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I am working on this exercise:

If $E$ is an intermediate field of an extension $F/K$ of fields. Suppose $F/E$ and $E/K$ are Galois extensions, and every $\sigma\in Gal(E/K)$ is extendible to an automorphism of $F$, then show that $F/K$ is Galois.

I can see that any $\sigma$ extended over $F$ fixes elements in $K$ but not in $E-K$. But how to show it doesn't fix elements in $F-E$?

Hints only please, this is homework.

p.s. we use Kaplansky, he doesn't require Galois extensions to be finite dimensional.

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Is your definition of Galois that the fixed field of $\operatorname{Aut}(L/k)$ is $k$? If so, then it seems helpful to note that $\operatorname{Gal}(F/E) \subset \operatorname{Aut}(F/K)$. –  Dylan Moreland Feb 10 '12 at 14:19
    
Thank you Dylan. –  Allan Feb 10 '12 at 15:22
    
i solved it. You can remoe this question. –  Allan Feb 10 '12 at 15:27
    
You have the power to remove it, I think, but I don't think there's any harm in leaving it up. –  Dylan Moreland Feb 10 '12 at 15:30
14  
@Allan: Instead of removing the question, post your solution as an answer! –  Arturo Magidin Feb 10 '12 at 16:56

1 Answer 1

Here is a full solution of the problem. Since the OP already solved it, I think there is no harm in writing a full solution of it here.

We denote by $Aut(K/k)$ the group of automorphisms of a field extension $K/k$. Let $K/k$ be a not necessarily finite dimensional algebraic extension field. If $K$ is normal and separable over $k$, we say $K/k$ is Galois. If $K/k$ is Galois, we write $G(K/k)$ instead of $Aut(K/k)$.

We need the following characterization of a Galois extension field.

Lemma An algebraic extension field $K/k$ is Galois if and only if the fixed subfield of $K$ by $Aut(K/k)$ is $k$.

Proof. Suppose $K/k$ is Galois. Let $\alpha \in K - k$. Let $f(X)$ be the minimal polynomial of $\alpha$ over $k$. Since $K/k$ is normal and separable, there exists a root $\beta$ of $f(X)$ such that $\alpha \ne \beta$ and $\beta \in K$. Let $\sigma\colon k(\alpha) \rightarrow k(\beta)$ be the unique isomorphism such that $\sigma(\alpha) = \beta$. Since $K/k$ is normal, $\sigma$ can be extended to an automorphism $\sigma'$ of $K/k$. Since $\sigma'(\alpha) = \beta$, we are done.

Conversely suppose the fixed subfield of $K$ by $G = Aut(K/k)$ is $k$. Let $\alpha$ be an element of $K$. Let $f(X)$ be the minimal polynomial of $\alpha$ over $k$. Since $\sigma(\alpha)$ is a root of $f(X)$ for every $\sigma \in G$, the set $S = \{\sigma(\alpha)\mid \sigma \in G\}$ is finite. Let $\sigma_1, \cdots, \sigma_m$ be elements of $G$ such that $\sigma_1(\alpha), \cdots, \sigma_m(\alpha)$ are pairwise distinct and $S = \{\sigma_1(\alpha), \cdots, \sigma_m(\alpha) \}$. Let $g(X) = (X - \sigma_1(\alpha))\cdots (X - \sigma_m(\alpha))$. Since every coefficient of $g(X)$ is fixed by $G$, $g(X) \in k[X]$. Since $g(\alpha) = 0$, $g(X)$ is divisible by $f(X)$. Since each $\sigma_i(\alpha)$ is a root of $f(X)$, $g(X) = f(X)$. Hence $\alpha$ is separable over $k$ and $K/k$ is normal. This completes the proof of the lemma.

Now let $F/K, E/K$ be as in the problem. By the lemma, it suffices to prove that the fixed subfield of $F$ by $Aut(F/K)$ is $K$. Suppose $\alpha \in F$ is fixed by $Aut(F/K)$. Since $G(F/E) \subset Aut(F/K)$, $\alpha$ is fixed by $G(F/E)$. Hence $\alpha \in E$ by the lemma. Since every element of $G(E/K)$ is extended to an element of $Aut(F/K)$, $\alpha$ is fixed $G(E/K)$. Hence $\alpha \in K$ by the lemma. This completes the proof.

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