Consider the following 'wrong' question.
Let $f(x) \in F[x]$ be an irreducible polynomial in a polynomial ring of a field $F$. Let $L$ be the splitting field of $f(x)$ over $F$. Assume that $L$ is a Galois extension over $F$. Let $\alpha \in L$ be a root of $f(x)$. Consider an intermediate field $L-K-F$. Let $g(x) \in K[x]$ be the minimal polynomial of $\alpha$. Let $M$ be a subfield of $L$ which is the splitting field of $g(x)$. Do we have $M=L$? (Wrong)
Suppose that $L=F(α)$ is a normal separable extension of F. Then it follows that for any intermediate field $L−K−F$, $K(α)=L$. I'm trying to generalize this. Of course if $f(x)$ in question is factorized as $g(x)h(x)$. It may be the case that in the splitting field of $g(x)$ over $K$, $h(x)$ remains irreducible or at least not factored into linear polynomials. As pointed by comment, the answer is no and there is a trivial counter example. What I'm interested is the condition to have $M=L$. So let me ask the question.
Is there plausible conditions to have $M=L$? What if $K$ is a galois extension of $F$? How about the case that the galois group of $K$ over $F$ were abelian?
Thank you for your attention.
PS. In the above question, the last condition that I have intended is '... of $L$ over $F$ were abelian?'