# Every algebraically closed field is infinite

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?

• No, if $F=\Bbb F_2$, for example, $P(X)=X^2+X+1$. Actually, the degree of $P$ is $|F|$. – ajotatxe Oct 9 '18 at 14:05
• It takes the value $1$ at every element of $F,$ but as a polynomial, its degree is the cardinality of $F$. – saulspatz Oct 9 '18 at 14:09
• 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 – Alessandro Codenotti Oct 9 '18 at 14:10
• So, we do not say that $P\left(x\right)=Q\left(x\right)$ for all $x\in F$ implies $P=Q$? – Joshua Tilley Oct 9 '18 at 14:16
• @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 $|F|$. – Andrés E. Caicedo Oct 9 '18 at 14:23

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.