TITLE: Inverse limit of modules and tensor product
QUESTION [23 upvotes]: Let $(M_n)_n$ be an inverse system of finitely generated modules over a commutative ring $A$ and $I\subset A$ an ideal.
When is the canonical homomorphism
$$\left(\varprojlim\nolimits_n M_n\right)\otimes_A A/I \rightarrow \varprojlim\nolimits_n \left(M_n \otimes_A A/I\right)$$
an isomorphism?
What does one need? E.g. all $M_n$ flat over $A$ or special conditions about $A$ and $I$?

REPLY [3 votes]: Let us show that the canonical morphism
$$
\left(\varprojlim\nolimits_n M_n\right)\otimes_A A/I \rightarrow \varprojlim\nolimits_n \left(M_n \otimes_A A/I\right)
$$
in the question is not always an isomorphism. (This fact was implicit in the question and in the accepted answer, but, as far as I can see, it was not proved.)
As in this answer, let $K$ be a field and $A$ the commutative $K$-algebra with one generated by the symbols $a$ and $b$ subject to the relations $ab^2=ab$ and $a^2=0$. 
Set $M_n:=(b^n)$ and $I:=(a)$. 
The above morphism becomes the obvious morphism 
$$
\left(\bigcap_n\ (b^n)\right)\Bigg/a\left(\bigcap_n\ (b^n)\right)\to\varprojlim\nolimits_n ((b^n)/a(b^n)).
$$ 
But we have 
$$
\bigcap_n\ (b^n)=(ab),\quad(ab)/a(ab)=(ab)/(0)\simeq(ab)\ne(0),
$$ 
and, for $n>0$, 
$$
(b^n)/a(b^n)=(b^n)/(b^n)=0,
$$ 
so that we get the morphism 
$$
0\ne(ab)\to0.
$$