# Subrings of Algebraic Extentions

• Mar 14th 2012, 08:57 AM
leshields
Subrings of Algebraic Extentions
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!!
• Mar 15th 2012, 05:46 AM
Deveno
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 $\displaystyle \mathbb{Q}[\pi]$, which is isomorphic to $\displaystyle \mathbb{Q}[x]$. this is certainly a subring of $\displaystyle \mathbb{R}$, which contains $\displaystyle \mathbb{Q}$. however, it is NOT a field, because (for one), it doesn't contain $\displaystyle \frac{1}{\pi}$
• Mar 17th 2012, 02:05 PM
leshields
Re: Subrings of Algebraic Extentions
Thanks, that's great!