- Prove that a finite field cannot be algebraically closed.
Solution
1. Let \(F\) be a finite field and define\[ g(T)=1+\prod_{a\in F}(T-a)\in F[T]. \]Then \(g\) is nonconstant.
If \(|F|=q\), the product has degree \(q\), so \(g\) also has degree \(q\ge1\).
2. The polynomial \(g\) has no root in \(F\).
For any \(b\in F\), one factor in the product is \(b-b=0\). Hence \[ g(b)=1+0=1\neq0. \] Thus no element of \(F\) is a zero of \(g\).
3. Therefore \(F\) is not algebraically closed.
An algebraically closed field has a root for every nonconstant polynomial in one variable. By 1 and 2, \(g\in F[T]\) is nonconstant and has no root in \(F\). Hence \(F\) is not algebraically closed.