Intermediate ring between a field and an algebraic extension. 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.
 A: 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.
A: Well, you know that $R$ is contained in an algebraic extension of $F$, so you should use it somehow. It directly implies that there is a polynomial (of minimal degree, say) with coefficients in $F$ which annihilates $r$. Can you manufacture an inverse for $r$ using this polynomial?
A: Let $F\subseteq R\subseteq E$, with $E$ algebraic over $F$, $R$ a ring.
As you note, the key is to show that any nonzero element of $r$ has an inverse in $R$.
So let $r\in R$, $r\neq 0$. Since $R\subseteq E$, then $r$ is algebraic over $F$. Therefore, there is a monic irreducible polynomial $p(x)$ with coefficients in $E$ with $p(r)=0$. Write
$$p(x) = x^n + a_{n-1}x^{n-1}+\cdots + a_1x + a_0.$$
Note that $a_0\neq 0$, since we are taking $p(x)$ to be the monic irreducible. We have
$$\begin{align*}
p(r) &= 0\\
r^n + a_{n-1}r^{n-1}+\cdots + a_1r + a_0 & = 0\\
r^n + a_{n-1}r^{n-1}+\cdots +a_1r &= -a_0\\
r(r^{n-1}+a_{n-1}r^{n-2}+\cdots + a_1) &= -a_0\\
r\left(-\frac{1}{a_0}\right)(r^{n-1}+a_{n-1}r^{n-2}+\cdots + a_1) &= 1.
\end{align*}$$
