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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.

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    Yes. But you only care about one element of $R$ at a time and any such element is contained in a finite extension of $F.$ Replace $E$ by this extension and $R$ by the intersection of $R$ and $E$2011-06-11

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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?

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    Got it, thanks a lot!2018-07-22
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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 $\begin{align*} p(r) &= 0\\ r^n + a_{n-1}r^{n-1}+\cdots + a_1r + a_0 & = 0\\ r^n + a_{n-1}r^{n-1}+\cdots +a_1r &= -a_0\\ r(r^{n-1}+a_{n-1}r^{n-2}+\cdots + a_1) &= -a_0\\ r\left(-\frac{1}{a_0}\right)(r^{n-1}+a_{n-1}r^{n-2}+\cdots + a_1) &= 1. \end{align*}$

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    There is [some meta discussion](http://meta.math.stackexchange.com/a/24840/242) on this overruling of author self-deletion, hints, etc.2016-08-19
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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.

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    @Wel Here's a further hint: $\rm\:1/r = g(r)\iff g(r)\ r = 1\:.\ $ Scale the minimal polynomial so it has this form, i.e. $\rm\ f(x) = g(x)\ x - 1\:.$2011-06-10