finite field cannot be algebraically closed

Theorem.

A finite field cannot be algebraically closed.

Proof.

The proof proceeds by the method of contradiction. Assume that a field $F$ is both finite and algebraically closed. Consider the polynomial $p(x)=x^{2}-x$ as a function from $F$ to $F$ . There are two elements which any field (in particular, $F$ ) must have — the additive identity $0$ and the multiplicative identity $1$ . The polynomial $p$ maps both of these elements to $0$ . Since $F$ is finite and the function $p\colon F\to F$ is not one-to-one, the function cannot map onto $F$ either, so there must exist an element $a$ of $F$ such that $x^{2}-x\not=a$ for all $x\in F$ . In other words, the polynomial $x^{2}-x-a$ has no root in $F$ , so $F$ could not be algebraically closed. ∎

Title	finite field cannot be algebraically closed
Canonical name	FiniteFieldCannotBeAlgebraicallyClosed
Date of creation	2013-03-22 16:29:09
Last modified on	2013-03-22 16:29:09
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	6
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 12F05
Related topic	AlgebraicClosureOfAFiniteField