This seminar by Kasadkad took place on 23rd November 2008 20:00 UTC, in #mathematics channel of irc.freenode.net.

The timestamps displayed in the IRC text are UTC.

Any explanation or errors regarding this seminar should be recorded in the discussion page.

## Seminar Edit

Nov 23 13:31:33 <Kasadkad> anyway ok Nov 23 13:31:33 <Kasadkad> might as well start Nov 23 13:31:41 <Kasadkad> today I'm going to give a traditional application of Galois theory, showing that there's no quintic equation: that is there are polynomials whose roots can't be written in terms of radicals Nov 23 13:32:09 * ChanServ has changed the topic to: Seminar in progress. If you have a question, say '!' and wait to be called on. Nov 23 13:33:26 <Kasadkad> first thing to do is define what "written in terms of radicals" means Nov 23 13:34:26 <Kasadkad> a field extension K/F is called a radical extension if you can write F = F_0 <= F_1 <= ... <= F_k = K, where F_{i+1} = F_i(a_i^{1/n_i}) for each Nov 23 13:34:36 <Kasadkad> with a_i in F_i and n_i some positive integers Nov 23 13:34:58 <Kasadkad> in other words, a tower of extensions where at each step you throw in an nth root of something from the previous step Nov 23 13:35:37 <Kasadkad> (you can't just throw in a bunch of roots at once since you want stuff like sqrt(3 + 7^(1/3)) Nov 23 13:36:34 <Kasadkad> incidentally, when I say "a_i^{1/n_i}", that just means "any (n_i)th root of a_i", there's no silly fusses about branch cuts or anything Nov 23 13:36:57 <Kasadkad> a polynomial f(x) in F[x] is called solvable by radicals if its roots lie in a radical extension Nov 23 13:37:01 <Kasadkad> (over F) Nov 23 13:37:56 <Kasadkad> a better way to phrase that is this Nov 23 13:38:20 <Kasadkad> given f(x) in F[x], you can form the splitting field K of F, which is just F(r_1, ..., r_n) where r_1, ..., r_n are all the roots of F Nov 23 13:38:26 <Kasadkad> *all the roots of f Nov 23 13:38:46 <Kasadkad> then f(x) is solvable by radicals if its splitting field is contained in a radical extension over F Nov 23 13:40:19 <Kasadkad> we're going to use Galois theory to translate "f(x) is solvable by radicals" into a statement about the Galois group of the splitting field of f Nov 23 13:41:18 <Kasadkad> if we're going to have any hope of figuring out the Galois group of a radical extension, we should start by figuring out Gal(F(a^{1/n})/F) Nov 23 13:41:49 <Kasadkad> unfortunately, F(a^{1/n})/F doesn't even have to be Galois!@ Nov 23 13:42:24 <Kasadkad> e.g. Q(2^(1/3))/Q is not Galois, because it doesn't contain all the conjugates of 2^(1/3), namely 2^(1/3) (-1 +- sqrt(-3))/2 Nov 23 13:43:16 <Kasadkad> the fix is that if z is a primitive nth root of unity, then F(a^{1/n}, z)/F is Galois Nov 23 13:43:46 <Kasadkad> more or less: you also need the characteristic of F not to divide n, but no problem if F = Q Nov 23 13:45:00 <Kasadkad> that extension is Galois because the conjugates of z are z^i, which are in it, and the conjugates of a^{1/n} are z^i a^{1/n}, which are in it Nov 23 13:45:17 <Kasadkad> whoops Nov 23 13:45:28 <Kasadkad> it's true that that extension isn't Galois but it isn't what I should be talking about Nov 23 13:45:52 <Kasadkad> we saw