Basic Field Theory (towards Galois Theory)

Seminar
 20:00:16 Kasadkad: ok my clock says 12:00 20:00:22 ChanServ changed the topic of #mathematics to: SEMINAR IN PROGRESS. If you want to ask a question, say ! and wait to be called

20:02:00 Kasadkad: this is the first of probably three seminars leading up to galois theory; since I figured if you've seen the field theory necessary to get there you've probably seen galois theory anyway, it seemed silly not to go over some of that first 20:02:45 Kasadkad: eventually we'll get to the sort of de rigueur showoff applications like the unsolvability of the quintic equation in terms of radicals 20:03:08 Kasadkad: so I'm going to assume people know what a field is 20:03:28 Kasadkad: and very basic linear algebra and group theory 20:04:19 Kasadkad: if you don't know what a field is just pretend when I say &quot;a field F&quot; that I'm saying &quot;the rational numbers Q&quot; (but really if you don't know what a                  field is you probably won't get much out of this) 20:04:20 Kasadkad: anyway 20:04:52 Kasadkad: we're going to be interested for a while in how one field behaves with a                  smaller field contained in it 20:05:28 Kasadkad: if F and K are fields with F a subset of K, we say that K is an extension of F, and talk about &quot;the extension K/F&quot; 20:05:49 Libster: ! 20:06:10 Kasadkad: we'll deal with especially nice sorts of extensions, algebraic extensions, but let me tell you what algebraic elements are first 20:06:11 Kasadkad: yes? 20:06:18 Libster: so it's essentially like a &quot;subfield?&quot; 20:06:23 Kasadkad: heh, yeah 20:06:28 Libster: with the operations +,* induced from K 20:06:31 Kasadkad: right 20:06:51 Kasadkad: I guess I should say that if we have a homomorphism from F into K then we'll                  also consider K to be an extension of F 20:07:02 Kasadkad: (since field homomorphisms are injective) 20:07:11 Kasadkad: but that's not a big deal 20:07:21 Kasadkad: ok so 20:07:44 Kasadkad: people may be familiar with what &quot;algebraic numbers&quot; and &quot;transcendental numbers&quot; are, and the general definition is the same 20:08:19 Kasadkad: given a field extension K/F, and c in K, we say that c is algebraic over F                  if there's a polynomial f(x) with coefficients in F such that f(c) = 0 20:08:53 Kasadkad: so like, if K = C the complex numbers and F = Q the rational numbers, then 2^(1/3) is algebraic over Q since it's a root of x^3 - 2, which has rational coefficients 20:09:49 Kasadkad: if c in K is not algebraic over F, then it's called transcendental over F 20:10:06 Kasadkad: e.g. pi in C is transcendental over Q (don't ask me to prove it) 20:11:20 Kasadkad: Now an extension K/F such that every c in K is algebraic over F is called an algebraic extension 20:11:57 Kasadkad: and if K/F isn't algebraic then it's transcendental, but we won't think about such things 20:12:59 Kasadkad: It would be nice to give an example of an algebraic extension but we need to                  do a little more work to get one 20:13:06 Kasadkad: for example 20:13:23 Kasadkad: given an extension K/F and c in K, you can consider the field F(c) 20:14:11 Kasadkad: this is the &quot;smallest subfield of K containing c&quot;: any field containing F                  and c had better contain all polynomials in c with coefficients in F, and in fact all rational functions in c with coefficients in F 20:14:25 Kasadkad: and it turns out that that actually does form a field, which we call F(c) 20:14:44 vixey: ! 