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If $a_1,...,a_n$ is a maximal $R$-sequence in an ideal $I$ of a noetherian commutative ring $R$ then $I⊆∪P$, where the union is taken over all the associated prime ideals of the $n$-generated ideal $(a_1,...,a_n)$.

Does this result also hold for non-noetherian rings?

Thanks for any help!

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Essentially you are asking if the set of zero divisors is the union of associated primes in the non-noetherian case. (This is the case $n=0$ of your question.) This is false as the following simple example shows. Set $$R=K[X_1,\dots,X_n,\dots]/(X_1^2,\dots,X_n^2,\dots),$$ and let $I=(x_1,\dots,x_n,\dots)$ be the maximal ideal of $R$. Then every element of $I$ is a zero-divisor (in fact, nilpotent), but $\operatorname{Ass}R=\emptyset$.

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