algebraic extension proof

morganfor · #1 $Old$ December 4th, 2008, 12:56 AM

I need help with this proof. Let E be an algebraic extension of field F. If R is a ring and F is contained in R is contained in F, show that R must be a field.
Thanks!

NonCommAlg · #2 $Old$ December 4th, 2008, 01:19 AM

Quote:

Originally Posted by morganfor $View Post$

I need help with this proof. Let E be an algebraic extension of field F. If R is a ring and F is contained in R is contained in E, show that R must be a field.
Thanks!

let $0 \neq r \in R$ and $p(x)=x^n + a_1x^{n-1} + \cdots + a_{n-1}x+a_n \in F[x]$ be the minimal polynomial of $r.$ then $a_n \neq 0$ and thus: $r(-a_n^{-1}r^{n-1}-a_n^{-1}a_1r^{n-2} - \cdots - a_n^{-1}a_{n-1})=1. \ \Box$

morganfor · #3 $Old$ December 4th, 2008, 11:28 AM

Thanks! I get it all the way up until the last part. How did you get the final polynomial and how is it that when you multiply it by r you get 1?

ThePerfectHacker · #4 $Old$ December 4th, 2008, 01:30 PM

Quote:

Originally Posted by morganfor $View Post$

Thanks! I get it all the way up until the last part. How did you get the final polynomial and how is it that when you multiply it by r you get 1?

By moving $a_n$ over to the other side and dividing by $a_n$ both sides.