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In a question related to introductory Galois theory, I was asked to given an example of a tower of fields $F \subset K \subset E$ such that $E$ is a splitting field for some polynomial $f(x) \in F[x]$, but that $K$ need not be a splitting field. The example that comes to mind for me is $\mathbb Q \subset \mathbb Q(\sqrt[3]{2}) \subset \mathbb Q(\omega_3, \sqrt[3]{2})$.

So I know that the minimal polynomial for $\mathbb Q(\omega_3, \sqrt[3]{2})$ over $\mathbb Q$ is $x^3-2$. This polynomial does not split in the intermediate field $\mathbb Q(\sqrt[3]{2})$.

I wanted to make sure that I'm justifying that this intermediate field is not a splitting field correctly. Now I may be confused about the books terminology. When we say something is not a splitting field, we always have to refer to a specific polynomial correct? I've read a term called a normal extension which I think refers to one in which every polynomial splits, but my book doesn't mention these.

If splitting field means with respect to the polynomial $x^3-2$, then I've already demonstrated that $\mathbb Q(\sqrt[3]{2})$ is not a splitting field. Also, it couldn't be a normal extension either because that polynomial doesn't split.

Does it sound like I'm understanding this correctly or are there some subtleties I might be missing?

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Splitting fields are normal extensions, which means that every irreducible polynomial with coefficients in the base field which has one root in the extension splits in the extension. Now, $ X^3 - 2 $ is irreducible by Eisenstein, and clearly has a root in $\mathbb{Q}(2^{1/3})/\mathbb{Q} $, however the other two roots are complex, and are not contained in this extension. Therefore, the extension is not normal, and cannot be a splitting field.

Your understanding is correct, although you have to prove the fact that every splitting field is necessarily a normal extension for your argument to work. Here is a simple proof: assume that $ L/K $ is a splitting field of the polynomial $ g \in K[X] $, and let $ f $ be an irreducible polynomial which has a root $ \alpha $ in $ L/K $. Our proof strategy is to show that if $\beta $ is a root of $ f $, then $[L(\alpha):L] = [L(\beta):L]$. Now, note that we have the isomorphisms $ K(\alpha) \cong K[X]/(f) \cong K(\beta) $. Then, we have a $K$-isomorphism $\phi : K(\alpha) \to K(\beta) $, and by extension of isomorphisms to splitting fields, and the fact that $ L(\alpha) $ is the splitting field of $ g $ over $ K(\alpha) $, it follows that this isomorphism extends to a $K$-isomorphism $ \phi' : L(\alpha) \to L(\beta) $ so that $ [L(\alpha):L] = [L(\beta):L] $. Now, since $ \alpha \in L $, we must have $ \beta \in L $ as well, completing the proof.

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  • $\begingroup$ Of course - fixed that. $\endgroup$ – Starfall May 6 '16 at 20:34
  • $\begingroup$ Oh ok. Well I should probably avoid the normal extension stuff for now (though I will probably read more later). For my specific situation can I argue something like this? Since a splitting field is defined to be the smallest field in which a polynomial splits, $\mathbb Q(\sqrt[3]{2})$ is not the splitting field of any polynomial which splits in $\mathbb Q$ already, because it is too large of a field. Therefore, any polynomial which splits in $\mathbb Q\sqrt[3]{2})$ must have $\sqrt[3]{2}$ as one of its roots. But the minimal polynomial for $\sqrt[3]{2}$ is $x^3-2$ (to be cont.) $\endgroup$ – user1236 May 6 '16 at 20:41
  • $\begingroup$ And any polynomial having $\sqrt[3]{2}$ as a root is divisible by the minimal polynomial, and so would have the roots not found in $\mathbb Q(\sqrt[3]2)$ as well. Thus no polynomial in $\mathbb Q[x]$ splits in $\mathbb Q(\sqrt[3]2)$. $\endgroup$ – user1236 May 6 '16 at 20:43
  • $\begingroup$ I do not see how you see that any polynomial which splits in $ \mathbb{Q}(2^{1/3}) $ must have $ 2^{1/3} $ as a root. Consider $ \mathbb{Q}(\sqrt{2})/\mathbb{Q} $, then the polynomial $ X^2 - 2X - 1 $ splits in this extension, but it doesn't have $ \sqrt{2} $ as a root. $\endgroup$ – Starfall May 6 '16 at 20:50
  • $\begingroup$ Thanks Starfall for your answer, it is very clear to me, and I was searching for such a proof. $\endgroup$ – reuns May 6 '16 at 20:58

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