Subring of a field extension is a subfield For the first part, 

I am trying to show that any subring of $E/F$ is a subfield where $E$ is an extension of field $F$ and all elements of $E$ are algebraic over $F$. 

My solution is to just show an element of a subring has an inverse and we will be done. Let $R$ be the subring and $r\in R$. Since $r$ is algebraic over $F$ we can find a polynomial $p(x)=a_0+a_1x+a_2x+\cdots + a_nx^n$ such that $p(r)=0$ where $a_i \in F$. Using this expression I explicitly computed what an inverse of $r$ must look like. 

However, the second part of the question asks to prove that any subring of a finite dimensional extension field $E/F$ is a subfield.

Is the second part of the question redundant? I didn't use whether the extension was finite or infinite for my solution in the first part. Will both the parts have the same solution that I wrote above?

Let $R$ be a ring such that $F\subset R\subset E$ and $\alpha\neq 0$ an element of $R$. We can as well suppose $R=F[\alpha]$. Then multiplication by $\alpha$ is an endomorphism of the finite dimensional $F$-vector space $R$. This endomorphism is injective because $R$ is an integral domain (subring of a field), hence it is surjective. In particular, $1$ is attained, i.e. there exists an element $\beta\in R$ such that $\beta\alpha=1$ – in other words, $\alpha$ has an inverse in $F[\alpha]$.
