# Splitting field of an intermediate field

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?

PS. In the above question, the last condition that I have intended is '... of $$L$$ over $$F$$ were abelian?'

• If $K=F(\alpha)$, then $g(x)=x-\alpha$, so $M=K$. This shows that $M$ can be much smaller than $L$. – GH from MO Sep 10 '19 at 8:05
• @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? – seoneo Sep 10 '19 at 8:12
• The answer is negative even to the strongest form of your question. See my response below. – 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 $$\{1\} such that $$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$$.

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

• 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. – seoneo Sep 10 '19 at 8:36
• No thanks, I just added a comment to your new question. – tanjia Sep 10 '19 at 8:43
• 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??? – seoneo Sep 10 '19 at 9:00
• yeah, you are right. I have modified my comment. – tanjia Sep 10 '19 at 11:42
• Aha! But your previous argument was suffices to apply for abelian galois group case. It was very helpful. – seoneo Sep 10 '19 at 12:19