Let $K \supset F$ a normal extension (finite). Show in details that, for every $\alpha \in K$, the polynomial

$$f(x)=\prod_{\sigma \in Aut_F(K)}(x-\sigma(\alpha))\in F[x];$$

My attempt:

Since $K$ is a finite extension of $F$, $K$ is an algebraic extension, so $\alpha$ is algebraic over $F$.

If $\alpha \in F$, then $\sigma(\alpha)=\alpha$ for every $\sigma \in$ Aut$_F(K)$, and, obviously, $f(x)=(x-\alpha)^k \in F[x]$, where $k$ $=$ |Aut$_F(K)$|.

If $\alpha \notin F$, we know that $\alpha$ is algebraic over $F$, so there is $p(x)=irr(\alpha,F) \in F$. Since, the extension is normal, every root of $p(x)$ is in $K$ and, it's know that any automorphism in Aut$_F(K)$ will take a root of an irreducible polynomial into another, so if $\{\alpha=\alpha_1,\ldots,\alpha_m\}$ is the set of rotos of $p(x)$, any automorphism will permute these roots.

We can write $p(x)=(x-\alpha_1)^{n_1}\ldots(x-\alpha_m)^{n_m} \in K[x]$.

I can't go any further than that... My initial idea was to show that $f(x)=p(x)$, however I see that it's not true, because the number of automorphisms in Aut$_F(K)$ can be bigger than $p$'s degree, so I can't see how to guarantee that $f(x) \in F[x]$.

I know that $f(x)=(x-\alpha_1)^{n'_1}\ldots(x-\alpha_m)^{n'_m}$ , $0\leq n'_i\leq k=$|Aut$_F(K)$|. So if $n'_i \geq n_i$, we will have that $f(x)=p(x)g(x)$, $g(x)=(x-\alpha_1)^{n'_1-n_1}\ldots(x-\alpha_m)^{n'_m-n_m}$. So I'd have to show that $g(x)\in F[x]$, but I can't see how to do that... And, even if I knew, it doesn't solve the entire problem..

Does someone have a better idea to solve this?

Thanks :)

  • $\begingroup$ Do you know anything about symmetric polynomials? $\endgroup$ – Kevin Arlin Sep 15 '14 at 1:24
  • $\begingroup$ @KevinCarlson No, I don't.. :( $\endgroup$ – Anna Sep 15 '14 at 1:33
  • $\begingroup$ Ah, it often gets covered in this course but isn't critical here. What you want to try to show is that this $f$ is a power of the minimal polynomial, namely it's $p(\alpha)^{[K:S]}$, where by $S:=S(p(\alpha))$ I mean the splitting field. To do this you want to write each automorphism $\sigma$ as an automorphism of $S$ composed with an automorphism of $K$ over $S$. Do you see how to do that? $\endgroup$ – Kevin Arlin Sep 15 '14 at 1:40
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    $\begingroup$ Ah, and as happens, I notice the proof I was going for was harder than necessary. Apologies. $\endgroup$ – Kevin Arlin Sep 15 '14 at 2:02
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    $\begingroup$ @KevinCarlson Thanks! I talked to the professor. He meant fields with characteristic 0, just forgot to write it haha :) $\endgroup$ – Anna Sep 19 '14 at 1:59

The necessary datum is the following:

Lemma Let $L/k$ be a normal and separable extension. Then the intersection of the fixed fields of automorphisms of $L/k$ is exactly $k$, that is, $k=\{x\in L:\sigma x=x\forall\sigma\in\text{Aut}(L/k)\}$.

From this we simply point out that $f$ is fixed by automorphisms of $L$, so that all of its coefficients are as well, and they're thus in $k$.

Proof of lemma Let $k$ and $L$ be embedded in a common algebraic closure $\bar k$. By definition, all automorphisms of $L$ over $k$ fix $k$. Then let $\alpha\in L\setminus k$, and $\bar\alpha\neq \alpha$ be another root of $\alpha$'s minimal polynomial. Then $k(\alpha)$ is isomorphic to $k(\bar\alpha)$ by $\alpha\mapsto \bar\alpha$. This automorphism extends to $\bar k$, so that not all automorphisms of $\bar k/k$ fix $\alpha$, and we just need to check we can restrict this automorphism to $L$. But one of the standard formulations of the condition that $L/k$ be normal is that every automorphism of $\bar k$ over $k$ map $L$ into itself.

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