last time that Q(2^(1/3), z)/Q has Galois group S_3 Nov 23 13:46:08 <Kasadkad> which is terrible, abelian groups are better Nov 23 13:46:56 <Kasadkad> you can get that by adjoining the nth roots of unity to the base field instead Nov 23 13:47:13 <Kasadkad> i.e. consider F(a^{1/n}, z)/F(z) Nov 23 13:47:18 <Kasadkad> that's Galois by what I said before Nov 23 13:47:25 <Kasadkad> let's see what its Galois group Nov 23 13:47:50 <Kasadkad> a member of the Galois group is entirely determined by what it does to a^{1/n} Nov 23 13:48:21 <Kasadkad> it must send it to some conjugate z^i a^{1/n} Nov 23 13:49:30 <Kasadkad> and if s(a^{1/n}) = z^i a^{1/n} and t(a^{1/n}) = z^j a^{1/n}, then st(a^{1/n}) = z^{i+j} a^{1/n} Nov 23 13:50:05 <Kasadkad> in other words (since z^n = 1), multiplication in the Galois group looks exactly like addition mod n Nov 23 13:50:18 <Kasadkad> so the Galois group is cyclic Nov 23 13:50:29 <Kasadkad> (maybe not of order n, but of some order dividing n) Nov 23 13:50:52 <Kasadkad> so let's use that to look at the Galois group of a radical extension Nov 23 13:51:05 <Kasadkad> we do need to assume that our base field has all the necessary roots of unity Nov 23 13:52:21 <Kasadkad> but that's not really a problem, because roots of unity are already radicals 1^{1/n}, so throwing them in to the base field is no problem for talking about solvability by radicals Nov 23 13:53:03 <Kasadkad> so let's say we have a radical extension F = F_0 <= F_1 <= ... <= F_k = K, where F_{i+1} = F_i(a_i^{1/n_i}) Nov 23 13:53:13 <Kasadkad> and F contains all the (n_i)th roots of unity for each i Nov 23 13:53:40 <Kasadkad> and let's say K/F is Galois Nov 23 13:54:06 <Kasadkad> the Galois correspondence says that there is a corresponding tower of subgroups in Gal(K/F) Nov 23 13:54:32 <Kasadkad> G_0 = Gal(K/F) >= G_1 >= ... >= G_k = {1} Nov 23 13:54:52 <Kasadkad> where G_i is the fixing subgroup of F_i Nov 23 13:55:53 <Kasadkad> here's something I mentioned last time: if you have F <= E <= E' <= K, with K/F Galois, and G_E' and G_E the fixing subgroups of E' and E resp., then E'/E is Galois exactly when G_E' is normal in G_E Nov 23 13:56:08 <Kasadkad> and, no surprise, if it is Galois then Gal(E'/E) =~ G_E/G_E' Nov 23 13:56:39 <Kasadkad> so here, we know that each F_{i+1}/F_i is Galois, with cyclic Galois group Nov 23 13:57:23 <Kasadkad> translating that with the Galois correspondence, G_{i+1} is normal in G_i for each i, and G_i/G_{i+1} is cyclic Nov 23 13:58:07 <Kasadkad> There's a special name for such groups: solvable Nov 23 13:58:28 <Kasadkad> inspired by this correspondence Nov 23 13:59:07 <Kasadkad> i.e. G is solvable if it has a tower of subgroups {1} = H_k <= ... <= H_1 <= H_0 = G, such that H_{i+1} is normal in H_i and H_i/H_{i+1} is cyclic Nov 23 13:59:17 <Kasadkad> for example, S_3 is a solvable group Nov 23 13:59:29 <Kasadkad> we have the series {1} <= A_3 <= S_3 Nov 23 13:59:45 <Kasadkad> A_3/{1} = A_3 =~ Z/3Z is cyclic Nov 23 13:59:57 <Kasadkad> and so is S_3/A_3 =~ Z/2Z Nov 23 14:00:07 <Kasadkad> on the other hand, A_5 is not solvable Nov 23 14:00:29 <Kasadkad> because it's simple: it has no nontrivial normal subgroups at all, so it certainly can't