0

This appears in a bigger context, but in summary, I have $K\subset L$ extension of finite fields and $\alpha\in L$ such that $L=K(\alpha)$. Also, $G=Aut_K L$. Now, I need to prove the following:

$$f_K^{\alpha}= \prod_{\sigma\in G} (X-\sigma(\alpha)) $$

But I think that’s pretty easy once you prove $g= \prod_{\sigma\in G} (X-\sigma(\alpha))\in K[X] $. How should I do it?

cut
  • 347
  • Not true in general: $K=\mathbb Q, L=\mathbb Q(\sqrt[3]2)$. Perhaps you mean $L/K$ Galois? – lhf Mar 13 '23 at 16:33
  • 1
    You used the tag [tag:finite-fields]. Do we know that $K$ is finite? It plays a role, because the objection raised by lhf becomes moot, if $K$ and $L$ are finite. Anyway, you should not rely on the choice of tags alone to convey such facts. – Jyrki Lahtonen Mar 13 '23 at 16:46
  • Yes, sorry. $K$ is finite and $L$ is a finite extension. I’m editing the question. – cut Mar 13 '23 at 16:59
  • Ok. Thanks for the clarification. A solution needs to explain, why the assumption $L=K(\alpha)$ is necessary? In other words, what goes "wrong" (or at least "differently"), if $K(\alpha)$ is a proper subfield of $L$? Otherwise $g$ won't be irreducible over $K$. – Jyrki Lahtonen Mar 14 '23 at 05:27
  • Yeah, I think that is what it means. But I do not know what the corresponding statement would be. – cut Mar 14 '23 at 09:29

1 Answers1

0

For simplicity, denote by $f$ the minimal polynomial of $\alpha$ over $K$. Note that your statement is equivalent to that the set of all roots of $f$ is exactly the set $\{\sigma(\alpha):\sigma\in\text{Aut}_K(L)\}$.

First, for any $\sigma\in\text{Aut}_K(L)$, we have $f(\sigma(\alpha))=\sigma(f(\alpha))=0$ because $\sigma$ fixes $K$. Conversely, let $\beta$ be a root of $f$ in an algebraic closure of $K$ containing $L$. Since $\alpha$ and $\beta$ are conjugates, there is an isomorphism $\varphi:L=K(\alpha)\to K(\beta)$ fixing $K$. This shows that $K(\beta)\cong L$. Now, since any extension between finite fields is a normal extension, the fact that $\alpha\in L$ implies that $\beta\in L$, so we have $K(\beta)\le L$. Since $K(\beta)$ and $L$ are both finite-dimensional vector spaces, we then have $K(\beta)=L$. We have now shown that $\varphi$ is in fact an automorphism of $L$ fixing $K$, and $\varphi(\alpha)=\beta$, as desired.

durianice
  • 2,624
  • Actually, since $K\subset L$ is always a Galois extension and $g$ is fixed by all automorphisms in $\text{Gal}(K/L)$, we can, by comparing coefficients, conclude that the coefficients of $g$ are fixed by those automorphisms as well. Then the coefficients of $g$ must lie in $K$ because they lie in the fixed field for the whole group $\text{Gal}(K/L)$. – durianice Apr 12 '23 at 05:26