′, (f g)′ = f g′ + f ′g. (ii) Assume f ̸= 0 and L is a splitting field of f. Then f has a repeated root in L if and only if f and f ′ have a common (non-constant) irreducible factor in K[t] (if and only if f and f ′ have a common root in L). This will allow us to show when irreducible polynomials are separable. Proof. 22 2 Field extensions II Galois Theory (i) (f + g)′ = f ′ + g′ is true by linearity. To show that (f g)′ = f g′ + f ′g, we use linearity to reduce to the case where f = tn, g = tm. Then both sides are (n + m)tn+m−1. So this holds. (ii) First assume that f has a repeated root
. So let f = (t − α)2h ∈ L[t] where α ∈ L. Then f ′ = 2(t − α)h + (t − α)2h′ = (t − α)(2h + (t − α)h′). So f (α) = f ′(α) = 0. So f and f ′ have common roots. However, we want a common irreducible factor in K[t], not L[t]. So we let Pα be the minimal polynomial of α over K. Then Pα \| f and Pα \| f ′. So done. Conversely, suppose e is a common irreducible factor of f and f ′ in K[t], with deg e > 0. Pick α ∈ Roote(L). Then α ∈ Rootf (L) ∩ Rootf ′(L). Since α is a root of f, we can write f = (t − α)q ∈ L[t] for some q. Then f ′ = (t − α)q′ + q. Since (t − α) \| f ′, we must have (t − α) \| q. So (t − α)2 \| f. Recall that the characteristic of a field char K is the minimum p such that If no such p exists, we say char K = 0. For example, Q has p · 1K = 0. characteristic 0 while Zp has characteristic p. Corollary. Let K be a field, f ∈ K[t] non-zero irreducible. Then (i) If char K = 0, then f is separable. (ii) If char K = p > 0, then f is not separable iff deg f > 0 and f ∈ K[tp]. For example, t2p + 3tp + 1 is not separable. Proof. By definition, for irreducible f, f is not separable iff f has a repeated root. So by our previous lemma, f is not separable if and only if f and f ′ have a common irreducible factor of positive degree in K[t]. However, since f is irreducible, its only factors are 1 and itself. So this can happen if and only if f ′ = 0. To make it more explicit, we can
write Then we can write f = antn + · · · + a1t + a0. f ′ = nantn−1 + · · · + a1. Now f ′ = 0 if and only if all coefficients iai = 0 for all i. (i) Suppose char K = 0, then if deg f = 0, then f is trivially separable. If deg f > 0, then f is not separable iff f ′ = 0 iff iai = 0 for all i iff ai = 0 for i ≥ 1. But we cannot have a polynomial of positive degree with all its coefficients zero (apart from the constant term). So f must be separable. (ii) If deg f = 0, then f is trivially separable. So assume deg f > 0. Then f is not separable ⇔ f ′ = 0 ⇔ iai = 0 for i ≥ 0 ⇔ ai = 0 for all i ≥ 1 not multiples of p ⇔ f ∈ K[tp]. Using this, it should be easy to find lots of examples of separable polynomials. 23 2 Field extensions II Galois Theory Definition (Separable elements and extensions). Let K ⊆ L be an algebraic field extension. We say α ∈ L is separable over K if Pα is separable, where Pα is the minimal polynomial of α over K. We say L is separable over K (or K ⊆ L is separable) if all α ∈ L are separable. Example. – The extensions Q ⊆ Q( √ 2) and R ⊆ C are separable because char Q = char R = 0. So we can apply our previous corollary. – Let L = Fp(s) be the field of rational functions in s over Fp (which is the fraction field of Fp[s]), and K = Fp(sp). We have K ⊆ L, and L = K(s). Since sp ∈ K, s is a root of tp − sp ∈ K[t]. So s is algebraic over K and hence L is algebraic over K. In fact Ps = tp − sp is the minimal polynomial of s over K. Now tp−sp = (t−s)p since the field has characteristic p.
So Roottp−sp (L) = {s}. So Ps is not separable. As mentioned in the beginning, separable extensions are nice, or at least non-weird. One particular nice result about separable extensions is that all finite separable extensions are simple, i.e. if K ⊆ L is finite separable, then L = K(α) for some α ∈ L. This is what we will be working towards for the remaining of the section. Example. Consider Q ⊆ Q( 2, we should be able to generate Q( √ fact, we can use α = 3). This is a separable finite extension. So 2, 3) by just one element, not just two. In 3, since we have 2 + √ √ √ √ √ √ α3 = 11 √ 3 = 2 √ 2 + 9α. 2 + 9 √ So since α3 ∈ Q(α), we know that 2 ∈ Q(α). So we also have √ 3 ∈ Q(α). In general, it is not easy to find an α that works, but we our later result will show that such an α exists. Before that, we will prove some results about the K-homomorphisms. Lemma. Let L/F/K be finite extensions, and E/K be a field extension. Then for all α ∈ L, we have \| HomK(F (α), E)\| ≤ [F (α) : F ]\| HomK(F, E)\|. Note that if Pα is the minimal polynomial of α over F, then [F (α) : F ] = deg Pα. So we can interpret this intuitively as follows: for each ψ ∈ HomK(F, E), we can obtain a K-homomorphism in HomK(F (α), E) by sending things in F according to ψ, and then send α to any root of Pα. Then there are at most [F (α) : F ] K-homomorphisms generated this way. Moreover, each Khomomorphism in HomK(F (α), E) can be created this way. So we get this result. Proof. We show that for each ψ ∈ HomK(F, E), there are at most [F (α) : F ] K-isomorphisms in Hom
K(F (α), E) that restrict to ψ in F. Since each Kisomorphism in HomK(F (α), E) has to restrict to something, it follows that there are at most [F (α) : F ]\| HomK(F, E)\| K-homomorphisms from F (α) to E. 24 2 Field extensions II Galois Theory Now let Pα be the minimal polynomial for α in F, and let ψ ∈ HomK(F, E). To extend ψ to a morphism F (α) → E, we need to decide where to send α. So there should be some sort of correspondence RootPα (E) ←→ {ϕ ∈ HomK(F (α), E) : ϕ\|F = ψ}. Except that the previous sentence makes no sense, since Pα ∈ F [t] but we are not told that F is a subfield of E. So we use our ψ to “move” our things to E. We let M = ψ(F ) ⊆ E, and q ∈ M [t] be the image of Pα under the homomorphism F [t] → M [t] induced by ψ. As we have previously shown, there is a one-to-one correspondence Rootq(E) ←→ HomM (M [t]/⟨q⟩, E). What we really want to show is the correspondence between Rootq(E) and the K-homomorphisms F [t]/⟨Pα⟩ → E that restrict to ψ on F. Let’s ignore the quotient for the moment and think: what does it mean for ϕ ∈ HomK(F [t], E) to restrict to ψ on F? We know that any ϕ ∈ HomL(F [t], E) is uniquely determined by the values it takes on F and t. Hence if ϕ\|F = ψ, then our ϕ must send F to ψ(F ) = M, and can send t to anything in E. This corresponds exactly to the M -homomorphisms M [t] → E that does nothing to M and sends t to that “anything” in E. The situation does not change when we put back the quotient. Changing from M [
t] → E to M [t]/⟨q⟩ → E just requires that the image of t must be a root of q. On the other hand, using F [t]/⟨Pα⟩ instead of F [t] requires that ϕ(Pα(t)) = 0. But we know that ϕ(Pα) = ψ(Pα) = q. So this just requires q(t) = 0 as well. So we get the one-to-one correspondence HomM (M [t]/⟨q⟩, E) ←→ {ϕ ∈ HomK(F [t]/⟨Pα⟩, E) : ϕ\|F = ψ}. Since F [t]/⟨Pα⟩ = F (α), there is a one-to-one correspondence Rootq(E) ←→ {ϕ ∈ HomK(F (α), E) : ϕ\|F = ψ}. So done. Theorem. Let L/K and E/K be field extensions. Then (i) \| HomK(L, E)\| ≤ [L : K]. In particular, \| AutK(L)\| ≤ [L : K]. (ii) If equality holds in (i), then for any intermediate field K ⊆ F ⊆ L: (a) We also have \| HomK(F, E)\| = [F : K]. (b) The map HomK(L, E) → HomK(F, E) by restriction is surjective. Proof. (i) We have previously shown we can find a sequence of field extensions K = F0 ⊆ F1 ⊆ · · · ⊆ Fn = L 25 2 Field extensions II Galois Theory such that for each i, there is some αi such that Fi = Fi−1(αi). Then by our previous lemma, we have \| HomK(L, E)\| ≤ [Fn : Fn−1]\| HomK(Fn−1, E)\| ≤ [Fn : Fn−1][Fn−1 : Fn−2]\| HomK(Fn−2, E)\|... ≤ [Fn : Fn−1][Fn−1 : Fn−2] · · · [F1 : F
0]\| HomK(F0, E)\| = [Fn : F0] = [L : K] (ii) (a) If equality holds in (i), then every inequality in the proof above has to an equality. Instead of directly decomposing K ⊆ L as a chain above, we can first decompose K ⊆ F, then F ⊆ L, then join them together. Then we can assume that F = Fi for some i. Then we get \| HomK(L, E)\| = [L : F ]\| HomK(F, E)\| = [L : K]. Then the tower law says \| HomK(F, E)\| = [F : K]. (b) By the proof of the lemma, for each ψ ∈ HomK(F, E), we know that {ϕ : HomK(L, E) : ϕ\|F = ψ} ≤ [L : F ]. (∗) As we know that \| HomK(F, E)\| = [F : K], \| HomK(L, E)\| = [L : K] we must have had equality in (∗), or else we won’t have enough elements. So in particular {ϕ : HomK(L, E) : ϕ\|F = ψ} ≥ 1. So the map is surjective. With this result, we can prove prove the following result characterizing separable extensions. Theorem. Let L/K be a finite field extension. Then the following are equivalent: (i) There is some extension E of K such that \| HomK(L, E)\| = [L : K]. (ii) L/K is separable. (iii) L = K(α1, · · ·, αn) such that Pαi, the minimal polynomial of αi over K, is separable for all i. (iv) L = K(α1, · · ·, αn) such that Rαi, the minimal polynomial of αi over K(α1, · · ·, αi−1) is separable for all i. Proof. 26 2 Field extensions II Galois Theory – (i) ⇒ (ii): For all α ∈ L, if Pα is the minimal polynomial of α over K, then since K(α) is a subfield of
L, by our previous theorem, we have \| HomK(K(α), E)\| = [K(α) : K]. We also know that \| RootPα (E)\| = \| HomK(K(α), E)\|, and that [K(α) : K] = deg Pα. So we know that Pα has no repeated roots in any splitting field. So Pα is a separable. So L/K is a separable extension. – (ii) ⇒ (iii): Obvious from definition – (iii) ⇒ (iv): Since Rαi is a minimal polynomial in K(α1, · · ·, αi−1), we know that Rαi \| Pαi. So Rαi is separable as Pαi is separable. – (iv) ⇒ (i): Let E be the splitting field of Pα1, · · ·, Pαn. We do induction on n to show that this satisfies the properties we want. If n = 1, then L = K(α1). Then we have \| HomK(L, E)\| = \| RootPαi (E)\| = deg Pα1 = [K(α1) : K] = [L : K]. We now induct on n. So we can assume that (iv) ⇒ (i) holds for smaller number of generators. For convenience, we write Ki = K(α1, · · ·, αi). Then we have \| HomK(Kn−1, E)\| = [Kn−1 : K]. We also know that \| HomK(Kn, E)\| ≤ [Kn : Kn−1]\| HomK(Kn−1, E)\|. What we actually want is equality. We now re-do (parts of) the proof of this result, and see that separability guarantees that equality holds. If we pick ψ ∈ HomK(Kn−1, E), then there is a one-to-one correspondence between {ϕ ∈ HomK(Kn, E) : ϕ\|Kn−1 = ψ} and Rootq(E), where q ∈ M [t] is defined as the image of Rαn under Kn−1[t] → M [t], and M is the image of ψ. Since Pαn ∈ K[t] and Rαn \| Pαn,
then q \| Pαn. So q splits over E. By separability assumption, we get that \| Rootq(E)\| = deg q = deg Rαn = [Kn : Kn−1]. Hence we know that \| HomK(L, E)\| = [Kn : Kn−1]\| HomK(Kn−1, E)\| = [Kn : Kn−1][Kn−1 : K] = [Kn : K]. So done. Before we finally get to the primitive element theorem, we prove the following lemma. This will enable us to prove the trivial case of the primitive element theorem, and will also be very useful later on. Lemma. Let L be a field, L∗ = L \ {0} be the multiplicative group of L. If G is a finite subgroup of L∗, then G is cyclic. 27 2 Field extensions II Galois Theory Proof. Since L∗ is abelian, G is also abelian. Then by the structure theorem on finite abelian groups, G ∼= Z ⟨n1⟩ × · · · × Z, ⟨nr⟩ for some ni ∈ N. Let m be the least common multiple of n1, · · ·, nr, and let f = tm − 1. If α ∈ G, then αm = 1. So f (α) = 0 for all α ∈ G. Therefore \|G\| = n1 · · · nr ≤ \| Rootf (L)\| ≤ deg f = m. Since m is the least common multiple of n1, · · ·, nr, we must have m = n1 · · · nr and thus (ni, nj) = 1 for all i ̸= j. Then by the Chinese remainder theorem, we have G ∼= So G is cyclic. Z ⟨n1nr⟩ ⟨n1 · · · nr⟩ We now come to the main theorem of the lecture: Theorem (Primitive element theorem). Assume L/K is a finite and separable extension. Then L/K is simple, i.e. there is some α ∈ L such that L = K(α). Proof. At some point in our proof, we will require that L is infinite. So we first do the finite case first. If K is finite, then L
is also finite, which in turns implies L∗ is finite too. So by the lemma, L∗ is a cyclic group (since it is a finite subgroup of itself). So there is some α ∈ L∗ such that every element in L∗ is a power of α. So L = K(α). So focus on the case where K is infinite. Also, assume K ̸= L. Then since L/K is a finite extension, there is some intermediate field K ⊆ F ⊊ L such that L = F (β) for some β. Now L/K is separable. So F/K is also separable, and [F : K] < [L : K]. Then by induction on degree of extension, we can assume F/K is simple. In other words, there is some λ ∈ F such that F = K(λ). Now L = K(λ, β). In the rest of the proof, we will try to replace the two generators λ, β with just a single generator. Unsurprisingly, the generator of L will be chosen to be a linear combination of β and λ. We set α = β + aλ for some a ∈ K to be chosen later. We will show that K(α) = L. Actually, almost any choice of a will do, but at the end of the proof, we will see which ones are the bad ones. Let Pβ and Pλ be the minimal polynomial of β and λ over K respectively. Consider the polynomial f = Pβ(α − at) ∈ K(α)[t]. Then we have f (λ) = Pβ(α − aλ) = Pβ(β) = 0. On the other hand, Pλ(λ) = 0. So λ is a common root of Pλ and f. We now want to pick an a such that λ is the only common root of f and Pλ (in E). If so, then the gcd of f and Pα in K(α) must only have λ as a root. But since Pλ is separable, it has no double roots. So the gcd must be t − λ. In particular, we must have λ ∈ K(α). Since α = β + aλ, it follows that β ∈ K(α) as well,
