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 elementswhich any field (in particular, $F$ ) must have — the additive identity $0$ and the multiplicative identity $1$ . The polynomial $p$ maps both ofthese 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, sothere must exist an element $a$ of $F$ such that $x^{2}-x\ot=a$ forall $x\\in F$ . In other words, the polynomial $x^{2}-x-a$ has no rootin $F$ , so $F$ could not be algebraically closed.∎

单词	FiniteFieldCannotBeAlgebraicallyClosed
释义	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 elementswhich any field (in particular, $F$ ) must have — the additive identity $0$ and the multiplicative identity $1$ . The polynomial $p$ maps both ofthese 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, sothere must exist an element $a$ of $F$ such that $x^{2}-x\ot=a$ forall $x\\in F$ . In other words, the polynomial $x^{2}-x-a$ has no rootin $F$ , so $F$ could not be algebraically closed.∎
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