# Algebraic Closure.

• Feb 18th 2006, 03:53 PM
ThePerfectHacker
Algebraic Closure.
Does there exists a finite field such is algebraically closed?

I tried of course the most obvious example $\displaystyle \mathbb{Z}_2$ but it is not algebraically closed for $\displaystyle x^2+x+1\in F[\mathbb{Z}_2]$ has no zero.

This problem seems strongly connected with number theory because it invloves prime numbers because all finite fields have order $\displaystyle p^n$.
• Feb 18th 2006, 10:38 PM
rgep
No finite field is algebraically closed. Consider the field of q elements (q some prime power). Then every field element satisfies the equation $\displaystyle X^q = X$, and so the polynomial $\displaystyle X^q - X - 1$ has no roots in the field.
• Feb 19th 2006, 09:55 AM
ThePerfectHacker
(For simplicity sake I am working with $\displaystyle \mathbb{Z}_p$ only)

1)Every finite field has form $\displaystyle \mathbb{Z}_{p}$.
2)Assume that $\displaystyle \mathbb{Z}_{p}$ is closed.
3)Then $\displaystyle x^2-a\in \mathbb{Z}_{p}[x],\forall a\in\mathbb{Z}_{p}$
4)Thus, $\displaystyle x^2-a\equiv 0 \mod (p)$.
5)Thus, $\displaystyle x^2\equiv a \mod (p)$.
6)Thus, $\displaystyle (a/p)=1$ for all $\displaystyle a\in\mathbb{Z}_p, a\not =0$ (Legendre symbol).
7)Thus, $\displaystyle \sum^{p-1}_{k=1}(k/p)=p-1$.
8)But, $\displaystyle \sum^{p-1}_{k=1}(k/p)=0$ (a theorem)
10)Thus, no finite field of form $\displaystyle \mathbb{Z}_{p}$ is algebraically closed.
Now perhaps it can be completed for all field of $\displaystyle \mathbb{Z}_{p^n}$