# 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 FiniteFieldCannotBeAlgebraicallyClosed 2013-03-22 16:29:09 2013-03-22 16:29:09 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Theorem msc 12F05 AlgebraicClosureOfAFiniteField
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