# Different notions of associated prime (in the non Noetherian case)

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.

Definitions

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.

Questions

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?

• 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! 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 $(I:x)=A\neq\mathcal M$]
• 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. :-) Mar 2 '10 at 21:21