and so K(α) = L. 28 2 Field extensions II Galois Theory Thus, it remains to choose an a such that there are no other common roots. We work in a splitting field of PβPλ, and write Pβ = (t − β1) · · · (t − βm) Pλ = (t − λ1) · · · (t − λn). We wlog β1 = β and λ1 = λ. Now suppose θ is a common root of f and Pλ. Then f (θ) = 0 Pλ(θ) = 0 ⇒ Pβ(α − aθ) = 0 Pλ(θ) = 0 ⇒ α − aθ = βi θ = λj for some i, j. Then we know that α = βi + aλj. However, by definition, we also know that α = β + aλ Now we see how we need to choose a. We need to choose a such that the elements for all i, j. But if they were equal, then we have β + aλ ̸= βi + aλj a = λ − λj βi − β, and there are only finitely many elements of this form. So we just have to pick an a not in this list. Corollary. Any finite extension L/K of field of characteristic 0 is simple, i.e. L = K(α) for some α ∈ L. Proof. This follows from the fact that all extensions of fields of characteristic zero are separable. We have previously seen that Q( 3)/Q is a simple extension, but that 2, is of course true from this theorem. A more interesting example would be one in which this fails. We will need a field with non-zero characteristic. √ √ Example. Let L = Fp(s, u), the fraction field of Fp[s, u]. Let K = Fp(sp, up). We have L/K. We want to show this is not simple. If α ∈ L, then αp ∈ K. So α is a root of tp − αp ∈ K[t]. Thus the minimal polynomial Pα has degree at most p. So [K(α) : K] = deg Pα ≤ p. On the other hand, we
have [L : K] = p2, since {siuj : 0 ≤ i, j < p} is a basis. So for any α, we have K(α) ̸= L. So L/K is not a simple extension. This then implies L/K is not separable. At this point, one might suspect that all fields with positive characteristic are not separable. This is not true by considering a rather silly example. 29 2 Field extensions II Galois Theory Example. Consider K = F2 and L = F2[s]/⟨s2 + s + 1⟩. We can check manually that s2 + s + 1 has no roots and hence irreducible. So L is a field. So L/F2 is a finite extension. Note that L only has 4 elements. Now if α ∈ L \ F2, and Pα is the minimal polynomial of α over F2, then Pα \| t2 + t + 1. So Pα is separable as a polynomial. So L/F2 is separable. In fact, we have Proposition. Let L/K be an extension of finite fields. Then the extension is separable. Proof. Let the characteristic of the fields be p. Suppose the extension were not separable. Then there is some non-separable element α ∈ L. Then its minimal polynomial must be of the form Pα = aitpi. Now note that the map K → K given by x → xp is injective, hence surjective. So we can write ai = bp i for all i. Then we have Pα = aitpi = bitip, and so Pα is not irreducible, which is a contradiction. 2.7 Normal extensions We are almost there. We will now move on to study normal extensions. Normal extensions are very closely related to Galois extensions. In fact, we will show that if an extension is normal and separable, then it is Galois. The advantage of introducing the idea of normality is that normality is a much more concrete definition to work with. It is much easier to check if an extension is normal than to check if \| AutK(L)\| = [K : L]. In particular, we will shortly prove that the splitting field of any polynomial is normal. This is an important result, since we are going to use the splitting field
to study the roots of a polynomial, and since we mostly care about polynomials over Q, this means all these splitting fields are automatically Galois extensions of Q. It is not immediately obvious why these extensions are called “normal” (just like most other names in Galois theory). We will later see that normal extensions are extensions that correspond to normal subgroups, in some precise sense given by the fundamental theorem of Galois theory. Definition (Normal extension). Let K ⊆ L be an algebraic extension. We say L/K is normal if for all α ∈ L, the minimal polynomial of α over K splits over L. In other words, given any minimal polynomial, L should have all its roots. √ Example. The extension Q( 3 √ t3 − 2 does not split over Q( 3 2)/Q is not normal since the minimal polynomial 2). In some sense, extensions that are not “normal” are missing something. This is somewhat similar to how Galois extensions work. Before we go deeper into this, we need a lemma. Lemma. Let L/F/K be finite extensions, and ¯K is the algebraic closure of K. Then any ψ ∈ HomK(F, ¯K) extends to some ϕ ∈ HomK(L, ¯K). 30 2 Field extensions II Galois Theory Proof. Let ψ ∈ HomK(F, ¯K). If F = L, then the statement is trivial. So assume L ̸= F. Pick α ∈ L \ F. Let qα ∈ F [t] be the minimal polynomial of α over F. Consider ψ(qα) ∈ ¯K[t]. Let β be any root of qα, which exists since ¯K is algebraically closed. Then as before, we can extend ψ to F (α) by sending α to β. More explicitly, we send aiαi → ψ(ai)βi, N i=0 which is well-defined since any polynomial relation satisfied by α in F is also satisfied by β. Repeat this process finitely many times to get some element in HomK(L, ¯K). We will use this lemma to characterize normal extensions. Theorem. Let L/K be a finite extension. Then L/K is a normal extension if and only if L is
the splitting field of some f ∈ K[t]. Proof. Suppose L/K is normal. Since L is finite, let L = K(α1, · · ·, αn) for some αi ∈ L. Let Pαi be the minimal polynomial of αi over K. Take f = Pα1 · · · Pαn. Since L/K is normal, each Pαi splits over L. So f splits over L, and L is a splitting field of f. For the other direction, suppose that L is the splitting field of some f ∈ K[t]. First we wlog assume L ⊆ ¯K. This is possible since the natural injection K → ¯K extends to some ϕ : L → ¯K by our previous lemma, and we can replace L with ϕ(L). Now suppose β ∈ L, and let Pβ be its minimal polynomial. Let β′ be another root. We want to show it lives in L. Now consider K(β). By the proof of the lemma, we can produce an embedding ι : K(β) → ¯K that sends β to β′. By the lemma again, this extends to an embedding of L into ¯K. But any such embedding must send a root of f to a root of f. So it must send L to L. In particular, ι(β) = β′ ∈ L. So Pβ splits over L. This allows us to identify normal extensions easily. The following theorem then allows us to identify Galois extensions using this convenient tool. Theorem. Let L/K be a finite extension. Then the following are equivalent: (i) L/K is a Galois extension. (ii) L/K is separable and normal. (iii) L = K(α1, · · ·, αn) and Pαi, the minimal polynomial of αi over K, is separable and splits over L for all i. Proof. – (i) ⇒ (ii): Suppose L/K is a Galois extension. Then by definition, this means \| HomK(L, L)\| = \| AutK(L)\| = [L : K]. 31 2 Field extensions II Galois Theory To show that L/K is separable, recall that we proved that an extension is separable if and only if there is some E such
that \| HomK(L, E)\| = [L : K]. In this case, just pick E = L. Then we know that the extension is separable. To check normality, let α ∈ L, and let Pα be its minimal polynomial over K. We know that \| RootPα(L)\| = \| HomK(K[t]/⟨Pα⟩, L)\| = \| HomK(K(α), L)\|. But since \| HomK(L, L)\| = [L : K] and K(α) is a subfield of L, this implies \| HomK(K(α), L)\| = [K(α) : K] = deg Pα. Hence we know that So Pα splits over L. \| RootPα (L)\| = deg Pα. – (ii) ⇒ (iii): Just pick α1, · · ·, αn such that L = K(α1, · · ·, αn). Then these polynomials are separable since the extension is separable, and they split since L/K is normal. In fact, by the primitive element theorem, we can pick these such that n = 1. – (iii) ⇒ (i): Since L = K(α1, · · ·, αn) and the minimal polynomials Pαi over K are separable, by a previous theorem, there are some extension E of K such that \| HomK(L, E)\| = [L : K]. To simplify notation, we first replace L with its image inside E under some K-homomorphism L → E, which exists since \| HomK(L, E)\| = [L : K] > 0. So we can assume L ⊆ E. We now claim that the inclusion HomK(L, L) → HomK(L, E) is a surjection, hence a bijection. Indeed, if ϕ : L → E, then ϕ takes αi to ϕ(αi), which is a root of Pαi. Since Pαi splits over L, we know ϕ(αi) ∈ L for all i. Since L is generated by these αi, it follows that ϕ(L) ⊆ L. Thus, we have [L : K] = \| HomK(L, E)\| = \| Hom
K(L, L)\|, and the extension is Galois. From this, it follows that if L/K is Galois, and we have an intermediate field K ⊆ F ⊆ L, then L/F is also Galois. Corollary. Let K be a field and f ∈ K[t] be a separable polynomial. Then the splitting field of f is Galois. This is one of the most crucial examples. 32 2 Field extensions II Galois Theory 2.8 The fundamental theorem of Galois theory Finally, we can get to the fundamental theorem of Galois theory. Roughly, given a Galois extension K ⊆ L, the fundamental theorem tell us there is a one-to-one correspondence between intermediate field extensions K ⊆ F ⊆ L and subgroups of the automorphism group Gal(L/K). Given an intermediate field F, we can obtain a subgroup of Gal(L/K) by looking at the automorphisms that fix F. To go the other way round, given a subgroup H ≤ Gal(L/K), we can obtain a corresponding field by looking at the field of elements that are fixed by everything in H. This is known as the fixed field, and can in general be defined even for non-Galois extensions. Definition (Fixed field). Let L/K be a field extension, H ≤ AutK(L) a subgroup. We define the fixed field of H as LH = {α ∈ L : ϕ(α) = α for all ϕ ∈ H}. It is easy to see that LH is an intermediate field K ⊆ LH ⊆ L. Before we get to the fundamental theorem, we first prove Artin’s lemma. This in fact proves part of the results in the fundamental theorem, but is also useful on its own right. Lemma (Artin’s lemma). Let L/K be a field extension and H ≤ AutK(L) a finite subgroup. Then L/LH is a Galois extension with AutLH (L) = H. Note that we are not assuming that L/K is Galois, or even finite! Proof. Pick any α ∈ L. We set {α1, · · ·, αn} = {ϕ(α) : ϕ ∈ H}, where αi are distinct. Here we are
allowing for the possibility that ϕ(α) = ψ(α) for some distinct ϕ, ψ ∈ H. By definition, we clearly have n < \|H\|. Let n (t − αi) ∈ L[t]. f = 1 We know that any ϕ ∈ H gives an homomorphism L[t] → L[t], and any such map fixes f because ϕ just permutes the αi. Thus, the coefficients of f are in LH, and thus f ∈ LH [t]. Since id ∈ H, we know that f (α) = 0. So α is algebraic over LH. Moreover, if qα is the minimal polynomial of α over LH, then qα \| f in LH [t]. Hence [LH (α) : LH ] = deg qα ≤ deg f ≤ \|H\|. Further, we know that f has distinct roots. So qα is separable, and so α is separable. So it follows that L/LH is a separable extension. We next show that L/LH is simple. This doesn’t immediately follow from the primitive element theorem, because we don’t know it is a finite extension yet, but we can still apply the theorem cleverly. Pick α ∈ L such that [LH (α) : LH ] is maximal. This is possible since [LH (α) : LH ] is bounded by \|H\|. The claim is that L = LH (α). 33 2 Field extensions II Galois Theory We pick an arbitrary β ∈ L, and will show that this is in LH (α). By the above arguments, LH ⊆ LH (α, β) is a finite separable extension. So by the primitive element theorem, there is some λ ∈ L such that LH (α, β) = LH (λ). Note that we must have [LH (λ) : LH ] ≥ [LH (α) : LH ]. By maximality of [LH (α) : LH ], we must have equality. So LH (λ) = LH (α). So β ∈ LH (α). So L = LH (α). Finally, we show it is a Galois extension. Let L = LH (α). Then [L : LH ] = [LH (α) : LH ] ≤ \|H\| ≤ \| AutLH
(L)\| Recall that we have previously shown that for any extension L/LH, we have \| AutLH (L)\| ≤ [L : LH ]. Hence we must have equality above. So [L : LH ] = \| AutLH (L)\|. So the extension is Galois. Also, since we know that H ⊆ AutLH (L), we must have H = AutLH (L). Theorem. Let L/K be a finite field extension. Then L/K is Galois if and only if LH = K, where H = AutK(L). Proof. (⇒) Suppose L/K is a Galois extension. We want to show LH = K. Using Artin’s lemma (and the definition of H), we have [L : K] = \| AutK(L)\| = \|H\| = \| AutLH (L)\| = [L : LH ] So [L : K] = [L : LH ]. So we must have LH = K. (⇐) By the lemma, K = LH ⊆ L is Galois. This is an important theorem. Given a Galois extension L/K, this gives us a very useful test of when elements of α ∈ L are in fact in K. We will use this a lot. Finally, we get to the fundamental theorem. Theorem (Fundamental theorem of Galois theory). Assume L/K is a (finite) Galois extension. Then (i) There is a one-to-one correspondence H ≤ AutK(L) ←→ intermediate fields K ⊆ F ⊆ L. This is given by the maps H → LH and F → AutF (L) respectively. Moreover, \| AutK(L) : H\| = [LH : K]. (ii) H ≤ AutK(L) is normal (as a subgroup) if and only if LH /K is a normal extension if and only if LH /K is a Galois extension. (iii) If H ◁ AutK(L), then the map AutK(L) → AutK(LH ) by the restriction map is well-defined and surjective with kernel isomorphic to H, i.e. AutK(L) H = AutK(LH ). 34 2 Field extensions II Galois Theory Proof. Note that since L/K is a
Galois extension, we know \| AutK(L)\| = \| HomK(L, L)\| = [L : K], By a previous theorem, for any intermediate field K ⊆ F ⊆ L, we know \| HomK(F, L)\| = [F : K] and the restriction map HomK(L, L) → HomK(F, L) is surjective. (i) The maps are already well-defined, so we just have to show that the maps are inverses to each other. By Artin’s lemma, we know that H = AutLH (L), and since L/F is a Galois extension, the previous theorem tells that LAutF (L) = F. So they are indeed inverses. The formula relating the index and the degree follows from Artin’s lemma. (ii) Note that for every ϕ ∈ AutK(L), we have that LϕHϕ−1 = ϕLH, since iff ϕ(ψ(ϕ−1(α))) = α for all ψ ∈ H iff ψ(ϕ−1(α)) = ϕ−1(α) α ∈ LϕHϕ−1 for all ψ ∈ H iff α ∈ ϕLH. Hence H is a normal subgroup if and only if ϕ(LH ) = LH for all ϕ ∈ AutK(L). (∗) Assume (∗). We want to first show that HomK(LH, LH ) = HomK(LH, L). Let ψ ∈ HomK(LH, L). Then by the surjectivity of the restriction map HomK(L, L) → HomK(LH, L), ψ must be the restriction of some ˜ψ ∈ HomK(L, L). So ˜ψ fixes LH by (∗). So ψ sends LH to LH. So ψ ∈ HomK(LH, LH ). So we have \| AutK(LH )\| = \| HomK(LH, LH )\| = \| HomK(LH, L)\| = [LH : K]. So LH /K is Galois, and hence normal. Now suppose LH /K is a normal extension. We want to