have this solvable series Nov 23 14:01:19 <Kasadkad> that suggests what to do: come up with a polynomial with Galois group say, A_5; if it was solvable by radicals, the Galois group would be solvable, but it isn't Nov 23 14:01:35 <Kasadkad> there are some technical details, though Nov 23 14:02:13 <Kasadkad> namely, if f(x) is solvable by radicals, we know that the splitting field K of f(x) is contained in some radical extension K'/F Nov 23 14:02:40 <Kasadkad> But who knows if K'/F is Galois Nov 23 14:03:16 <Kasadkad> also we're missing roots of unity, and it's not a big deal to add them in, but it has to be done in such a way that you still have a radical extension Nov 23 14:03:29 <Kasadkad> I won't go over the details of this Nov 23 14:04:04 <Kasadkad> Given a radical extension F = F_0 <= ... <= F_k = K, you can put F_i' = F_i(necessary roots of unity) Nov 23 14:05:10 <Kasadkad> Then F_{i+1}'/F_i' is now Galois with cyclic Galois group ("is a cyclic extension"), and each step is still just adjoining an nth root Nov 23 14:05:50 <Kasadkad> the only issue is that you got F <= F_0' by adding a bunch of roots of unity at once, so that's not really a radical extension, but you can just adjoin them one at a time so that it is Nov 23 14:06:20 <Kasadkad> Then you have F <= F_0' <= ... <= F_k' = K' where each subextension is cyclic Nov 23 14:06:26 <Kasadkad> Unfortunately K'/F doesn't have to be Galois itself Nov 23 14:07:06 <Kasadkad> which can be fixed straightforwardly by adjoining all the conjugates of all the a_i^{1/n_i} one at a time Nov 23 14:07:30 <Kasadkad> but it is kind of a mess to say exactly what that means, so I won't Nov 23 14:07:44 <Kasadkad> anyway, point is that any radical extension K/F is contained in a Galois radical extension K'/F Nov 23 14:09:23 <Kasadkad> now the situation is, f(x) has splitting field E/F which is contained in a Galois radical extension K'/F Nov 23 14:09:34 <Kasadkad> we know Gal(K'/F) is solvable, and want to see that Gal(E/F) is solvable Nov 23 14:10:12 <Kasadkad> well, by what I said earlier, Gal(E/F) =~ Gal(K'/F)/Gal(K'/E) Nov 23 14:10:23 <Kasadkad> so the necessary fact is that a quotient of a solvable group is solvable Nov 23 14:11:08 <Kasadkad> which is true Nov 23 14:11:17 <Kasadkad> you can check if you like Nov 23 14:11:29 <Kasadkad> now we're in the situation we want to be in Nov 23 14:11:46 <Kasadkad> all that's left is to illustrate a polynomial with non-solvable Galois group, and that polynomial will not be solvable by radicals Nov 23 14:12:20 <Kasadkad> so here is a completely random polynomial: x^5 - 3x^2 + 1 in Q[x] Nov 23 14:12:43 <Kasadkad> we have no idea what the roots are, so it's not clear how to figure out what the Galois group is Nov 23 14:13:09 <Kasadkad> but we can anyway Nov 23 14:13:37 <Kasadkad> check that it's irreducible (reducing mod 2 and checking there will work), so that the Galois group has order divisible by 5 Nov 23 14:14:10 <Kasadkad> (because the splitting field contains a root of the polynomial, and a root of an irreducible polynomial of degree n has degree n) Nov 23 14:14:36 <Kasadkad> so the Galois group contains an element of order 5 Nov 23 14:15:02 <Kasadkad> if we look