The Galois Correspondence

Seminar
 [12:02] * ChanServ changes topic to 'SEMINAR IN PROGRESS. If you want to ask a                   question, say ! and wait to be called' [12:03] &lt;Kasadkad&gt; so let's finally get to defining kind of the main object of Galois theory, the automorphism group of a field extension K/F [12:04] &lt;Kasadkad&gt; for K a field, a function phi : K -&gt; K is an automorphism if it's 1)                   a bijection and 2) a homomorphism (i.e. phi(x+y) = phi(x) + phi(y)                    and phi(xy) = phi(x)phi(y), for all x,y in K) [12:05] &lt;Kasadkad&gt; e.g. if K = C, complex conjugation is an automorphism [12:06] &lt;Kasadkad&gt; general automorphisms are too much, though; there are tons of bizarre discontinuous automorphisms of C [12:06] &lt;Kasadkad&gt; so what we'll look at is automorphisms of K that fix the base field F [12:06] &lt;Kasadkad&gt; i.e. phi(x) = x for all x in F [12:06] &lt;Kasadkad&gt; notice that complex conjugation is also an automorphism of the extension C/R in this sense, since it fixes real numbers [12:08] &lt;Kasadkad&gt; here's an important property of automorphisms of K/F: if c in K is a                   root of p(x), then phi(c) is also a root of p(x) [12:08] &lt;Kasadkad&gt; and that's just from the homomorphism property: if p(x) = a_0 + a_1 x                   + ... + a_n x^n, then p(phi(c)) = a_0 + a_1 phi(c) + ... + a_n phi(c)^n = phi(p(c)) = 0 [12:09] &lt;Kasadkad&gt; well [12:09] &lt;Kasadkad&gt; sorry I messed that up a little [12:09] &lt;Kasadkad&gt; if c is a root of p(x) in F[x] [12:09] &lt;Kasadkad&gt; because the whole point is that we need phi to fix the coefficients a_i of p(x) [12:10] &lt;Kasadkad&gt; so, automorphisms of K/F take field elements to conjugates [12:10] &lt;Kasadkad&gt; (remember that if c in K is algebraic over F, it has a minimal                   polynomial m(x) over F, and the conjugates of c are the roots of                    m(x)) [12:11] &lt;Kasadkad&gt; so already that puts a big restriction on how many automorphisms we                   can have of a finite extension [12:12] &lt;Kasadkad&gt; for example if phi is an automorphism of C/R, then phi(i) has to be                   either +-i, and then phi(a+bi) = a +- bi, so phi is either the identity map or complex conjugation [12:12] &lt;Kasadkad&gt; there's a converse to the fact that phi(c) is a conjugate of c [12:13] &lt;Kasadkad&gt; namely, if c and d in K satisfy the same irreducible polynomial in                   F[x], then there's an automorphism phi of K/F such that phi(c) = d [12:14] &lt;Kasadkad&gt; err [12:14] &lt;Kasadkad&gt; let me state it a little more generally, and a little less wrong [12:15] &lt;Kasadkad&gt; given an automorphism phi : F -&gt; F, and c, d in K which are conjugate, there's a unique isomorphism phi' : F(c) -&gt; F(d) which agrees with phi on F [12:17] &lt;Kasadkad&gt; if phi is the identity, you can see that easily because {1, c, ..., c^{n-1}} is a basis of F(c) and {1, d, ..., d^{n-1}} is a basis of                   F(d), so you get a unique bijective linear map F(c) -&gt; F(d) sending c^k to d^k, and that does it [12:18] &lt;Kasadkad&gt; so let me give a little informal motivation for the Galois correspondence [12:18] &lt;Kasadkad&gt; well, first [12:18] &lt;Kasadkad&gt; notice that the automorphisms of K/F form a group [12:19] &lt;Kasadkad&gt; if phi_1, phi_2 are two automorphisms of K/F, their composition is                   definitely an automorphism of K and fixes F, their inverses also fix F, and we have the identity map[ [12:19] &lt;Kasadkad&gt; this