Let $A$ be a commutative ring, $M$ be an $A$-module and $N \leq M$. There are two definitions of primary modules:
1) $M/N$ is coprimary (i.e., every zero divisor is nilpotent);
2) $\text{Ann}_A(N)$ is a primary ideal of $A$.
The first definition is the usual one, however the second one appears in some linear algebra books (for instance, Roman's "Advanced Linear Algebra").
Of course, definition 1) is just for submodules of $M$ while definition 2) is for any module $N$. So what I'm really asking is if there is any reasonable way of matching these two definitions in the sense that $N$ satisfies 2 iff there exists some (special) module $M$ (not any $M$) associated to $N$ that makes 1) true.
I'm interested mainly in the case of $A$ Noetherian or even the case of $A$ being a PID.
Thanks in advance.
I can't comment otherwise I would. But I think it must be a typo, in the sense that (2) is supposed to be: $\mathrm{Ann}_A(M/N)$ is a primary ideal of $A$.
Then it seems rather tautological to prove that (1) implies (2) supposed to be that $R$ is noetherian and $M$ is finitely generated. Indeed, we may assume that $N=0$. Let $\mathfrak{p}$ be the unique associated prime of $M$. Then $\mathfrak{p}$ is minimal over $I=\mathrm{Ann}_A(M)$, in particular an associated prime of $A/I$. To prove that $\mathfrak{p}$ is the unique associated prime of $A/I$ it suffices to prove that any $x\notin\mathfrak{p}$ is a nonzero divisor on $A/I$. But suppose that $y\in A\setminus I$ such that $xy\in I$. Then we can find $m\neq 0$ such that $ym\neq 0$. But clearly $xym=0$ which implies that $x$ is a nonzero divisor on $M$ - a contradition to the fact that $M$ is coprimary.
The other implication "(2) implies (1)" does not always hold put it holds if in addition to the previous assumptions $M$ has no embedded primes. Indeed, let $\mathfrak{p}$ be the unique associated prime of $A/I$ then $\mathfrak{p}$ is also associated to $M$. Suppose that $x\notin\mathfrak{p}$ with $xm=0$ for some $m\neq 0$. Then $x\in\mathfrak{q}$ for some associated prime $\mathfrak{q}$ of $M$ different from $\mathfrak{p}$. But since $\mathfrak{p}=\sqrt{I}\subseteq\mathfrak{q}$ it follows that $\mathfrak{q}$ is an embedded prime. Therefore such an $x$ does not exists and it follows that every element not in $\mathfrak{p}$ is a nonzerodivisor on $M$ which means that $M$ is coprimary.
Even if $A$ is a PID the implication "(2) implies (1)" is false. For instance let $A=\mathbb{Z}$ and $M=\mathbb{Z}\oplus\mathbb{Z}/(p)$ for some prime $p$. Then $\mathrm{Ann}_A(M)=0$ and $A$ is coprimary. But $M$ has associated primes $(0)$ and $(p)$ and is therefore not coprimary.