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Let $R$ be a commutative Noetherian ring with $\mathfrak p$ a nilpotent prime ideal, that is $\mathfrak p^r=0$ for some positive integer. If $\oplus_{i=0}^{r-1}\mathfrak p^i/\mathfrak p^{i+1}$ is a flat $R/\mathfrak p$-module, how to prove that an $R/\mathfrak p$-sequence is also an $R$-sequence?

nick
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Since the $R/\mathfrak p$-modules $\mathfrak p^i/\mathfrak p^{i+1}$ are flat, any $R/\mathfrak p$-sequence is also an $\mathfrak p^i/\mathfrak p^{i+1}$-sequence. Now consider the exact sequences $0\to \mathfrak p^i/\mathfrak p^{i+1}\to R/\mathfrak p^{i+1}\to R/\mathfrak p^i\to 0$ for $i\ge 0$ and observe that an $R/\mathfrak p^i$-sequence which is also an $\mathfrak p^i/\mathfrak p^{i+1}$-sequence is an $R/\mathfrak p^{i+1}$-sequence. Since $\mathfrak p$ is nilpotent, at some moment you get that the initial $R/\mathfrak p$-sequence must be an $R$-sequence.

  • Dear YACP, thank you for your reply. I think the main point is that, for an exact sequence of R-modules $0\rightarrow N \rightarrow M \rightarrow Q \rightarrow 0$, if $x_1,\cdots,x_n$ is an Q-sequence, let $I=(x_1,\cdots,x_n)$, then $0\rightarrow N/IN \rightarrow M/IM \rightarrow Q/IQ \rightarrow 0$ is an exact sequence. – nick Apr 28 '13 at 14:48