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I am looking for an example of a module $M$, a ring $A$, and a prime ideal $\mathfrak p$ such that $\operatorname{depth}_{\mathfrak p} M < \operatorname{depth}_{A_{\mathfrak p}} M_{\mathfrak p}$. How can I find such an example?

If $A$ is local with maximal ideal $\mathfrak m$, and $M = A$, then if $\mathfrak m$ is associated we have $\operatorname{depth}_A M=0$. On the other hand, if I can choose a prime $\mathfrak p$ that is not an associated prime of $A$, we get $\operatorname{depth}_{A_{\mathfrak p}} M_{\mathfrak p} \ge 1$. How can I look for this example?

Let's take $A = k[x,y]/(xy,y^2)$. Then $\operatorname{depth}_A A = 0$. Take $\mathfrak p = (y)$. Is it true however that $\operatorname{depth}_{A_{\mathfrak p}} A_{\mathfrak p} \geq 1$?

user26857
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Lynn16
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  • No, it is not true that $\operatorname{depth}{A{\mathfrak p}} A_{\mathfrak p} \geq 1$ for some obvious reason: $\mathfrak p^2=0$. – user26857 Oct 27 '14 at 09:02

1 Answers1

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Hint. Try a similar example for $A$, that is, a quotient ring of a polynomial ring in three indeterminates.

user26857
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  • Ok. But is my ring and module correct now? The problem is that the only prime I can see containing $(xy,y^2,z^2)$ is of course $(x,y,z)$ which is maximal. Unless I now take $M = k[x,y,z]/(xy,y^2,yz)$. – Lynn16 Oct 26 '14 at 23:33
  • @Lynn16 Finally you made the right choice for $M$! What $p$ do you suggest now? (BTW, $(y,z)$ also contains $(xy,y^2,z^2)$, and is not maximal.) – user26857 Oct 26 '14 at 23:35
  • Ok, so $A = k[x,y,z]$, $p = (y,z)$ and $M = k[x,y,z]/(xy,y^2,yz)$. We clearly have $depth_(y,z) M = 0$ since the ideal acts by zero divisors on $M$. When we localize at $M$, $M_{(y,z)} = k[x,y,z]/(xy,y^2,yz){(y,z)} = k[x,y,z]/(y,y^2,yz){(y,z)} = k[x,z]_{(z)}$ and so the depth is now greater than or equal to 1 yes? – Lynn16 Oct 26 '14 at 23:46