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?
Best Answer
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.