Inverse limit of modules and tensor product

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 $I$?

Solution 1:

It's not true in general that tensor product commute with projective limits.

E.g. consider $\mathbb Z_p := \projlim_n \mathbb Z/p^n.$ We have that $\mathbb Z_p \otimes_{\mathbb Z} \mathbb Q$ is non-zero; it is the field $\mathbb Q_p$.

On the other hand $\mathbb Z/p^n \otimes_{\mathbb Z} \mathbb Q = 0$ for each value of $n$.

On the other hand, suppose that the modules $M_n$ are of finite length, and that $N$ is finitely presented. Then $(\varprojlim_n M_n)\otimes_A N \to \varprojlim_{n} M_n\otimes N$ is an isomorphism.

To see this, choose a finite presentation $A^r \to A^s \to N \to 0$ of $N$.

Then we have to show that the cokernel of $\varprojlim_n M_n^r \to \varprojlim_n M_n^s$ is isomorphic to the projective limit of the cokernels of the maps $M_n^r \to M_n^s$. This follows from the finite length assumption, which shows (applying Mittag--Leffler) that the projective limit of the cokernels is indeed the cokernel of the projective limits.

Now suppose that $I$ is finitely generated (e.g. assume $A$ is Noetherian). Then $A/I$ is finitely presented, and so if the $M_n$ are furthermore of finite length, the natural map you ask about is an isomorphism.

Solution 2:

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 $$ 0\ne(ab)\to0. $$