Smallest subfield containing $F$ and $\alpha$

Question

Let $F$ be a field and let $K$ be an extension of $F$. Show that if $\alpha\in K$ is algebraic over $F$, $F[\alpha]=\{p(\alpha)\mid p(x)\in F[x]\}$ is the smallest subfield of $K$ containing $F$ and $\alpha$.

Answer 1

You have to show that (1) $F[\alpha]$ is a field, and (2) that whenever $K$ is a field with $F \subseteq K$ and $\alpha \in K$, then $F[\alpha] \subseteq K$.

Once you have shown (1), then (2) is not too hard: Assume that $K$ is a field with $F \subseteq K$ and $\alpha \in K$. Then surely all elements of the form $$a_0 + a_1 \alpha + a_2 \alpha^2 + \cdots + a_n \alpha^n$$ with $a_i \in F$ and $n \in \mathbb{N}$ must be elements of $K$, i.e. $F[\alpha] \subseteq K$.

Thus it remains to show (1). The proof relies on the fact that $\alpha$ is algebraic over $F$. Let $p(x)$ denote the minimal polynomial of $\alpha$ over $F$. Then in fact the ring $F[x]/(p(x))$ will be isomorphic to $F[\alpha]$ (as rings). If you can show this, and that $F[x]/(p(x))$ is a field, then you're done.