$\rm Spec$ of an infinite product

Question

I know that the spectrum of an infinite product of rings is not an infinite disjoint union of spaces. I always see this fact being proven using compactness, but I would like to understand where the argument of the finite case fails.

I started reasoning under the assumption that the ideals in a product of rings $R:=\prod _{i\in I}R_i$ , where the cardinality of $I$ is not finite, are exactly the subsets $\prod _{i\in I}J_i$, with $J_i\subset R_i$ ideals; but at this point the argument for the finite case seems to work fine, so probably the initial assumption is wrong. I'm not sure though: the product of ideals is an ideal in $R$, so I just need to know if any ideal in $R$ is of this form. Let $J\subset R$ be an ideal, and let $J_i\subset R_i$ be the image of $J$ under the corresponding projection. Clearly $J\subset J':=\prod _{i\in I}J_i$. Let $j:=(j_i)_{i\in I}$ be an element of $J'$: then for every $i$, the element of $J'$ whose coordinates are all $0$, except for the $i$-th one which is $j_i$, is contained in $J$. Can we say that $j\in J$ then? I'm not convinced, because the sum of an infinite number of terms does not make sense in general, but here maybe it does: the sum is defined coordinatewise, and at every coordinate we have a sum of all zeros but a term, that can be computed.

Answer 1

As KReiser says in the comments, there are already some questions and answers here discussing what $\text{Spec}$ of an infinite product looks like; I don't consider this question a duplicate of those because the question here is specifically why the obvious argument does not work. This is the step that doesn't work:

Let $j:=(j_i)_{i\in I}$ be an element of $J'$: then for every $i$, the element of $J'$ whose coordinates are all $0$, except for the $i$-th one which is $j_i$, is contained in $J$. Can we say that $j\in J$ then?

No. If $R = \prod_i R_i$ is any infinite product of nonzero rings, it has an ideal $J$ consisting of all elements of $R$ whose coordinates vanish except at finitely many entries. The projection $J_i$ of $J$ to every factor $R_i$ is all of $R_i$ by construction, so $J' = (1)$ is the unit ideal. But $J$ is never the unit ideal itself, because it doesn't contain any element whose coordinates are nonzero infinitely often, and in particular never contains the identity. It is true that one can make sense of the "infinite sum" $\sum j_i$ here but ideals are simply not required to be closed under such sums.

The prime ideals containing $J$ are where the "interesting" part of $\text{Spec } R$ is, involving ultrafilters and so on as explained in the linked answers.

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There is also a topological analogy here here which may be helpful motivation. Let $X$ be a topological space and consider the ring $C_b(X, \mathbb{C})$ of bounded continuous functions $X \to \mathbb{C}$. This ring is naturally a commutative unital $C^{\ast}$-algebra, so by the commutative Gelfand-Naimark theorem it can be identified with the ring of continuous functions on its spectrum, which is a compact Hausdorff space. The "obvious" part of the spectrum consists of the maximal ideals corresponding to evaluation at each $x \in X$; if $X$ is noncompact then there must be other points in the spectrum to guarantee compactness. And indeed the spectrum turns out to be exactly the Stone-Cech compactification $\beta X$.

In particular, if $X$ is a discrete space then $C_b(X, \mathbb{C})$ is naturally a subring of the infinite product $\prod_{x \in X} \mathbb{C}$ (the two turn out to have the same spectrum), and $\beta X$ turns out to be exactly the space of ultrafilters on $X$.