show this implies (∗). Pick any α ∈ LH and ϕ ∈ AutK(L). Let Pα be the minimal polynomial of α over K. So ϕ(α) is a root of Pα (since ϕ fixes Pα ∈ K, and hence maps roots to roots). Since LH /K is normal, Pα splits over LH. This implies that ϕ(α) ∈ LH. So ϕ(LH ) = LH. Hence, H is a normal subgroup if and only if ϕ(LH ) = LH if and only if LH /K is a Galois extension. (iii) Suppose H is normal. We know that AutK(L) = HomK(L, L) restricts to HomK(LH, L) surjectively. To show that we in fact have restriction to AutK(LH ), by the proof above, we know that ϕ(LH ) = LH for all ϕ ∈ AutK(LH ). So this does restrict to an automorphism of LH. In other words, the map AutK(L) → AutK(LH ) is well-defined. It is easy to see this is a group homomorphism. Finally, we have to calculate the kernel of this homomorphism. Let E be the kernel. Then by definition, E ⊇ H. So it suffices to show that \|E\| = \|H\|. By surjectivity of the map and the first isomorphism theorem of groups, we have \| AutK(L)\| \|E\| = \| AutK(LH )\| = [LH : K] = [L : K] [L : LH ] = \| AutK(L)\| \|H\|, noting that LH /K and L/K are both Galois extensions, and \|H\| = [LH : K] by Artin’s lemma. So \|E\| = \|H\|. So we must have E = H. 35 2 Field extensions II Galois Theory √ √ √ Example. Let p be an odd prime, and ζp be a primitive pth root of unity. Given n ∈ Q(ζp) if and a (square-free) integer n, when is only if Q( n) is a quadratic extension. √ n in Q(ζp)? We know
that √ n) : Q] = 2, i.e. Q( n) ⊆ Q(ζp). Moreover, [Q( We will later show that Gal(Q(ζp)/Q) ∼= (Z/pZ)∗ ∼= Cp−1. Then by the fundamental theorem of Galois theory, quadratic extensions contained in Q(ζp) correspond to index 2-subgroups of Gal(Q(ζp)/Q). By general group theory, there is exactly one such subgroup. So there is exactly one square-free n such that Q( n)), given by the fixed field of the index 2 subgroup of (Z/pZ)∗. n) ⊆ Q(ζp) (since all quadratic extensions are of the form Q( √ √ Now we shall try to find some square root lying in Q(ζp). We will not fully justify the derivation, since we can just square the resulting number to see that it is correct. We know the general element of Q(ζp) looks like p−1 k=0 ckζ k p. We know Gal(Q(ζp)/Q) ∼= (Z/pZ)∗ acts by sending ζp → ζ n p for each n ∈ (Z/pZ)∗, and the index 2 subgroup consists of the quadratic residues. Thus, if an element is fixed under the action of the quadratic residues, the quadratic residue powers all have the same coefficient, and similarly for the non-residue powers. If we wanted this to be a square root, then the action of the remaining elements of Gal(Q(ζp)/Q) should negate this object. Since these elements swap the residues and the non-residues, we would want to have something like ck = 1 if k is a quadratic residue, and −1 if it is a non-residue, which is just the Legendre symbol! So we are led to try to square τ = p−1 k=1 k p ζ k p. It is an exercise in the Number Theory example sheet to show that the square of this is in fact τ 2 = −1 p p. So we have √ p ∈ Q(ζp) if p ≡ 1 (mod 4
), and √ −p ∈ Q(ζp) if p ≡ 3 (mod 4). 2.9 Finite fields We’ll have a slight digression and look at finite fields. We adopt the notation where p is always a prime number, and Zp = Z/⟨p⟩. It turns out finite fields are rather simple, as described in the lemma below: Lemma. Let K be a finite field with q = \|K\| element. Then (i) q = pd for some d ∈ N, where p = char K > 0. (ii) Let f = tq − t. Then f (α) = 0 for all α ∈ K. Moreover, K is the splitting field of f over Fp. This means that a finite field is completely determined by the number of elements. 36 2 Field extensions II Galois Theory Proof. (i) Consider the set {m · 1K}m∈Z, where 1K is the unit in K and m· represents repeated addition. We can identify this with Fp. So we have the extension Fp ⊆ K. Let d = [K : Fp]. Then q = \|K\| = pd. (ii) Note that K ∗ = K \ {0} is a finite multiplicative group with order q − 1. Then by Lagrange’s theorem, αq−1 = 1 for all α ∈ K ∗. So αq − α = 0 for all α ̸= 0. The α = 0 case is trivial. Now every element in K is a root of f. So we need to check that all roots of f are in K. Note that the derivative f ′ = qtq−1 − 1 = −1 (since q is a power of the characteristic). So f ′(α) = −1 ̸= 0 for all α ∈ K. So f and f ′ have no common roots. So f has no repeated roots. So K contains q distinct roots of f. So K is a splitting field. Lemma. Let q = pd, q′ = pd′, where d, d′ ∈ N. Then (i) There is a finite field K with exactly q elements, which is unique up to isomorphism. We write this as Fq. (ii) We can embed Fq ⊆ F
q′ iff d \| d′. Proof. (i) Let f = tq − t, and let K be a splitting field of f over Fp. Let L = Rootf (K). The objective is to show that L = K. Then we will have \|K\| = \|L\| = \| Rootf (K)\| = deg f = q, because the proof of the previous lemma shows that f has no repeated roots. To show that L = K, by definition, we have L ⊆ K. So we need to show every element in K is in L. We do so by showing that L itself is a field. Then since L contains all the roots of f and is a subfield of the splitting field K, we must have K = L. It is straightforward to show that L is a field: if α, β ∈ L, then (α + β)q = αq + βq = α + β. So α + β ∈ L. Similarly, we have (αβ)q = αqβq = αβ. So αβ ∈ L. Also, we have (α−1)q = (αq)−1 = α−1. So α−1 ∈ L. So L is in fact a field. Since any field of size q is a splitting field of f, and splitting fields are unique to isomorphism, we know that K is unique. (ii) Suppose Fq ⊆ Fq′. Then let n = [Fq′ : Fq]. So q′ = qn. So d′ = nd. So d \| d′. On the other hand, suppose d \| d′. Let d′ = dn. We let f = tq′ for any α ∈ Fq, we have − t. Then f (α) = αq′ − α = αqn − α = (· · · ((αq)q)q · · · )q − α = α − α = 0. Since Fq′ is the splitting field of f, all roots of f are in Fq′. So we know that Fq ⊆ F′ q. 37 2 Field extensions II Galois Theory Note that if ¯Fp is the algebraic closure of Fp, then Fq ⊆ ¯Fp for every q = pd. We then have Fpk = ¯Fp,
k∈N because any α ∈ ¯Fp is algebraic over Fp, and so belongs to some Fq. Definition. Consider the extension Fqn /Fq, where q is a power of p. The Frobenius Frq : Fqn → Fqn is defined by α → αq. This is a homomorphism precisely because the field is of characteristic zero. In fact, Frq ∈ AutFq (Fqn ), since αq = α for all α ∈ Fq. The following two theorems tells us why we care about the Frobenius. Theorem. Consider Fqn /Fq. Then Frq is an element of order n as an element of AutFq (Fqn ). Proof. For all α ∈ Fqn, we have Frn n. q (α) = αqn If m \| n, then the set = α. So the order of Frq divides {α ∈ Fqn : Frm q (α) = α} = {α ∈ Fqn : αqm = α} = Fqm. So if m is the order of Frq, then Fqm = Fqn. So m = n. Theorem. The extension Fqn/Fq is Galois with Galois group Gal(Fqn /Fq) = AutFq (Fqn ) ∼= Z/nZ, generated by Frq. qn = Fqn \ {0} is finite. We have previously Proof. The multiplicative group F∗ seen that multiplicative groups of finite fields are cyclic. So let α be a generator of this group. Then Fqn = Fq(α). Let Pα be the minimal polynomial of α over Fq. Then since AutFq (Fqn ) has an element of order n, we get n ≤ \| AutFq (Fqn)\| = \| HomFq (Fq(α), Fqn )\|. Since Fq(α) is generated by one element, we know \| HomFq (Fq(α), Fqn )\| = \| RootPα(Fqn )\| So we have n ≤ \| RootPα (Fqn )\| ≤ deg Pα = [Fqn : Fq] = n. So we know that \| Aut
Fq (Fqn )\| = [Fqn : Fq] = n. So Fqn /Fq is a Galois extension. Since \| AutFq (Fqn )\|, it has to be generated by Frq, since this has order n. In particular, this group is cyclic. We see that finite fields are rather nice — there is exactly one field of order pd for each d and prime p, and these are all of the finite fields. All extensions are Galois and the Galois group is a simple cyclic group. 38 2 Field extensions II Galois Theory Example. Consider F4/F2. We can write F2 = {0, 1} ⊆ F4 = {0, 1, α, α2}, where α is a generator of F∗ 4. Define ϕ ∈ AutF2(F4) by ϕ(α) = α2. Then AutF2(F4) = {id, ϕ} since it has order 2. Note that we can also define the Frobenius Frp : ¯Fp → ¯Fp, where α → αp. p. So we can recover this subfield by Then Fpd is the elements of ¯Fp fixed by Frd just looking at the Frobenius. 39 3 Solutions to polynomial equations II Galois Theory 3 Solutions to polynomial equations We have now proved the fundamental theorem of Galois theory, and this gives a one-to-one correspondence between (intermediate) field extensions and subgroups of the Galois group. That is our first goal achieved. Our next big goal is to use this Galois correspondence to show that, in general, polynomials of degree 5 or more cannot be solved by radicals. First of all, we want to make this notion of “solving by radicals” precise. We all know what this means if we are working over Q, but we need to be very precise when working with arbitrary fields. For example, we know that the polynomial f = t3 − 5 ∈ Q[t] can be “solved by radicals”. In this case, we have √ Rootf (C) = { 3 √ 5, µ 3 √ 5, µ2 3 5}, where µ3 = 1, µ ̸= 1. In general fields, we want to properly define the analogues
√ of µ and 3 5. These will correspond to two different concepts. The first is cyclotomic extensions, where the extension adds the analogues of µ, and the second is √ Kummer extensions, where we add things like 3 5. Then, we would say a polynomial is soluble by radicals if the splitting field of the polynomial can be obtained by repeatedly taking cyclotomic and Kummer extensions. 3.1 Cyclotomic extensions Definition (Cyclotomic extension). For a field K, we define the nth cyclotomic extension to be the splitting field of tn − 1. Note that if K is a field and L is the nth cyclotomic extension, then Roottn−1(L) is a subgroup of multiplicative group L∗ = L \ {0}. Since this is a finite subgroup of L∗, it is a cyclic group. Moreover, if char K = 0 or 0 < char K ∤ n, then (tn − 1)′ = ntn−1 and this has no common roots with tn − 1. So tn − 1 has no repeated roots. In other words, tn − 1 has n distinct roots. So as a group, Roottn−1(L) ∼= Z/nZ. In particular, this group has at least one element µ of order n. Definition (Primitive root of unity). The nth primitive root of unity is an element of order n in Roottn−1(L). These elements correspond to the elements of the multiplicative group of units in Z/nZ, written (Z/nZ)×. The next theorem tells us some interesting information about these roots and some related polynomials. Theorem. For each d ∈ N, there exists a dth cyclotomic monic polynomial ϕd ∈ Z[t] satisfying: (i) For each n ∈ N, we have tn − 1 = ϕd. d\|n 40 3 Solutions to polynomial equations II Galois Theory (ii) Assume char K = 0 or 0 < char K ∤ n. Then Rootϕn (L) = {nth primitive roots of unity}. Note that here we have an abuse of notation, since ϕn is a polynomial in Z[t], not K[t], but we can just use the canonical map Z[t] → K
[t] mapping 1 to 1 and t to t. Proof. We do induction on n to construct ϕn. When n = 1, let ϕ1 = t − 1. Then (i) and (ii) hold in this case, trivially. Assume now that (i) and (ii) hold for smaller values of n. Let f = ϕd. d\|n,d<n By induction, f ∈ Z[t]. Moreover, if d \| n and d < n, then ϕd \| (tn − 1) because (td − 1) \| (tn − 1). We would like to say that f also divides tn − 1. However, we have to be careful, since to make this conclusion, we need to show that ϕd and ϕd′ have no common roots for distinct d, d′ \| n (and d,, d′ < n). Indeed, by induction, ϕd and ϕ′ d have no common roots because Rootϕd (L) = {dth primitive roots of unity}, Rootϕd′ (L) = {d′th primitive roots of unity}, and these two sets are disjoint (or else the roots would not be primitive). Therefore ϕd and ϕd′ have no common irreducible factors. Hence f \| tn − 1. So we can write tn − 1 = f ϕn, where ϕn ∈ Q[t]. Since f is monic, ϕn has integer coefficients. So indeed ϕn ∈ Z[t]. So the first part is proven. To prove the second part, note that by induction, Rootf (L) = {non-primitive nth roots of unit}, since all nth roots of unity are dth primitive roots of unity for some smaller d. Since f ϕn = tn − 1, ϕn contains the remaining, primitive nth roots of unit. Since tn − 1 has no repeated roots, we know that ϕn does not contain any extra roots. So Rootϕn (L) = {nth primitive roots of unity}. These ϕn are what we use to “build up” the polynomials tn − 1. These will later serve as a technical tool to characterize the Galois group of the nth cyclotomic extension of Q. Before we an reach that
, we first take a tiny step, and prove something that works for arbitrary fields first. Theorem. Let K be a field with char K = 0 or 0 < char K ∤ n. Let L be the nth cyclotomic extension of K. Then L/K is a Galois extension, and there is an injective homomorphism θ : Gal(L/K) → (Z/nZ)×. In addition, every irreducible factor of ϕn (in K[t]) has degree [L : K]. 41 3 Solutions to polynomial equations II Galois Theory The important thing about our theorem is the homomorphism θ : Gal(L/K) → (Z/nZ)×. In general, we don’t necessarily know much about Gal(L/K), but the group (Z/nZ)× is well-understood. In particular, we now know that Gal(L/K) is abelian. Proof. Let µ be an nth primitive root of unity. Then Roottn−1(L) = {1, µ, µ2, · · ·, µn−1} is a cyclic group of order n generated by µ. We first construct the homomorphism θ : AutK(L) → (Z/nZ)× as follows: for each ϕ ∈ AutK(L), ϕ is completely determined by the value of ϕ(µ) since L = K(µ). Since ϕ is an automorphism, it must take an nth primitive root of unity to another nth primitive root of unity. So ϕ(µ) = µi for some i such that (i, n) = 1. Now let θ(ϕ) = ¯i ∈ (Z/nZ)×. Note that this is well-defined since if µi = µj, then i − j has to be a multiple of n. Now it is easy to see that if ϕ, ψ ∈ AutK(L) are given by ϕ(µ) = µi, and ψ(µ) = µj, then ϕ ◦ ψ(µ) = ϕ(µj) = µij. So θ(ϕψ) = ¯ij = θ(ϕ)θ(ψ).
