# Necessarily a field between a field and its 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.

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Consider the case where $E/F$ is finite and think about $R$ an $F$ vector space and multiplication by $r$ as an endomorphism of $R$. Show that this map is surjective. Hint: Think about dimension. – jspecter Jun 9 '11 at 17:35
F extends E but R contains F and is contained in E? Are you sure it's not a typo? – Gadi A Jun 9 '11 at 17:36
You mean: Let $E$ be an algebraic extension of $F$. – lhf Jun 9 '11 at 17:47
@lhf, Gadi: thanks. Corrected. – Weltschmerz Jun 9 '11 at 17:49
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 $E$ – jspecter Jun 11 '11 at 4:46

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 $x$. Can you manufacture an inverse for $r$ using this polynomial?

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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 $\rm\:F\:.$

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 is a 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).

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@Bill: I don't really get that. Could you please point me to where the tool you mention is explained? Thank you. – Weltschmerz Jun 10 '11 at 19:05
@Wel I'm happy to elaborate if you tell me which part you'd like explained further. – Bill Dubuque Jun 10 '11 at 20:31
@Bill: I understand that the problem is writing $r^{-1}$ in terms of an expression involving powers of $r$ and elements of $F$. I don't know about the identity you mention and the notation therein, or about rationalizing to F the denominator of $1/r$. A small example would be enough, I'll try to work on the general case. Thanks. – Weltschmerz Jun 10 '11 at 21:16
@Wel To elaborate on those points requires giving the complete solution hinted in the first sentence. Since you said you desired only a hint, is it ok to do such? – Bill Dubuque Jun 10 '11 at 21:26
@Bill: alright, I'll keep thinking about it. If I don't solve it I shall bother you again :) Thanks. – Weltschmerz Jun 10 '11 at 21:29