Timeline for A question about the assassinator (={associated primes}) and the support of a module.

8 events

when toggle format	what		by	license	comment
Jun 3, 2010 at 20:13	comment	added	ressing		@Karl Schwede: It does, yes. I should have been clearer in my comment in response to Hailong's.
Jun 3, 2010 at 8:24	comment	added	Karl Schwede		So Long's comment answers the question, right?
Jun 2, 2010 at 20:12	history	edited	Harry Gindi	CC BY-SA 2.5	edited title
Jun 2, 2010 at 20:05	history	edited	ressing	CC BY-SA 2.5	deleted 3 characters in body
Jun 2, 2010 at 20:01	comment	added	ressing		Thank you, you're right. $\mathfrak{p} \in Ass(N) \implies 0 \rightarrow A/\mathfrak{p} \rightarrow N$ exact $\implies 0 \rightarrow A/\mathfrak{p} \otimes A_{\mathfrak{p}} \rightarrow N \otimes A_{\mathfrak{p}}$ exact $\implies 0 \rightarrow A_{\mathfrak{p}}/{\mathfrak{p}A_{\mathfrak{p}}} \rightarrow N_{\mathfrak{p}}$ exact. Thus $N_{\mathfrak{p}}$ contains a nonzero submodule and so $\mathfrak{p} \in Supp(N)$.
Jun 2, 2010 at 19:45	comment	added	Hailong Dao		Not sure I understand. $Ass(M/xM) \subseteq Supp(M/xM)$, right?
Jun 2, 2010 at 19:40	history	edited	ressing	CC BY-SA 2.5	Changed latex formatting in last line; deleted 7 characters in body; deleted 2 characters in body
Jun 2, 2010 at 19:24	history	asked	ressing	CC BY-SA 2.5