# Further explanation on proof that associated primes are precisely those belonging to primary modules in reduced decomposition of $0$.

Consider the following theorem in Chapter X Noetherian Rings and Modules from Lang's Algebra (page 423, third edition):

Theorem 3.5. Let $$A$$ and $$M \neq 0$$ be Noetherian. The associated primes of $$M$$ are precisely the primes which belong to the primary submodules in a reduced primary decomposition of $$0$$ in $$M$$. In particular, the set of associated primes of $$M$$ is finite.

Proof: Let $$0=Q_1\cap\cdots\cap Q_r$$ be a reduced primary decomposition of $$0$$ in $$M$$. There is an injective homomorphism $$M\to\bigoplus_{i=1}^r M/Q_i.$$ Then every associated prime of $$M$$ belongs to some $$Q_i$$. (1)

Conversely, let $$N=Q_2\cap\cdots\cap Q_r$$. Then $$N\neq 0$$ because the decomposition is reduced. Then $$N=N/(N\cap Q_1)\approx (N+Q_1)/Q_1\subset M/Q_1.$$ Hence $$N$$ is isomorphic to a submodule of $$M/Q_1$$, and consequently has an associated prime which can be none other than the prime $$p_1$$ belonging to $$Q_1$$. (2)

1. How does one conclude from the injective homomorphism that every associated prime of $$M$$ belongs to some $$Q_i$$? I'm aware that a submodule $$Q$$ of $$M$$ is primary iff $$M/Q$$ has exactly one associated prime $$p$$, in which case $$p$$ belongs to $$Q$$. I also know that for a submodule $$N$$ of $$M$$, an associated prime of $$M$$ is associated with $$N$$ or with $$M/N$$, but don't know how to tie it together.

2. Why does $$N$$ being isomorphic to a submodule of $$M/Q_1$$ implies that $$p_1$$ is its associated prime?

• For 1: You could use the fact that the set of associated primes of a finite sum is the union of the associated primes of the summands. Feb 21, 2012 at 9:59
• For 2: If $\phi: M\rightarrow N$ is an imbedding of $A$ modules and $P\subset A$ is the annihilator of $m\in M$ then $P$ is also the annihilator of $\phi (m)\in N$ so $Ass_A(M)\subset Ass_A(N)$. Feb 21, 2012 at 10:07
• Tim kinsella, of Cap'n Jazz? Feb 21, 2012 at 10:23
• haha! i never thought anyone on MSE would recognize my alias. alas it is only an alias Feb 21, 2012 at 10:25

I am not sure how Lang intends the reader to use the fact that we have an injective homomorphism $$M \to \bigoplus_{i=1}^r M/Q_i$$. Here is how I would argue that every associated prime of $$M$$ belongs to some $$Q_i$$:

Since each $$Q_i$$ is a primary submodule of $$M$$, by Proposition 3.4 we know that each $$M/Q_i$$ has exactly one associated prime $$\mathfrak{p}_i$$, which in fact belongs to $$Q_i$$, so that $$Q_i$$ is $$\mathfrak{p}_i$$-primary.

Hence, it suffices to show that if $$\mathfrak{p}$$ is an associated prime of $$M$$, then $$\mathfrak{p}$$ is an associated prime of $$M/Q_i$$ for some $$i$$, because from the previous statement we can conclude that $$\mathfrak{p}$$ must be $$\mathfrak{p}_i$$ and so $$\mathfrak{p}$$ must belong to $$Q_i$$.

Now, let $$\mathfrak{p}$$ be an associated prime of $$M$$ and let $$x$$ be an element of $$M$$ such that $$\mathfrak{p} = \mathrm{ann}_A(x)$$. Examining the proof of Proposition 2.12, we see that $$\mathfrak{p}$$ is an associated prime of $$M/Q_i$$ if and only if $$Ax \cap Q_i = 0$$. Suppose, for the sake of contradiction, that $$Ax \cap Q_i \neq 0$$ for all $$i$$. Let $$s_i \in A$$ be such that $$s_i x$$ is a nonzero element of $$Ax \cap Q_i$$ for each $$i$$. Consider $$s = s_1 \dotsm s_r \in A$$. We have $$sx \in Ax \cap Q_i$$ for all $$i$$, so $$sx \in A \cap (Q_1 \cap \dotsb \cap Q_r) = A \cap 0 = 0$$. Thus, $$sx = 0$$, which implies that $$s \in \mathrm{ann}_A(x)$$, that is, $$s_1 \dotsm s_r \in \mathfrak{p}$$. Hence, $$s_i \in \mathfrak{p}$$ for some $$i$$. But, this implies that $$s_i x = 0$$, which is a contradiction.

Hence, there exists some $$i$$ such that $$Ax \cap Q_i = 0$$, and the proof of Proposition 2.12 shows that $$\mathfrak{p}$$ is an associated prime of $$M/Q_i$$ for this $$i$$.

Let $$\mathfrak{p}_1$$ be the prime belonging to $$Q_1$$. To show that $$\mathfrak{p}_1$$ is an associated prime of $$M$$, it suffices to show that it is an associated prime of some submodule of $$M$$, by Proposition 2.12. Specifically, we will show that $$\mathfrak{p}_1$$ is an associated prime of the nonzero submodule $$N = Q_2 \cap \dotsb \cap Q_r$$ of $$M$$.
Now, by Proposition 3.4, we know that $$M/Q_1$$ has a unique associated prime, namely $$\mathfrak{p}_1$$. So, again by Proposition 2.12, if $$\mathfrak{q}$$ is an associated prime of any submodule of $$M/Q_1$$, then $$\mathfrak{q}$$ is also an associated prime of $$M/Q_1$$, and hence $$\mathfrak{q} = \mathfrak{p}_1$$. Hence, if we show that $$N$$ is isomorphic to a submodule of $$M/Q_1$$, then we will have shown that $$\mathfrak{p}_1$$ is an associated prime of $$N$$, and so we will be done.
$$A$$ is a commutative ring with $$1 \neq 0$$, all modules are $$A$$-modules and all homomorphisms are $$A$$-homomorphisms.
Proposition 2.12. Let $$N$$ be a submodule of $$M$$. Every associated prime of $$N$$ is associated with $$M$$ also. An associated prime of $$M$$ is associated with $$N$$ or with $$M/N$$.
Proposition 3.4. Let $$A$$ and $$M$$ be Noetherian, $$M \neq 0$$. A submodule $$Q \neq M$$ of $$M$$ is primary if and only if $$M/Q$$ has exactly one associated prime $$\mathfrak{p}$$, and in that case, $$\mathfrak{p}$$ belongs to $$Q$$, i.e. $$Q$$ is $$\mathfrak{p}$$-primary.