# Ideals of infinite product rings

(Rewritten following earlier feedback...)

(1) What are the ideals of ${\mathbb Z}^{\mathbb N}$? That is, take the ring which is the product of a countable infinity of copies of $\mathbb Z$; is there a simple explicit way to describe its ideals?

More generally, suppose $R = \prod_{i \in I} R_i$ is a product of rings (unital, not necessarily commutative). We'd like to describe the left ideals of $R$ in terms of $R_i$.

I see that in a finite product all the ideals are the products of ideals in the factors. For example, if $J$ is an ideal of $R_0 \times R_1$, define $J_i = \pi_i(J)$; then $J \subseteq J_0 \times J_1$, but also if $r_i \in J_i$, then we can get $s_i \in J \cap \pi_i^{-1}(r_i)$ and we have $(r_0, r_1) = (1_{R_0}, 0)s_0 + (0, 1_{R_1})s_1 \in J$.

So only the infinite products are interesting. There we have ideals such as the elements which are zero except at finitely many coordinates. Here are the ones I can see. Suppose each $R_i$ has left ideals $J_{i0} \subseteq J_{i1}$, and $K \subseteq 2^I$ is an ideal in the power set, in the sense that it's closed under subset and finite union. Then take $J$ to be the set of elements whose coordinates are in the small ideals and occasionally in the large ideals: $J = \{r \in R: r_i \in J_{i1}, \{i: r_i \notin J_{i0}\} \in K\}$. This seems to be a left ideal in $R$.

(2) Are there any other left ideals in $R$?

• The better way to look at this problem is to find all ideals of $\prod \mathbb Z$. In this context your question is equivalent to: Is every ideal in $\prod \mathbb Z$ product of ideals in $\mathbb Z$? This is not true for infinite products. – user52045 Dec 19 '13 at 13:12
• For example, every filter on $\mathbb{N}$ gives rise to an ideal. – Martin Brandenburg Dec 19 '13 at 13:49
• @MartinBrandenburg: Is it true tho that every ideal comes from filter on index set? – user52045 Dec 19 '13 at 13:59
• No, $\prod_i n_i \mathbb{Z}$ doesn't come from a filter. In a product of fields, ideals correspond to filters on the index set. The situation of $\mathbb{Z}$ is much more complicated and probably not solvable. – Martin Brandenburg Dec 19 '13 at 14:00
• – Martin Brandenburg Nov 17 '15 at 10:01