This seminar by Kasadkad took place on 9th November 2008 20:00 UTC, in #mathematics channel of

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Seminar Edit

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 "a field F" that
                   I'm saying "the rational numbers Q" (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 "the extension K/F"
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 "subfield?"
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 "algebraic numbers" and "transcendental
                   numbers" 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
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 "smallest subfield of K containing c": 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 "smallest subfield of K containing F and c"
20:16:17 Kasadkad: or "subfield generated by F and c"
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 > [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
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
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 >= 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
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 < 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) < 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) < 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 "symmetries");
                   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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