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?'

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    $\begingroup$ If $K=F(\alpha)$, then $g(x)=x-\alpha$, so $M=K$. This shows that $M$ can be much smaller than $L$. $\endgroup$ – GH from MO Sep 10 '19 at 8:05
  • $\begingroup$ @GHfromMO Wow. What a nice counter example. Your counter example proves my first question is actually wrong. OK. But what if $K$ is another Galois extension of $F$? How about the case that the galois group of $L$ over $F$ is abelian? $\endgroup$ – seoneo Sep 10 '19 at 8:12
  • $\begingroup$ The answer is negative even to the strongest form of your question. See my response below. $\endgroup$ – GH from MO Sep 10 '19 at 9:26

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 $K$.

  • $\begingroup$ Ahhhh I ment G(L/F) abelian... But yours is a nice ce. $\endgroup$ – seoneo Sep 10 '19 at 16:36
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    $\begingroup$ @seoneo: If $L/F$ is abelian, then any intermediate field extension is Galois over $F$. This forces that $L=F(\alpha)=K(\alpha)$, hence $L=M$. $\endgroup$ – GH from MO Sep 10 '19 at 23:36

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 $M=L$.

  • $\begingroup$ Many many thanks to your response. It is actually correct answer to my 'original' question. However, what I have in mind is the 'condition' to have $M=L$. And so after realizing the incorrectness of my original question, I have edited the post. $\endgroup$ – seoneo Sep 10 '19 at 8:36
  • $\begingroup$ No thanks, I just added a comment to your new question. $\endgroup$ – tanjia Sep 10 '19 at 8:43
  • $\begingroup$ Thanks again for your quick response. How do we conclude that $M/F$ is Galois from the fact that $M/K$ and $K/F$ are Galois??? $\endgroup$ – seoneo Sep 10 '19 at 9:00
  • $\begingroup$ yeah, you are right. I have modified my comment. $\endgroup$ – tanjia Sep 10 '19 at 11:42
  • $\begingroup$ Aha! But your previous argument was suffices to apply for abelian galois group case. It was very helpful. $\endgroup$ – seoneo Sep 10 '19 at 12:19

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