Let $\displaystyle L:K$ be an algebraic extension. Suppose that $\displaystyle R$ is a subring of $\displaystyle L$that contains $\displaystyle K$. Show that $\displaystyle R$ is a field. Give an example to show that the algebraic restriction is necessary.

Take an $\displaystyle r\in R, r \neq 0$ and $\displaystyle r \notin K$. As $\displaystyle R$ is a subring only need to show that $\displaystyle r$ has an inverse also in $\displaystyle R$ to show $\displaystyle R$ is a field. As $\displaystyle r$ is algebraic over $\displaystyle K$, there exists a minimal poly, $\displaystyle p$ say which is irreducible over $\displaystyle K$. So $\displaystyle p(r)= a_{n}r^n+...+ar_{1}+a_{0} = 0 $ with $\displaystyle a_{i}\in K$.

As $\displaystyle p$ is irreducible $\displaystyle a_{0}\neq 0$ otherwise $\displaystyle p(r)= r(a_{n}r^{n-1}+...+a_{1})=0$ but as $\displaystyle r \neq 0 \Rightarrow a_{n}r^{n-1}+...+a_{1}=0$ so $\displaystyle p$ would not be the minimal polynomial.

Now $\displaystyle p(r) \Rightarrow 1= r( -a_{0}^{-1}) (a_{n}r^{n-1}+...+a_{1})$ and so $\displaystyle -a_{0}^{-1} (a_{n}r^{n-1}+...+a_{1}) \in R$ is the inverse of $\displaystyle r$ thus $\displaystyle R$ is a field.

I think this is all OK and correct but an (annoying) friend of mine was saying oh but you haven't defined how many degrees p has. He's annoying as thats all he said and refused to tell me how to work it out, surely as nothing is known about degrees of $\displaystyle K$ you can't really know?

Also I'm struggling to think of an example...

I'm assuming I'm missing something simple but if anyone could help with these two wee bits or show I've gone wrong somewhere else I'd appreciate it!!