Most books I have treat primary decomposition only in the Noetherian case. Atyiah-MacDonald goes a step further and prover the uniqueness theorems of primary decomposition without the Noetherian hypothesis. But it seems to me they get a slight different result from the usual one.


Recall that a prime $P$ is said to be associated to the $A$-module $M$ if there exists $m \in M$ such that $P = Ann(m)$; equivalently $A/P$ injects into $M$. I denote by $Ass(M)$ the set of associated primes. If $A$ and $M$ are Noetherian, this is always not empty.

For an ideal $I$ we let $Ass(I) = Ass(A/I)$. So a prime $P$ is associated to $I$ if and only if $P$ is of the form $(I : x)$ for some $x \in A$.

For the purposes of this question let me say that $P$ belongs to $I$ if and only if $P$ is of the form $\sqrt{(I : x)}$ for some $x \in A$. We call $Bel(I)$ the set of primes belonging to $I$.

Then the result of Atyiah-MacDonald shows that if $I$ has a (minimal) primary decomposition $I = \bigcap Q_i$, and if we let $P_i = \sqrt{Q_i}$, the set of $P_i$ which appear is exactly $Bel(I)$. The usual formulation gives instead that for $A$ Noetherian this set is $Ass(I)$.

The problem

I want to understand the relationship between $Ass(I)$ and $Bel(I)$. Clearly, since prime ideals are radical, $Ass(I) \subset Bel(I)$. In general I see no reason why the opposite inclusion should be true.

Let us see how to go proving the opposite inclusion in a special case. Assume $I$ is decomposable. Then by the result of Atyiah-MacDonald it is enough to show that if we have a minimal primary decomposition $I = \bigcap Q_i$, and if we let $P_i = \sqrt{Q_i}$, then $P_i \in Ass(I)$.

Let us do this for $P_1$ and call $R = Q_2 \cap \cdots \cap Q_n$. I also call $P = P_1$, $Q = Q_1$, so $I = Q \cap R$.

Then observe that $R/I = R/(R \cap Q) \cong (R + Q) /Q \subset A/Q$. Since $Q$ is $P$-primary, $Ass(A/Q) = P$. So $Ass(R/I) \subset \{ P \}$.

If moreover $A$ is Noetherian this set has to be non empty, so $Ass(R/I) = \{ P \}$ and a fortiori it follows that $P \in Ass(A/I)$.

I don't see how to do this without the Noetherian hypothesis, though.


Is $Ass(I) = Bel(I)$ always, even if $A$ is not Noetherian?

Is $Ass(I) = Bel(I)$ if we assume that $A$ is not Noetherian, but at least $I$ is decomposable?

  • $\begingroup$ Very interesting question! I did use your development (and the answer Georges Elencwajg provided) in my notes), it helps understand primary decompositions, even in the noetherian case! $\endgroup$ Oct 22 '16 at 20:17

Dear Andrea, let $A=K[X_1,X_2,\ldots ,X_n,\ldots]$, the polynomial ring in countably many variables and $I$ be the ideal $I=(X_1^2, X_2^2,\ldots )\subset A $. Then

$$\mathcal M=(X_1,X_2,\ldots)\in Bel(I) \setminus Ass(I)$$

Indeed, $(I:1)=I$ has as radical $\sqrt I=\mathcal M$, hence $\mathcal M \in Bel(I)$. But there is no polynomial $x=P(X_1,X_2,\ldots ,X_N)\notin I$ such that $(I:x)=\mathcal M$ because $X_M$ will not satisfy $X_M.x\in I$ for $M>N$ [Of course if $x\in I$, we have $(I:x)=A\neq\mathcal M$]

  • $\begingroup$ Thank you very much! Since the radical of $I$ is maximal, $I$ is actually primary, so equality does not hold even in the decomposable case. So I guess this settles everything. :-) $\endgroup$ Mar 2 '10 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.