at the Galois group as a permutation group, permuting the 5 roots of x^5 - 3x^2 + 1 Nov 23 14:15:12 <Kasadkad> that element of order 5 must be a 5-cycle Nov 23 14:15:24 <Kasadkad> just because the elements of order 5 in S_5 are exactly the 5-cycles Nov 23 14:15:26 <maxote> what time is now? :s Nov 23 14:15:39 <Kasadkad> 21:15 UT Nov 23 14:15:39 <Kasadkad> C Nov 23 14:15:49 <Kasadkad> so the Galois group has a 5-cycle in it Nov 23 14:15:52 <maxote> ufff Nov 23 14:16:03 <Kasadkad> Let's also see that it has a transposition in it Nov 23 14:16:20 <Kasadkad> You can use the intermediate value theorem to see that f(x) = x^5 - 3x^2 + 1 has exactly 3 real roots Nov 23 14:16:28 <Kasadkad> so 2 complex roots, which are conjugates Nov 23 14:16:48 <Kasadkad> so complex conjugation, restricted to the splitting field of f, is a member of the Galois group of f Nov 23 14:17:01 <Kasadkad> and it interchanges the two complex roots, while fixing the three real roots Nov 23 14:17:04 <Kasadkad> that is, it's a transposition Nov 23 14:17:13 <Kasadkad> now the Galois group contains a 5-cycle and a transposition Nov 23 14:17:24 <Kasadkad> and you can check that a subgroup of S_5 containing a 5-cycle and a transposition must be all of S_5 Nov 23 14:17:29 <Kasadkad> so the Galois group is S_5 Nov 23 14:17:35 <Kasadkad> do we know that's not solvable? Nov 23 14:17:43 <Kasadkad> we know A_5 isn't solvable because it's simple Nov 23 14:18:12 <Kasadkad> If S_5 was solvable, it would have a solvable series like {1} <= H_k <= ... <= H_1 <= H_0 = S_5 Nov 23 14:18:15 <Kasadkad> where each is normal in the last Nov 23 14:18:44 <Kasadkad> but the only proper nontrivial subgroup of S_5 is A_5, which you can check by intersecting such a subgroup with A_5 to get a normal subgroup of A_5, and going from there Nov 23 14:18:57 <Kasadkad> so a solvable series would have to look like {1} <= ... <= A_5 <= S_5 Nov 23 14:19:04 <Kasadkad> but since A_5 is simple there's no way to fill in the rest Nov 23 14:19:07 <Kasadkad> so, S_5 is not solvable Nov 23 14:19:16 <Kasadkad> and so x^5 - 3x^2 + 1 is not solvable in terms of radicals Nov 23 14:19:52 <Kasadkad> exciting Nov 23 14:20:00 <|Steve|> ! Nov 23 14:20:04 <Kasadkad> yes Nov 23 14:20:17 <|Steve|> Do you mean the only proper, nontrivial, normal subgroup of S_5 is A_5? Nov 23 14:20:55 <Kasadkad> oh, heh Nov 23 14:20:56 <Kasadkad> Yes Nov 23 14:21:41 <Kasadkad> There's a converse to this, btw, that if f(x) has solvable Galois group then it's solvable by radicals Nov 23 14:22:13 <Kasadkad> the key point is proving that if K/F is cyclic of degree n, and F contains the nth roots of unity (and char F doesn't divide n), then K = F(a^{1/n}) for some a in F Nov 23 14:22:30 <Kasadkad> which is actually constructive, and so you can use it to solve polynomials Nov 23 14:22:40 <Kasadkad> e.g., to come up with the cubic equation from a solvable series of S_3 Nov 23 14:23:02 <Kasadkad> (I'm not going to do so) Nov 23 14:23:11 <Pthing> why Nov 23 14:23:23 <Kasadkad> it would take time Nov 23 14:23:24 <|Steve|> Computing Galois groups is hard. Nov 23 14:23:30 <Kasadkad> Yeah Nov 23 14:23:46 <Kasadkad> I was going to say a word about one technique you can use