group is called Aut(K/F) [12:20] &lt;Kasadkad&gt; so, suppose we have some element c of K; we can look at the subset H                   of G = Aut(K/F) consisting of those elements which fix c [12:20] &lt;Kasadkad&gt; in fact it's easy to see that H is a subgroup of G [12:21] &lt;Kasadkad&gt; now we want to look at the conjugates of c in K [12:21] &lt;Kasadkad&gt; we know that they're exactly {phi(c)} as phi runs over the group Aut(K/F) [12:22] &lt;Kasadkad&gt; I sort of already pointed it out, but if K/F is finite, then so is                   Aut(K/F) [12:23] &lt;Kasadkad&gt; since K = F(c_1, ..., c_n), and any automorphism of K sends each c_i to one of its finitely many conjugates, and once you know what it                   does to the c_i it's completely determined [12:23] &lt;Kasadkad&gt; so back to H [12:24] &lt;Kasadkad&gt; we know that the conjugates of c are {phi(c)}, but they might be                   repeated as phi runs over all the automorphisms of K/F [12:24] &lt;Kasadkad&gt; certainly they'll be repeated |H| times as we run over the subgroup H, for example [12:24] &lt;Kasadkad&gt; but notice if we mod out by H, i.e. if phi_1 H, ..., phi_n H are the distinct cosets of H in G = Aut(K/F), then we do get each one exactly once [12:25] &lt;Kasadkad&gt; that is, phi_1(c), ..., phi_n(c) are exactly the distinct conjugates of c in K [12:25] &lt;Kasadkad&gt; where n = [G:H] [12:25] &lt;Kasadkad&gt; so that's something [12:26] &lt;Kasadkad&gt; suppose for the moment that K contains ALL the conjugates of c; that is, the minimal polynomial factors completely in K [12:26] &lt;Kasadkad&gt; then this tells us the minimal polynomial of c over F must look like (x - phi_1(c))^k_1 ... (x - phi_n(c))^k_n, for some integers k_i &gt;= 1 [12:26] &lt;Kasadkad&gt; since the phi_i(c)'s are exactly the roots of thise polynomial [12:27] &lt;Kasadkad&gt; also remember that the minimal polynomial is irreducible over F [12:28] &lt;Kasadkad&gt; it might seem like an irreducible polynomial should not be able to                   have multiple roots, and fortunately that's true for most field extensions (including any extension of a field of characteristic zero                   or a finite field) [12:28] &lt;Kasadkad&gt; so the minimal polynomial of c over F is (x - phi_1(c))...(x -                   phi_n(c)) [12:28] &lt;Kasadkad&gt; in other words, the degree of c over F is n = [G:H] [12:28] &lt;Kasadkad&gt; and that's the same as the degree [F(c) : F] [12:29] &lt;Kasadkad&gt; so we've managed to show, given a field F &lt;= F(c) &lt;= K, that there's                   a subgroup H of Aut(K/F) such that [G:H] = [F(c):F] [12:31] &lt;Kasadkad&gt; let's see about the few assumptions we made in that argument [12:31] &lt;Kasadkad&gt; the first one was that given c in K, the minimal polynomial of c over F factors completely in K [12:31] &lt;Kasadkad&gt; i.e. &quot;all the conjugates of c are in K&quot; [12:32] &lt;Kasadkad&gt; this certainly doesn't have to happen in general [12:32] &lt;Kasadkad&gt; for example the extension Q(2^(1/3))/Q; the minimal polynomial of                   3^(1/3) over Q is x^3 - 2, but that doesn't factor over Q(2^(1/3)) because we don't have the elements 2^(1/3) (-1 +- sqrt(-3))/2 [12:33] &lt;Kasadkad&gt; when it does happen, we say that K/F is a normal extension [12:34] &lt;Kasadkad&gt; normal extensions sound a little hard to recognize at first, but there's sort of a surprising equivalence [12:35] &lt;Kasadkad&gt; if p(x) is a polynomial in F[x], and it factors completely in K                   containing F, with roots r_1, ..., r_n, then