Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $E|F$ be an algebraic field extension and a ring $K$ such that $F\subseteq K\subseteq E$. It is true that $K$ is a field?

share|cite|improve this question
It might be useful to note that Alex's solution extends to the following stronger version: If $K$ is an integral domain that is finitely generated as a vector space over a field $L$, then $K$ is a field. – Rijul Saini Dec 8 '14 at 18:12

3 Answers 3

up vote 13 down vote accepted

Yes. Suppose $0\ne k\in K$. Since $k\in E$ and $E/F$ is algebraic, we have some minimal polynomial $x^n+a_{n-1}x^{n-1}+\cdots+a_0$ with coefficients in $F$ that is satisfied by $k$. By minimality the coefficient $a_0$ must be nonzero, so it has an inverse $a_0^{-1}$ in $F$. Then $k(-a_0^{-1})(k^{n-1}+a_{n-1}k^{n-2}+\cdots+a_1)=1$, so $k^{-1}=(-a_0^{-1})(k^{n-1}+a_{n-1}k^{n-2}+\cdots+a_1)$. Since $k$ and each $a_i$ are in $K$, it follows that $k^{-1}$ is in $K$. Thus $K$ is a field.

share|cite|improve this answer
Note how the Cayley–Hamilton theorem proves in the same way as above that a matrix over a field is invertible iff its determinant is not zero. – lhf May 21 '13 at 2:25
@lhf Eh, all the proofs of C-H that I can think of use he fact that $A A^\dagger = (\det A) I = A^\dagger A$ for a certain matrix $A^\dagger$. And this already implies what you said. Though it's true that this argument is worth remembering. – Ryan Reich May 21 '13 at 3:27
Standard, but strong argument! (+1) – i707107 May 21 '13 at 4:24
@i707107 I never claimed it was original :) – Alex Becker May 21 '13 at 4:27

Yes. Let $a \in E$ with $a\ne0$.

Since $a \in E$ is algebraic over $F$, we have that $F[a]$ is a finite-dimensional $F$-algebra because $F[a]$ is by definition the $F$-vector space generated by the powers of $a$ and powers having exponent larger than the degree of $a$ can be replaced by linear combinations of lower powers by using a monic polynomial equation satisfied by $a$.

Since $F[a]$ is a finite-dimensional $F$-algebra, the map $x \mapsto ax$ on $F[a]$ is $F$-linear and injective and hence surjective. Therefore, $1$ is in the image and $a$ is invertible in $F[a]$.

Since $F[a]$ is by definition the smallest ring containing $F$ and $a$, we have that $a$ is invertible in every subring $K$ of $E$ that contains $a$. In particular, every nonzero element of $K$ is invertible, and $K$ is a field.

Actually, the following are equivalent (proof left as an exercise):

  • $a$ is algebraic over $F$
  • $F[a]$ is a finite-dimensional $F$-algebra
  • $F[a]$ is a field
  • $F[a]=F(a)$
share|cite|improve this answer
Compare the proof above with the proof that "every finite domain is a field" or that "every finite cancellation monoid is a group" which use the same map; finite dimension replaces finiteness here. – lhf May 21 '13 at 3:39

A more general argument can be given: let $A\subset B$ be two integral domains. An element $b\in B$ is said to be integral over $A$ if there exists a monic polynomial $p(t)\in A[t]$ with $p(b)=0$. The ring $B$ is said to be integral over $A$ is every element in $B$ is integral over $A$.

Theorem: Let $A\subset B$ be I.D.'s , with $B$ integral over $A$. Then, $A$ is a field iff $B$ is a field.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.