Here is a question I think should be elementary but I am having trouble thinking about it.

Let $A$ be a commutative, unital ring, and let $I,J,K$ be ideals of $A$.

We have an inclusion $I\cap K + J \subset (I + J)\cap (K+J)$, which in general can be proper, though they will have the same radical.

On the other hand, if $I\subset K$, then $I\cap K + J = I + J = (I+J)\cap (K+J)$.

What if in place of $I,K$ we have a countable descending sequence of ideals $I_1\supset I_2\supset I_3\supset\dots$?

Does $J + \bigcap I_i = \bigcap (J + I_i)$?

Another way to ask the question is this: does taking the intersection of a descending sequence of ideals $(I_i)$ commute with taking the quotient by an arbitrary ideal $J$? I.e. does the image of $\bigcap I_i$ in $A/J$ equal the intersection in $A/J$ of the images of each $I_i$?

If it's true, what's the proof? If it's not true, is it true under some reasonable hypotheses (e.g. $A$ is noetherian)?


(A partial answer in case my thinking on additional conditinons doesn't pan out quickly.)


Let $R=\prod_{i=1}^\infty F$ for a field $F$, and let $I_i$ be the ideal which is the set of elements which is zero on coordinates $j$ for $1\leq j\leq i$. This creates a descending chain of ideals, each of which is a finite intersection of maximal ideals of the form $M_i$ given by the set of elements zero on coordinate $i$.

Let $J$ be a maximal ideal containing $\bigoplus_{i=1}^\infty F$. It's well known this is a maximal ideal distinct from any of the $M_i$. We claim that $I_i\nsubseteq J$ for any $i$. If it were true, then $\prod_{1\leq j\leq i} M_j\subseteq I_i\subseteq J$ would imply that one of the $M_i\subseteq J$, but this is not possible since they are distinct maximal ideals.

Then $J=J + \bigcap I_i \neq \bigcap (J + I_i)=R$

  • 1
    $\begingroup$ I was trying to figure out if modularity helps if the chain stabilizes. Also, if Noetherian, maybe primary decomposition helps? I'm not handy enough with it to see. $\endgroup$ – rschwieb Feb 21 '17 at 17:57
  • $\begingroup$ By $\prod_{i\leq j} M_j$ do you mean $\prod_{j\leq i} M_j$? $\endgroup$ – Ben Blum-Smith Feb 21 '17 at 17:59
  • $\begingroup$ @BenBlum-Smith Yes! sorry. $\endgroup$ – rschwieb Feb 21 '17 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.