we say the extension F(r_1, ..., r_n)/F is a splitting field for p [12:35] &lt;Kasadkad&gt; it's the &quot;smallest field in which p factors completely&quot; [12:35] &lt;Kasadkad&gt; certainly if a splitting field contains a root of p(x) then it                   contains all the conjugates of that root, by definition [12:36] &lt;Kasadkad&gt; the surprising thing is that a splitting field is in fact normal, and that finite normal extensions are splitting fields! [12:37] &lt;Kasadkad&gt; so something like Q(2^(1/3), sqrt(-3))/Q is a normal extension, because it's a splitting field for x^3 - 2 [12:37] &lt;Kasadkad&gt; the other assumption we made was that an irreducible polynomial over F cannot have multiple roots in K [12:38] &lt;Kasadkad&gt; I'm not going to talk about this much: if that's true, then K/F is                   called separable, and it's not hard to check that extensions of                    characteristic zero fields and of finite fields are separable [12:39] &lt;Kasadkad&gt; the trick is that f(x) has a multiple root in some extension iff f(x) and f'(x) have a common root (that's the derivative of f) [12:39] &lt;Kasadkad&gt; so you look at gcd(f(x), f'(x)) [12:40] &lt;Kasadkad&gt; in the motivation I gave earlier we saw that if K/F is finite, normal, and separable, then we get a neat way to associate subgroups of Aut(K/F) with subfields F(c) of K containing F [12:40] &lt;Kasadkad&gt; so there's a special name for such an extension: K/F is called a                   Galois extension when it is finite, normal, and separable [12:41] &lt;Kasadkad&gt; let's introduce a little notation [12:41] &lt;Kasadkad&gt; if K/F is Galois then we write Gal(K/F) instead of Aut(K/F), and call it the Galois group of K/F [12:42] &lt;Kasadkad&gt; if E is a subfield of K containing F, then E is called an                   intermediate field of K/F [12:42] &lt;Kasadkad&gt; as before we can look at the subgroup of G = Gal(K/F) which fix E,                   and we'll denote that G_E [12:43] &lt;Kasadkad&gt; we saw that, at least, [F(c) : F] = [G : G_F(c)] [12:43] &lt;Kasadkad&gt; when K/F is Galois [12:43] &lt;Kasadkad&gt; and this is true for more general intermediate fields E: [E : F] = [G : G_E] [12:44] &lt;Kasadkad&gt; (in fact, it turns out that any finite separable extension K/F has                   the form K = F(c) for some c, which proves what I just said, but I'm                    not going to prove it and you can do things without assuming that) [12:45] &lt;Kasadkad&gt; so we have this association E |-&gt; G_E of intermediate fields to                   subgroups of G [12:45] &lt;Kasadkad&gt; let's see that this is one-to-one [12:45] &lt;Kasadkad&gt; i.e. if E', E are intermediate fields and G_E = G_E', then E = E' [12:45] &lt;Kasadkad&gt; the thing to consider is the field generated by E' and E within K [12:46] &lt;Kasadkad&gt; we've seen the field F(c_1, ..., c_n) generated by F and c_1, ..., c_n, and this is the same construction [12:47] &lt;Kasadkad&gt; e.g. if E = F(a_1, ..., a_m), E' = F(b_1, ..., b_n) then the composite EE' is just F(a_1, ..., a_m, b_1, ..., b_n) [12:47] &lt;Kasadkad&gt; &quot;the smallest subfield of K containing both E and E'&quot; [12:47] &lt;Kasadkad&gt; the fact that E and E' generate EE', together with G_E = G_E', means that G_{EE'} = G_E also [12:48] &lt;Kasadkad&gt; since if an automorphism fixes E and E', then it also fixes EE', and if it fixes EE', that contains E and E', so it definitely fixes those too [12:48] &lt;Kasadkad&gt; but now we can exploit our formula for the size of the group G_E [12:49] &lt;Kasadkad&gt; that is, [EE' : F] = [G : G_{EE'}] = [G : G_E] = [E : F] [12:49] &lt;Kasadkad&gt; but EE' contains E, so by finite dimensionality, [EE' : F] = [E : F]                   is only possible if EE' = E [12:49] &lt;Kasadkad&gt; and that implies that E' was contained in E to start with [12:50] &lt;Kasadkad&gt; (and then just do this argument with E replaced by E' to see that E                   is also contained in E) [12:50] &lt;Kasadkad&gt; so, the association of intermediate fields E to their fixing subgroups G_E is one-to-one! [12:50] &lt;Kasadkad&gt; that already tells us something surprising: there are only finitely many intermediate fields of a Galois extension K/F [12:50] &lt;Kasadkad&gt; since there are only finitely many subgroups of the finite group Gal(K/F) [12:52] &lt;Kasadkad&gt; it would be nice to see that this association was also onto; that is, any subgroup of Gal(K/F) is the fixing subgroup of some intermediate field of K/F [12:52] &lt;Kasadkad&gt; well, there's at least an easy candidate for such a field [12:53] &lt;Kasadkad&gt; namely, given H a subgroup of Gal(K/F), the fixed field K^H of H is                   the set of elements x in K such that phi(x) = x for all x in H [12:53] &lt;Kasadkad&gt; (it's easy to check that K^H really is a field) [12:54] &lt;Kasadkad&gt; what's less easy to check is that this actually works: that is, that the fixing subgroup G_{K^H} of K^H is exactly H again [12:54] &lt;Kasadkad&gt; (not that it's that difficult to prove, but you probably wouldn't                   come up with how to do it out of nowhere) [12:55] &lt;Kasadkad&gt; although we can prove it in the case that K has the form K^H(c) for some c [12:55] &lt;Kasadkad&gt; which as I said before is actually true in the setting we're in, so [12:56] &lt;Kasadkad&gt; notice that H &lt;= G_{K^H} is trivial from the definitions; certainly everything in H fixes everything in K^H [12:56] &lt;Kasadkad&gt; but maybe there are things that fix K^H that weren't in the original group H? [12:57] &lt;Kasadkad&gt; so let's prove that [K : K^H] = |H| [12:59] &lt;Kasadkad&gt; that will show that H = G_{K^H}, because we know [G : G_{K^H}] = [K^H : F] [13:00] &lt;Kasadkad&gt; and also |G| = [K:F], so dividing gives us |G_{K^H}| = [K : K^H] = |H|, so finiteness forces G_{K^H} = H [13:00] &lt;Kasadkad&gt; so, suppose H = {h_1, ..., h_n}, where n = |H| [13:01] &lt;Kasadkad&gt; look at the polynomial (x - h_1(c))...(x - h_n(c)) [13:01] &lt;Kasadkad&gt; notice that if you apply a member of H to this polynomial (i.e. its                   coefficients), it remains unchanged, because {hh_1, ..., hh_n} is                    again the group H, for any h in H [13:02] &lt;Kasadkad&gt; so doing that just interchanges the factors (x - h_i(c)), which doesn't change the polynomial [13:02] &lt;Kasadkad&gt; so by definition, the coefficients of the polynomial are in K^H, since they're fixed by H [13:03] &lt;Kasadkad&gt; that is, we've exhibited a polynomial of degree n in K^H[x] with c as                   a root, so we must have [K : K^H] = [K^H(c) : K^H] &lt;= n = |H| [13:03] &lt;Kasadkad&gt; which is exactly the other direction of the inequality that was needed [13:03] &lt;Kasadkad&gt; so! [13:03] &lt;Kasadkad&gt; let's collect what we have now [13:04] &lt;Kasadkad&gt; there's a bijection between intermediate fields of K/F and subgroups of Gal(K/F) [13:04] &lt;Kasadkad&gt; given by sending an intermediate field E to its fixing subgroup G_E, and in the other direction, a subgroup H to its fixed field K^H [13:05] &lt;Kasadkad&gt; there are a few other