So θ is a group homomorphism. Now we check that θ is injective. If θ(ϕ) = ¯1 (note that (Z/nZ)× is a multiplicative group with unit 1), then ϕ(µ) = µ. So ϕ = id. Now we show that L/K is Galois. Recall that L = K(µ), and let Pµ be a minimal polynomial of µ over K. Since µ is a root of tn − 1, we know that Pµ \| tn − 1. Since tn − 1 has no repeated roots, Pµ has no repeated roots. So Pµ is separable. Moreover, Pµ splits over L as tn − 1 splits over L. So the extension is separable and normal, and hence Galois. Applying the previous theorem, each irreducible factor g of ϕn is a minimal polynomial of some nth primitive root of unity, say λ. Then L = K(λ). So deg g = deg Pλ = [K(λ) : K] = [L : K]. Example. We can calculate the following in Q[t]. (i) ϕ1 = t − 1. (ii) ϕ2 = t + 1 since t2 − 1 = ϕ1ϕ2. (iii) ϕ3 = t2 + t + 1. (iv) ϕ4 = t2 + 1. These are rather expected. Now take K = F2. Then 1 = −1. So we might be able to further decompose these polynomials. For example, t + 1 = t − 1 in F2. So we have ϕ4 = t2 + 1 = t2 − 1 = ϕ1ϕ2. So in F2, ϕ4 is not irreducible. Similarly, if we have too much time, we can show that ϕ15 = (t4 + t + 1)(t4 + t3 + 1). So ϕ15 is not irreducible. However, they are irreducible over the rationals, as we will soon see. 42 3 Solutions to polynomial equations II Galois Theory So far, we know Gal(L/K) is an abelian group, isomorphic to a subgroup of (Z/nZ)×
. However, we are greedy and we want to know more. The following lemma tells us when this θ is an isomorphism. Lemma. Under the notation and assumptions of the previous theorem, ϕn is irreducible in K[t] if and only if θ is an isomorphism. Proof. (⇒) Suppose ϕn is irreducible. Recall that Rootϕn(L) is exactly the nth primitive roots of unity. So if µ is an nth primitive root of unity, then Pµ, the minimal polynomial of µ over K is ϕn. In particular, if λ is also an nth primitive root of unity, then Pµ = Pλ. This implies that there is some ϕλ ∈ AutK(L) such that ϕλ(µ) = λ. Now if ¯i ∈ (Z/nZ)×, then taking λ = µi, this shows that we have ϕλ ∈ AutK(L) such that θ(ϕλ) = ¯i. So θ is surjective, and hence an isomorphism. (⇔) Suppose that θ is an isomorphism. We will reverse the above argument and show that all roots have the same minimal polynomial. Let µ be a nth primitive root of unity, and pick ¯i ∈ (Z/nZ)×, and let λ = µi. Since θ is an isomorphism, there is some ϕλ ∈ AutK(L) such that θ(ϕλ) = ¯i, i.e. ϕλ(µ) = µi = λ. Then we must have Pµ = Pλ. Since every nth primitive root of unity is of the form µi (with (i, n) = 1), this implies that all nth primitive roots have the same minimal polynomial. Since the roots of ϕn are all the nth primitive roots of unity, its irreducible factors are exactly the minimal polynomials of the primitive roots. Moreover, ϕn does not have repeated roots. So ϕn = Pµ. In particular, ϕn is irreducible. We want to apply this lemma to the case of rational numbers. We want to show
that θ is an isomorphism. So we have to show that ϕn is irreducible in Q[t]. Theorem. ϕn is irreducible in Q[t]. In particular, it is also irreducible in Z[t]. Proof. As before, this can be achieved by showing that all nth primitive roots have the same minimal polynomial. Moreover, let µ be our favorite nth primitive root. Then all other primitive roots λ are of the form λ = µi, where (i, n) = 1. By the fundamental theorem of arithmetic, we can write i as a product i = q1 · · · qn. Hence it suffices to show that for all primes q ∤ n, we have Pµ = Pµq. Noting that µq is also an nth primitive root, this gives Pµ = Pµq1 = P(µq1 )q2 = Pµq1q2 = · · · = Pµq1 ···qr = Pµi. So we now let µ be an nth primitive root, Pµ be its minimal polynomial. Since µ is a root of ϕn, we can write Pµ \| ϕn inside Q[t]. So we can write ϕn = PµR, Since ϕn and Pµ are monic, R is also monic. By Gauss’ lemma, we must have Pµ, R ∈ Z[t]. Note that showing Pµ = Pµq is the same as showing µq is a root of Pµ, since deg Pµ = deg Pµq. So suppose it’s not. Since µq is an nth primitive root of unity, it is a root of ϕn. So µq must be a root of R. Now let S = R(tq). Then µ is a root of S, and so Pµ \| S. We now reduce mod q. For any polynomial f ∈ Z[t], we write the result of. Since reducing the coefficients mod q as ¯f. Then we have ¯S = R(tq) = R(t) q 43 3 Solutions to polynomial equations II Galois Theory ¯Pµ divides ¯S (by Gauss
’ lemma), we know ¯Pµ and R(t) have common roots. But ¯ϕn = ¯Pµ ¯R, and so this implies ¯ϕn has repeated roots. This is impossible since ¯ϕn divides tn − 1, and since q ∤ n, we know the derivative of tn − 1 does not vanish at the roots. So we are done. Corollary. Let K = Q and L be the nth cyclotomic extension of Q. Then the injection θ : Gal(L/Q) → (Z/nZ)× is an isomorphism. Example. Let p be a prime number, and q = pd, d ∈ N. Consider Fq, a field with q elements, and let L be the nth cyclotomic extension of Fq (where p ∤ n). Then we have a homomorphism θ : Gal(L/Fq) → (Z/nZ)×. We have previously shown that Gal(L/Fq) must be a cyclic group. So if (Z/nZ)× is non-cyclic, then θ is not an isomorphism, and ϕn is not irreducible in Fq[t]. For example, take p = q = 7 and n = 8. Then (Z/8Z)× = {¯1, ¯3, ¯5, ¯7} is not cyclic, because manual checking shows that there is no element of order 4. Hence θ : Gal(L/F7) → (Z/8Z)× is not an isomorphism, and ϕ8 is not irreducible in F7[t]. 3.2 Kummer extensions We shall now consider a more general case, and study the splitting field of tn − λ ∈ K[t]. As we have previously seen, we will need to make use of the nth primitive roots of unity. The definition of a Kummer extension will involve a bit more that it being the splitting field of tn − λ. So before we reach the definition, we first studying some properties of an arbitrary splitting field of tn − λ, and use this to motivate the definition of a Kummer extension. Definition (Cyclic extension). We say a Galois extension L/K is cyclic is Gal(L/K) is a cycl
ic group. Theorem. Let K be a field, λ ∈ K non-zero, n ∈ N, char K = 0 or 0 < char K ∤ n. Let L be the splitting field of tn − λ. Then (i) L contains an nth primitive root of unity, say µ. (ii) L/K(µ) is a cyclic (and in particular Galois) extension with degree [L : K(µ)] \| n. (iii) [L : K(µ)] = n if and only if tn − λ is irreducible in K(µ)[t]. Proof. (i) Under our assumptions, tn − λ and (tn − λ)′ = ntn−1 have no common roots in L. So tn − λ has distinct roots in L, say α1, · · ·, αn ∈ L. It then follows by direct computation that α1α−1 are distinct roots of unity, i.e. roots of tn − 1. Then one of these, say µ must be an nth primitive root of unity. 1, · · ·, αnα−1 1 1, α2α−1 44 3 Solutions to polynomial equations II Galois Theory (ii) We know L/K(µ) is a Galois extension because it is the splitting field of the separable polynomial tn − λ. To understand the Galois group, we need to know how this field exactly looks like. We let α be any root of tn − λ. Then the set of all roots can be written as {α, µα, µ2α, · · ·, µn−1α} Then L = K(α1, · · ·, αn) = K(µ, α) = K(µ)(α). Thus, any element of Gal(L/K(µ)) is uniquely determined by what it sends α to, and any homomorphism must send α to one of the other roots of tn − λ, namely µiα for some i. Define a homomorphism σ : Gal(L/K(µ)) → Z/nZ that sends ϕ to the corresponding i (as an element of Z/nZ, so that it is well-defined). It is easy to see
that σ is an injective group homomorphism. So we know Gal(L/K(µ)) is isomorphic to a subgroup of Z/nZ. Since the subgroup of any cyclic group is cyclic, we know that Gal(L/K(µ)) is cyclic, and its size is a factor of n by Lagrange’s theorem. Since \| Gal(L/K(µ))\| = [L : K(µ)] by definition of a Galois extension, it follows that [L : K(µ)] divides n. (iii) We know that [L : K(µ)] = [K(µ, α) : K(µ)] = deg qα. So [L : K(µ)] = n if and only if deg qα = n. Since qα is a factor of tn − λ, deg qα = n if and only if qα = tn − λ. This is true if and only if tn − λ is irreducible K(µ)[t]. So done. Example. Consider t4 + 2 ∈ Q[t]. Let µ = of unity. Now √ −1, which is a 4th primitive root t4 + 2 = (t − α)(t + α)(t − µα)(t + µα), √ where α = 4 Q ⊆ Q(µ) ⊆ Q(µ, α), where Q(µ, α) is a splitting field of t4 + 2. −2 is one of the roots of t4 + 2. Then we have the field extension Since −2 ̸∈ Q(µ), we know that t4 + 2 is irreducible in Q(µ)[t] by looking at the factorization above. So by our theorem, Q(µ) ⊆ Q(µ, α) is a cyclic extension of degree exactly 4. √ Definition (Kummer extension). Let K be a field, λ ∈ K non-zero, n ∈ N, char K = 0 or 0 < char K ∤ n. Suppose K contains an nth primitive root of unity, and L is a splitting field of tn − λ. If deg[L : K] = n, we say L/K is a Kummer
extension. Note that we used to have extensions K ⊆ K(µ) ⊆ L. But if K already contains a primitive root of unity, then K = K(µ). So we are left with the cyclic extension K ⊆ L. To following technical lemma will be useful: Lemma. Assume L/K is a field extension. Then HomK(L, L) is linearly independent. More concretely, let λ1, · · ·, λn ∈ L and ϕ1, · · ·, ϕn ∈ HomK(L, L) distinct. Suppose for all α ∈ L, we have λ1ϕ1(α) + · · · + λnϕn(α) = 0. Then λi = 0 for all i. 45 3 Solutions to polynomial equations II Galois Theory Proof. We perform induction on n. Suppose we have some λi ∈ L and ϕi ∈ HomK(L, L) such that λ1ϕ1(α) + · · · + λnϕn(α) = 0. The n = 1 case is trivial, since λ1ϕ1 = 0 implies λ1 = 0 (the zero homomorphism does not fix K). Otherwise, since the homomorphisms are distinct, pick β ∈ L such that ϕ1(β) ̸= ϕn(β). Then we know that λ1ϕ1(αβ) + · · · + λnϕn(αβ) = 0 for all α ∈ L. Since ϕi are homomorphisms, we can write this as λ1ϕ1(α)ϕ1(β) + · · · + λnϕn(α)ϕn(β) = 0. On the other hand, by just multiplying the original equation by ϕn(β), we get λ1ϕ1(α)ϕn(β) + · · · + λnϕn(α)ϕn(β) = 0. Subtracting the equations gives λ1ϕ1(α)(ϕ1(β) − ϕn(β)) + · · · + λn−1ϕn−1(
α)(ϕn−1(β) − ϕn(β)) = 0 for all α ∈ L. By induction, λi(ϕi(β) − ϕn(β)) = 0 for all 1 ≤ i ≤ n − 1. In particular, since ϕ1(β) − ϕn(β) ̸= 0, we have λ1 = 0. Then we are left with λ2ϕ2(α) + · · · + λnϕn(α) = 0. Then by induction again, we know that all coefficients are zero. Theorem. Let K be a field, n ∈ N, char K = 0 or 0 < char K ∤ n. Suppose K contains an nth primitive root of unity, and L/K is a cyclic extension of degree [L : K] = n. Then L/K is a Kummer extension. This is a rather useful result. If we look at the splitting field of a polynomial tn − λ, even if the ground field includes the right roots of unity, a priori, this doesn’t have to be a Kummer extension if it doesn’t have degree n. But we previously showed that the extension must be cyclic. And so this theorem shows that it is still a Kummer extension of some sort. This is perhaps not too surprising. For example, if, say, n = 4 and λ is secretly a square, then the splitting field of t4 − λ is just the splitting field of t2 − λ. √ Proof. Our objective here is to find a clever λ ∈ K such that L is the splitting field of tn − λ. To do so, we will have to hunt for a root β of tn − λ in L. Pick ϕ a generator of Gal(L/K). We know that if β were a root of tn − λ, then ϕ(β) = µ−1β for some primitive nth root of unity µ. Thus, we want to find an element that satisfies such a property. By the previous lemma, we can find some α ∈ L such that β = α + µϕ(α) + µ2ϕ2(α) + · · · + µn−1ϕn−1(α) ̸= 0
. 46 3 Solutions to polynomial equations II Galois Theory Then, noting that ϕn is the identity and ϕ fixes µ ∈ K, we see that β trivially satisfies ϕ(β) = ϕ(α) + µϕ2α + · · · + µn−1ϕn(α) = µ−1β, In particular, we know that ϕ(β) ∈ K(β). Now pick λ = βn. Then ϕ(βn) = µ−nβn = βn. So ϕ fixes βn. Since ϕ generates Gal(L/K), we know all automorphisms of L/K fixes βn. So βn ∈ K. Now the roots of tn − λ are β, µβ, · · ·, µn−1β. Since these are all in β, we know K(β) is the splitting field of tn − λ. Finally, to show that K(β) = L, we observe that id, ϕ\|K(β),..., ϕn\|K(β) are distinct elements of AutK(K(β)) since they do different things to β. Recall our previous theorem that [K(β) : K] ≥ \| AutK(K(β))\|. So we know that n = [L : K] = [K(β) : K]. So L = K(β). So done. Example. Consider t3 − 2 ∈ Q[t], and µ a third primitive root of unity. Then √ 2). Then Q ⊆ Q(µ) is a cyclotomic we have the extension Q ⊆ Q(µ) ⊆ Q(µ, 3 √ extension of degree 2, and Q(µ) ⊆ Q(µ, 3 2) is a Kummer extension of degree 3. 3.3 Radical extensions We are going to put these together and look at radical extensions, which allows us to characterize what it means to “solve a polynomial with radicals”. Definition (Radical extension). A field extension L/K is radical if there is some further extension E/L and with a sequence K = E0 ⊆ E1 ⊆ · · · ⊆ Er = E, such that each Ei ⊆ Ei
