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

Consider the following theorem:

Let $A$ and $M$ be Noetherian. The associated primes of $M$ are precisely the prime which belong to the primary modules 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)

Convsersely, 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 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 $Ass_A(M)\subset Ass_A(N)$. – Tim kinsella Feb 21 '12 at 10:07