# If $R$ is a Noetherian ring, $I,J,Q$ are its ideals, $Q$ is primary and $IJ \subseteq Q$ , then $I \subseteq Q$ or $J^n \subseteq Q$ for some $n$

We consider a Noetherian commutative ring $$R$$ and $$I,J,Q$$ ideals of $$R$$.
If $$Q$$ is a primary ideal and $$IJ \subseteq Q$$, then we need to show that either $$I \subseteq Q$$ or that there is a positive integer $$n$$ such that $$J^n \subseteq Q$$.
I considered $$x \in I$$ and $$y \in J$$. Then $$xy \in IJ \subseteq Q$$. So, we have two cases: either $$x \in Q$$ (so $$I \subseteq Q$$), or $$x \notin Q$$.
I don't know where to go from there.
I encountered this as an exercise in my course of commutative algebra.
The definition of primary ideal that I've been taught is the following:
An ideal $$I \subset R$$ is primary, if $$xy \in I \ \ \text{and} \ \ x \notin I \Rightarrow y^n \in I$$ for some n
Any help would be greatly appreciated.

• $x\in Q$ does not imply $I\subseteq Q$, because you are not assuming $I=(x)$. In any case... what is your definition of "primary ideal"? Commented Apr 6, 2021 at 22:10
• @ArturoMagidin Edit: I've added the definition of primary ideal. Commented Apr 6, 2021 at 22:15
• Hint: J is finitely generated. If $Ι \not\subseteq Q$ then some power of all the generators of $J$ will belong to $Q$. Find the generators of $J^n$ and select some appropriate $n$. Commented Apr 6, 2021 at 22:19

In a commutative ring $$A$$, an ideal $$Q$$ is primary if and only if whenever $$xy\in Q$$, either $$x\in Q$$ or there exists $$n\gt 0$$ such that $$y^n\in Q$$.

You assume that $$IJ\subseteq Q$$. We want to show that either $$I\subseteq Q$$ or else that there exists $$n\gt 0$$ such that $$J^n\subseteq Q$$.

Your original assertion (now removed) that for $$x\in I$$ and $$y\in J$$, $$x\in Q$$ implies $$I\subseteq Q$$ is unwarranted: you only know $$x\in Q$$, and you are not assuming $$I=(x)$$, so you cannot just conclude that $$I\subseteq Q$$ from the fact that $$x\in Q$$. For all you know, sometimes you get $$x\in I$$ (for some choices of $$x$$ and $$y$$), and sometimes you'll get $$y^n\in Q$$ (for other choices), so that you don't always get that the first factor is in $$Q$$. So you cannot argue like that.

Instead, let's consider two possibilities: either $$I\subseteq Q$$, or $$I\not\subseteq Q$$. If $$I\subseteq Q$$ then we are done. So we may assume that $$I\not\subseteq Q$$. Let $$x\in I$$ be such that $$x\notin Q$$.

Since $$R$$ is noetherian, $$J$$ is finitely generated; say $$J=(y_1,\ldots,y_m)$$.

Now, each of $$xy_i\in Q$$, but $$x\notin Q$$. Therefore, since $$Q$$ is primary, there exists $$n_i\gt 0$$ such that $$y_i^{n_i}\in Q$$. Let $$N=n_1+\cdots+n_m+1$$. I claim that $$J^N\subseteq Q$$.

To prove that, consider an arbitrary element $$r_k$$ of $$J$$, which is of the form $$r_k=a_{1k}y_1+\cdots + a_{mk}y_m$$ with $$a_{ik}\in R$$. Now look at the element $$r_1\cdots r_N$$.

Prove that each summand in the obvious expression of $$r_1\cdots r_N$$ will contain some $$y_j^{n_j}$$, and hence be an element of $$Q$$. Conclude that $$J^N\subseteq Q$$, as desired.