Timeline for Reducedness of a ring with prime nilradical

6 events

when toggle format	what		by	license	comment
Jan 22, 2015 at 13:16	comment	added	MooS		This clarifies everything. Thanks a lot!
Jan 22, 2015 at 13:16	vote	accept	MooS
Jan 22, 2015 at 10:32	comment	added	Jesse Elliott		If you're looking for a condition on $\mathfrak{q}$, then one such necessary and sufficient condition is that $\mathfrak{q}$ be primary. (Indeed, $\mathfrak{q}$ is primary if and only if the map $A/\mathfrak{q} \longrightarrow (A/\mathfrak{q})_{\mathfrak{p}}$ is injective, which since the image is a field implies that $\mathfrak{q}$ is prime.) Under the assumption that $A/\mathfrak{q}$ is $(R_0)$, the ideal $\mathfrak{q}$ is prime iff it is primary iff it is radical. I'm not sure what else can be said.
Jan 22, 2015 at 8:09	comment	added	MooS		Thank You for your answer. Actually I am aware of this criteria for a Noetherian ring to be reduced and it was precisely what I had in mind when giving the proof. $(R_0)$ is just another term for "generically reduced" and $(S_1)$ is kind a weaker version of Cohen-Macaulay. Actually in my counterexample $A/\mathfrak q$ was one-dimensional, hence Cohen-Macaulay and $(S_1)$ coincide in this ring. Maybe this is the remaining the question now: Is there another sufficient condition in terms of the ideal $\mathfrak q$, that suffices to show that $A/\mathfrak q$ is reduced?
Jan 22, 2015 at 6:26	history	edited	Jesse Elliott	CC BY-SA 3.0	added 167 characters in body
Jan 22, 2015 at 6:20	history	answered	Jesse Elliott	CC BY-SA 3.0