When a subfield of a splitting field is a splitting field 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?
 A: 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).
