The complex numbers as a quotient of $\prod_{p\in\mathcal{P}}\bar{\mathbb{F}}_p$

Question

I saw the following problem proposed in an open notebook from my institute.

Let's denote $\mathcal{P}$ the set of prime ideals and $\bar{\mathbb{F}}_p$ an algebraic closure of $\mathbb{F}_p$.

On the ring $\displaystyle \prod_{p\in\mathcal{P}}\bar{\mathbb{F}}_p$ consider a maximal ideal $M$ containing the ideal of elements $(a_p)_{p\in\mathcal{P}}$ that are $0$ for all but finitely many $p$. Prove that $\displaystyle \prod_{p\in\mathcal{P}}\bar{\mathbb{F}}_p/M\cong \mathbb{C}$

I have not been able to solve it and I am interested to see a solution. Any comments about the origin or implications of the problem will be welcome.

Answer 1

Such a field is an ultraproduct of the fields $(\overline{\mathbb{F}}_p)_{p\in \mathcal{P}}$ by a non-principal ultrafilter. So you can argue as follows:

1. Show that any maximal ideal in the product ring has the form $M_\mathcal{U} = \{(a_p)_{p\in \mathcal{P}}\mid a_p = 0\text{ for all } p\in \mathcal{U}\}$, where $\mathcal{U}$ is an ultrafilter on $\mathcal{P}$, and the condition about containing the finite support ideal is equivalent to $\mathcal{U}$ being non-principal.

2. Check that the quotient $\prod_{p\in \mathcal{P}}\overline{\mathbb{F}}_p/M_{\mathcal{U}}$ is the same as the ultraproduct $\prod_{p\in \mathcal{P}}\overline{\mathbb{F}}_p/\mathcal{U}$.

3. By Łos's Theorem, the ultraproduct is an algebraically closed field of characteristic $0$.

4. It follows from the Steinitz classification of algebraically closed fields by characteristic and transcendence degree that $\mathbb{C}$ is the unique algebraically closed field of characteristic $0$ of size $2^{\aleph_0}$ up to isomorphism.

5. So we can finish by using the fact that an ultraproduct of countable structures by a non-principal ultrafilter on a countable set always has cardinality $2^{\aleph_0}$.

It's surely possible to give an argument that looks more like algebra and less like logic in this case. But the point is that once you're familiar with these facts from logic, the result is pretty immediate.

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Edit: Giving more details on point 5. as requested in the comments. The basic task is to find a continuum-sized family of functions $(f_\alpha\colon I\to J)_{\alpha\in 2^{\aleph_0}}$, where $I$ and $J$ are countable sets, such that if $\alpha\neq \beta$, then $\{i\in I\mid f_\alpha(i) = f_\beta(i)\}$ is finite. Then putting $\mathcal{P}$ in bijection with $I$ and each $\overline{\mathbb{F}}_p$ in bijection with $J$, our functions $f_\alpha$ turn into distinct elements of the ultraproduct.

And there's a nice trick for solving this problem: Take $I = \mathbb{N}$ and $J = \mathbb{Q}$. For each $r\in \mathbb{R}$, let $f_r\colon \mathbb{N}\to \mathbb{Q}$ be a sequence of rational numbers converging to $r$. Then when $r\neq s$, $f_r(n)\neq f_s(n)$ for large enough $n$.

For more information on cardinalities of ultraproducts, there is a classic paper by Keisler.