20:14:49 Kasadkad: yes? 20:14:52 vixey: F(c) doesn't depend on F at all, only K? 20:15:06 Kasadkad: It does depend on F 20:15:13 Kasadkad: It's rational functions in the variable c with coefficients in F 20:15:51 Kasadkad: e.g. Q(sqrt(2)) = {a + b sqrt(2) : a,b in Q} is different from R(sqrt(2)), which is just R 20:15:53 vixey: sorry I don't see why it is smallest subfield of K 20:16:08 Kasadkad: oh, I should have said &quot;smallest subfield of K containing F and c&quot; 20:16:17 Kasadkad: or &quot;subfield generated by F and c&quot; 20:16:26 vixey: ok I follow, thanks 20:16:30 Kasadkad: (I don't intend to qualify those with anything rigorous) 20:16:31 Kasadkad: ok 20:16:53 Kasadkad: it turns out that, if c is algebraic, F(c) is actually just polynomials in c                  with coefficients in F 20:18:17 Kasadkad: maybe that's not entirely obvious, but for example if a_0 + a_1 c + ... + a_n c^n = 0 where a_0 != 0 and a_i in F, then we can multiply through by c^{-1} to get a formula for c^{-1} as a polynomial in c 20:19:38 Kasadkad: so given something like 1/(3sqrt(2) - 4) we can rewrite that as (3sqrt(2) + 4)/2, a polynomial in sqrt(2) 20:19:40 Kasadkad: anyway 20:19:47 Kasadkad: I digressed a little 20:20:01 Kasadkad: the point is, given c algebraic over F, we'd like to say that F(c)/F is an                  algebraic extension 20:20:28 Kasadkad: but it's not really clear that if c is algebraic over F, then so is                  say 7c^3 - c + 4 20:20:46 Kasadkad: and we need all elements of F(c)/F to be algebraic over F 20:20:50 Libster: ! 20:20:52 Kasadkad: yes 20:20:59 Libster: sorry little confused on notation 20:21:10 Libster: you said K/F before 20:21:15 Libster: where F is a subfield of K 20:21:17 Kasadkad: yeah 20:21:26 Libster: isn't F(c) a subfield of... oh nevermind 20:21:28 Libster: sorry 20:21:33 Libster: it makes sense now 20:21:34 Kasadkad: F(c) is a subfield of K, F is a subfield of F 20:21:39 Kasadkad: err F is a subfield of F(c) 20:21:39 Libster: right 20:21:45 Kasadkad: ok 20:22:34 Kasadkad: I should mention that we can also have things like F(c_1, ..., c_n), and it's what you're probably imagining 20:22:55 Kasadkad: rational functions in c_1, ..., c_n with coefficients in F, and just polynomials if all the c_i's are algebraic 20:23:18 Kasadkad: alternatively we can describe F(c_1, ..., c_n) inductively as F(c_1, ..., c_{n-1})(c_n) 20:23:26 Kasadkad: we just adjoin the c_i's one at a time 20:23:38 Kasadkad: so 20:23:41 Kasadkad: here's an important definition 20:23:52 Kasadkad: notice that if K/F is a field extension, then K is actually a vector space over F 20:24:30 Kasadkad: e.g. Q(sqrt(2)) = {a + b sqrt(2) : a,b in Q} looks like a vector space over Q with basis {1, sqrt(2)} 20:25:07 Kasadkad: the dimension of K as an F-vector space is called the degree of K/F, and is written [K:F] 20:26:08 Kasadkad: other examples are [Q(2^(1/3)) : Q] = 3; a basis of Q(2^(1/3)) is {1, 2^(1/3), 2^(2/3)} 20:26:37 Kasadkad: (though we can't really prove that easily yet) 20:27:17 Kasadkad: The situation we'll be interested in is when [K:F] is finite, in which case we say K/F is a finite extension 20:27:34 Kasadkad: you can do galois theory when K/F isn't finite but it's harder 20:28:38 Kasadkad: now the idea we'll use to show some extensions are