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 $n$

Question

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.

Answer 1

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.