Show that no finite field is algebraically closed.
There are just so many ways this can be proven. Here's one:
Consider a finite field $ \scriptstyle F $ with elements $ \scriptstyle a_1,a_2,\ldots,a_n $. Because the polynomial $ \scriptstyle(x-a_1)(x-a_2)\cdots(x-a_n)+1 $ is irreducible in $ \scriptstyle F $, there exists a splitting field $ \scriptstyle E $ for the polynomial over $ \scriptstyle F $ by Theorem 20.2. That is, $ \scriptstyle E $ is a proper algebraic extension of $ \scriptstyle F $, so $ \scriptstyle F $ cannot be algebraically closed.
- --Nick Rupley 00:40, 30 April 2009 (UTC)
