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I want to show that

$$\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)=\operatorname{Ass}_R\left(\dfrac{R}{x^3R}\right)$$

where $x\in R$ is a non-zero-divisor and non-unit element and $R$ is a Noetherian ring. To proof this exercise, I've been used these two short exact sequences:

$$0\to\dfrac{R}{x^2R}\to\dfrac{R}{x^3R}\to\dfrac{R}{xR}\to0$$

and

$$0\to\dfrac{R}{xR}\to\dfrac{R}{x^3R}\to\dfrac{R}{x^2R}\to0.$$

From the first sequence we have:

$$\operatorname{Ass}_R\left(\dfrac{R}{x^2R}\right)\subseteq\operatorname{Ass}_R\left(\dfrac{R}{x^3R}\right)\subseteq\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)\cup\operatorname{Ass}_R\left(\dfrac{R}{x^2R}\right)=\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)$$

and second one gives us:

$$\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)\subseteq\operatorname{Ass}_R\left(\dfrac{R}{x^3R}\right)\subseteq\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)\cup\operatorname{Ass}_R\left(\dfrac{R}{x^2R}\right)=\operatorname{Ass}_R\left(\dfrac{R}{x^2R}\right).$$ So we have the equality.

Santana Afton
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Alireza Shamsian
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  • The first sequence is correct by the help of Mutsumura's book, Commutative Algebra, where he says we have this short exact sequence for any $n\in\mathbb{N}$: $$0\to\dfrac{R}{x^{n-1}R}\to\dfrac{R}{x^nR}\to\dfrac{R}{xR}\to0,$$ but I'm not sure about the second one. – Alireza Shamsian Aug 06 '19 at 23:50
  • The second one is also true. $R/xR$ is isomorphic to $x^2R/x^3R=\ker(R/x^3R\rightarrow R/x^2R)$ through the map $a+xR\mapsto x^2a+x^3R$ provided $x$ is nonzero divisor. – Rafael Aug 07 '19 at 00:15
  • Thank you. You're right. – Alireza Shamsian Aug 07 '19 at 11:12

1 Answers1

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If you see Page 289 of the book, by the third isomorphism theorem, $xR/x^nR\cong R/x^{n-1}R$, $0→R/x^{n-1}R→R/x^nR→R/xR→0$.

From the sequence, you can get the result inductively. Let $n=2$. Then $0→R/xR→R/x^2R→R/xR→0$ and $Ass_{R}(R/xR)\subset Ass_{R}(R/x^2R)\subset Ass_{R}(R/xR)∪ Ass_{R}(R/xR)=Ass_{R}(R/xR)$. And let $n=3$....