algebraic is that: if K/F is finite then it's algebraic 20:29:07 Kasadkad: the proof is short 20:29:19 Kasadkad: suppose d = [K:F] is finite 20:29:35 Kasadkad: we need to pick any c in K and show it's algebraic over F 20:29:41 Kasadkad: so, look at the set {1, c, ..., c^d} 20:29:54 Kasadkad: this has d+1 &gt; [K:F] elements, so it must be linearly dependent over F 20:30:23 Kasadkad: but a linear dependence of {1, c, ..., c^d} over F is exactly the same as a                  polynomial a_0 + a_1 c + ... + a^d c^d = 0, where each a_i is in F 20:30:33 Kasadkad: so c is algebraic over F, and hence K/F is algebraic 20:31:38 Kasadkad: in particular if we're given c algebraic over F, this shows that F(c)/F is                  algebraic 20:32:13 Kasadkad: c is a root of some a_0 + a_1 x + ... + a_n x^n, and then {1, c, ..., c^n} must be an F-basis of F(c) 20:32:20 Kasadkad: so F(c)/F is finite and algebraic 20:32:50 Kasadkad: the converse is false, by the way 20:33:07 Kasadkad: something like Q(sqrt(2), sqrt(3), sqrt(5), ...)/Q is algebraic but isn't                  finite 20:33:15 vixey: ! 20:33:16 Kasadkad: or (algebraic closure of Q)/Q 20:33:17 Kasadkad: yes? 20:33:36 vixey: is the statement, F(c)/F is algebraic  the same as the statement   F(c) is                algebraic  ? 20:33:52 Kasadkad: well 20:34:01 Kasadkad: really you should specify the extension 20:34:09 Kasadkad: it might happen that K/F is algebraic but K/E isn't 20:34:24 vixey: ok 20:34:51 Kasadkad: (e.g. R(pi i)/R is algebraic but R(pi i)/Q isn't) 20:36:16 Kasadkad: not all fields will be given to us as F(c), so we'd like to at least see that F(c_1, ..., c_n)/F is also algebraic when the c_i's are 20:36:36 Kasadkad: but it's not so clear that that's even finite, a priori 20:36:47 Kasadkad: there's an important dimension formula that tells you it is 20:37:15 Kasadkad: if F/E and K/F are extensions, so if K/E, and we have [K:E] = [K:F][F:E] 20:38:09 Kasadkad: I won't prove this, but the idea is that if {a_1, ..., a_m} is a basis of F                  over E and {b_1, ..., b_n} is a basis of K over F, then {a_i b_j} is a basis of K over E; it's not hard to check that 20:38:55 Kasadkad: now we can see why F(c_1, ..., c_n)/F is algebraic 20:40:01 Kasadkad: the degree is                  [F(c_1, ..., c_n) : F] = [F(c_1, ..., c_n) : F(c_1, ..., c_{n-1})] ... [F(c_1,c_2) : F(c_1)][F(c_1) : F] 20:40:28 Kasadkad: each of these is finite because c_i is clearly algebraic over F(c_1, ..., c_{i-1}) if it's algebraic over F 20:40:57 Kasadkad: so these are really all the algebraic extensions we'll need 20:41:22 Kasadkad: in fact, any finite extension K/F has the form F(c_1, ..., c_n)/F with the c_i's algebraic: just take {c_1, ..., c_n} a basis of K over F 20:42:26 Kasadkad: the next thing to do is, given c algebraic, get our hands on a distinguished polynomial that c is the root of 20:43:36 Kasadkad: so remember how we got a polynomial that c is a root of when F(c)/F is                  finite, by looking at {1, c, ..., c^d} 20:43:48 Kasadkad: where d = [F(c):F] 20:44:23 Kasadkad: suppose c is a root of some b_0 + b_1 x + ... + b_m x^m in F[x] 20:44:39 Kasadkad: (btw F[x] is polynomials in x with coefficients in F, if you didn't know) 20:44:51 Kasadkad: and b_m != 0 20:45:19 Kasadkad: then we can write c^m = -b_m^{-1}(b_0 + b_1 c + ... + b_{m-1} c^{m-1}) 20:45:54 Kasadkad: and that says that the set {1, c, ..., c^{m-1}} is an