+1 is a cyclotomic or Kummer extension, i.e. Ei+1 is a splitting field of tn − λi+1 over Ei for some λi+1 ∈ Ei. Informally, we say Ei+1 is obtained by adding the roots “ nλi+1” to Ei. Hence we interpret a radical extension as an extension that only adds radicals. Definition (Solubility by radicals). Let K be a field, and f ∈ K[t]. f. We say f is soluble by radicals if the splitting field of f is a radical extension of K. √ This means that f can be solved by radicals of the form n Let’s go back to our first lecture and describe what we’ve done in the language λi. we’ve developed in the course. Example. As we have shown in lecture 1, any polynomial f ∈ Q[t] of degree at most 4 can be solved by radicals. For example, assume deg f = 3. So f = t3 + at2 + bt + c. Let L be the splitting field of f. Recall we reduced the problem of “solving” f to the case 3. Then we found our β, γ ∈ C such that a = 0 by the substitution x → x − a each root αi can be written as a linear combination of β and γ (and µ), i.e. L ⊆ Q(β, γ, µ). Then we showed that {β3, γ3} = −27c ± (27c)2 + 4 × 27b3 2. 47 3 Solutions to polynomial equations II Galois Theory We now let λ = (27c)2 + 4 × 27b3. Then we have the extensions Q ⊆ Q(λ) ⊆ Q(λ, µ) ⊆ Q(λ, µ, β), and also Q ⊆ L ⊆ Q(λ, µ, β). Note that the first extension Q ⊆ Q(λ) is a Kummer extension since it is a splitting field of t2 − λ2. Then Q(λ) ⊆ Q(λ, µ) is the third cyclotomic extension. Finally, Q(λ, µ) ⊆ Q(λ, µβ)
is a Kummer extension, the splitting field of t3 − β3. So Q ⊆ L is a radical extension. Let’s go back to the definition of a radical extension. We said L/K is radical if there is a further extension E/L that satisfies certain nice properties. It would be great if E/K is actually a Galois extensions. To show this, we first need a technical lemma. Lemma. Let L/K be a Galois extension, char K = 0, γ ∈ L and F the splitting field of tn − γ over L. Then there exists a further extension E/F such that E/L is radical and E/K is Galois. Here we have the inclusions K ⊆ L ⊆ F ⊆ E, where K, L and F are given and E is what we need to find. The idea of the proof is that we just add in the “missing roots” to obtain E so that E/K is Galois, and doing so only requires performing cyclotomic and Kummer extensions. Proof. Since we know that L/K is Galois, we would rather work in K than in L. However, our γ is in L, not K. Hence we will employ a trick we’ve used before, where we introduce a new polynomial f, and show that its coefficients are fixed by Gal(L/K), and hence in K. Then we can look at the splitting field of f or its close relatives. Let f = (tn − ϕ(γ)). ϕ∈Gal(L/K) Each ϕ ∈ Gal(L/K) induces a homomorphism L[t] → L[t]. Since each ϕ ∈ Gal(L/K) just rotates the roots of f around, we know that this induced homomorphism fixes f. Since all automorphisms in Gal(L/K) fix the coefficients of f, the coefficients must all be in K. So f ∈ K[t]. Now since L/K is Galois, we know that L/K is normal. So L is the splitting field of some g ∈ K[t]. Let E be the splitting field of f g over K. Then K ⊆ E is normal. Since the characteristic is zero, this is automatically separable. So the extension K ⊆ E is
Galois. We have to show that L ⊆ E is a radical extension. We pick our fields as follows: – E0 = L – E1 = splitting field of tn − 1 over E0 48 3 Solutions to polynomial equations II Galois Theory – E2 = splitting field of tn − γ over E1 – E3 = splitting field of tn − ϕ1(γ) over E2 –... – Er = E, where we enumerate Gal(L/K) as {id, ϕ1, ϕ2, · · · }. We then have the sequence of extensions L = E0 ⊆ E1 ⊆ E2 ⊆ · · · ⊆ Er Here E0 ⊆ E1 is a cyclotomic extension, and E1 ⊆ E2, E2 ⊆ E3 etc. are Kummer extensions since they contain enough roots of unity and are cyclic. By construction, F ⊆ E2. So F ⊆ E. Theorem. Suppose L/K is a radical extension and char K = 0. Then there is an extension E/L such that E/K is Galois and there is a sequence K = E0 ⊆ E1 ⊆ · · · ⊆ E, where Ei ⊆ Ei+1 is cyclotomic or Kummer. Proof. Note that this is equivalent to proving the following statement: Let K = L0 ⊆ L1 ⊆ · · · Ls be a sequence of cyclotomic or Kummer extensions. Then there exists an extension Ls ⊆ E such that K ⊆ E is Galois and can be written as a sequence of cyclotomic or Kummer extensions. We perform induction on s. The s = 0 case is trivial. If s > 0, then by induction, there is an extension M/Ls−1 such that M/K is Galois and is a sequence of cyclotomic and Kummer extensions. Now Ls is a splitting field of tn − γ over Ls−1 for some γ ∈ Ls−1. Let F be the splitting field of tn − γ over M. Then by the lemma and its proof, there exists an extension E/M that is a sequence of cyclotomic or Kummer extensions, and E/K is
Galois. √ Ls = Ls−1( n γ) K Ls−1 √ F = M ( n γ) E M However, we already know that M/K is a sequence of cyclotomic and Kummer extensions. So E/K is a sequence of cyclotomic and Kummer extension. So done. 49 3 Solutions to polynomial equations II Galois Theory 3.4 Solubility of groups, extensions and polynomials Let f ∈ K[t]. We defined the notion of solubility of f in terms of radical extensions. However, can we decide whether f is soluble or not without resorting to the definition? In particular, is it possible to decide whether its soluble by just looking at Gal(L/K), where L is the splitting field of f over K? It would be great if we could do so, since groups are easier to understand than fields. The answer is yes. It turns out the solubility of f corresponds to the solubility of Gal(L/K). Of course, we will have to first define what it means for a group to be soluble. After that, we will find examples of polynomials f of degree at least 5 such that Gal(L/K) is not soluble. In other words, there are polynomials that cannot be solved by radicals. Definition (Soluble group). A finite group G is soluble if there exists a sequence of subgroups Gr = {1} ◁ · · · ◁ G1 ◁ G0 = G, where Gi+1 is normal in Gi and Gi/Gi+1 is cyclic. Example. Any finite abelian group is solvable by the structure theorem of finite abelian groups: G ∼= Z ⟨n1⟩ × · · · × Z. ⟨nr⟩ Example. Let Sn be the symmetric group of permutations of n letters. We know that G3 is soluble since {1} ◁ A3 ◁ S3, where S3/A3 ∼= Z/⟨2⟩ and A3/{0} ∼= Z/⟨3⟩. S4 is also soluble by {1} ◁ {e, (1 2)(3 4)} ◁ {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2
3)} ◁ A4 ◁ S4. We can show that the quotients are Z/⟨2⟩, Z/⟨3⟩, Z/⟨2⟩ and Z/⟨2⟩ respectively. How about Sn for higher n? It turns out they are no longer soluble for n ≥ 5. To prove this, we first need a lemma. Lemma. Let G be a finite group. Then (i) If G is soluble, then any subgroup of G is soluble. (ii) If A ◁ G is a normal subgroup, then G is soluble if and only if A and G/A are both soluble. Proof. (i) If G is soluble, then by definition, there is a sequence Gr = {1} ◁ · · · ◁ G1 ◁ G0 = G, such that Gi+1 is normal in Gi and Gi/Gi+1 is cyclic. Let Hi = H ∩ Gi. Note that Hi+1 is just the kernel of the obvious homomorphism Hi → Gi/Gi+1. So Hi+1 ◁ Hi. Also, by the first isomorphism theorem, this gives an injective homomorphism Hi/Hi+1 → Gi/Gi+1. So Hi/Hi+1 is cyclic. So H is soluble. 50 3 Solutions to polynomial equations II Galois Theory (ii) (⇒) By (i), we know that A is solvable. To show the quotient is soluble, by assumption, we have the sequence Gr = {1} ◁ · · · ◁ G1 ◁ G0 = G, such that Gi+1 is normal in Gi and Gi/Gi+1 is cyclic. We construct the sequence for the quotient in the obvious way. We want to define Ei as the quotient Gi/A, but since A is not necessarily a subgroup of E, we instead define Ei to be the image of quotient map Gi → G/A. Then we have a sequence Er = {1} ◁ · · · ◁ E0 = G/A. The quotient map induces a surjective homomorphism Gi/Gi+1 → Ei/Ei+1, showing that Ei/Ei+1 are cyclic. (⇐) From the assumptions, we get
the sequences Am = {1} ◁ · · · ◁ A0 = A Fn = A ◁ · · · ◁ F0 = G where each quotient is cyclic. So we get a sequence Am = {1} ◁ A1 ◁ · · · ◁ A0 = Fn ◁ Fn−1 ◁ · · · ◁ F0 = G, and each quotient is cyclic. So done. Example. We want to show that Sn is not soluble if n ≥ 5. It is a well-known fact that An is a simple non-abelian group, i.e. it has no non-trivial subgroup. So An is not soluble. So Sn is not soluble. The key observation in Galois theory is that solubility of polynomials is related to solubility of the Galois group. Definition (Soluble extension). A finite field extension L/K is soluble if there is some extension L ⊆ E such that K ⊆ E is Galois and Gal(E/K) is soluble. Note that this definition is rather like the definition of a radical extension, since we do not require the extension itself to be “nice”, but just for there to be a further extension that is nice. In fact, we will soon see they are the same. Lemma. Let L/K be a Galois extension. Then L/K is soluble if and only if Gal(L/K) is soluble. This means that the whole purpose of extending to E is just to make it a Galois group, and it isn’t used to introduce extra solubility. Proof. (⇐) is clear from definition. (⇒) By definition, there is some E ⊆ L such that E/K is Galois and Gal(E/K) is soluble. By the fundamental theorem of Galois theory, Gal(L/K) is a quotient of Gal(E/K). So by our previous lemma, Gal(L/K) is also soluble. We now come to the main result of the lecture: Theorem. Let K be a field with char K = 0, and L/K is a radical extension. Then L/K is a soluble extension. 51 3 Solutions to polynomial equations II Galois Theory Proof. We have already shown that if we have a radical extension L/K, then
there is a finite extension K ⊆ E such that K ⊆ E is a Galois extension, and there is a sequence of cyclotomic or Kummer extensions E0 = K ⊆ E1 ⊆ · · · ⊆ Er = E. Let Gi = Gal(E/Ei). By the fundamental theorem of Galois theory, inclusion of subfields induces an inclusion of subgroups G0 = Gal(E/K) ≥ G1 ≥ · · · ≥ Gr = {1}. In fact, Gi ▷ Gi+1 because Ei ⊆ Ei+1 are Galois (since cyclotomic and Kummer extensions are). So in fact we have G0 = Gal(E/K) ▷ G1 ▷ · · · ▷ Gr = {1}. Finally, note that by the fundamental theorem of Galois theory, Gi/Gi+1 = Gal(Ei+1/Ei). We also know that the Galois groups of cyclotomic and Kummer extensions are abelian. Since abelian groups are soluble, our previous lemma implies that L/K is soluble. In fact, we will later show that the converse is also true. So an extension is soluble if and only if it is radical. Corollary. Let K be a field with char K = 0, and f ∈ K[t]. If f can be solved by radicals, then Gal(L/K) is soluble, where L is the splitting field of f over K. Again, we will later show that the converse is also true. However, to construct polynomials that cannot be solved by radicals, this suffices. In fact, this corollary is all we really need. Proof. We have seen that L/K is a Galois extension. By assumption, L/K is thus a radical extension. By the theorem, L/K is also a soluble extension. So Gal(L/K) is soluble. To find an f ∈ Q[t] which cannot be solved by radicals, it suffices to find an f such that the Galois group is not soluble. We don’t know many non-soluble groups so far. So in fact, we will find an f such that Gal(L/Q) = S5. To do so, we want to relate Galois groups to permutation groups. Lemma. Let
