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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    $\begingroup$ 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. $\endgroup$
    – jspecter
    Jun 9, 2011 at 17:35
  • $\begingroup$ F extends E but R contains F and is contained in E? Are you sure it's not a typo? $\endgroup$
    – Gadi A
    Jun 9, 2011 at 17:36
  • $\begingroup$ You mean: Let $E$ be an algebraic extension of $F$. $\endgroup$
    – lhf
    Jun 9, 2011 at 17:47
  • $\begingroup$ @lhf, Gadi: thanks. Corrected. $\endgroup$ Jun 9, 2011 at 17:49
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    $\begingroup$ 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 $E$ $\endgroup$
    – jspecter
    Jun 11, 2011 at 4:46

3 Answers 3


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?

  • $\begingroup$ Hi, I am a bit confused about the statement that "there is a polynomial (of minimal degree, say) with coefficients in $F$ which annihilates $r$". We know that for every $f \in F$, there is a unique monic, irreducible polynomial in $E[x]$ with $f$ being its root. But how does this imply that for every $e \in E$, there is a polynomial in $F[x]$ with $e$ being a root? Couldn't it be the case that it does not take all elements in $E$ to make every $f \in F$ a root of some polynomial? Thanks a lot! $\endgroup$
    – msd15213
    Jul 22, 2018 at 16:13
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    $\begingroup$ @michaelshiyu I'm not sure I understand your question. What you phrased in your question is exactly the definition of an algebraic field extension. The other claim - that an element $f$ of $F$ is the root of a polynomial in $F[x]$ (or $E[x]$) - is trivial: $x-f$ is such a polynomial. $\endgroup$
    – Mark
    Jul 22, 2018 at 19:18
  • $\begingroup$ sorry for being vague, I meant to ask why is every element in $E$ (instead of $F$) a root of a polynomial in $F[x]$? I understand that by the definition of algebraic extension, every element in $F$ is a root for some polynomial in $E[x]$, and I also understand that that polynomial can be viewed as a polynomial of some element in $E$ with coefficients in $F$. So, some elements in $E$ are roots of some polynomials in $F[x]$, but why is this true for every element in $E$? Hope this question makes sense now. Thanks. $\endgroup$
    – msd15213
    Jul 22, 2018 at 20:21
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    $\begingroup$ Because, as I said, this is exactly the definition of $E$ being an algebraic extension of $F$. $\endgroup$
    – Mark
    Jul 22, 2018 at 22:10
  • $\begingroup$ Got it, thanks a lot! $\endgroup$
    – msd15213
    Jul 22, 2018 at 22:18

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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    $\begingroup$ Arturo: the OP asked for a "not-too-weak, not-too-strong hint". (To my taste this was what @Mark Schwarzmann provided.) But what you've written looks like a complete answer... $\endgroup$ Jun 9, 2011 at 18:52
  • $\begingroup$ @Pete: Yeah, you're right. Missed that in the post. I should really read more carefully even if I'm late for lunch. $\endgroup$ Jun 9, 2011 at 18:53
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    $\begingroup$ For the record, this post had been (self-)deleted shortly after posting to comply with the request for hints only in the question. It was undeleted, by three users including me, around the writing of this comment, as the concern seemed by now obsolete. (The motivation for the self-deletion was recorded in two comments, one by the answerer, which may not be visible in the future anymore; they are visible at the time of writing the comment.) $\endgroup$
    – quid
    Aug 19, 2016 at 14:18
  • $\begingroup$ There is some meta discussion on this overruling of author self-deletion, hints, etc. $\endgroup$ Aug 19, 2016 at 14:40
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    $\begingroup$ @Jneven: algebraic does not imply finite. If the extension is not algebraic, then it contains an element $x\in E$ that is transcendental over $F$, and then $F[x]$ is an intermediate ring that is not field. $\endgroup$ Jan 6 at 18:20

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.

  • $\begingroup$ @Bill: I don't really get that. Could you please point me to where the tool you mention is explained? Thank you. $\endgroup$ Jun 10, 2011 at 19:05
  • $\begingroup$ @Wel I'm happy to elaborate if you tell me which part you'd like explained further. $\endgroup$ Jun 10, 2011 at 20:31
  • $\begingroup$ @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. $\endgroup$ Jun 10, 2011 at 21:16
  • $\begingroup$ @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? $\endgroup$ Jun 10, 2011 at 21:26
  • $\begingroup$ @Bill: alright, I'll keep thinking about it. If I don't solve it I shall bother you again :) Thanks. $\endgroup$ Jun 10, 2011 at 21:29

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