Let $R$ and $S$ be fields. Consider the ring $R \times S$.

Since $R$ (resp. $S$) is a field, the ideals of $R$ (resp. $S$) are $\{0_R\}$ and $R$ (resp. $\{0_S\}$ and $S$). Since every ideal of the direct product of rings is the direct product of ideals of these rings, we get the following result using induction:

For fields $R_1,...,R_k$ and $S \subset \{1,...,k\}$, let $I(S) = \{(r_1,...,r_k) \in R_1 \times ... \times R_k : r_s = 0$ for all $s \in S \}$. Then an ideal $I$ of $R_1 \times ... R_k$ equals to $I(S)$ for some $S \subset \{1,...,k\}$.

Now what happens if we consider an infinite product of fields ?

I believe that the set $\{ (r_1, r_2,...,r_n,0,0,...) \in \Pi_{i=0}^{\infty} R_i : \exists n \in \mathbb{N}$ and $r_m = 0$ for all $m > n \}$ is an ideal of $\Pi_{i=0}^{\infty} R_i$ and is not of the form $I(S)$ for some $S \subset \mathbb{N}$.

Are there other examples of ideals of $\Pi_{i=0}^{\infty} R_i$ that are not of the form $S(I)$ for some $S \subset \mathbb{N}$?

Thank you very much for your help!

  • 1
    $\begingroup$ There are certainly other ideals. For any subset $S$ of $\mathbb{N}$, you can take the intersection of $I\left(S\right)$ with your "finitely supported" ideal, and get a new ideal (at least when $S$ is infinite). Or take the sum instead of the intersection. $\endgroup$ Commented Jan 15, 2019 at 21:21
  • $\begingroup$ Thank you ! But I wonder if my example is actually working. Obviously $S$ is not finite here but it could be an infinite subset of $\mathbb{N}$ no? $\endgroup$
    – riri92
    Commented Jan 15, 2019 at 21:23
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    $\begingroup$ @riri92 - You can take $S$ to be infinite. It's a minor tweak to my answer poster below. $\endgroup$ Commented Jan 15, 2019 at 21:45
  • 1
    $\begingroup$ @riri92: Yes, your example is actually working. $\endgroup$ Commented Jan 15, 2019 at 21:52

2 Answers 2


Your example is correct. More generally, ideals in an arbitrary product $R=\prod_{i\in J} R_i$ of fields correspond to what are called (Boolean) ideals (of sets) on the index set $J$. An ideal on a set $J$ is a collection $D$ of subsets of $J$ with the following properties:

  • $\emptyset\in D$.
  • If $A,B\in D$ then $A\cup B\in D$.
  • If $A\in D$ and $B\subseteq A$, then $B\in D$.

Intuitively, the idea is that the elements of $D$ are sets that are "small" in some sense.

Given an ideal $D$ on the set $J$, you can define an ideal $I(D)$ in the ring $R$ which consists of all elements whose set of nonzero coordinates is an element of $D$. For example, if $D$ is the collection of all finite subsets of $J=\mathbb{N}$, then the ideal $I(D)$ is exactly your example. But there are many other ideals of sets: you can take some random infinite collection of subsets of $J$ and consider the ideal they generate (i.e., their closure under the three conditions above). Or, more systematically, many reasonable notions of "smallness" give ideals of sets. You can find several examples on Wikipedia.

Conversely, using the fact that the $R_i$ are fields, you can show that every ideal in $R$ is of the form $I(D)$ for some ideal $D$ on $J$. Namely, you can take $D$ to be the set of subsets $A\subseteq J$ such that some element of your ideal is nonzero on every coordinate in $A$.

When $J$ is finite, given any ideal $D$ on $J$ you can just take the union of all the elements of $D$ to get a set $S\in D$ which contains all the elements of $D$. (We have $S\in D$ since $D$ is closed under finite unions.) So, in this case $I(D)$ is just what you called $I(S)$. The difference when $J$ is infinite is that an ideal $D$ on $J$ may not be closed under arbitrary unions, so there may be no single element of $D$ that contains all other elements of $D$.


You can expand on your example to build one. Let $R_i = \mathbb{R}$ for all positive integers $i.$ Let $s = \prod_i R_i.$ Then $s$ is the set of all sequences of real numbers. Moreover, $s$ is a ring under coordinate-wise addition and multiplication. Let $F$ be the subset of $s$ consisting of those sequences with all but finitely many terms equal to $0.$ It is not hard to show that $F$ is an ideal in $s.$ A little thought shows that $F= \oplus_i R_i.$ If I'm awake, I believe that the ideals of which you speak are of the form $I(S)$ for some subset $S$ of the positive integers. If $S=\{n_1, n_2, ... , n_k\}$ for distinct integers $n_i,$ then $I=\prod_iJ_i,$ where $J_i = R_i$ if $i\in S,$ and $J_i=0$ otherwise, is an ideal of $s.$ In fact, each of these ideals is contained in $F,$ but none is actually equal to $F.$


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