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Take the following proof that every algebraically closed field is infinite.

Suppose we have a finite, algebraically closed field $F$. The polynomial $$P\left(X\right)=\prod _{f\in F}\left(X-f\right)+1$$ has no roots. It is clear that $P\in F\left[X\right]$, and that $P$ has no roots. However, to show that $F$ is not algebraically closed, we need a polynomial with degree at least $1$. Isn't this polynomial degree $0$? It is equal to $1$ identically?

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  • $\begingroup$ No, if $F=\Bbb F_2$, for example, $P(X)=X^2+X+1$. Actually, the degree of $P$ is $|F|$. $\endgroup$ – ajotatxe Oct 9 '18 at 14:05
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    $\begingroup$ It takes the value $1$ at every element of $F,$ but as a polynomial, its degree is the cardinality of $F$. $\endgroup$ – saulspatz Oct 9 '18 at 14:09
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    $\begingroup$ Especially for finite fields you should not confuse an element of $F[X]$, which is a formal sequence of coefficents, with the polynomial function we often think about instead! For example over $\Bbb F_2$ the polynomial $X^4+X^2$ evaluates to zero at all elements of the field, but it is not the zero polynomial $\endgroup$ – Alessandro Codenotti Oct 9 '18 at 14:10
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    $\begingroup$ So, we do not say that $P\left(x\right)=Q\left(x\right)$ for all $x\in F$ implies $P=Q$? $\endgroup$ – Joshua Tilley Oct 9 '18 at 14:16
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    $\begingroup$ @Joshua No, but almost. The implication holds if $P(x)=Q(x)$ for more values of $x$ than the degrees of $P,Q$. In the example in the question, $Q$ is 1, of degree 0, but $P$ is of degree $|F|$. $\endgroup$ – Andrés E. Caicedo Oct 9 '18 at 14:23
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Alternatively, if a field $F$ is finite and has $n$ elements, then all elements of $F$ are roots of $x^n-x$ and so the polynomial $x^n-x+1$ has no roots in $F$. Thus, $F$ is not algebraically closed.

This is essentially the same proof, since $x^n-x=\prod _{f\in F} (x-f)$, but perhaps it's psychologically clearer.

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