to do so Nov 23 14:23:57 <Kasadkad> that people who have seen Galois theory might not have seen Nov 23 14:25:41 <Kasadkad> let G be the Galois group of f(x) in Q[x], and G' the Galois group of f(x) mod p over F_p Nov 23 14:26:15 <Kasadkad> G' is cyclic, as any Galois group of a finite extension of a finite field is Nov 23 14:26:17 <Kasadkad> if p doesn't divide the discriminant of f(x), then G' injects into G Nov 23 14:26:42 <Kasadkad> which is okay, but the nicer thing is that the injection preserves cycle structure Nov 23 14:27:09 <Kasadkad> e.g. Nov 23 14:27:51 <Kasadkad> well you might think that just shifts the problem to compute Galois groups over F_p which isn't any better, but it is Nov 23 14:28:00 <Kasadkad> take x^5 + 4x^4 + x + 3 Nov 23 14:28:20 <Kasadkad> the discriminant is 5*127*11087, so forget about those primes Nov 23 14:28:24 <Kasadkad> mod 7, that's irreducible Nov 23 14:29:39 <Kasadkad> the important fact is that the Galois group of a polynomial is transitive on its roots Nov 23 14:30:11 <Kasadkad> so if the Galois group of x^5 + 4x^4 + x + 3 mod 7 is generated by some g, then g had better be a 7-cycle Nov 23 14:30:33 <Kasadkad> otherwise no power of g will send a root in one cycle to a root in another cycle Nov 23 14:30:38 <Kasadkad> err, a 5-cycle Nov 23 14:30:40 <Kasadkad> 5 roots Nov 23 14:30:57 <Kasadkad> so that tells us right away that the Galois group over Q also contains a 5-cycle Nov 23 14:31:48 <Kasadkad> how about we factor mod 2 Nov 23 14:31:56 <Kasadkad> that's (x^3 + x^2 + 1)(x^2 + x + 1) Nov 23 14:32:15 <Kasadkad> by the same argument, the Galois group over F_2 is cyclic generated by some (2,3)-cycle Nov 23 14:32:30 <Kasadkad> uhh, sorry Nov 23 14:32:38 <Kasadkad> <Kasadkad> the important fact is that the Galois group of a polynomial is transitive on its roots <-- an *irreducible* polynomial Nov 23 14:32:58 <|Steve|> What's a (2,3)-cycle? Nov 23 14:33:06 <Kasadkad> a member of the Galois group can only send something to another roots of its minimal polynomial Nov 23 14:33:26 <Kasadkad> cycle type (2,3), (a b)(c d e) Nov 23 14:33:55 <Kasadkad> the Galois group sends roots of x^2 + x + 1 to roots of x^2 + x + 1, and roots of x^3 + x^2 + 1 to roots of x^3 + x^2 + 1 Nov 23 14:34:11 <Kasadkad> and it's transitive on the roots of each one Nov 23 14:34:15 <Kasadkad> and is cyclic Nov 23 14:34:21 <Kasadkad> so it must be generated by a (2,3)-cycle Nov 23 14:34:40 <Kasadkad> and so the Galois group over Q also contains a (2,3)-cycle Nov 23 14:34:52 <Kasadkad> reducing mod 23 will give you a 4-cycle Nov 23 14:35:10 <Kasadkad> so we have a subgroup of S_5 containing a 5-cycle, a 4-cycle, and a (2,3)-cycle Nov 23 14:35:39 <Kasadkad> it better have order >= 60 by Lagrange, but also it contains an odd permutation, the 4-cycle Nov 23 14:35:42 <Kasadkad> so it's all of S_5 Nov 23 14:36:31 <Kasadkad> it works best when the Galois group is largish Nov 23 14:36:51 <Kasadkad> since it can force it to be all of S_n or A_n or something Nov 23 14:36:58 <|Steve|> That is pretty neat, but it does require computing potentially many Galois groups as well as the discriminant of the polynomial which is nontrivial (resultant of p and p' or