# $M^\wedge_I \to N^\wedge_I$ an isomorphism if $S_P^{-1}M^\wedge_P \to S_P^{-1}N^\wedge_P$ is an isomorphism for all primes $P$ containing $I$

Let $$R$$ be a Noetherian ring, $$I \subseteq R$$ an ideal, and $$S \subseteq R$$ a multiplicative set.

Lemma 2.3 of Adam, Haeberly, Jackowski, and May's paper A generalisation of the Segal conjecture states that if $$M = \{M_\alpha\}$$ is a pro-module and $$S_P^{-1}M^\wedge_P$$ is pro-zero for all primes $$P$$ disjoint from $$S$$ and containing and ideal $$I$$, then $$S^{-1}M^\wedge_I$$ is pro-zero.

I'd like the following generalisation. If $$M = \{M_\alpha\} \to N = \{N_\beta\}$$ is a homomorphism of pro-groups such that the induced map $$S_P^{-1}M^\wedge_P \to S_P^{-1}N^\wedge_P$$ is an isomorphism for all primes $$P$$ containing $$I$$, then $$S^{-1}M^\wedge_I \to S^{-1}N^\wedge_I$$ is an isomorphism.

Is this already stated somewhere in the literature? I'm from an algebraic topology background, so I'm not really sure where I should be looking. It seems like it might be possible to generalise Adam et al's proof to this case, but if the result is already known then that would be better.

• I look at their paper, and they are assuming that this is a pro-system of finitely generated modules. Would you assume the same?
– Z. M
Jan 12 at 18:51
• @Z.M yes, I'd assume the same Jan 13 at 7:54
• In that case, the completion is an exact functor, thus commutes with taking kernel and cokernel, and you reduce to the case in that paper (btw, you misspelled Adams).
– Z. M
Jan 13 at 10:03
• Thanks! It's simple in hindsight. If you want to write this as an answer to the question I'll choose it as the accepted answer. Jan 13 at 10:34
• I am not satisfied with this generality. For example, what happens if we remove the Noetherianness, and replace classical completion by derived completion (which is the completion in stable homotopy theory), say. The proof in that paper does not seem to be conceptual enough, either.
– Z. M
Jan 13 at 10:40