This seminar by Kasadkad took place on 16th November 2008 20:00 UTC, in #mathematics channel of

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Seminar Edit

[12:02] * ChanServ changes topic to 'SEMINAR IN PROGRESS.  If you want to ask a 
                   question, say ! and wait to be called'
[12:03] <Kasadkad> 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] <Kasadkad> for K a field, a function phi : K -> 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] <Kasadkad> e.g. if K = C, complex conjugation is an automorphism
[12:06] <Kasadkad> general automorphisms are too much, though; there are tons of bizarre 
                   discontinuous automorphisms of C
[12:06] <Kasadkad> so what we'll look at is automorphisms of K that fix the base field F
[12:06] <Kasadkad> i.e. phi(x) = x for all x in F
[12:06] <Kasadkad> notice that complex conjugation is also an automorphism of the 
                   extension C/R in this sense, since it fixes real numbers
[12:08] <Kasadkad> 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] <Kasadkad> 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] <Kasadkad> well
[12:09] <Kasadkad> sorry I messed that up a little
[12:09] <Kasadkad> if c is a root of p(x) in F[x]
[12:09] <Kasadkad> because the whole point is that we need phi to fix the coefficients 
                   a_i of p(x)
[12:10] <Kasadkad> so, automorphisms of K/F take field elements to conjugates
[12:10] <Kasadkad> (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 
[12:11] <Kasadkad> so already that puts a big restriction on how many automorphisms we 
                   can have of a finite extension
[12:12] <Kasadkad> 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] <Kasadkad> there's a converse to the fact that phi(c) is a conjugate of c
[12:13] <Kasadkad> 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] <Kasadkad> err
[12:14] <Kasadkad> let me state it a little more generally, and a little less wrong
[12:15] <Kasadkad> given an automorphism phi : F -> F, and c, d in K which are 
                   conjugate, there's a unique isomorphism phi' : F(c) -> F(d) which 
                   agrees with phi on F
[12:17] <Kasadkad> 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) -> F(d) sending 
                   c^k to d^k, and that does it
[12:18] <Kasadkad> so let me give a little informal motivation for the Galois 
[12:18] <Kasadkad> well, first
[12:18] <Kasadkad> notice that the automorphisms of K/F form a group
[12:19] <Kasadkad> 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] <Kasadkad> this group is called Aut(K/F)
[12:20] <Kasadkad> 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] <Kasadkad> in fact it's easy to see that H is a subgroup of G
[12:21] <Kasadkad> now we want to look at the conjugates of c in K
[12:21] <Kasadkad> we know that they're exactly {phi(c)} as phi runs over the group 
[12:22] <Kasadkad> I sort of already pointed it out, but if K/F is finite, then so is 
[12:23] <Kasadkad> 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] <Kasadkad> so back to H
[12:24] <Kasadkad> 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] <Kasadkad> certainly they'll be repeated |H| times as we run over the subgroup 
                   H, for example
[12:24] <Kasadkad> 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 
[12:25] <Kasadkad> that is, phi_1(c), ..., phi_n(c) are exactly the distinct conjugates 
                   of c in K
[12:25] <Kasadkad> where n = [G:H]
[12:25] <Kasadkad> so that's something
[12:26] <Kasadkad> suppose for the moment that K contains ALL the conjugates of c; that 
                   is, the minimal polynomial factors completely in K
[12:26] <Kasadkad> 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 >= 1
[12:26] <Kasadkad> since the phi_i(c)'s are exactly the roots of thise polynomial
[12:27] <Kasadkad> also remember that the minimal polynomial is irreducible over F
[12:28] <Kasadkad> 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] <Kasadkad> so the minimal polynomial of c over F is (x - phi_1(c))...(x - 
[12:28] <Kasadkad> in other words, the degree of c over F is n = [G:H]
[12:28] <Kasadkad> and that's the same as the degree [F(c) : F]
[12:29] <Kasadkad> so we've managed to show, given a field F <= F(c) <= K, that there's 
                   a subgroup H of Aut(K/F) such that [G:H] = [F(c):F]
[12:31] <Kasadkad> let's see about the few assumptions we made in that argument
[12:31] <Kasadkad> the first one was that given c in K, the minimal polynomial of c over 
                   F factors completely in K
[12:31] <Kasadkad> i.e. "all the conjugates of c are in K"
[12:32] <Kasadkad> this certainly doesn't have to happen in general
[12:32] <Kasadkad> 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] <Kasadkad> when it does happen, we say that K/F is a normal extension
[12:34] <Kasadkad> normal extensions sound a little hard to recognize at first, but 
                   there's sort of a surprising equivalence
[12:35] <Kasadkad> 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] <Kasadkad> it's the "smallest field in which p factors completely"
[12:35] <Kasadkad> certainly if a splitting field contains a root of p(x) then it 
                   contains all the conjugates of that root, by definition
[12:36] <Kasadkad> the surprising thing is that a splitting field is in fact normal, and 
                   that finite normal extensions are splitting fields!
