I remember a result that states that $\Bbb C$ as a field can be realized as a quotient of the ring $\displaystyle \prod_{p~\text{prime}} \Bbb F_p$ or something similar. Does anyone have the precise statement and/or references?


1 Answer 1


The true statement is the following : let $\mathcal{U}$ be a free ultrafilter on the set of primes $P$. Then $\mathbb{C}$ is isomorphic to the quotient of $R=\displaystyle\prod_{p\in P}\overline{\mathbb{F}_p}$ by the ideal $I=\{x\in R\mid \{i\in P\mid x_i=0\}\in\mathcal{U}\}$; where $\overline{\mathbb{F}_p}$ is the algebraic closure of $\mathbb{F}_p$. This quotient is also denoted $\displaystyle\prod_{p\in P}\overline{\mathbb{F}_p}/\mathcal{U}$.

Indeed it's easy to prove, by Los's theorem that this quotient is an algebraically closed field of characteristic $0$; and a bit of combinatorics shows that this has cardinal $2^{\aleph_0}$.

Now a theorem of model theory ensures that any two algebraically closed fields of characteristic $0$ and of the same uncountable cardinality are isomorphic; so the conclusion follows.

I'm adding a little unnecessary note because of another answer to this question (that was deleted): the structure of the maximal ideals of an infinite poduct of rings is very hard in general, but very easy for fields: if $k_i, i\in I$ are fields, then the maximal ideals of $\displaystyle\prod_{i\in I}k_i$ are precisely those given by some ultrafilter on $I$ as in what precedes. Hence there are $|I|$ many maximal ideals for a finite $I$ (the quotients are the $k_i$) and $2^{2^{|I|}}$ for an infinite $I$. This also works for bilateral ideals of division rings


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