nice properties that I haven't stated but are easy to check [13:05] &lt;Kasadkad&gt; like, this bijection is inclusion-reversing [13:05] &lt;Kasadkad&gt; if E &lt;= E', then G_{E'} &lt;= G_E, and if H &lt;= H', then K^{H'} &lt;= K^H [13:05] &lt;Kasadkad&gt; those are just straight from the definitions [13:06] &lt;Kasadkad&gt; also, the normal subgroups H of G correspond exactly the subextensions K^H/F of K/F which are themselves Galois [13:06] &lt;Kasadkad&gt; but I won't say any more about that now [13:07] &lt;Kasadkad&gt; this bijection is called the Galois correspondence, btw [13:07] &lt;Kasadkad&gt; so let's actually compute one of these Galois groups and then I'll                   stop [13:08] &lt;Kasadkad&gt; set z = (-1 + sqrt(-3))/2; this is a primitive cube root of unity [13:09] &lt;Kasadkad&gt; let's do K = Q(3^(1/3), z) and compute Gal(K/Q) [13:09] &lt;Kasadkad&gt; notice that K/Q is indeed a Galois extension, because it's exactly the extension of Q generated by the roots of x^3 - 2 (it's a                   splitting field) [13:10] &lt;Kasadkad&gt; well, the thing to do is to look at what the automorphisms do to the generators 3^(1/3), z [13:10] &lt;Kasadkad&gt; once we know how they act on those, we know how they act on the whole field [13:10] &lt;Kasadkad&gt; there are six possibilities [13:10] &lt;Kasadkad&gt; 3^(1/3) must map to one of its three conjugates, 3^(1/3), z 3^(1/3), or z^2 3^(1/3) [13:11] &lt;Kasadkad&gt; and z must map to one of its two conjugates, z or z^2 [13:11] &lt;Kasadkad&gt; we can't conclude from this that these automorphisms form the Galois group of K/F, though [13:11] &lt;Kasadkad&gt; in fact we can't even conclude that they're automorphisms! [13:11] &lt;Kasadkad&gt; or anything at all [13:11] &lt;Kasadkad&gt; there's nothing telling us that defining automorphisms on generators like this will give us something well-defined [13:12] &lt;Kasadkad&gt; but fortunately there's a trick to get around that [13:13] &lt;Kasadkad&gt; namely, we already know what |Gal(K/F)| = |G| has to be, because it's                   just [G : {identity}] = [K^{identity} : F] = [K : F] [13:13] &lt;Kasadkad&gt; that is, |Gal(K/F)| = [K:F] when K/F is Galois [13:13] &lt;Kasadkad&gt; and here [K:F] is 6 [13:14] &lt;Kasadkad&gt; err, K/Q I guess [13:14] &lt;Kasadkad&gt; that's not too hard to see in this case: [K:Q] = [Q(3^(1/3), z) : Q(3^(1/3))][Q(3^(1/3)) : Q], since degrees multiply [13:15] &lt;Kasadkad&gt; [Q(3^(1/3)) : Q] = 3 is easy because x^3 - 3 is irreducible [13:15] &lt;Kasadkad&gt; [Q(3^(1/3), z) : Q(3^(1/3))] is also easy because x^2 + x + 1 is                   irreducible over Q(3^(1/3)); otherwise it would have a root in                    Q(3^(1/3)), but it can't because its roots are both non-real [13:15] &lt;Kasadkad&gt; uh [Q(3^(1/3), z) : Q(3^(1/3))] = 2 [13:15] &lt;Kasadkad&gt; so [K:F] = 6 and hence |Gal(K/F)| = 6 [13:16] &lt;Kasadkad&gt; we already saw that there are at most 6 possible automorphisms in                   Gal(K/F), and we listed what they would have to be [13:16] &lt;Kasadkad&gt; so this tells us that there really are 6, and that the ones we listed are indeed well-defined [13:16] &lt;Kasadkad&gt; that doesn't tell us what group Gal(K/F) is, though [13:17] &lt;Kasadkad&gt; it could be either Z/6Z or S_3, the two groups of order 6 [13:17] &lt;Kasadkad&gt; an easy way to distinguish them is to produce a pair of elements that don't commute [13:18] &lt;Kasadkad&gt; for example if s sends 3^(1/3) to z 3^(1/3) and z to z, and t sends 3^(1/3) to 3^(1/3) and z to z^2 [13:19] &lt;Kasadkad&gt; then