# Infinite product of fields

The main source of inspiration for this question is this excerpt

Recall: An ultrafilter on the set X gives you a maximal ideal in the ring of all real-valued functions, and these are the only prime ideals.

from this MathOverflow answer of Tom Goodwillie. (In the same thread there is an outstanding answer of our friend Georges Elencwajg.)

Let $X$ be an infinite set, for each $x$ in $X$ let $K_x$ be a field, and let $A$ be the product of the $K_x$.

Here is what I'd like to know:

$(1)$ What are the maximal ideals of $A$, and what are the corresponding residue fields?

$(2)$ What are the non-maximal prime ideals of $A$, and what are the corresponding residue domains?

$(3)$ What are the non-prime primary ideals of $A$, and what are the corresponding residue rings?

$(4)$ How can one describe the Zariski topology on the prime spectrum of $A$? In particular, does this topology depend only on $X$, or does it really depend on the $K_x$?

For $f$ in $A$ and $x$ in $X$, write $f(x)$ for the $x$ component of $f$, and $f^{-1}(0)$ for the set of all $x$ in $X$ such that $f(x)=0$.

Here are the few things I believe I know:

$(5)$ To each $x$ in $X$ is attached an obvious maximal ideal $\mathfrak m_x$ with residue field $K_x$: the set of all $f$ vanishing at $x$.

$(6)$ More generally, to each ultrafilter $U$ on $X$ is attached the set, which is in fact a prime ideal, of all $f$ such that $f^{-1}(0)$ is in $U$. Then (5) corresponds to the principal ultrafilters. But I have no idea about the residue domain associated with a non-principal ultrafilter.

I believe that the answer to (2) is that all primes are maximal.

I (perhaps naively) expect all primary ideals to be prime, and thus maximal.

So, as you see, I need tons of help! Thank you very much in advance!

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Here is a generalization of one of the subquestions: Does the spectrum of a product of a family of (commutative) rings depend only (up to homeomorphism) on the spectra of the members? – Pierre-Yves Gaillard Dec 3 '11 at 12:35
Excerpt from Mumford's Red Book (p.74): "Example G. $\text{Spec}(\prod_{i=1}^\infty k)$, $k$ a field. Those familiar with ultrafilters and similar far-out mysteries will have no trouble proving that this topological space is the Stone-Cech compactification of $\mathbb Z_+$. Logicians assure us that we can prove more theorems if we use these outrageous spaces." – Pierre-Yves Gaillard Dec 3 '11 at 18:52
I was just about to ask the same question myself! To answer (6), the one uses Łoś's theorem to characterise the residue field after quotienting by the maximal ideal corresponding to an ultrafilter. It is an example of an ultraproduct. – Zhen Lin Mar 28 '12 at 1:05
Dear @ZhenLin: Thanks for your comment. I think it would make a great answer... – Pierre-Yves Gaillard Mar 28 '12 at 3:54

You are absolutely right when you write "I believe that the answer to (2) is that all primes are maximal", Pierre-Yves :

Proposition 1 The ring $A$ is zero-dimensional.
Proof Recall that a commutative ring $R$ is called von Neumann if for all $r\in R$ there exists an $s\in R$ such that $r=sr^2$.
It is clear that a field is von Neumann and that a product of any family of von Neumann rings is von Neumann, so that your ring $A$ is von Neumann.
But a von Neumann ring has Krull dimension zero: indeed, if ${\mathfrak p } \subset A$ is prime, the quotient $D:=A/{\mathfrak p }$ is clearly von Neumann.
And to end the proof just note that a von Neumann domain $D$ is a field: if $d\neq 0\in D$, the relation $d=rd^2$ implies $1=rd$ so that $d$ is invertible.

