$F(\alpha,\beta)$ coincides with $F[\alpha,\beta]$. Let $\alpha$, $\beta$ be algebraic over $K$. Show the ring $K[\alpha,\beta]$ coincides with the field $K(\alpha,\beta)$ .
 A: Proposition: Let $L | K$ be an algebraic field extension, then every intermediate ring $ L \supset R \supset K$ is a field.
Proof: Let $ L \supset R \supset K$ be an intermediate ring and $x \in R$, $x \neq 0$, then as $x \in L$, $x$ is algebraic over $K$, we have a polynomial equation $0=a_nx^n + a_{n-1}x^{n-1}+ \dots +a_1x+a_0$ with $a_i \in K$ We may assume that $a_0 \neq 0$, if not, divide by a suitable power of $x$. Then we have $1=\frac{a_nx^n+\dots+a_1x}{-a_0}$, so $x^{-1} = \frac{a_nx^{n-1}+\dots+a_2x+a_1}{-a_0} \in R$. As $x$ was arbitrary, $R$ is field.
Now we show that $K[\alpha,\beta]= K(\alpha, \beta)$: As $K(\alpha, \beta)$ is generated by elements algebraic over $K$, $K(\alpha,\beta) | K$ is an algebraic extension. Now $K(\alpha,\beta) \supset K[\alpha,\beta] \supset K$ is an intermediate ring, so by the proposition, $K[\alpha,\beta]$ is a field. Thus $K[\alpha,\beta]$ is equal to its own field of fractions, which is $K(\alpha,\beta)$.
A: This results from this very simple lemma from commutative algebra:

Let $K$ be a field. If a  finite $K$-algebra $S$ is an integral domain, $S$ is a field.

Indeed, let $s\ne 0$ be an element of $S$. Multiplication $S$ is a $K$-linear endomorphism of $S$, which is injective  since $S$ is an integral domain. Now, on finite dimensional vector spaces, injective endomorphisms are bijective. 
Therefore $1$ is attained, i.e. there exists an element $s'\in S$ such that $ss'=1$. In other words, any non-zero element in $S$ has an inverse.
