# Under Noetherian assumptions, the associated primes equal the primes belonging to the zero submodule (Exercise 7.18 of Atiyah, Macdonald)

In Atiyah, Macdonald, Introduction to Commutative Algebra, Chapter 7, we find:

7.18. Let $$A$$ be a Noetherian ring, $$\mathfrak{p}$$ a prime ideal of $$A$$, and $$M$$ a finitely generated $$A$$-module. Show that the following are equivalent:

1. $$\mathfrak{p}$$ belongs to $$0$$ in $$M$$;
2. there exists $$x \in M$$ such that $$\operatorname{Ann}(x)=\mathfrak{p}$$;
3. there exists a submodule of $$M$$ isomorphic to $$A / \mathfrak{p}$$.

Deduce that there exists a chain of submodules $$0=M_0 \subset M_1 \subset \ldots \subset M_r=M$$ such that each quotient $$M_{i} / M_{i-1}$$ is of the form $$A / \mathfrak{p}_{i}$$, where $$\mathfrak{p}_{i}$$ is a prime ideal of $$A$$.

How does one proceed?

• Last part is 00L0. Commented Apr 29 at 7:49

$$\def\p{\mathfrak{p}} \def\ann{\operatorname{Ann}}$$The equivalence 1-2-3 works without the Noetherian assumption on $$A$$. One must only assume $$M$$ is a Noetherian $$A$$-module and $$\p$$ is a finitely generated prime ideal. Let's see how. Firstly note that part 1 makes sense: By [AM, Exercise 7.17] (which asks to show that every submodule of $$M$$ is the intersection of finitely many irreducible submodules and that every irreducible submodule of $$M$$ is primary), the zero submodule of $$M$$ has a primary decomposition. By [MSE, Lemma 4.3⁎], there is a minimal primary decomposition $$0=\bigcap_{i=1}^nQ_i$$, and $$\{r_M(Q_i)\}$$ is the set of primes belonging to $$0\subset M$$.

2$$\Leftrightarrow$$3. Easy.

2$$\Rightarrow$$1. In particular, $$\p=r(0:x)$$. The result follows from [MSE, Proposition 4.5⁎].

1$$\Rightarrow$$2. Let's generalize the proof of [AM, Proposition 7.17]. We have to show that every finitely generated prime belonging to $$0\subset M$$ is of the form $$\ann x$$, for some $$x\in M$$. Suppose $$0=\bigcap_{i=1}^nQ_i$$ is a minimal primary decomposition and denote $$\p_i=r_M(Q_i)$$. Let $$N_i=\bigcap_{j\neq i}Q_j\neq 0$$. From the proof of [MSE, Proposition 4.5⁎], we have $$r(\ann x)=\p_i$$ for any $$x\neq 0$$ in $$N_i$$, so that $$\ann x\subset\p_i$$.

We have $$\p_i=r(Q_i:M)$$. Now assume $$\p_i$$ is finitely generated (for some $$i$$). Then there is an integer $$m$$ such that $$\p_i^m\subset(Q_i:M)$$ and therefore $$\p_i^mN_i\subset \p_i^mM\cap N_i\subset(Q_i:M)M\cap N_i\subset Q_i\cap N_i=0$$. Let $$m\geq 1$$ be the smallest integer such that $$\p_i^mN_i=0$$, and let $$x$$ be a non-zero element in $$\p_i^{m-1}N_i$$. Then $$\p_ix=0$$, therefore for such an $$x$$ we have $$\ann x\supset\p_i$$, and hence $$\ann x=\p_i$$.

This finishes the proof of the equivalence 1-2-3. To show the last claim, we need to further assume $$A$$ is Noetherian. The proof is done in [Y, Exercise 7.18].

#### References

AM. Atiyah, Macdonald, Introduction to Commutative algebra.

MSE. Mathematics Stack Exchange, 1st and 2nd uniqueness theorems for primary decomposition of modules.

Y. B. Yu, Supplement and Solution Manual for Introduction to Commutative algebra. Also saved in the Wayback Machine.

• Note this result was asked as well here, but the proof is different there. Commented Apr 2 at 15:21