Corollary For every prime (=maximal) ideal $\mathfrak p\subset A$ the residue field is $\kappa (\mathfrak p)=A_{\mathfrak p}$
Proof Since the ring $A$ is reduced, so is its localization, the zero-dimensional (cf.Proposition) ring $A_{\mathfrak p}$.
But then ${\mathfrak p} A_{\mathfrak p}$ is the nilpotent radical of that reduced ring $A_{\mathfrak p}$ and so must be zero: ${\mathfrak p} A_{\mathfrak p}=0$ .
Hence we get the (rather unusual !) formula $\kappa (\mathfrak p)=A_{\mathfrak p}/ \mathfrak pA_{\mathfrak p}=A_{\mathfrak p}/0=A_{\mathfrak p}$ .

And your other hope is true too:
Proposition 2 Every primary ideal of $A$ is maximal.
Indeed , since the ring $A$ is von Neumann it is absolutely flat (Atiyah-Macdonald, Introduction to Commutative Algebra, Chapter 2, Exercise 27 ) and thus every primary ideal of $A$ is maximal. (Same reference, Chapter 4, Exercise 3)

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Mind-blowing!!! Thanks a lot! – Pierre-Yves Gaillard Dec 3 '11 at 12:29

For (1), there will be variety results depended on the fields.

（1）If $A=\prod_{i\in \mathbb{N}}\mathbb{F}_p$ then all the residue fields are equal to $\mathbb{F}_p$. (since $x^p=x$)

(2) If $A=\prod_{p}\mathbb{F}_p$, then for any prime $\mathfrak{m}$ containing $\oplus_p\mathbb{F}_p$ , then $\mathbb{Q}\subseteq A/\mathfrak{m}$ and the cardinality of $A/\mathfrak{m}$ is $\aleph$, thus $A/\mathfrak{m}$ is not an algebraic extension. It has transcendental elements over $\mathbb{Q}$.

Proof about the cardinality: define a map $\varphi$ from $\{0,1\}^{\mathbb{N}}$ to $A/\mathfrak{m}$ by sending $(a_i)_{i=1}^{\infty}$ to $(x_k)$, $x_k=\sum_{j=1}^na_j2^{j-1}$ if $2^n-1<p_k\leq2^{n+1}-1$. Here index $k$ denotes $k$-th coordinate and $p_k$ denotes the $k$-th prime number. Then $\varphi$ is injective.

In this case, we want to ask if $A$ is a purely transcendental extension over $\mathbb{Q}$ ?

It is not true at least for some maximal ideal!

Consider the equation $x^2+1=0$. Then it can be solved in $\mathbb{F}_p$ if only if $p=1\mod 4$ or $p=2$. Since those primes are infinite, let $a=(a_{p_k})\in A$, $a_{p_k}=1$ if $p_k=1 \mod 4$, then we find a maximal ideal $\mathfrak{m}$ containing $\oplus\mathbb{F}_p$ such that $1-a\in \mathfrak{m}$. Hence $x^2+1=0$ can be solvable in $A/\mathfrak{m}$.

But also we can find some maximal ideal $\mathfrak{m}$ such that the $x^2+1=0$ can not be solvable in $A/\mathfrak{m}$.

The residue fields are always not purely algebraic extension. Since $(x^2+1)(x^2+2)(x^2-2)$ can be always solvable.

(3) If $A=\prod_{i\in \mathbb{N}}\mathbb{Q}$, the discussion about the cardinality of the residue field is similar.

Those questions relate to my recent thread the ring of Cauchy sequence.

Another way to show that $\dim A=0$:

suppose we have a proper inclusion between prime ideals, say $\mathfrak{p}\subsetneq\mathfrak{m}$, then you can find an $x\in \mathfrak{m}-\mathfrak{p}$ such that the coordinate of $x$ is either $1$ or $0$, then we must have $1-x\in\mathfrak{p}\subset \mathfrak{m}$, contradiction!

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Dear wxu: Thank you for your nice answer! – Pierre-Yves Gaillard Dec 3 '11 at 18:35