st(3^(1/3)) = s(3^(1/3)) = z 3^(1/3), while ts(3^(1/3)) = t(z                   3^(1/3)) = z^2 3^(1/3) [13:19] &lt;Kasadkad&gt; so s and t don't commute, and we conclude that Gal(K/F) =~ S_3 [13:19] &lt;Kasadkad&gt; I keep writing K/F instead of K/Q :( [13:20] &lt;Kasadkad&gt; an even easier way to see that is that normal subgroups of Gal(K/Q)                   correspond to intermediate fields of K/Q that are Galois over Q [13:20] &lt;Kasadkad&gt; if Gal(K/Q) were abelian, all its intermediate fields would be Galois                    over Q [13:20] &lt;Kasadkad&gt; but they aren't, namely Q(3^(1/3))/Q isn't [13:20] &lt;Kasadkad&gt; so now with Gal(K/Q) we can list all the intermediate fields of K/Q [13:21] &lt;Kasadkad&gt; notice that Gal(K/Q) is generated by the s and t I gave above; it's                    {1, s, s^2, t, ts, ts^2} [13:21] &lt;Kasadkad&gt; (s is a &quot;rotation&quot; and t is a &quot;reflection&quot;, thinking of D_3) [13:22] &lt;Kasadkad&gt; and the subgroups of Gal(K/Q) =~ S_3 are: {1}, {1, t}, {1, ts}, {1,                    ts^2}, and {1, s, s^2} [13:22] &lt;Kasadkad&gt; so what's the fixed field of, say, {1, t}? [13:22] &lt;Kasadkad&gt; well we can forget about the 1, it's just the identity map [13:23] &lt;Kasadkad&gt; t sends 3^(1/3) to 3^(1/3) and z to z^2 [13:23] &lt;Kasadkad&gt; so one thing t definitely fixes is 3^(1/3) [13:23] &lt;Kasadkad&gt; and then t at least fixes the whole field Q(3^(1/3)) [13:23] &lt;Kasadkad&gt; but that's actually the whole fixed field [13:23] &lt;Kasadkad&gt; because [Q(3^(1/3)) : Q] = 3 = [Gal(K/Q) : {1, t}] [13:24] &lt;Kasadkad&gt; you can do this similarly for the other subgroups to see that the intermediate fields of K/Q are exactly: [13:25] &lt;Kasadkad&gt; Q, Q(3^(1/3)), Q(z 3^(1/3)), Q(z^2 3^(1/3)), Q(z), and K itself [13:26] &lt;Kasadkad&gt; this intermediate field stuff might seem a little boring, but it can actually tell you stuff about the extension K/F [13:26] &lt;Kasadkad&gt; we'll see next week that if K/F is a &quot;radical extension&quot;, i.e. one whose elements can be written in terms of roots of elements of F,                   then Gal(K/F) must have a special property [13:28] &lt;Kasadkad&gt; and that if K is the splitting field of a certain polynomial of                   degree 5 over Q, maybe x^5 - 3x^2 + 1 [13:28] &lt;Kasadkad&gt; then Gal(K/F) doesn't have that property [13:28] &lt;Kasadkad&gt; so that the roots of x^4 - 3x^2 + 1 can't be written in terms of                   radicals [13:28] &lt;Kasadkad&gt; i.e. there's no &quot;quartic equation&quot; [13:28] &lt;Kasadkad&gt; uh &quot;quintic' [13:28] &lt;Kasadkad&gt; okay that's it for me [13:28] &lt;Kasadkad&gt; today [13:28] &lt;thermoplyae&gt; excellent [13:28] &lt;breeden&gt; thank you Kasadkad :) [13:29] &lt;thermoplyae&gt; do you plan to continue next week? [13:29] &lt;Kasadkad&gt; yeah [13:29] &lt;Kasadkad&gt; it won't be as long as this one since I just want to go over the                   insolubility of the quintic as an application [13:30] &lt;thermoplyae&gt; do you have a topic handy? [13:30] &lt;Kasadkad&gt; oh, now I see why I said &quot;quartic equation&quot; [13:30] &lt;thermoplyae&gt; ah, ok [13:30] &lt;Kasadkad&gt; x^5 - 3x^2 + 1, of course [13:30] &lt;Kasadkad&gt; &quot;The Insolubility of the Quintic&quot; seems appropriate [13:30] &lt;Kasadkad&gt; or like [13:31] &lt;Kasadkad&gt; yeah that's fine [13:31] &lt;Kasadkad&gt; probably there'll be time to spare, and I can talk a little about                    something else like how to actually compute Galois groups [13:31] &lt;Dudicon&gt; wooo