some such). Nov 23 14:37:14 <Kasadkad> Mmwell Nov 23 14:37:34 <Kasadkad> All you have to do is factor mod p, though Nov 23 14:37:48 <Kasadkad> Which is not so hard Nov 23 14:38:00 <Kasadkad> Also you don't even have to compute the discriminant Nov 23 14:38:15 <Kasadkad> if p divides the discriminant, then there will be multiple roots mod p which you'll see in the factorization Nov 23 14:39:24 <maxote> ! Nov 23 14:39:27 <Kasadkad> yes Nov 23 14:39:35 <maxote> what's the max. value of p? Nov 23 14:39:40 <Kasadkad> there is none Nov 23 14:39:49 <Kasadkad> this works for any prime p which doesn't divide the discriminant of f Nov 23 14:39:58 <Kasadkad> only finitely many primes can divide that discriminant, so Nov 23 14:40:32 <Kasadkad> One issue is if the Galois group is small, like Z/5Z or something Nov 23 14:41:09 <Kasadkad> so you keep getting 5-cycles when you reduce mod p, but maybe there's a 4-cycle out when p = 104729 Nov 23 14:41:31 <Kasadkad> but even then all isn't lost Nov 23 14:41:43 <maxote> of course, a good p that is practical Nov 23 14:42:05 <Kasadkad> remarkably, the proportion of primes p for which you get a particular cycle type C when reducing mod p is exactly the same as the proportion of elements of the Galois group over Q with cycle type C Nov 23 14:43:04 <Kasadkad> so at least you can give yourself high confidence that the Galois group is what you think Nov 23 14:44:35 <Kasadkad> this whole business is difficult to prove, by the way Nov 23 14:45:10 <|Steve|> How are you defining a proportion of an infinite set? Nov 23 14:45:15 <Kasadkad> vaguely, thank you Nov 23 14:45:55 <Kasadkad> Apparently it's http://en.wikipedia.org/wiki/Natural_density Nov 23 14:46:08 <Kasadkad> The theorem is http://en.wikipedia.org/wiki/Chebotarev's_density_theorem , and I have no idea how it's proved Nov 23 14:47:08 <Kasadkad> Just to get the right homomorphism from the Galois group mod p into the Galois group over Q takes some doing, you have to know a little about factorization of prime ideals in Dedekind domains Nov 23 14:47:32 <Kasadkad> and then showing it's injective if p doesn't divide the discriminant takes the Dedekind discriminant theorem Nov 23 14:47:38 <Kasadkad> which is more algebraic number theory Nov 23 14:50:02 <Kasadkad> p-adics, etc Nov 23 14:50:28 <|Steve|> (Beyond me. Maybe you should give a seminar on p-adic numbers.) Nov 23 14:50:29 <maxote> ! Nov 23 14:51:55 <Kasadkad> |Steve|: I read enough algebraic number theory to see the proof of the theorem for almost all primes not dividing the discriminant, but I never actually got around to reading the real version with p-adics Nov 23 14:51:56 <Kasadkad> so Nov 23 14:51:58 <Kasadkad> maxote, yeah Nov 23 14:52:05 <maxote> the proportions, are they equals either mathematically or statistically? Nov 23 14:52:16 <Kasadkad> they're equal Nov 23 14:52:17 <Kasadkad> as in Nov 23 14:52:18 <Kasadkad> numbers Nov 23 14:52:34 <Kasadkad> http://en.wikipedia.org/wiki/Natural_density is the right definition of "proportion of primes p such ..." Nov 23 14:52:44 <Kasadkad> just a number between 0 and 1 Nov 23 14:52:56 <Kasadkad> and the Galois group is a finite group, so the proportion of elements such