TITLE: Intermediate ring between a field and an algebraic extension.
QUESTION [31 upvotes]: This is an exercise in some textbooks.

Let $E$ be an algebraic extension of $F$. Suppose $R$ is ring that contains $F$ and is contained in $E$. Prove that $R$ is a field.

The trouble is really with the inverse of $r$, where $r\in R$. How to prove that $r^{-1}\in R$, in apparent lack of a characterization of $R$.
It occurred to me to use the smallest field containing $R$ ($R$ is easily shown to be an integral domain), that's the field of quotients, and proving that it's $R$ itself, but I don't really know how to proceed.
A not-too-weak, not-too-strong hint will be much appreciated. 
Beware $ $ Readers seeking only hints should  beware that there is now a complete answer.

REPLY [4 votes]: Hint $\ $ Read off an inverse of $\rm\:r \ne 0\:$ from a minimal polynomial $\rm\:f(x)\in F[x]\:$ of $\rm\:r\:$ over $\rm\:F,\,$ e.g. $\rm\,ax^2+bx = c \,\Rightarrow\, x\,(ax+b)/c = 1,\,$  where $\rm\,c\neq 0\,$ by minimality (else we could cancel $\rm x)$
Note: this is a special case of using the Euclidean algorithm to compute inverses via the Bezout identity, viz. $\rm\: (x,f(x)) = 1\ \Rightarrow\ a(x)\ x + b(x)\ f(x) = 1\ $ so $\rm\: a(r)\ r = 1\ $ by evaluating at $\rm\:x = r.\:$ 
This can be viewed as  generalization of rationalizing denominators in low-degree extensions, i.e. one may compute an inverse of $\rm\:r\:$ by "rationalizing" (to $\rm\:F)\,$ the denominator of $\rm\:1/r\:$ (which can also be done using norms, resultants, etc). See here for much further discussion.