K be a field, f ∈ K[t] of degree n with no repeated roots. Let L be the splitting field of f over K. Then L/K is Galois and there exist an injective group homomorphism Gal(L/K) → Sn. Proof. Let Rootf (L) = {α1, · · ·, αn}. Let Pαi be the minimal polynomial of αi over K. Then Pαi \| f implies that Pαi is separable and splits over L. So L/K is Galois. Now each ϕ ∈ Gal(L/K) permutes the αi, which gives a map Gal(L/K) → Sn. It is easy to show this is an injective group homomorphism. 52 3 Solutions to polynomial equations II Galois Theory Note that there is not a unique or naturally-defined injective group homomorphism to Sn. This homomorphism, obviously, depends on how we decide to number our roots. Example. Let f = (t2 − 2)(t2 − 3) ∈ Q[t]. Let L be the splitting field of f over Q. Then the roots are Rootf (L) = { √ √ √ 2, 2, − 3, − √ 3}. We label these roots as α1, α2, α3, α4 in order. Now note that L = Q( 3), and thus [L : Q] = 4. Hence \| Gal(L/Q)\| = 4 as well. We can let Gal(L/Q) = {id, φ, ψ, λ}, where 2, √ √ √ √ √ √ id( 2) = φ( ψ( 2) = − √ 2) = λ( 2) = − √ 2 √ 2 √ √ √ √ √ id( φ( ψ( λ( 2 2 √ √ 3 3) = 3) = 3) = − 3) = − 3 √ √ 3 3 Then the image of Gal(L/Q) → S4 is given by {e, (1 2), (3 4), (1 2)(3 4)}. What we really want to know is if there are polynomials in which this map is in fact an isomorphism, i.e
. the Galois group is the symmetric group. If so, then we can use this to produce a polynomial that is not soluble by polynomials. To find this, we first note a group-theoretic fact. Lemma. Let p be a prime, and σ ∈ Sp have order p. Then σ is a p-cycle. Proof. By IA Groups, we can decompose σ into a product of disjoint cycles: σ = σ1 · · · σr. Let σi have order mi > 1. Again by IA Groups, we know that p = order of σ = lcm(m1, · · ·, mr). Since p is a prime number, we know that p = mi for all i. Hence we must have r = 1, since the cycles are disjoint and there are only p elements. So σ = σ1. Hence σ is indeed an p cycle. We will use these to find an example where the Galois group is the symmetric group. The conditions for this to happen are slightly awkward, but the necessity of these will become apparent in the proof. Theorem. Let f ∈ Q[t] be irreducible and deg f = p prime. Let L ⊆ C be the splitting field of f over Q. Let Rootf (L) = {α1, α2, · · ·, αp−2, αp−1, αp}. Suppose that α1, α2, · · ·, αp−2 are all real numbers, but αp−1 and αp are not. In particular, αp−1 = ¯αp. Then the homomorphism β : Gal(L/Q) → Sn is an isomorphism. 53 3 Solutions to polynomial equations II Galois Theory Proof. From IA groups, we know that the cycles (1 2 · · · p) and (p − 1 p) generate the whole of Sn. So we show that these two are both in the image of β. As f is irreducible, we know that f = Pα1, the minimal polynomial of α1 over Q. Then p = deg Pαi = [Q(α1) : Q]. By the tower law, this divides [L : Q], which is equal to \| Gal(L/Q)\| since the extension is
Galois. Since p divides the order of Gal(L/Q), by Cauchy’s theorem of groups, there must be an element of Gal(L/Q) that is of order p. This maps to an element σ ∈ im β of order exactly p. So σ is a p-cycle. On the other hand, the isomorphism C → C given by z → ¯z restricted to L gives an automorphism in Gal(L/Q). This simply permutes αp−1 and αp, since it fixes the real numbers and αp−1 and αp must be complex conjugate pairs. So τ = (p − 1 p) ∈ im β. Now for every 1 ≤ i < p, we know that σi again has order p, and hence is a p-cycle. So if we change the labels of the roots α1, · · ·, αp and replace σ with σi, and then waffle something about combinatorics, we can assume σ = (1 2 · · · p − 1 p). So done. Example. Let t5 − 4t + 2 ∈ Q[t]. Let L be the splitting field of f over Q. First note that deg f = 5 is a prime. Also, by Eisenstein’s criterion, f is irreducible. By finding the local maximum and minimum points, we find that f has exactly three real roots. So by the theorem, there is an isomorphism Gal(L/Q) → S5. In other words, Gal(L/Q) ∼= S5. We know S5 is not a soluble group. So f cannot be solved by radicals. After spending 19 lectures, we have found an example of a polynomial that cannot be solved by radicals. Finally. Note that there are, of course, many examples of f ∈ Q[t] irreducible of degree at least 5 that can be solved by radicals, such as f = t5 − 2. 3.5 Insolubility of general equations of degree 5 or more We now want to do something more interesting. Instead of looking at a particular example, we want to say there is no general formula for solving polynomial equations of degree 5 or above. First we want to define certain helpful notions. Definition (Field of symmetric rational functions). Let K be a field, L = K(
x1, · · ·, xn), the field of rational functions over K. Then there is an injective homomorphism Sn → AutK(L) given by permutations of xi. We define the field of symmetric rational functions F = LSn to be the fixed field of Sn. There are a few important symmetric rational functions that we care about more. Definition (Elementary symmetric polynomials). The elementary symmetric polynomials are e1, e2, · · ·, en defined by ei = xℓ1 · · · xℓi. 1≤l1<l2<···<li≤n 54 3 Solutions to polynomial equations II Galois Theory It is easy to see that e1 = x1 + x2 + · · · + xn e2 = x1x2 + x1x3 + · · · + xn−1xn en = x1 · · · xn. Obviously, e1, · · ·, en ∈ F. Theorem (Symmetric rational function theorem). Let K be a field, L = K(x1, · · ·, xn). Let F the field fixed by the automorphisms that permute the xi. Then (i) L is the splitting field of f = tn − e1tn−1 + · · · + (−1)nen over F. (ii) F = LSn ⊆ L is a Galois group with Gal(L/F ) isomorphic to Sn. (iii) F = K(e1, · · ·, en). Proof. (i) In L[t], we have f = (t − x1) · · · (t − xn). So L is the splitting field of f over F. (ii) By Artin’s lemma, L/K is Galois and Gal(L/F ) ∼= Sn. (iii) Let E = K(e1, · · ·, en). Clearly, E ⊆ F. Now E ⊆ L is a Galois extension, since L is the splitting field of f over E and f has no repeated roots. By the fundamental theorem of Galois theory, since we have the Galois extensions E ⊆ F ⊆ L, we have Gal(L/F ) ≤ Gal(L/E).
So Sn ≤ Gal(L/E). However, we also know that Gal(L/E) is a subgroup of Sn, we must have Gal(L/E) = Gal(L/F ) = Sn. So we must have E = F. Definition (General polynomial). Let K be a field, u1, · · ·, un variables. The general polynomial over K of degree n is f = tn + u1tn−1 + · · · + un. Technically, this is a polynomial in the polynomial ring K(u1, · · ·, un)[t]. However, we say this is the general polynomial over K be cause we tend to think of these ui as representing actual elements of K. We say the general polynomial over K of degree n can be solved by radicals if f can be solved by radicals over K(u1, · · ·, un). Example. The general polynomial of degree 2 over Q is This can be solved by radicals because its roots are t2 + u1t + u2. −u1 ± u2 1 − 4u2 2 55. 3 Solutions to polynomial equations II Galois Theory Theorem. Let K be a field with char K = 0. Then the general polynomial polynomial over K of degree n cannot be solved by radicals if n ≥ 5. Proof. Let f = tn + u1tn−1 + · · · + un. be our general polynomial of degree n ≥ 5. Let N be a splitting field of f over K(u1, · · ·, un). Let Rootf (N ) = {α1, · · ·, αn}. We know the roots are distinct because f is irreducible and the field has characteristic 0. So we can write f = (t − α1) · · · (t − αn) ∈ N [t]. We can expand this to get u1 = −(α1 + · · · + αn) u2 = α1α2 + α1α3 + · · · + αn−1αn... ui = (−1)i(ith elementary symmetric polynomial in α1, · · ·, αn). Now let x1, · · ·, xn be new variables, and ei the ith elementary symmetric polynomial
in x1, · · ·, xn. Let L = K(x1, · · ·, xn), and F = K(e1, · · ·, en). We know that F ⊆ L is a Galois extension with Galois group isomorphic to Sn. We define a ring homomorphism θ : K[u1, · · ·, un] → K[e1, · · ·, en] ⊆ K[x1, · · ·, xn] ui → (−1)iei. This is our equations of ui in terms αi, but with xi instead of αi. We want to show that θ is an isomorphism. Note that since the homomorphism just renames ui into ei, the fact that θ is an isomorphism means there are no “hidden relations” between the ei. It is clear that θ is a surjection. So it suffices to show θ is injective. Suppose θ(h) = 0. Then h(−e1, · · ·, (−1)nen) = 0. Since the xi are just arbitrary variables, we now replace xi with αi. So we get h(−e1(α1, · · ·, αn), · · ·, (−1)n(en(α1, · · ·, αn))) = 0. Using our expressions for ui in terms of ei, we have h(u1, · · ·, un) = 0, But h(u1, · · ·, un) is just h itself. So h = 0. Hence θ is injective. So it is an isomorphism. This in turns gives an isomorphism between K(u1, · · ·, un) → K(e1, · · ·, en) = F. We can extend this to their polynomial rings to get isomorphisms between K(u1, · · ·, un)[t] → F [t]. 56 3 Solutions to polynomial equations II Galois Theory In particular, this map sends our original f to f → tn − e1tn−1 + · · · + (−1)nen = g. Thus, we get an isomorphism between the splitting field of f over K(u1, · · ·, un) and
the splitting field g over F. The splitting field of f over K(u1, · · ·, un) is just N by definition. From the symmetric rational function theorem, we know that the splitting field of g over F is just L, and So N ∼= L. So we have an isomorphism Gal(N/K(u1, · · ·, un)) → Gal(L/F ) ∼= Sn. Since Sn is not soluble, f is not soluble. This is our second main goal of the course, to prove that general polynomials of degree 5 or more are not soluble by radicals. Recall that we proved that all radical extensions are soluble. We now prove the converse. Theorem. Let K be a field with char K = 0. If L/K is a soluble extension, then it is a radical extension. Proof. Let L ⊆ E be such that K ⊆ E is Galois and Gal(E/K) is soluble. We can replace L with E, and assume that in fact L/K is a soluble Galois extension. So there is a sequence of groups {0} = Gr ◁ · · · ◁ G1 ◁ G0 = Gal(L/K) such that Gi/Gi+1 is cyclic. By the fundamental theorem of Galois theory, we get a sequence of field extension given by Li = LGi: K = L0 ⊆ · · · ⊆ Lr = L. Moreover, we know that Li ⊆ Li+1 is a Galois extension with Galois group Gal(Li+1/Li) ∼= Gi/Gi+1. So Gal(Li+1/Li) is cyclic. Let n = [L : K]. Recall that we proved a previous theorem that if Gal(Li+1/Li) is cyclic, and Li contains a primitive kth root of unity (with k = [Li+1 : Li]), then Li ⊆ Li+1 is a Kummer extension. However, we do not know of Li contains the right root of unity. Hence, the trick here is to add an nth primitive root of unity to each field in the sequence. Let µ an nth primitive root of unity. Then if we add the nth primitive root to each item of the sequence, we have L0(µ) ⊆ · · · ⊆
Li(µ) ⊆ Li+1(µ) ⊆ · · · ⊆ Lr(µ) ⊆ K = L0 ⊆ · · · ⊆ ⊆ Li ⊆ ⊆ ⊆ Li+1 ⊆ · · · ⊆ Lr = L We know that L0 ⊆ L0(µ) is a cyclotomic extension by definition. We will now show that Li(µ) ⊆ Li+1(µ) is a Kummer extension for all i. Then L/K is radical since L ⊆ Lr(µ). 57 3 Solutions to polynomial equations II Galois Theory Before we do anything, we have to show Li(µ) ⊆ Li+1(µ) is a Galois extension. To show this, it suffices to show Li ⊆ Li+1(µ) is a Galois extension. Since Li ⊆ Li+1 is Galois, Li ⊆ Li+1 is normal. So Li+1 is the splitting of some h over Li. Then Li+1(µ) is just the splitting field of (tn − 1)h. So Li ⊆ Li+1(µ) is normal. Also, Li ⊆ Li+1(µ) is separable since char K = char Li = 0. Hence Li ⊆ Li+1(µ) is Galois, which implies that Li(µ) ⊆ Li+1(µ) is Galois. We define a homomorphism of groups Gal(Li+1(µ)/Li(µ)) → Gal(Li+1/Li) by restriction. This is well-defined because Li+1 is the splitting field of some h over Li, and hence any automorphism of Li+1(µ) must send roots of h to roots of h, i.e. Li+1 to Li+1. Moreover, we can see that this homomorphism is injective. If ϕ → ϕ\|Li+1 = id, then it fixes everything in Li+1. Also, since it is in Gal(Li+1(µ)/Li(µ)), it fixes Li(µ). In particular, it fixes µ. So ϕ must fix the
