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?'



Splitting field of an intermediate field
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](https://groupprops.subwiki.org/wiki/Subgroup_structure_of_dihedral_group: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 $K$.


