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Let $M$ be $A$-module, $A$ commutative ring, and $N$ submodule and let $$N=Q_1\cap\dots\cap Q_r=Q'_1\cap \dots \cap Q'_s$$ be reduced primary decompositions of $N$. Then $r=s$. The set of primes belonging to $Q_1,\dots,Q_r$ and $Q'_1,\dots,Q'_r$ is the same.

It is part of theorem 3.2 in Chapter X of Lang's Algebra, and he said that proof follows from this theorem (3.5):

Let $A$ and $M$ be Noetherian. The associated primes of $M$ are precisely the primes belonging to the primary modules in a reduced primary decomposition of $0$ in $M$.

I don't see how it follow from this proof since we need both module and ring to be Noetherian in second theorem? Is there some direct proof for first statement (I tried to prove it myself but without success).

Just some definitions for clarity:

  • Submodule $Q$ of $M$ is primary if for every $a\in A$ function $a_{M/Q}:M\rightarrow M$, $a_{M/Q}(x)=ax$ is either injective or nilpotent. And set of all $a\in A$ for which that function is nilpotent is prime ideal which belongs to $Q$.
  • Primary decomposition is reduced if primes that belong to different submodules in decompositions are distinct.
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If $N=Q_1\cap\cdots\cap Q_n$ is a reduced primary decomposition, and $P_i=\sqrt{Q_i}$, then $\operatorname{Ass}(M/N)=\{P_1,\dots,P_n\}$.

We have an exact sequence $0\to M/N\to\bigoplus_{i=1}^n M/Q_i$. Then $\operatorname{Ass}(M/N)\subseteq\operatorname{Ass}(\bigoplus_{i=1}^n M/Q_i)$. But $\operatorname{Ass}(\bigoplus_{i=1}^n M/Q_i)=\bigcup_{i=1}^n\operatorname{Ass}(M/Q_i)=\{P_1,\dots,P_n\}$.

For the converse, pick up an $x_i\in\bigcap_{j\ne i}Q_j\setminus Q_i$. Now notice that the submodule $Ax_i$ of $M/N$ is isomorphic to the submodule $Ax_i$ of $M/Q_i$ and hence they have the same associated prime ideals. But on one hand $\operatorname{Ass}(Ax_i)\subseteq\operatorname{Ass}(M/N)$, and on the other hand $\operatorname{Ass}(Ax_i)\subseteq\operatorname{Ass}(M/Q_i)=\{P_i\}$. In order to conclude now that $P_i\in\operatorname{Ass}(M/N)$ we have to be sure that $\operatorname{Ass}(Ax_i)\neq\emptyset$, and this requires the assumption $A$ noetherian (no assumption on $M$ is needed!).

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    $\begingroup$ Hi, I have some questions regarding your answer, I hope you can clarify them for me: (1) by the notation $P_i = \sqrt{Q_i}$, do you mean that $P_i$ is the prime belonging to $Q_i$, i.e. $Q_i$ is $P_i$-primary? I have only seen the notation $\sqrt{I}$ used when referring to the radical of the ideal $I$, so I am a bit confused. $\endgroup$ – Brahadeesh Jun 16 at 6:59
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    $\begingroup$ (2) Lang proves that if $A$ and $M \neq 0$ are Noetherian, then the submodule $Q \neq M$ is primary if and only if $M/Q$ has a unique associated prime (this is Proposition X.3.4). In your answer, you do not assume a priori that $A$ and $M$ are Noetherian, but you write $\mathrm{Ass}(M/Q_i) = \{ P_i \}$. Does this mean you are working under the additional hypothesis that $M/Q_i$ has a unique associated prime for each $i$? $\endgroup$ – Brahadeesh Jun 16 at 7:01
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    $\begingroup$ (3) I see that $\mathrm{Ass}(\bigoplus M/Q_i) \supset \bigcup \mathrm{Ass}(M/Q_i)$, but how do I prove the reverse inclusion? Sorry about the multiple comments, I would really appreciate some clarification on these points. $\endgroup$ – Brahadeesh Jun 16 at 7:02
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    $\begingroup$ (1) The notation is standard and your guess is right. $\endgroup$ – user26857 Jun 16 at 9:57
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    $\begingroup$ (2) If $Q$ is $P$-primary then $\mathrm{Ass}(M/Q)=\{P\}$ without any extra conditions. $\endgroup$ – user26857 Jun 16 at 10:00

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