# Thread: Subrings of Algebraic Extentions

1. ## Subrings of Algebraic Extentions

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

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

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

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

2. ## Re: Subrings of Algebraic Extentions

your proof seems fine to me. one caveat, there is nothing saying L might not EQUAL K, so you need a one-line dispatch for that. the degree of L over K isn't relevant, and without it given, you can't say much about the degree of the minimal polynomial of r.

as for a counter-example, consider $\mathbb{Q}[\pi]$, which is isomorphic to $\mathbb{Q}[x]$. this is certainly a subring of $\mathbb{R}$, which contains $\mathbb{Q}$. however, it is NOT a field, because (for one), it doesn't contain $\frac{1}{\pi}$

3. ## Re: Subrings of Algebraic Extentions

Thanks, that's great!