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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?

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marked as duplicate by user26857 abstract-algebra Jan 6 at 8:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new 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
up vote 14 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.

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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
  • $F[a]=F(a)$
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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.

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