TITLE: Show that an algebraically closed field must be infinite.
QUESTION [24 upvotes]: Show that an algebraically closed field must be infinite. 

Answer
If F is a finite field with elements $a_1, ... , a_n$ the polynomial $f(X)=1 + \prod_{i=1}^n (X - a_i)$ has no root in F, so F cannot be algebraically closed.
My Question
Could we not use the same argument if F was countably infinite? Couldn't we say that if F was a field with elements $a_1, a_2, ... $ then the polynomial $f(X) = 1 + \prod_{i=1}^{\infty} (X - a_i)$ does not split over F?
Thank you in advance

REPLY [10 votes]: Since there are a countable number of polynomials over $\mathbb Q$, the roots of those polynomials are also countable (countable union of finite sets is countable).
The sum and product of elements algebraic over $\mathbb Q$ are both algebraic over $\mathbb Q$.
An element which is algebraic over an algebraic extension of $\mathbb Q$ is algebraic over $\mathbb Q$.
Therefore the algebraic closure of $\mathbb Q$ (which we can prove exists and is unique up to isomorphism - eg realised within $\mathbb C$) is countable.
The same is true of any countable field, for similar reasons. Since a countable algebraically closed field exists, one cannot prove that it does not.