Algebraic Closure.

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February 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 $\mathbb{Z}_2$ but it is not algebraically closed for $x^2+x+1\in F[\mathbb{Z}_2]$ has no zero.

This problem seems strongly connected with number theory because it invloves prime numbers because all finite fields have order $p^n$ .
February 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 $X^q = X$ , and so the polynomial $X^q - X - 1$ has no roots in the field.
February 19th 2006, 09:55 AM
ThePerfectHacker

I was thinking about this today and I have a different approach.
(For simplicity sake I am working with $\mathbb{Z}_p$ only)

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

Now perhaps it can be completed for all field of $\mathbb{Z}_{p^n}$