## Subrings of an algebraic field extension which contain the base field are subfields

Let $K/F$ be an algebraic field extension and let $R \subseteq K$ be a subring containing $F$. Show that $R$ is a subfield of $K$.

It suffices to show that $R$ is closed under inversion. To this end, let $\alpha \in R$. Since $K$ is algebraic over $F$, every element of $K$ is algebraic over $F$. Let $p(x)$ be the minimal polynomial of $\alpha$ over $F$; if $p(x) = \sum c_ix^i$, then we have $\sum c_i\alpha^i = 0$. Note that $c_0 \neq 0$ since $p$ is irreducible. Rearranging, we see that $\alpha (\frac{-1}{c_0} \sum_{i \neq 0} c_i \alpha^{i-1}) = 1$, and $\frac{-1}{c_0} \sum_{i \neq 0} c_i \alpha^{i-1} \in R$. So $\alpha^{-1} \in R$, and $R$ is closed under inversion. Thus $R$ is a subfield of $K$.