3
$\begingroup$

Look at the following proposition:

Let $K\subset L\subset M$ be three fields. If $M$ is a splitting field over $K$ of a polinomial in $K[X]$ and moreover if for every $\sigma\in G=Gal(M/K)$ we have that $\sigma(L)\subseteq L$, then $L$ is a splitting field over $K$.

My attempt of proof is the following, but I have problems in the last part:

if $M=K(a_1,\ldots, a_n)$ where $a_1,a_2,\ldots, a_n$ are the roots of $f\in K[X]$, then (rilabelling the roots if it is necessary) we have that $L=K(a_1,\ldots,a_t)$ with $t<n$. Now if $g=(X-a_1)\ldots(X-a_n)$, by the property of $L$, we can argue that $g\in Fix(G)[X]$, and so $L$ is the splitting field of $g$ over $Fix(G)[X]$ but not over $K$. In characteristic $0$ we could conclude that $Fix(G)=K$, but what about the general case?

$\endgroup$
1
$\begingroup$

I think the following may work:

First, note that if $\sigma \in \operatorname{Gal}(M/K)$ and $\sigma(L) \subseteq L$, then since $L/K$ is algebraic, $\sigma|_L$ is actually an automorphism $L\to L$ (Lang's Algebra, Lemma 2.1, p. 230).

I assume based on your other questions that you are familiar with normal extensions, but just for reference (this is taken almost verbatim from Morandi's Field and Galois Theory):

Let $L/K$ be an algebraic extension. The following are equivalent: (1) $L/K$ is normal (i.e., $L$ is a splitting field over $K$). (2) If $N$ is an algebraic closure of $L$ and if $\tau: L \to N$ is a $K$-homomorphism, then $\tau(L)=L$. (3) If $K \subseteq F \subseteq L \subseteq N$ are fields and if $\tau: F \to N$ is a $K$-homomorphism, then $\tau(F) \subseteq L$, and there is a $\sigma\in \operatorname{Gal}(L/K)$ with $\sigma|_F = \tau$.

We shall show $L/K$ satisfies (2). Let $N$ be an algebraic closure of $M$; then since $N/M$ and $M/L$ are both algebraic, $N$ is also an algebraic closure of $L$. Let $\tau: L \to N$ be a $K$-homomorphism. Now, $M/K$ is normal, so we may apply (3): There is a $\sigma \in \operatorname{Gal}(M/K)$ such that $\sigma|_L=\tau$ . In particular, we have $\tau(L) = \sigma(L)= L$ (by the first paragraph).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.