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I would like the functor $$(-\otimes_{\mathbb Z} F)\hat{}: \mathbb{Z}[x_1,\dots,x_r]\text{-Mod}_{\mathrm{f.g.}}\longrightarrow \mathbb{Z}[x_1,\dots,x_r]\text{-Mod}$$ to be exact, where completion is w.r.t. the ideal $\langle x_1,\dots,x_r\rangle$ and $F$ is a free $\mathbb{Z}$-module that is not finitely generated. Since $\mathbb{Z}[x_1,\dots,x_r]$ is Noetherian, I know this would work if $F$ was finitely generated.

A typical counterexample for failure of exactness for completion in the infinitely generated case is the sequence of $\mathbb{Z}[x]$-modules $$\bigoplus_{n\in\mathbb N} \mathbb Z [x] \xrightarrow{\mathrm{diag}(1,x,x^2,\dots)} \bigoplus_{n\in\mathbb N} \mathbb Z [x] \xrightarrow{\mathrm{proj}} \bigoplus_{n\in\mathbb N} \mathbb Z [x]/(x^n) ,$$ while my situation looks like $$ \bigoplus_{n\in\mathbb N} M \xrightarrow{\mathrm{diag}(f,f,f,\dots)} \bigoplus_{n\in\mathbb N} N \xrightarrow{\mathrm{diag}(g,g,g,\dots)} \bigoplus_{n\in\mathbb N} P ,$$ so it cannot possibly run into the same issue. I'm willing throw in the additional assumption that $M$, $N$ and $P$ are free over $\mathbb Z$ (but not $\mathbb{Z}[x_1,\dots,x_r]$), but that may not help in any way. Thanks!

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  • $\begingroup$ $$ (-\otimes_{\mathbb Z} F)\hat{}: \mathbb{Z}[x_1,\dots,x_r]\mathrm{-Mod}_{\mathrm{f.g.}}\longrightarrow \mathbb{Z}[x_1,\dots,x_r]\mathrm{-Mod} $$ $$ (-\otimes_{\mathbb Z} F)\hat{}: \mathbb{Z}[x_1,\dots,x_r]\text{-Mod}_{\mathrm{f.g.}}\longrightarrow \mathbb{Z}[x_1,\dots,x_r]\text{-Mod} $$ I edited to change the first display above to the second. I wonder whether people write things like the first version because they don't know how to do the second, or they don't care, or they somehow don't notice, or they prefer the first, or what. $\endgroup$ Feb 17, 2022 at 18:19
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    $\begingroup$ Thanks, that looks better! $\endgroup$
    – user109300
    Feb 17, 2022 at 20:17
  • $\begingroup$ You might consult the concept of weak pro-regularity in, say, Yekutieli's paper arxiv.org/pdf/2002.04901.pdf $\endgroup$
    – Z. M
    Feb 19, 2022 at 22:10

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