Splitting field of an intermediate field

Question

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?'

Answer 1

Even when $K/F$ is quadratic (hence galois and abelian), it can happen that $M$ is smaller than $L$.

For a straightforward counterexample, take a chain of groups $\{1\}<I<H<D_8$ such that $I$ is not normal (cf. subgroups of D8). Let $L/F$ be a $D_8$ extension, and let $K$ (resp. $M$) be the fixed field of $H$ (resp. $I$). Then $K/F$ and $M/K$ are quadratic extensions, while the quartic extension $M/F$ is not normal. So if $M=F(\alpha)$, then $L$ is the splitting field of $\alpha$ over $F$, while $M=K(\alpha)$ is the splitting field of $\alpha$ over $K$.

Answer 2

Here is a direct counterexample. $L=\mathbb{Q}(\sqrt[3]{2},\xi_3)$,$F=\mathbb{Q}$, $\alpha=\sqrt[3]{2}$, $K=\mathbb{Q}(\alpha)$ and $f=x^3-2$. We have $M=K\subsetneqq L$.

This is not true even when $K/F$ is galoisian. A slight modification of the counterexample above gives a new counterexample. Lets take $F=\mathbb{Q},K=\mathbb{Q}(\sqrt{2}),M=\mathbb{Q}(\sqrt[4]{2}), L=\mathbb{Q}(\sqrt[4]{2},\xi_4),\alpha=\sqrt[4]{2},f=x^4-2$ and $g=x^2-\sqrt{2}$. $L$ is the splitting field of $f$ over $F$. $\alpha$ is a root of $f$. The extension $K/F$ is galoisian. The minimal polynomial of $\alpha$ over $K$ is $g$ and the splitting field of $g$ over $K$ is $M$, but we don't have $M=L$.