Extending Automorphisms, Isomorphisms

Question

Suppose we have an isomorphism $\varphi: K_1 \to K_2$ where there is a base field $F$ and a Galois extension field $L$, ie, $F \subsetneq K_1, K_2 \subsetneq L$, $\varphi|_F = Id$ and $|Gal(L/F)| < \infty$. One case we consider is that where $K_1 = K_2$ and thus $\varphi$ is an automorphism of $K_1$ fixing $F$. Does there exist an automorphism $\overline{\varphi} \in Gal(L/F)$ such that $\overline{\varphi}|_{K_1} = \varphi$?

I propose the following proof that such an automorphism exists, though I am not certain as to its validity.

Suppose $\varphi:K_1 \to K_2$ as above. Since $L$ is Galois over $K_1$ we have by the Primitive Element Theorem that $L = K_1(\alpha)$. Thus, $1, \alpha, ... , \alpha^{n-1}$ is a basis for $L$ over $K_1$. Let $m_{\alpha, K_1} \in K_1[x]$ denote the minimal polynomial of $\alpha \in L$ over $K_1$. Let $\beta$ be a root of $\varphi(m_{\alpha, K_1})$ where we extend $\varphi$ naturally to a ring homomorphism from $K_1[x]$ to $K_2[x]$. Then since $K_1 \cong K_2$, $K_1(\alpha) \cong K_2(\beta)$. Since $L = K_1(\alpha) \cong K_2(\beta)$, I claim there exists $\beta \in L$ such that $\varphi(m_{\alpha, K_1})(\beta) = 0$. Thus, $1, \beta, ... , \beta^{n-1}$ is a basis for $L$ over $K_2$ and if we let $\overline{\varphi} |_{K_1} = \varphi$ and $\overline{\varphi}: \alpha \mapsto \beta$, then $\overline{\varphi}$ is the desired automorphism of $L$.

I believe the weakest claim is that $\beta \in K_1(\alpha) = L$, and I am not sure how to shore it up. I would love feedback on the rest of the proof as well.

Answer 1

I guess the crucial point of your proof is the following.

$m_{\alpha,K_{1}}$ divides $m_{\alpha,F}$.

Now

• $L/F$ is Galois,
• $m_{\alpha,F}$ is irreducible in $F[x]$, and
• $m_{\alpha,F}$ has a root $\alpha$ in $L$.

This yields that $m_{\alpha,F}$ splits completely in $L$.

Now $\varphi(m_{\alpha, K_1})$ also divides $m_{\alpha,F}$, as $\varphi$ fixes $F$ pointwise, so $\varphi(m_{\alpha, K_1})$ also splits completely in $L$, and there's your $\beta$.

Answer 2

I think your proof works fine (provided you can show that $\beta$ exists). Here is an alternative:

Since you are using Dummit and Foote, take a look at theorem 27 on page 541 (of the 3rd edition). Since $L/F$ is a finite Galois extension, then it is the splitting field of some $f\in F[x]$, then $L/K_1$ and $L/K_2$ are splitting fields for $f\in K_1[x]$ and $f\in K_2[x]$ respectively. Then applying the theorem gives exactly the extended automorphism you are looking for.