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It seems pretty clear that this statement should be true, but I can't seem to figure out how to prove it.

What I aim to prove is this: Given an irreducible polynomial $p(x)$ in $F[x]$, if $E$ is the splitting field of $p(x)$ in $F$, then for any two zeroes $a_1$ and $a_2$ of $p(x)$ there is an automorphism on $E$ that sends $a_1$ to $a_2$. (*edited due to point made by commentor).

I tried and I think I found a way to prove that if $a_1$ and $a_2$ are two zeroes I can create an automorphism on $F[a_1,a_2]$ that sends $a_1$ to $a_2$. The problem here is there's no guarantee that $a_1$ and $a_2$ split $p(x)$. And I can't seem to extend the automorphism so it works on all of $E$.

This feels like something that should be easy but I just can't seem to get it. This is for homework, but it isn't the full problem (there's probably other ways to do the problem, but my inability to prove this theorem is really bothersome).

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Hm. Perhaps you're right. Maybe it just isn't true. That might explain why it's so tough to prove. But this is actually a little more than I needed. I just need the fact that for any two zeroes, there is an automorphism on $E$ taking one to the other. I guess this doesn't necessarily imply that I get all permutations. My bad. I suppose I could phrase it as prove the group of permutations you can get out of the zeroes of p(x) is transitive. –  Rioghasarig Mar 21 '13 at 5:16
    
There generally wont be automorphisms of $F[a_1,a_2]$. –  Hurkyl Mar 21 '13 at 5:41

1 Answer 1

Let $\alpha$ and $\beta$ be two roots of $p$ in $E$. Then $F(\alpha) \cong F(\beta)$ as field extensions of $F$. If this isn't obvious, note they are both isomorphic to $F[x] / p(x)$.

The embedding $F(\alpha) \hookrightarrow E$ makes $E$ the splitting field of $p$ over $F(\alpha)$. The embedding $F(\alpha) \to F(\beta) \hookrightarrow E$ also makes $E$ the splitting field of $p$ over $F(\alpha)$. Because the splitting field of a polynomial over a field is unique, these two different extensions of $F(\alpha)$ must be isomorphic. This gives an isomorphism of $E$ that sends $\alpha$ to $\beta$.

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Incidentally, the way I think about the problem is that the splitting field is a series of fields built as quotients of polynomial rings; e.g. pick an irreducible factor $h(x,y)$ of $p(y)$ over $F[x]/p(x)$ and define the next extension $F[x,y]/(p(x),h(x,y))$, and so forth. Since I never make choose a root, it's clear that I can send $x$ to any of them. But stating a proof precisely along these lines would have been cumbersome. –  Hurkyl Mar 21 '13 at 5:56
    
+1. It is very instructive to translate the magical abstract argument that seems to make something from nothing, into the tower-of-factorizations construction of a splitting field. –  zyx Mar 21 '13 at 5:58
    
Ah, I see. Thank you. This helped a lot. –  Rioghasarig Mar 21 '13 at 6:10

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