[12:37] <Kasadkad> 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] <Kasadkad> the other assumption we made was that an irreducible polynomial over 
                   F cannot have multiple roots in K
[12:38] <Kasadkad> 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] <Kasadkad> 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] <Kasadkad> so you look at gcd(f(x), f'(x))
[12:40] <Kasadkad> 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] <Kasadkad> 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] <Kasadkad> let's introduce a little notation
[12:41] <Kasadkad> 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] <Kasadkad> if E is a subfield of K containing F, then E is called an 
                   intermediate field of K/F
[12:42] <Kasadkad> 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] <Kasadkad> we saw that, at least, [F(c) : F] = [G : G_F(c)]
[12:43] <Kasadkad> when K/F is Galois
[12:43] <Kasadkad> and this is true for more general intermediate fields E: [E : F] =
                   [G : G_E]
[12:44] <Kasadkad> (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] <Kasadkad> so we have this association E |-> G_E of intermediate fields to 
                   subgroups of G
[12:45] <Kasadkad> let's see that this is one-to-one
[12:45] <Kasadkad> i.e. if E', E are intermediate fields and G_E = G_E', then E = E'
[12:45] <Kasadkad> the thing to consider is the field generated by E' and E within K
[12:46] <Kasadkad> 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] <Kasadkad> 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] <Kasadkad> "the smallest subfield of K containing both E and E'"
[12:47] <Kasadkad> the fact that E and E' generate EE', together with G_E = G_E', means 
                   that G_{EE'} = G_E also
[12:48] <Kasadkad> 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 
[12:48] <Kasadkad> but now we can exploit our formula for the size of the group G_E
[12:49] <Kasadkad> that is, [EE' : F] = [G : G_{EE'}] = [G : G_E] = [E : F]
[12:49] <Kasadkad> but EE' contains E, so by finite dimensionality, [EE' : F] = [E : F] 
                   is only possible if EE' = E
[12:49] <Kasadkad> and that implies that E' was contained in E to start with
[12:50] <Kasadkad> (and then just do this argument with E replaced by E' to see that E 
                   is also contained in E)
[12:50] <Kasadkad> so, the association of intermediate fields E to their fixing 
                   subgroups G_E is one-to-one!
[12:50] <Kasadkad> that already tells us something surprising: there are only finitely 
                   many intermediate fields of a Galois extension K/F
[12:50] <Kasadkad> since there are only finitely many subgroups of the finite group 
[12:52] <Kasadkad> 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] <Kasadkad> well, there's at least an easy candidate for such a field
[12:53] <Kasadkad> 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] <Kasadkad> (it's easy to check that K^H really is a field)
[12:54] <Kasadkad> 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] <Kasadkad> (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] <Kasadkad> although we can prove it in the case that K has the form K^H(c) for 
                   some c
[12:55] <Kasadkad> which as I said before is actually true in the setting we're in, so
[12:56] <Kasadkad> notice that H <= G_{K^H} is trivial from the definitions; certainly 
                   everything in H fixes everything in K^H
[12:56] <Kasadkad> but maybe there are things that fix K^H that weren't in the original 
                   group H?
[12:57] <Kasadkad> so let's prove that [K : K^H] = |H|
[12:59] <Kasadkad> that will show that H = G_{K^H}, because we know [G : G_{K^H}] =
                   [K^H : F]
[13:00] <Kasadkad> 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] <Kasadkad> so, suppose H = {h_1, ..., h_n}, where n = |H|
[13:01] <Kasadkad> look at the polynomial (x - h_1(c))...(x - h_n(c))
[13:01] <Kasadkad> 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] <Kasadkad> so doing that just interchanges the factors (x - h_i(c)), which 
                   doesn't change the polynomial
[13:02] <Kasadkad> so by definition, the coefficients of the polynomial are in K^H, 
                   since they're fixed by H
[13:03] <Kasadkad> 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] <= n = |H|
[13:03] <Kasadkad> which is exactly the other direction of the inequality that was 
[13:03] <Kasadkad> so!
[13:03] <Kasadkad> let's collect what we have now
[13:04] <Kasadkad> there's a bijection between intermediate fields of K/F and subgroups 
                   of Gal(K/F)
[13:04] <Kasadkad> 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] <Kasadkad> there are a few other nice properties that I haven't stated but are 
                   easy to check
[13:05] <Kasadkad> like, this bijection is inclusion-reversing
[13:05] <Kasadkad> if E <= E', then G_{E'} <= G_E, and if H <= H', then K^{H'} <= K^H
[13:05] <Kasadkad> those are just straight from the definitions
[13:06] <Kasadkad> also, the normal subgroups H of G correspond exactly the 
                   subextensions K^H/F of K/F which are themselves Galois
[13:06] <Kasadkad> but I won't say any more about that now
[13:07] <Kasadkad> this bijection is called the Galois correspondence, btw
[13:07] <Kasadkad> so let's actually compute one of these Galois groups and then I'll 
[13:08] <Kasadkad> set z = (-1 + sqrt(-3))/2; this is a primitive cube root of unity
[13:09] <Kasadkad> let's do K = Q(3^(1/3), z) and compute Gal(K/Q)
[13:09] <Kasadkad> 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] <Kasadkad> well, the thing to do is to look at what the automorphisms do to the 
                   generators 3^(1/3), z
[13:10] <Kasadkad> once we know how they act on those, we know how they act on the whole 
[13:10] <Kasadkad> there are six possibilities
[13:10] <Kasadkad> 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] <Kasadkad> and z must map to one of its two conjugates, z or z^2
[13:11] <Kasadkad> we can't conclude from this that these automorphisms form the Galois 
                   group of K/F, though
[13:11] <Kasadkad> in fact we can't even conclude that they're automorphisms!
