In a commutative ring, can I say

An ideal $\mathfrak q$ in a ring $A$ is primary if $\mathfrak q \neq A $ and if $ xy \in \mathfrak q \Rightarrow $ either $ x \in \mathfrak q$ or $y^n \in \mathfrak q $ for some $n > 0$ or either $ y \in \mathfrak q$ or $x^m \in \mathfrak q $ for some $m > 0$.

Why I am asking this question is if I use the above definition, and use the ideal $P^2$ in the question Is each power of a prime ideal a primary ideal?, then $\bar{x}\bar{y} = \bar{z}^2 \in P^2 \text{ and } \bar x^2 \in P^2 $ gives a contradiction to the fact that $P^2$ is not primary.

Where my understanding went wrong?


You’ve misunderstood the effect of commutativity. It’s not either ... or: it’s both ... and. Let me repeat the definition:

A proper ideal $Q$ of a commutative ring $A$ is primary if whenever $xy\in Q$, then either $x\in Q$, or $y^n\in Q$ for some $n>0$.

Now suppose that some $xy\in Q$. $A$ is commutative, so $yx\in Q$ as well, and the definition then requires that either $y\in Q$, or $x^n\in Q$ for some $n>0$. In other words, the definition actually implies that $xy\in Q$ iff both

  • either $x\in Q$, or $y^n\in Q$ for some $n>0$, and
  • either $y\in Q$, or $x^n\in Q$ for some $n>0$.

An equivalent formulation that may be clearer: if $xy\in Q$, and $x,y\notin Q$, then there are integers $m,n>1$ such that $x^m,y^n\in Q$. $P^2$ fails this condition: $\bar x,\bar y\notin P^2$, and no power of $\bar y$ belongs to $P^2$.

  • $\begingroup$ so that should be $"and"$ this is what I missed and misunderstood, thanks for your answer Brian, thanks alot. $\endgroup$ – Ram Dec 25 '12 at 9:55
  • $\begingroup$ @Ram: You’re very welcome. $\endgroup$ – Brian M. Scott Dec 25 '12 at 9:58
  • $\begingroup$ @Scott, can you outline the proof how "and" is derived from the definition? And lets say I have an ideal $\mathfrak a $ such that $ xy \in \mathfrak a $ and $ x^n \notin \mathfrak a $ $\forall n > 0$ and $ y^m \in \mathfrak a$ for some $ m>0 $, this should not be primary ideal, I mean such a primary ideal doesn't exist, because of your statement above, how to prove that? can you give me some hints? $\endgroup$ – Ram Dec 30 '12 at 8:56
  • $\begingroup$ Sorry to trouble you Scott, now I am able to derive both conditions from the other definition i.e., all zero divisors are nilpotent. $\endgroup$ – Ram Dec 30 '12 at 11:33
  • $\begingroup$ @Ram: So it’s okay now? (I meant to get back to your other question sooner, but I forgot.) $\endgroup$ – Brian M. Scott Dec 30 '12 at 11:36

I am posting my explanation as an answer since I couldn't sum it up in a single comment.

@Brian, thanks for coming back, I think its okay now, let me explain.

@Scott, can you outline the proof how "and" is derived from the definition?

This is (and condition in your answer) is direct consequence of the definition, $\mathfrak a $ is primary iff $ A/ \mathfrak a \neq 0$ and every zero divisor in $ A/ \mathfrak a$ is nilpotent.

So, assume $\mathfrak a$ is primary and $ xy \in \mathfrak a$ $\Rightarrow$ $ \overline x$ is zerodivisor in $ A/ \mathfrak a$ since $ \overline {xy} = 0$, hence $ \overline x$ should be nilpotent so $ x^m \in \mathfrak a$ for some $ m \in \mathbb Z $ and similarly $ y^n \in \mathfrak a$ for some $ n \in \mathbb Z $ provided $ x,y \notin \mathfrak a$ and if either $ x $ or $ y \in \mathfrak a$ then both are no longer zero divisors in $A/\mathfrak a$ so it's not necessary that $ x^m $ should be in $\mathfrak a$ if $ y \in \mathfrak a$ or vice versa.

And coming to my question, if $ x^n \notin \mathfrak a$ for all $n > 0$ then $\mathfrak a$ is primary if $ y \in \mathfrak a$ else it is not primary.


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