Can someone Check my proof? It feels weird that I didn't really have to do anything.

Question

Since $D$ is a commutative ring with unity, it remains to show that every nonzero element of $D$ has a multiplicative inverse in $D$. Let $\alpha \in D \setminus \{0\}$ be arbitrary. Since $E$ is a finite extension of $F$, it is an algebraic extension of $F$, so $\alpha$ is algebraic over $F$. Let $n = \deg(\alpha, F)$. Since $\alpha \in D$, the ring axioms imply $1, \alpha, \alpha^2, \dots, \alpha^{n - 1} \in D$. The ring axioms thus imply that any linear combination of $1, \alpha, \alpha^2, \dots, a^{n - 1}$ with coefficients in $F$ is in $D$, that is, $F(\alpha) \subseteq D$. In particular, $\alpha^{-1} \in D$.

Answer 1

Depends what you call $F(\alpha)$, it seems to me that you already know it is a field.

If you don't know that, you need one more step: take the minimal polynomial of $\alpha$: $$\mu_{\alpha}(x)=a_0+a_1x+a_2x^2+\ldots+a_kx^k$$ with $a_i\in F$, and then: $$0=\mu_{\alpha}(\alpha)=a_0+a_1\alpha+a_2\alpha^2+\cdots+a_k\alpha^k$$

You know that $a_0\ne 0$ because $\alpha\ne 0$ (and so if $a_0=0$ you could find a smaller-degree polynomial that would vanish at $\alpha$). Therefore, you can cancel $a_0$, or equivalently assume $a_0=1$. Then: $$0=1+a_1\alpha+a_2\alpha^2+\ldots+a_k\alpha^k$$ which gives you $1=\alpha(-a_1-a_2\alpha-\ldots-a_k\alpha^{k-1})$ and so the element $-a_1-a_2\alpha-\ldots-a_k\alpha^{k-1}$ is your $\alpha^{-1}\in F(\alpha)$.