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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)?

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

Counterexample

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

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    $\begingroup$ 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

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