whole of Li+1(µ). So ϕ = id. By injectivity, we know that Gal(Li+1(µ)/Li(µ)) is isomorphic to a subgroup of Gal(Li+1/Li). Hence it is cyclic. By our previous theorem, it follows that Li(µ) ⊆ Li+1(µ) is a Kummer extension. So L/K is radical. Corollary. Let K be a field with char K = 0 and h ∈ K[t]. Let L be the splitting of h over K. Then h can be solved by radicals if and only if Gal(L/K) is soluble. Proof. (⇒) Proved before. (⇐) Since L/K is a Galois extension, L/K is a soluble extension. So it is a radial extension. So h can be solved by radicals. Corollary. Let K be a field with char K = 0. Let f ∈ K[t] have deg f ≤ 4. Then f can be solved by radicals. Proof. Exercise. Note that in the case where K = Q, we have proven this already by given explicit solutions in terms of radicals in the first lecture. 58 4 Computational techniques II Galois Theory 4 Computational techniques In the last three lectures, we will look at some techniques that allow us to actually compute the Galois group of polynomials (i.e. Galois groups of their splitting fields). 4.1 Reduction mod p The goal of this chapter is to see what happens when we reduce a polynomial f ∈ Z[t] to the corresponding polynomial ¯f ∈ Fp[t]. More precisely, suppose we have a polynomial f ∈ Z[t], and E is its splitting field over Q. We then reduce f to ¯f ∈ Fp[t] by reducing the coefficients mod p, and let ¯E be the splitting field of ¯f over Fp. The ultimate goal is to show that under mild assumptions, there is an injection Gal(E/Fp) → Gal(E/Q). To do this, we will go through a lot of algebraic fluff to obtain an alternative characterization of the Galois group, and obtain the result as an easy corollary. This section will be notationally heavy. First, in the background, we have a polynomial f
of degree n (whose field we shall specify later). Then we will have three distinct set of variables, namely (x1, · · ·, xn), (u1, · · ·, un), plus a t. They will play different roles. – The xi will be placeholders. After establishing our definitions, we will then map each xi to αi, a root of our f. – The ui will stay as “general coefficients” all the time. – t will be the actual variable we think our polynomial is in, i.e. all polynomials will be variables in t, and ui and xi will form part of the coefficients. To begin with, let L = Q(x1, · · ·, xn) F = Q(e1, · · ·, en). where xi are variables and ei are the symmetric polynomials in the x1, · · ·, xn. We have seen that Gal(L/F ) ∼= Sn. Now let B = Z[x1, · · ·, xn] A = Z[e1, · · ·, en]. It is an exercise on example sheet 4 to show that B ∩ F = A. (∗) We will for now take this for granted. We now add it new variables u1, · · ·, un, t. We previously mentioned that Sn can act on, say L[u1, · · ·, un, t] by permuting the variables. Here there are two ways in which this can happen — a permutation can either permute the xi, or permute the ui. We will have to keep this in mind. 59 4 Computational techniques II Galois Theory Now for each σ ∈ Sn, we define the linear polynomial Rσ = t − xσ(1)u1 − · · · − xσ(n)un. For example, we have R(1) = t − x1u1 − · · · − xnun. As mentioned, an element ρ ∈ Sn can act on Rρ in two ways: it either sends Rσ → Rρσ or Rσ → Rσρ−1. It should be clear that the first action permutes the xi. What the second action does is permute the ui. To see this, we can consider a simple case
where n = 2. Then the action ρ acting on R(1) sends t − x1u1 − x2u2 → t − xρ−1(1)u1 − xρ−2(2)u2 = t − x1uρ(1) − x2uρ(2). Finally, we define the following big scary polynomial: R = σ∈Sn Rσ ∈ B[u1, · · ·, un, t]. We see that this is fixed by any permutation in σ ∈ Sn under both actions. Considering the first action and using (∗), we see that in fact R ∈ A[u1, · · ·, un, t]. This is since if we view R as a polynomial over B in the variables u1, · · ·, un, t, then its coefficients will be invariant under permuting the xi. So the coefficients must be a function of the ei, i.e. lie in A. With these definitions in place, we can focus on a concrete polynomial. Let K be a field, and let f = tn + a1tn−1 + · · · + an ∈ K[t] have no repeated roots. We let E be the splitting field of f over K. Write Rootf (E) = {α1, · · ·, αn}. Note that this is the setting we had at the beginning of the chapter, but with an arbitrary field K instead of Q and Fp. We define a ring homomorphism θ : B → E by xi → αi. This extends to a ring homomorphism θ : B[u1, · · ·, un, t] → E[u1, · · ·, un, t]. Note that the ring homomorphism θ send ei → (−1)iai. So in particular, if we restrict the homomorphism θ to A, then the image is restricted to the field generated by ai. But we already have ai ∈ K. So θ(A) = K. In particular, since R ∈ A[u1, · · ·, un, t], we have θ(R) ∈ K[u1, · · ·, un, t]. Now let P be an irreducible factor of θ(R) in K
[u1, · · ·, un, t]. We want to say each such irreducible polynomial is related to the Galois group G = Gal(E/K). 60 4 Computational techniques II Galois Theory Since f has no repeated roots, we can consider G as a subgroup of Sn, where the elements of G are just the permutations of the roots αi. We will then show that each irreducible polynomial corresponds to a coset of G. Recall that at the beginning, we said Sn can act on our polynomial rings by permuting the xi or ui. However, once we have mapped the xi to the αi and focus on a specific field, Sn as a whole can no longer act on the αi, since there might be non-trivial relations between the αi. Instead, only the subgroup G ≤ Sn can act on αi. On the other hand, Sn can still act on ui. Recall that R is defined as a product of linear factors Rσ’s. So we can find a subset Λ ⊆ Sn such that P = σ∈Λ Rσ. We will later see this Λ is just a coset of the Galois group G. Pick σ ∈ Λ. Then by definition of P, Rσ \| P in E[u1, · · ·, un, t]. Now if ρ ∈ G, then we can let ρ act on both sides by permuting the xi (i.e. the αi). This does not change P because P has coefficients in K and the action of G has to fix K. Hence we have More generally, if we let Rρσ \| P. H = ρ∈G Rρσ ∈ E[u1, · · ·, un, t], then H \| P by the irreducibility of P. Since H is also invariant under the action of G, we know H ∈ K[u1, · · ·, un, t]. By the irreducibility of P, we know H = P. Hence, we know Λ = Gσ. We have thus proved that the irreducible factors of θ(R) in K[u1, · · ·, un, t] are in one-to-one correspondence with the cosets of G in Sn. In
particular, if P corresponds to G itself, then P = Rτ. In general, if P corresponds to a coset Gσ, we can let λ ∈ Sn act on P by permuting the ui’s. Then this sends τ ∈G P = ρ∈G Rρσ → Q = ρ∈G Rρσλ−1. So this corresponds to the coset Gσλ−1. In particular, P = Q if and only if Gσ = Gσλ−1. So we can use this to figure out what permutations preserve an irreducible factor. In particular, taking σ = (1), we have 61 4 Computational techniques II Galois Theory Theorem. G = {λ ∈ Sn : λ preserves the irreducible factor corresponding to G}. (†) This is the key result of this chapter, and we will apply this as follows: Theorem. Let f ∈ Z[t] be monic with no repeated roots. Let E be the splitting field of f over Q, and take ¯f ∈ Fp[t] be the obvious polynomial obtained by reducing the coefficients of f mod p. We also assume this has no repeated roots, and let ¯E be the splitting field of ¯f. Then there is an injective homomorphism ¯G = Gal( ¯E/Fp) → G = Gal(E/Q). Moreover, if ¯f factors as a product of irreducibles of length n1, n2, · · ·, nr, then Gal(f ) contains an element of cycle type (n1, · · ·, nr). Proof. We apply the previous theorem twice. First, we take K = Q. Then θ(R) ∈ Z[u1, · · ·, un, t]. Let P be the irreducible factor of θ(R) corresponding to the Galois group G. Applying Gauss’ lemma, we know P has integer coefficients. Applying the theorem again, taking K = Fp. Denote the ring homomorphism as ¯θ. Then ¯θ(R) ∈ Fp[u1, · · ·, un, t]. Now let Q be the irreducible factor ¯θ(R) corresponding to ¯G. Now note that θ(R(1
)) \| P and ¯θ(R(1)) \| Q, since the identity is in G and ¯G. Also, note that ¯θ(R) = θ(R), where the bar again denotes reduction mod p. So Q \| ¯P. Considering the second action of Sn (i.e. permuting the ui), we can show ¯G ⊆ G, using the characterization (†). Details are left as an exercise. This is incredibly useful for computing Galois groups, as it allows us to explicitly write down some cycles in Gal(E, Q). 4.2 Trace, norm and discriminant We are going to change direction a bit and look at traces and norms. These will help us understand the field better, and perhaps prove some useful facts from it. They will also lead to the notion of the discriminant, which is again another tool that can be used to compute Galois groups, amongst many other things. Definition (Trace). Let K be a field. If A = [aij] is an n × n matrix over K, we define the trace of A to be tr(A) = n i=1 aii, i.e. we take the sum of the diagonal terms. It is a well-known fact that if B is an invertible n × n matrix, then tr(B−1AB) = tr(A). Hence given a finite-dimensional vector space V over K and σ : V → V a K-linear map, then we can define the trace for the linear map as well. 62 4 Computational techniques II Galois Theory Definition (Trace of linear map). Let V be a finite-dimensional vector space over K, and σ : V → V a K-linear map. Then we can define tr(σ) = tr(any matrix representing σ). Definition (Trace of element). Let K ⊆ L be a finite field extension, and α ∈ L. Consider the K-linear map σ : L → L given by multiplication with α, i.e. β → αβ. Then we define the trace of α to be trL/K(α) = tr(σ). Similarly, we can consider the determinant, and obtain the norm. Definition (Norm of element). We define the norm of α to be where σ is, again, the multiplication-by-α map. NL/K(α) = det(
σ), This construction gives us two functions trL/K, NL/K : L → K. It is easy to see from definition that trL/K is additive while NL/K is multiplicative. Example. Let L/K be a finite field extension, and x ∈ K. Then the matrix of x is represented by xI, where I is the identity matrix. So NL/K(x) = x[L:K], trL/K(x) = [L : K]x. Example. Let K = Q, L = Q(i). Consider an element a + bi ∈ Q(i), and pick the basis {1, i} for Q(i). Then the matrix of a + bi is a −b a b. So we find that trL/K(a + bi) = 2a and N (a + bi) = a2 + b2 = \|a + bi\|2. √ −d) where d > 0 is square-free, then −d\|2. However, for other fields, the norm is In general, if K = Q and L = Q( −d) = a2 + b2d = \|a + b N (a + b not at all related to the absolute value. √ √ In general, computing norms and traces with the definition directly is not fun. It turns out we can easily find the trace and norm of α from the minimal polynomial of α, just like how we can find usual traces and determinants from the characteristic polynomial. To do so, we first prove the transitivity of trace and norm. Lemma. Let L/F/K be finite field extensions. Then trL/K = trF/K ◦ trL/F, NL/K = NF/K ◦ NL/F. To prove this directly is not difficult, but involves some confusing notation. Purely for the sake of notational convenience, we shall prove the following more general fact: Lemma. Let F/K be a field extension, and V an F -vector space. Let T : V → V be an F -linear map. Then it is in particular a K-linear map. Then detK T = NF/K(detF T ), trK T = trF/K(trF T ). 63 4 Computational techniques II Galois Theory Taking V to be L and T to be multiplication by α �
� F clearly gives the original intended result. Proof. For α ∈ F, we will write mα : F → F for multiplication by α map viewed as a K-linear map. By IB Groups, Rings and Modules, there exists a basis {ei} such that T is in rational canonical form, i.e. such that T is block diagonal with each diagonal looking like  0 1   0  ...   0 0 0 1... 0 · · · · · · · · ·... · · · a0 0 a1 0 a2 0...... 1 ar−1 .       Since the norm is multiplicative and trace is additive, and det A 0 0 B = det A det B, tr A 0 0 B = tr A + tr B, we may wlog T is represented by a single block as above. From the rational canonical form, we can read off detF T = (−1)r−1a0, trF T = ar−1. We now pick a basis {fj} of F over K, and then {eifj} is a basis for V over K. Then in this basis, the matrix of T over K is given by        0 0 · · · 0 ma0 1 0 · · · 0 ma1 0 1 · · · 0 ma2............ 0 0 · · · 1 mar−1...       . It is clear that this has trace trK(mar−1 ) = trF/K(ar−1) = trF/K(trF T ). Moreover, writing n = [L : K], we have detK        0 0 · · · 0 ma0 1 0 · · · 0 ma1 0 1 · · · 0 ma2............ 0 0 · · · 1 mar−1...        = (−1)n(r−1) detK �
