Show that if $F \subset D \subset E$ then D is a field

Question

Exercise: Let $E$ be a finite extension field of $F$, and $D$ an integral domain such that $F \subset D \subset E$. Show that $D$ is a field.

Attempt: All of the field axioms for D are inherited by the inclusion relationship and ring axioms except for the existence of multiplicative inverse. $\alpha \in D \cap F \implies \alpha^{-1} \in F \implies \alpha^{-1} \in D$, so consider $\beta \in D \backslash F$.

Since $E/F$ is finite it's also algebraic, but then $F[\beta] = F(\beta)$ and so $\beta^{-1} \in F(\beta) \subset D$.

I'm not really sure this proof is correct, but in case it is there are two things that bug me:

• If we just ask $D$ to be a ring then from $D \subset E$ follows $D$ is an integral domain.
• We only used the fact that $E/F$ is an algebraic extension in the proof.

Thanks for your help.

Answer 1

It's already an integral domain, all you're missing is inverses. Let $p_\beta(x) = a_0+a_1x+\ldots + a_nx^n$ be the minimal polynomial for $\beta$ over $F$. Then note

$$-a_0^{-1}(a_1+a_2\beta+a_3\beta^2+\ldots + a_n\beta^{n-1})\cdot\beta=1$$

So $\beta^{-1}\in D$, showing inverses. Here we use that all such $\beta\in E$ which is finite, so that there is a minimal polynomial for it over $F$.