[13:11] <Kasadkad> or anything at all
[13:11] <Kasadkad> there's nothing telling us that defining automorphisms on generators 
                   like this will give us something well-defined
[13:12] <Kasadkad> but fortunately there's a trick to get around that
[13:13] <Kasadkad> 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] <Kasadkad> that is, |Gal(K/F)| = [K:F] when K/F is Galois
[13:13] <Kasadkad> and here [K:F] is 6
[13:14] <Kasadkad> err, K/Q I guess
[13:14] <Kasadkad> 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] <Kasadkad> [Q(3^(1/3)) : Q] = 3 is easy because x^3 - 3 is irreducible
[13:15] <Kasadkad> [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] <Kasadkad> uh [Q(3^(1/3), z) : Q(3^(1/3))] = 2
[13:15] <Kasadkad> so [K:F] = 6 and hence |Gal(K/F)| = 6
[13:16] <Kasadkad> 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] <Kasadkad> so this tells us that there really are 6, and that the ones we listed 
                   are indeed well-defined
[13:16] <Kasadkad> that doesn't tell us what group Gal(K/F) is, though
[13:17] <Kasadkad> it could be either Z/6Z or S_3, the two groups of order 6
[13:17] <Kasadkad> an easy way to distinguish them is to produce a pair of elements that 
                   don't commute
[13:18] <Kasadkad> 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] <Kasadkad> 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] <Kasadkad> so s and t don't commute, and we conclude that Gal(K/F) =~ S_3
[13:19] <Kasadkad> I keep writing K/F instead of K/Q :(
[13:20] <Kasadkad> 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] <Kasadkad> if Gal(K/Q) were abelian, all its intermediate fields would be Galois 
                   over Q
[13:20] <Kasadkad> but they aren't, namely Q(3^(1/3))/Q isn't
[13:20] <Kasadkad> so now with Gal(K/Q) we can list all the intermediate fields of K/Q
[13:21] <Kasadkad> 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] <Kasadkad> (s is a "rotation" and t is a "reflection", thinking of D_3)
[13:22] <Kasadkad> 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] <Kasadkad> so what's the fixed field of, say, {1, t}?
[13:22] <Kasadkad> well we can forget about the 1, it's just the identity map
[13:23] <Kasadkad> t sends 3^(1/3) to 3^(1/3) and z to z^2
[13:23] <Kasadkad> so one thing t definitely fixes is 3^(1/3)
[13:23] <Kasadkad> and then t at least fixes the whole field Q(3^(1/3))
[13:23] <Kasadkad> but that's actually the whole fixed field
[13:23] <Kasadkad> because [Q(3^(1/3)) : Q] = 3 = [Gal(K/Q) : {1, t}]
[13:24] <Kasadkad> you can do this similarly for the other subgroups to see that the 
                   intermediate fields of K/Q are exactly:
[13:25] <Kasadkad> Q, Q(3^(1/3)), Q(z 3^(1/3)), Q(z^2 3^(1/3)), Q(z), and K itself
[13:26] <Kasadkad> this intermediate field stuff might seem a little boring, but it can 
                   actually tell you stuff about the extension K/F
[13:26] <Kasadkad> we'll see next week that if K/F is a "radical extension", 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] <Kasadkad> 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] <Kasadkad> then Gal(K/F) doesn't have that property
[13:28] <Kasadkad> so that the roots of x^4 - 3x^2 + 1 can't be written in terms of 
[13:28] <Kasadkad> i.e. there's no "quartic equation"
[13:28] <Kasadkad> uh "quintic'
[13:28] <Kasadkad> okay that's it for me
[13:28] <Kasadkad> today
[13:28] <thermoplyae> excellent
[13:28] <breeden> thank you Kasadkad :)
[13:29] <thermoplyae> do you plan to continue next week?
[13:29] <Kasadkad> yeah
[13:29] <Kasadkad> 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] <thermoplyae> do you have a topic handy?
[13:30] <Kasadkad> oh, now I see why I said "quartic equation"
[13:30] <thermoplyae> ah, ok
[13:30] <Kasadkad> x^5 - 3x^2 + 1, of course
[13:30] <Kasadkad> "The Insolubility of the Quintic" seems appropriate
[13:30] <Kasadkad> or like
[13:31] <Kasadkad> yeah that's fine
[13:31] <Kasadkad> 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] <Dudicon> wooo

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