The subring of a field extension contains the field, is a subfield of the field extension

Question

I am trying to prove that if $Q$ is an algebraic extension of $F$ ($F$ is a field), if $R$ is a subring of $Q$, and we have $F\subseteq R$, then $R$ is a subfield of $Q$.

My first approach was to show every element has an inverse. But there was no result.

Any help would be great.

Answer 1

Hint:

You just have to show that, for any $x\in Q$, the subring $F[x]\subset Q$ is a field.

Now observe that $F[x]$ is a finite-dimensional $F$-vector space, which is an integral domain.

Consider, for any $y\in F[x]$, $y\ne 0$, the multiplication-by-$y$ map on $F[x]$: this is an injective $F$-linear map. What can you conclude about this map?

Answer 2

If $a\in R$ and $f(a)=0$ with non-trivial $f\in F[X]$, then $f(X)=c(Xg(X)-1)$ for some $g\in F[X]$ and $c\in F^\times$. Then $b:=g(a)\in R$ and $c(ba-1)=f(a)=0$, so $ba=1$.