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Jul 28, 2022 at 19:12 comment added Johan When $n = 1$ you could argue like this: $M$ is $x = x_1$-torsion free, so completion is equal to derived completion. Then the short exact sequence $0 \to M \to M \to M/xM \to 0$ upon derived completion becomes the distinguished triangle $M^\wedge \to M^\wedge \to (M/xM)^\wedge$. But $(M/xM)^\wedge = M/xM$ so sits only in degree $0$ hence $x$ is regular on $M^\wedge$. For larger $n$ try induction on $n$ (using the same argument essentially). Cheers!
Jul 28, 2022 at 14:41 history asked Neil Strickland CC BY-SA 4.0