that blah blah makes sense Nov 23 14:55:24 <maxote> i'm confused if they are exactly equals or probably equals Nov 23 14:56:17 <Kasadkad> they are equal Nov 23 14:56:40 <maxote> ok Nov 23 14:59:58 <maxote> ! Nov 23 15:00:34 <maxote> i think it's your time Nov 23 15:03:07 <Kasadkad> ? Nov 23 15:03:09 <Kasadkad> my time? Nov 23 15:03:14 <maxote> 22:00 UTC Nov 23 15:03:28 <Kasadkad> what time is that Nov 23 15:03:33 <Kasadkad> I'm done, certainly Nov 23 15:03:35 <maxote> err, sorry, it was 20:00 Nov 23 15:04:28 <maxote> no problem Nov 23 15:05:22 <maxote> ! Nov 23 15:06:11 <maxote> does it require much computation to solve smaller problems? (e.g. 3rd-degree and 4th-degree polynomials) Nov 23 15:06:44 <Kasadkad> you don't have to do ! anymore btw Nov 23 15:06:48 <Kasadkad> the topic is lying Nov 23 15:06:58 <Kasadkad> 3rd degree polynomials are easy Nov 23 15:07:00 <maxote> thanks Nov 23 15:07:20 <maxote> in seconds? or in minutes? or in hourS? Nov 23 15:07:22 <Kasadkad> the Galois group of an irreducible cubic has to have order a multiple of 3 and be a subgroup of S_3, so it's A_3 or S_3 Nov 23 15:08:26 <Kasadkad> you can tell those two apart by looking at the discriminant of the polynomial Nov 23 15:08:35 <Kasadkad> if it's a square in the base field, the Galois group is A_3 Nov 23 15:08:37 <Kasadkad> if not, it's S_3 Nov 23 15:08:47 <Kasadkad> maybe you need the base field to have characteristic not 2, I don't remember Nov 23 15:08:54 <Kasadkad> over Q it's true Nov 23 15:10:09 <maxote> and is difficult to obtain S_4? Nov 23 15:11:20 <Kasadkad> probably not Nov 23 15:12:11 <maxote> so, for obtaining S_5, is it impossible or improbable? Nov 23 15:12:35 <Kasadkad> I gave two examples of polynomials with Galois group S_5 :P Nov 23 15:13:11 <maxote> but i was saying S_? that are solvables Nov 23 15:13:32 <Kasadkad> Ah Nov 23 15:13:37 <Kasadkad> S_n is not solvable for all n >= 5 Nov 23 15:13:45 <Kasadkad> because A_n is simple for all n >= 5 Nov 23 15:14:04 <Kasadkad> so yes, there are always polynomials of degree n that aren't solvable in radicals, if n >= 5 Nov 23 15:14:11 <maxote> i don't like this condition Nov 23 15:14:20 <Kasadkad> why not Nov 23 15:15:19 <maxote> remember the question that did Cardano to Tartaglia, more or less that are sides of perfect squares that are negatives, and Tartaglia didn't solve any of their problems. Nov 23 15:15:50 <maxote> /that/that there are Nov 23 15:16:08 <Kasadkad> ?? Nov 23 15:16:14 <Kasadkad> I don't know what you mean Nov 23 15:18:37 <maxote> well, it was the comment in the middleage Nov 23 15:18:57 <Kasadkad> how does it to relate to anything I said? Nov 23 15:19:10 <maxote> i've a question more Nov 23 15:19:26 <maxote> i think "S_n is not solvable for all n >= 5" <--- for what formulated problem is it? Nov 23 15:20:26 <Kasadkad> uhh Nov 23 15:20:34 <Kasadkad> oh, well Nov 23 15:20:37 <Kasadkad> "solvable" is a technical term Nov 23 15:20:40 <Kasadkad> which I defined earlier Nov 23 15:20:52 <maxote> or didn't find solution because there is not time to solve every possibilities Nov 23 15:22:35 <maxote> Kasadkad, a group is solvable if exists an Abelian group Nov 23 15:23:02 <Kasadkad> if it has a series