# 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 $I$?

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## 1 Answer

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.

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