@PeterTamaroff Automorphisms are bijections and another name for bijection is permutation. Anon's hint was that these are determined by considering permutations on just three elements. See where this is going?
A surjective homomorphism maps generators to generators, yes, so to find Aut(S3) it suffices to check which permutations of the transpositions extend to an automorphism of all of S3 (which is all of them here).
it is possible for a map that sends generators to generators to not extend to a homomorphism. for instance define $f:\{1,a,b\}\to\{1,a,b\}$ by $f(1)=a$, $f(a)=b$ and $f(b)=1$, where $1,a,b$ are generators of ${\bf Z}\oplus F\{a,b\}$ (the latter being the free group generated by $a$ and $b$)
I think we need our group to be abelian for my argument to work, actually, since there will only be one way to write each element in terms of generators in that case. In the nonabelian case there will potentially be multiple strings representing the same element.
It suffices to check that every permutation of the three transpositions (12),(23),(31) is induced by an inner automorphism. This is easy if you know $\sigma(s_1\cdots s_n)\sigma^{-1}=(\sigma(s_1)\cdots\sigma(s_n))$ intuitively (in cycle notation)
@Khromonkey no - I mentioned the Prufer decomposition of the group Q/Z being the number field analogue of partial fraction decomposition in function fields. (this is very much a number-theoretic thing.)
@PeterTamaroff Check that every permutation on the set {(12),(23),(32)} is a restriction of an automorphism on S3 given by $x\mapsto \sigma x\sigma^{-1}$ for some $\sigma\in S_3$ (such automorphisms are called "inner")
@Khromonkey I study number theory, groups and representations, diagrammatic algebras, and may look more into categories and combinatorics (ie species).
@PeterTamaroff It will tell you that each candidate for an automorphism (which we mean to be a permutation of the three transpositions) lifts to an actual automorphism, and we will know it does because we concretely describe the actual automorphisms (specifically as inner autos for known $\sigma$s).
@Khromonkey I mentioned "species" in combinatorics, a sort of categorical language for generatingfunctionology. You then said you were not interested in category theory applied to combinatorics, but the other way around.
The underlying set will be the direct product of G and Aut(G), but the multiplication will be different. You will have $(g,a)(h,b)=(ga(h),ab)$, where by $a(h)$ we mean the automorphism $a\in{\rm Aut}(G)$ applied to $h\in G$.
upvote those you thought were helpful to you or to other readers, or which you think deserve reputation or to be highlighted for others. accept whichever one helped you best, assuming you have understood it sufficiently first.
because this sends the message "I have to accept one of the answers to keep my acceptance rate up, but I do not actually appreciate any of the answers, I am just keeping up appearances"
but seriously, the standards for accepting an answer should be set higher than the standards for upvoting the answer, so if you do the former but not the latter your priorities are confusing to say the least
Why? couldn't I have $L(s,\chi)=o(\frac{1}{\zeta(s)})$, so that it still might vanish, but when multiplied by zeta(s), as s->1, the previous limit still works?
the definition of $L(s,\chi)=o(\zeta(s)^{-1})$ is $\frac{L(s,\chi)}{\zeta(s)^{-1}}\to0$, which is the same as $L(s,\chi)\zeta(s)\to0$, which is the same as $\frac{1}{L(s,\chi)\zeta(s)}\to\infty$
@Ethan, you can find a proof of nonvanishing at many places. For example Stein and Shakarchi Book 1, last chapter; Davenport, Multiplicative Number Theory; or some of Milne's notes.
You would probably appreciate the proof in the first two sources more than the last one, which is what anon's talking about.
@JacobSchlather With regards to your question, I haven't really studied separable extensions. My experience with field theory has been mainly with algebraic number fields.
Although it also has the strange note that although K(x^2) is strictly contained in K(x), x^2 is still transcendental over K so that K(x^2) \cong K(x).
@JacobSchlather right. Have you studied things like algebraic number theory before? It helps to know some when dealing with finite extensions of $\Bbb{Q}$.
@Sanchez Let $K(\alpha)/K$ be a separable extension let $L$ be the normal closure of $K(\alpha)$ now apply my argument. Maybe I should make that explicit.
@BenjaLim Yeah. I read that. I tried to get some info from the ANU site because some programs do take students in september, I could not find anything definitive about which programs though. If you get time, can you just verify? No rush though.
@Sanchez Yeah, I think the first time I proved it I did it for $L=K(\alpha)/K$ a Galois extension, then when I worked on the separable case I realized the previous proof generalized very nicely. Then it all fell apart for the inseparable case. I have some thoughts on writing an inseparable extension as K(\alpha)/K as K(\alpha)/L/K where L is the separable closure of K in K(alpha). But I'm trying to get this annoying paper on periodic generalized binomial coefficients written currently.
I don't thinks so, although I've never been entirely sure on inseparable extensions. This all seems right to me, it'll just take me a minute or two to digest
suffices to do the case where n = p, and for that, note that $\alpha = f(\beta)$, where $f$ is a rational function. Take $p$-th power on both sides give $\alpha^p = f(\beta)^p = f(\beta^p)$
This shows that $K(\alpha^p) \subset K(\beta^p)$. By symmetry they are equal.