of normal subgroups where each consecutive quotient is cyclic Nov 23 15:23:06 <Kasadkad> (or equivalently, abelian) Nov 23 15:23:40 <Kasadkad> brb Nov 23 15:24:46 <maxote> my question is, if no such Abelian group then "what means it?" why Abel declared it "unsolvable"? Nov 23 15:28:09 <maxote> i need to know the sufficient condition of the existence of solvable groups when a nth-degree polynomial has almost n roots Nov 23 15:34:04 <TheBlueWizard> ! Nov 23 15:34:46 <Kasadkad> the seminar isn't in progress, you don't have to do ! Nov 23 15:35:13 <TheBlueWizard> oh ok...so what's the next seminar? just curious Nov 23 15:35:36 <Kasadkad> No idea Nov 23 15:35:58 <TheBlueWizard> ok Nov 23 15:43:42 <smax> hi Nov 23 15:44:07 <smax> anyone here familiar with big-oh notataion? Nov 23 15:44:56 <Kaznak> smax, #math Nov 23 15:44:57 <smax> when choosing constant c, is it always safe, for ALL functions (even with a term of 2^n), to assign c = the sum of the absolute value of all terms coeficients? Nov 23 15:45:00 <smax> ok Nov 23 15:54:58 <maxote> i've another dude Nov 23 15:56:00 <maxote> how can i know that a radical (of finite operators) is expressed in the field of the Galois groups? Nov 23 15:59:30 <Kasadkad> uhh Nov 23 15:59:34 <Kasadkad> I don't know what that means Nov 23 16:01:45 <maxote> i don't see why the theorem says the no such radical (expressed with a finite number of operators) exists in all S_5 Nov 23 16:03:47 <maxote> my question is how much value is finite? Nov 23 16:04:06 <|Steve|> ? Nov 23 16:06:03 <maxote> i've not seen any proof for radicals of 20 operators e.g. Nov 23 16:06:24 <Kasadkad> "operators"? Nov 23 16:06:45 <maxote> +,*, -, /, sqrt, cubic root, and so on Nov 23 16:07:27 <Kasadkad> radicals of operators? Nov 23 16:07:37 <Kasadkad> why would there be 20? Nov 23 16:07:44 <Kasadkad> +, *, taking nth roots Nov 23 16:07:47 <Kasadkad> I don't know what else you have in mind Nov 23 16:08:30 <maxote> the solutions of the 3rd-degree poly. are radicals that have many operators Nov 23 16:09:12 <Kasadkad> the relevant notion here is "radical extension" Nov 23 16:10:44 <maxote> Kasadkad. 20 is a example of finite cardinality Nov 23 16:14:25 <maxote> i've the hopeness that quintic, sixtic, ... equations have radical roots Nov 23 16:14:49 <Kasadkad> they don't, in general Nov 23 16:14:57 <Kasadkad> i exhibit a quintic which doesn't have solvable Galois group Nov 23 16:15:05 <Kasadkad> so doesn't have roots expressable in terms of radicals Nov 23 16:15:11 <Kasadkad> exhibited Nov 23 16:15:27 <maxote> in what field? Nov 23 16:15:30 <Kasadkad> over Q Nov 23 16:15:45 <maxote> and why over Complex? Nov 23 16:15:46 <Kasadkad> x^5 - 3x^2 + 1 Nov 23 16:15:52 <maxote> and why NOT over Complex? Nov 23 16:16:11 <Kasadkad> it does have roots in the complex numbers Nov 23 16:16:44 <maxote> i will find the complex radicals for this x^5 - 3x^2 + 1 Nov 23 16:16:50 <Kasadkad> ok Nov 23 16:17:06 <Kasadkad> I'd be happy to see it Nov 23 16:17:30 <maxote> when i find the complex radicals then what? Nov 23 16:17:41 <Kasadkad> show me, I'd like to see Nov 23 16:18:13 <maxote> of course, i'll need time for the equations rewritings Nov 23 16:18:52 <Kasadkad> go ahead