��       ma0 ma1 ma2............... mar−1 0 0 · · · 1        So the result follows. = (−1)n(r−1) detK(ma0) = detK((−1)r−1ma0) = NF/K((−1)r−1a0) = NF/K(detF T ). 64 4 Computational techniques II Galois Theory As a corollary, we have the following very powerful tool for computing norms and traces. Corollary. Let L/K be a finite field extension, and α ∈ L. Let r = [L : K(α)] and let Pα be the minimal polynomial of α over K, say Pα = tn + an−1tn−1 + · · · + a0. with ai ∈ K. Then and trL/K(α) = −ran−1 NL/K(α) = (−1)nrar 0. Note how this resembles the relation between the characteristic polynomial and trace/determinants in linear algebra. Proof. We first consider the case r = 1. Write mα for the matrix representing multiplication by α. Then Pα is the minimal polynomial of mα. But since deg Pα = n = dimK K(α), it follows that this is also the characteristic polynomial. So the result follows. Now if r ̸= 1, we can consider the tower of extensions L/K(α)/K. Then we have NL/K(α) = NK(α)/K(NL/K(α)(α)) = NK(α)/K(αr) = (NK(α)/K(α))r = (−1)nrar 0. The computation for trace is similar. It is also instructive to prove this directly. In the case r = 1, we can pick the basis {1, α, α2, · · ·, αn−1} of L over K. Then the multiplication map sends So the matrix is just 1 → α α → α2... αn−1 → αn = −an−1αn−1 − · · · − a0 A = 
0 1   0  ...   0 0 0 1... 0 −a0 −a1 −a2... · · · · · · · · ·... · · · −an−1        The characteristic polynomial of this matrix is det(tI − A) = det        0 t t −1 0 −1...... 0 0 · · · · · · · · ·... · · ·        a0 a1 a2... t + an−1 65 4 Computational techniques II Galois Theory By adding ti multiples of the ith row to the first row for each i, this gives det(tI − A) = det        0 0 −1 t 0 −1...... 0 0 · · · · · · · · ·... · · ·        Pα a1 a2... t + an−1 = Pα. Then we notice that for r ̸= 1, in an appropriate choice of basis, the matrix looks like · · ·............ · · · A 0 0 .     Theorem. Let L/K be a finite but not separable extension. Then trL/K(α) = 0 for all α ∈ L. Proof. Pick β ∈ L such that Pβ, the minimal polynomial of β over K, is not separable. Then by the previous characterization of separable polynomials, we know p = char K > 0 with Pβ = q(tp) for some q ∈ K[t]. Now consider K ⊆ K(βp) ⊆ K(β) ⊆ L. To show trL/K = 0, by the previous proposition, it suffices to show trK(β)/K(βp) = 0. Note that the minimal
polynomial of βp over K is q because q(βp) = 0 and q is irreducible. Then [K(β) : K] = deg Pβ = p deg q and deg[K(βp) : K] = deg q. So [K(β) : K(βp)] = p. Now {1, β, β2, · · ·, βp−1} is a basis of K(β) over K(βp). Let Rβi be the minimal polynomial of βi over K(βp). Then Rβi = t − 1 tp − βir i = 0 i ̸= 0, We get the second case using the fact that p is a prime number, and hence K(βp)(βi) = K(β) if 1 ≤ i < p. So [K(βp)(βi) : K(βp)] = p and hence the minimal polynomial has degree p. Hence trK(β)/K(βp)(βi) = 0 for all i. Thus, trK(β)/K(βp) = 0. Hence trL/K = trK(βp)/K ◦ trK(β)/K(βp) ◦ trL/K(β) = 0. Note that if L/K is a finite extension, and char K = 0, then trL/K(1) = [L : K] ̸= 0. So trL/K ̸= 0. It is in fact true that all separable extensions have trL/K ̸= 0, not only when the field has characteristic 0. √ Example. We want to show 3 √ L = Q( 3 exists some a, b, c ∈ Q such that √ 3) = Q( 3 √ 3 ̸∈ Q( 3 2). Suppose not. Then we have 2), since both extensions of Q have degree 3. Then there √ √ 22. 66 4 Computational techniques II Galois Theory We now compute the traces over Q. The minimal polynomials over Q are P 3√ 3 = t3 − 3, P 3√ 2 = t3 − 2, P 3√ 4 = t3 − 4. So we have √ trL/Q( 3 √ 3)
= a trL/Q(1) + b trL/Q( 3 √ 2) + c trL/Q( 3 4). Since the minimal polynomials above do not have coefficients in t2, the traces of the cube roots are zero. So we need a = 0. Then we are left with √. √ We apply the same trick again. We multiply by 3 2 to obtain 4 + 2c. √ We note that the minimal polynomial of 3 √ 3 √ 6 = b 3 6 is t3 − 6. Taking the trace gives √ trL/Q( 3 √ 6) = b trL/Q( 3 4) + 6c. Again, the traces are zero. So c = 0. So we have √ 3 √ 3 = b 3 2. In other words, b3 = 3 2, √ which is clearly nonsense. This is a contradiction. So 3 √ 3 ̸∈ Q( 3 2). We can obtain another formula for the trace and norm as follows: Theorem. Let L/K be a finite separable extension. Pick a further extension E/L such that E/K is normal and \| HomK(L, E)\| = [L : K]. Write HomK(L, E) = {φ1, · · ·, φn}. Then trL/K(α) = n i=1 φi(α), NL/K(α) = n i=1 φi(α) for all α ∈ L. Proof. Let α ∈ L. Let Pα be the minimal polynomial of α over K. Then there is a one-to-one correspondence between HomK(K(α), E) ←→ RootPα (E) = {α1, · · ·, αd}. wlog we let α = α1. Also, since we get \| HomK(L, E)\| = [L : K], \| HomK(K(α), E)\| = [K(α) : K] = deg Pα. 67 4 Computational techniques II Galois Theory Moreover, the restriction map HomK(L, E) → HomK(K(α), E) (defined by φ → φ\|K(α)) is surjective and sends exactly [K(α) : K] elements to
any particular element in HomK(K(α), E). Therefore φi(α) = [L : K(α)] ψ(α) = [L : K(α)] ψ∈HomK (K(α),E) d i=1 αi. Moreover, we can read the sum of roots of a polynomial is the (negative of the) coefficient of td−1, where Pα = td + ad−1td−1 + · · · + a0. So φi(α) = [L : K(α)](−ad−1) = trL/K(α). Similarly, we have φi(α) =    [L:K(α)] ψ(α)  ψ∈HomK (K(α),E) [L:K(α)] d = αi i=1 = ((−1)da0)[L:K(α)] = NL/K(α). Corollary. Let L/K be a finite separable extension. Then there is some α ∈ L such that trL/K(α) ̸= 0. Proof. Using the notation of the previous theorem, we have trL/K(α) = φi(α). Similar to a previous lemma, we can show that φ1, · · ·, φn are “linearly independent” over E, and hence φi cannot be identically zero. Hence there is some α such that trL/K(α) = φi(α) ̸= 0. Example. Let K = Fq ⊆ L = Fqn, with q is a power of some prime number p. By a previous theorem on finite fields, we know L/K is Galois and Gal(L/K) = Z nZ and is generated by the Frobenius φ = Frq. To apply the theorem, we had to pick an E such that E/K is normal and HomK(L, E) = [L : K]. However, since L/K is Galois, we can simply pick E = L. 68 4 Computational techniques II Galois Theory Then we know trL/K(α) = ψ(α) ψ∈Gal(L/K) = n−
1 i=0 φi(α) = α + αq + αq2 + · · · + αqn−1. Similarly, the norm is NL/K(α) = n−1 i=0 φi(α) = α · αq · · · · · αqn−1. Recall that when solving quadratic equations f = t2 + bt + c, we defined the discriminant as b2 − 4c. This discriminant then determined the types of roots of f. In general, we can define the discriminant of a polynomial of any degree, in a scary way. Definition (Discriminant). Let K be a field and f ∈ K[t], L the splitting field of f over K. So we have f = a(t − α1) · · · (t − αn) for some a, α1, · · ·, αn ∈ L. We define ∆f = i<j (αi − αj), Df = ∆2 f = (−1)n(n−1)/2 (αi − αj). i̸=j We call Df the discriminant of f. Clearly, Df ̸= 0 if and only if f has no repeated roots. Theorem. Let K be a field and f ∈ K[t], L is the splitting field of f over K. Suppose Df ̸= 0 and char K ̸= 2. Then (i) Df ∈ K. (ii) Let G = Gal(L/K), and θ : G → Sn be the embedding given by the permutation of the roots. Then im θ ⊆ An if and only if ∆f ∈ K (if and only if Df is a square in K). Proof. (i) It is clear that Df is fixed by Gal(L/K) since it only permutes the roots. (ii) Consider a permutation σ ∈ Sn of the form σ = (ℓ m), and let it act on the roots. Then we claim that σ(∆f ) = −∆f. (†) So in general, odd elements in Sn negate ∆f while even elements fix it. Thus, ∆f ∈ K iff ∆f is fixed by Gal(L/K
) iff every element of Gal(L/K) is even. 69 4 Computational techniques II Galois Theory To prove (†), we have to painstakingly check all terms in the product. We wlog ℓ < m. If k < ℓ, m. Then this swaps (αk − αℓ) with αk − αm), which has no effect. The k > m case is similar. If ℓ < k < m, then this sends (αℓ − αk) → (αm − αk) and (αk − αm) → (αℓ − αm). This introduces two negative signs, which has no net effect. Finally, this sends (αk − αm) to its negation, and so introduces a negative sign. We will later use this result to compute certain Galois groups. Before that, we see how this discriminant is related to the norm. Theorem. Let K be a field, and f ∈ K[t] be an n-degree monic irreducible polynomial with no repeated roots. Let L be the splitting field of f over K, and let α ∈ RootF (L). Then Df = (−1)n(n−1)/2NK(α)/K(f ′(α)). Proof. Let HomK(K(α), L) = {φ1, · · ·, φn}. Recall these are in one-to-one correspondence with Rootf (L) = {α1, · · ·, αn}. Then we can compute (αi − αj) = (αi − αj). i̸=j i j̸=i Note that since f is just monic, we have f = (t − α1) · · · (t − αn). Computing the derivative directly, we find So we have (αi − αj) = f ′(αi). j̸=i (αi − αj) = i̸=j f ′(αi). i Now since the φi just maps α to αi, we have (αi − αj) = i̸=j i φi(f ′(α)) = NK(α)/K(f ′(α)). Finally, multiplying the factor of (−1)n(n
−1)/2 gives the desired result. Example. Let K be a field with char K ̸= 2, 3. Let f ∈ K[t] have degree 3, say f = t3 + bt + c where we have gotten rid of the t2 term as in the first lecture. We further assume f is irreducible with no repeated roots, and let L be the splitting field of f. We want to compute the discriminant of this polynomial. Let α ∈ Rootf (L). Then Then we can see β = f ′(α) = 3α2 + b. β = −2b − 3c α. 70 4 Computational techniques II Galois Theory Alternatively, we have α = −3c β + 2b. (∗) Putting (∗) into α3 + bα + c = 0, we find the minimal polynomial of β has constant term −4b3 − 27c2. This then gives us the norm, and we get Df = −NK(α)/K(β) = −4b3 − 27c2. This is the discriminant of a cubic. We can take a specific example, where f = t3 − 31t + 62. Then f is irreducible over Q. We can compute Df, and find that it is a square. So the previous theorem says the image of the Galois group Gal(L/K) is a subgroup of A3. However, we also know Gal(L/K) has three elements since deg f = 3. So we know Gal(L/K) ∼= A3. 71ic triangles are rather nice. They are so nice that we can actually prove something about them! Proposition. For every geodesic triangulation of S2 (and respectively T ) has e = 2 (respectively, e = 0). Of course, we know this is true for any triangulation, but it is diﬃcult to prove that without algebraic topology. Proof. For any triangulation τ, we denote the “faces” of ∆1, · · ·, ∆F, and write τi = αi + βi + γi for the sum of the interior angles of the triangles (with i = 1, · · ·, F ). Then we have τi = 2πV, since the total angle
around each vertex is 2π. Also, each triangle has three edges, and each edge is from two triangles. So 3F = 2E. We write this in a more convenient form: F = 2E − 2F. How we continue depends on whether we are on the sphere or the torus. 25 3 Triangulations and the Euler number IB Geometry – For the sphere, Gauss-Bonnet for the sphere says the area of ∆i is τi − π. Since the area of the sphere is 4π, we know 4π = = area(∆i) (τi − π) = 2πV − F π = 2πV − (2E − 2F )π = 2π(F − E + V ). So F − E + V = 2. – For the torus, we have τi = π for every face in ˚Q. So 2πV = τi = πF. 2V = F = 2E − 2F. 2(F − V + E) = 0, So So we get as required. Note that in the deﬁnition of triangulation, we decomposed X into topological triangles. We can also use decompositions by topological polygons, but they are slightly more complicated, since we have to worry about convexity. However, apart from this, everything works out well. In particular, the previous proposition also holds, and we have Euler’s formula for S2: V − E + F = 2 for any polygonal decomposition of S2. This is not hard to prove, and is left as an exercise on the example sheet. 26 4 Hyperbolic geometry IB Geometry 4 Hyperbolic geometry At the beginning of the course, we studied Euclidean geometry, which was not hard, because we already knew about it. Later on, we studied spherical geometry. That also wasn’t too bad, because we can think of S2 concretely as a subset of R3. We are next going to study hyperbolic geometry. Historically, hyperbolic geometry was created when people tried to prove Euclid’s parallel postulate (that given a line and a point P ∈, there exists a unique line containing P that does not intersect ). Instead of proving the parallel postulate, they managed to create a new geometry where this is false, and this is hyper

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