22
$\begingroup$

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

$\endgroup$
1
23
$\begingroup$

We can't, because $\prod_{i=1}^\infty (X-a_i)$ is not a polynomial. Infinite combinations of finitary operations are not defined; when we do talk about them, what is really going on is that we are taking some sort of limit, but to do that we need topology to be present, and even then, the infinitary operation will be subtle. A priori, one cannot talk about an infinite sum or an infinite product of elements.

$\endgroup$
1
  • 12
    $\begingroup$ It's not even just a thing which is not a polynomial, it's not a thing at all. The notation isn't well-defined yet; it doesn't even define a formal power series. $\endgroup$ – Qiaochu Yuan Jun 10 '13 at 19:42
8
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy