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Jun 8, 2018 at 20:11 comment added Luc Guyot @YCor Thanks for explaining this, and sorry for the trouble. The naive way (that is mine) to read your line about the correspondance is to assume that the correspondance is the Matlis duality $N \mapsto N' = \text{Hom}_{\hat{R}}(N, S)$. But what you mean is different, your are referring to another correspondance, namely $N \mapsto N^{\ast} = \ker(\hat{R}^n \twoheadrightarrow N')$. This is now clear for me.
Jun 8, 2018 at 17:32 comment added YCor Luc, I wrote on purpose "submodules of $\hat{R}^n$: it's a kind of "orthogonal" bijection, mapping a submodule to the set of homomorphisms into $S$ that vanish on the submodule. Indeed if it maps $N$ to $N'$, then the quotient by $N'$ is the Matlis dual of $N$. Actually to write properly, I should have written "correspondence between the set of submodules... and the set of submodules...".
Jun 8, 2018 at 17:28 history edited Luc Guyot CC BY-SA 4.0
Fixes typo: Hom reverses arrows
Jun 8, 2018 at 15:27 history edited Luc Guyot CC BY-SA 4.0
Fixes a typo
Jun 7, 2018 at 14:43 comment added Chris Leary This is an interesting observation, and I will see if this helps with what I am trying to do. Thank you.
Jun 7, 2018 at 9:09 history answered YCor CC BY-SA 4.0