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# 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. ∎

## Mathematics Subject Classification

12F05*no label found*

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