Ring Inside an Algebraic Field Extension 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?
 A: 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

*$F[a]=F(a)$

A: 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.
A: 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.
