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Let $A$ be a Noetherian ring and let $x \in A$ be an element which is neither a unit nor a zerodivisor. Why the ideals $xA$ and $x^{n}A$ (where $n \in \mathbb{N}$) have the same prime divisors? i.e. $$\operatorname{Ass}(A/xA) = \operatorname{Ass}(A/x^{n}A).$$ (Matsumura, Commutative Ring Theory, Exercise 6.3)

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since $x$ is not zero divisor, $(x:y)=(x^n:yx^{n-1})$. In a Noetherian ring, the set of associated prime ideals of an ideal is the set of primes which can be written as $(I:z)$.

Edit: I have not completed the proof. If $(x:y)$ is a prime then $(x:y)=(x^n:yx^{n-1})$ is a prime belonged to $x^nA$, conversely, if $(x^n:z)$ is a prime, we want to show $(x^n:z)$ is an associated prime to $A/xA$. If $(x^n:z)=(x:z)$, we are done. If not, suppose $y\in (x:z)$, but $y\notin (x^n:z)$, then $yx^{n-1}\in (x^n:z)$. If $y\notin(x^n:z)$, since $(x^n:z)$ is prime, we obtain that $z\in xA$, thus $(x^n:z)=(x^{n-1}:z^{\prime})$ for some $z^{\prime}$. By induction, we can finish the proof.

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If $0\to M'\to M\to M''\to 0$ is a short exact sequence of $A$-modules, then it is well known that $\operatorname{Ass}(M')\subseteq\operatorname{Ass}(M)\subseteq\operatorname{Ass}(M')\cup\operatorname{Ass}(M'')$.

Since $x$ is a non-zerodivisor we have $x^{n-1}A/x^nA\simeq A/xA$ (isomorphism of $A$-modules) and thus we have the following short exact sequence: $0\to A/xA\to A/x^nA\to A/x^{n-1}A\to 0$. Now apply the property above and induction on $n$.

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  • $\begingroup$ Where do you use the fact that $A$ is Noetherian? The proposition you cite is Theorem 6.3 in Matusmura, but he does not require that the modules are finitely generated. $\endgroup$ – Potato Jul 11 '13 at 18:33

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