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Jun 8, 2020 at 14:06 comment added Ben Webster Fair point. Luckily, D-modules on a smooth variety are always perfect.
Jun 7, 2020 at 9:21 comment added user20948 A remark: you need to assume that $M$ is a perfect complex to make $R\operatorname{Hom}(M,R)\otimes_R^{\mathbb L}N\simeq R\operatorname{Hom}(M,N)$ valid (see Lurie, Higher Algebra, Proposition 7.2.4.4, which says that this equivalent is true if and only if $M$ is a perfect complex).
Apr 3, 2018 at 13:45 vote accept Hang
Apr 2, 2018 at 18:26 history answered Ben Webster CC BY-SA 3.0