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!