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?


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.

  • $\begingroup$ Of course - fixed that. $\endgroup$
    – Ege Erdil
    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. $\endgroup$
    – Ege Erdil
    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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.