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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)

I have two questions about this proof I hope somebody can clarify.

  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?

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  • $\begingroup$ 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. $\endgroup$ Feb 21, 2012 at 9:59
  • $\begingroup$ 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 $Ass_A(M)\subset Ass_A(N)$. $\endgroup$ Feb 21, 2012 at 10:07
  • $\begingroup$ Tim kinsella, of Cap'n Jazz? $\endgroup$
    – Buble
    Feb 21, 2012 at 10:23
  • $\begingroup$ haha! i never thought anyone on MSE would recognize my alias. alas it is only an alias $\endgroup$ Feb 21, 2012 at 10:25

1 Answer 1

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Answer to (1):

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 $i$.


Answer to (2):

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.

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