F-basis of F(c), because we can always use c^m = etc. to replace high powers of c with ones in {1, ..., c^{m-1}} 20:46:28 Kasadkad: that's a basis with m elements, so we better have m &gt;= d = [F(c):F] 20:46:57 Kasadkad: that is, our construction earlier shows that there's a polynomial of degree d with c as a root, and that's the smallest degree possible for such a                  polynomial 20:47:33 Kasadkad: so let's say m(x) = a_0 + a_1 x + ... + a_d x^d has c as a root 20:47:43 Kasadkad: then m(x) is irreducible 20:48:33 Kasadkad: if not, if m(x) factors as f(x)g(x) where deg(f(x)) and deg(g(x)) are both &lt; d, then we'd have to have f(c) = 0 or g(c) = 0 20:48:35 Kasadkad: but that can't happen 20:49:06 Kasadkad: here's something even better 20:49:19 Kasadkad: if f(x) in F[x] has f(c) = 0, then m(x) divides f(x) 20:50:03 Kasadkad: this just comes from dividing f(x) by m(x), i.e. f(x) = q(x)m(x) + r(x), where q is the quotient and r the remainder, so deg(r) &lt; deg(m) = d                   or r = 0 20:50:21 Kasadkad: but 0 = f(c) = q(c)m(c) + r(c) and m(c) = 0 says r(c) = 0 20:50:39 Kasadkad: and we know that can't happen if deg(r) &lt; deg(m), so the only possibility is r = 0 20:50:44 Kasadkad: i.e. m(x) divides f(x) 20:50:54 Kasadkad: notice that this says m(x) is essentially unique! 20:51:17 Kasadkad: if g(x) is another polynomial of degree d = [F(c):F] with c as a root, then m(x) divides g(x) 20:51:27 Kasadkad: but they have the same degree so m(x) is a constant multiple of g(x) 20:51:47 Kasadkad: in particular, we can choose the leading term of m(x) to be monic, and that fixes it 20:52:07 Kasadkad: thus: given c in K algebraic over F, there's a unique monic irreducible polynomial in F[x] with c as a root 20:52:13 Kasadkad: called the minimal polynomial of c over F 20:52:51 Kasadkad: some examples 20:53:02 Kasadkad: the minimal polynomial of sqrt(2) over Q is just x^2 - 2 20:53:14 Kasadkad: it's irreducible because neither of its roots are rational 20:53:19 Kasadkad: and so it must be the minimal polynomial 20:54:13 Kasadkad: if you like you can work out that x^4 - 10x^2 + 1 is the minimal polynomial of sqrt(2) + sqrt(3) over Q 20:56:08 Kasadkad: given c algebraic over F and m(x) its minimal polynomial, the other roots of m are special 20:56:23 Kasadkad: if c' is also a root of m(x) we say that c' and c are conjugates (over F) 20:57:37 Kasadkad: at this point I can tell you about field automorphisms, but maybe I should stop before that 21:00:10 Kasadkad: the goal is sort of, look at the automorphisms of K/F (the &quot;symmetries&quot;); it turns out that automorphisms send c to a conjugate of c, and if we know something about the automorphisms, we can use that to figure something about c, like that it lies in a particular subfield of K 21:01:05 Kasadkad: the point being that we want to know something about the structure of K/F, and the subfields of K containing F are a good place to start 21:01:28 Kasadkad: so I'll stop there 21:02:19 Kasadkad: the next few seminars I'll go into less detail about proving stuff so that we can go faster 21:03:09 ~brett1479: Just my 2 cents ... I don't think you should change your speed at all. I                    think that was pretty perfect. Well done! 21:03:25 vixey: yeah, this is great, thanks Kasadkad 21:03:31 * shminux concurs 21:03:40 * Libster dissents j/k that was great

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