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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 $I$?

t.b.
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Cyril
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3 Answers3

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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.

user26857
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Matt E
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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 $$ 0\ne(ab)\to0. $$

Pierre-Yves Gaillard
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I would like to suggest another proof inspired by marlu's proof in direct product commutes with tensor product?. In the case of a surjective system $\{M_i\}$ and a flat finitely presented $A$-module $N$, we can relax the requirement that the members $M_i$ in the inverse system have to be of finite length in order for $(-)\otimes_AN$ to commute with the inverse limit. Alternatively, if we assume that $\{M_i\}$ is a surjective system of flat $A$-modules, then we can relax the flatness condition on $N$, as well. Then it suffices if $N$ is finitely presented over $A$. In the following, I consider both situations separately (Claim 1 and Claim 2).

Claim 1: Let A be a commutative ring. Let $\{M_i\}$ be a surjective system of $A$-modules. Let further $N$ be a finitely presented and flat $A$-module. Then, there is an isomorphism of $A$-modules $$(\lim_{\leftarrow i}M_i)\otimes_A N\cong\lim_{\leftarrow i}(M_i\otimes_AN).$$

Proof: Since $N$ is finitely presented, by definition, there are integers $m, n\geq1$ together with two surjective maps of $A$-modules $$\varphi: A^n\twoheadrightarrow N$$ and $$\psi:A^m\twoheadrightarrow\ker\varphi.$$ Both maps give rise to the short exact sequence of $A$-modules $$0\to A^m/\ker\psi\hookrightarrow A^n\twoheadrightarrow N\to0.~~~(1)$$
Since $A^n$ is free, hence flat over $A$ and $N$ is by assumption flat over $A$, the $A$-module $\ker\varphi\cong A^m/\ker\psi$ is flat over $A$, too. This means that $$\mathrm{Tor}^k(N, M)=\mathrm{Tor}^k(A^n, M)=\mathrm{Tor}^k(\ker\varphi, M)=0$$ for every $k\geq1$ and every $A$-module $M$. Hence, from the $\mathrm{Tor}$-long exact sequence, we deduce that for each $i$, there is a short exact sequence of $A$-modules $$0\to M_i\otimes_AA^m/\ker\psi\to M_i\otimes_AA^n\to M_i\otimes_AN\to0.$$ The right exactness of the functor $(-)\otimes_AA^m/\ker\psi$ implies that $\{M_i\otimes_AA^m/\ker\psi\}$ is a surjective system, too. Hence, by Proposition $10.2$ in Atiyah-MacDonald's book Introduction to Commutative Algebra, the sequence of $A$-modules $$0\to\lim_{\leftarrow i}(M_i\otimes_A\frac{A^m}{\ker\psi})\to\lim_{\leftarrow i}(M_i\otimes_AA^n)\to\lim_{\leftarrow i}(M_i\otimes_AN)\to0~~~(2)$$
is exact, too. For every $i$, there is a canonical isomorphism of quotient $A$-modules: $$\frac{M_i}{\{0\}}\otimes_A\frac{A^m}{\ker\psi}\cong \frac{M_i^{\oplus m}}{M_i\otimes\ker\psi}.$$ As the inverse system $\{M_i\otimes_A\ker\psi\}$ is surjective, Proposition $10.2$ in Atiyah-MacDonald’s book implies the following isomorphism: $$\lim_{\leftarrow i}(\frac{M_i^{\oplus m}}{M_i\otimes\ker\psi})\cong\frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}.$$ Thus, we can write short exact sequence $(2)$ in the form $$0\to\frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}\to\lim_{\leftarrow i}(M_i^{\oplus n})\to\lim_{\leftarrow i}(M_i\otimes_AN)\to0.~~~(3)$$
Applying the right exact functor $\lim_{\leftarrow i}(M_i)\otimes_A(-)$ on the short exact sequence $(1)$ yields the exact sequence $$\frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{(\lim_{\leftarrow i}M_i)\otimes_A\ker\psi} \to(\lim_{\leftarrow i}M_i)^{\oplus n}\to (\lim_{\leftarrow i}M_i)\otimes_AN\to0~~~(4)$$ The canonical map $$(\lim_{\leftarrow i}M_i)^{\oplus m}\to\lim_{\leftarrow i}(M_i^{\oplus m})~~~(5)$$ is an isomorphism by the fact that finite direct sums are limits and limits always commute with other limits. At the same time, that isomorphism maps the subspace $\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi$ to $\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)$. Hence, the canonical map $(5)$ descends to a surjective map $$\frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi}\twoheadrightarrow\frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}~~~(6)$$
Now, both exact sequences $(3)$ and $(4)$ fit in the following commutative diagram $\require{AMScd}$ \begin{CD} \frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi} @>>> (\lim_{\leftarrow i}M_i)^{\oplus n} @>>> (\lim_{\leftarrow i}M_i)\otimes_AN@>>>0@>>>0 \\ @VVV @VVV @VVV @VVV @VVV \\ \frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)} @>>> \lim_{\leftarrow i}(M_i^{\oplus n}) @>>> \lim_{\leftarrow i}(M_i\otimes_AN)@>>>0@>>>0. \end{CD} The top and bottom rows are exact. The vertical arrows are the canonical maps. The first vertical arrow is the surjective map $(6)$. The second one is an isomorphisms, as already explained, by the fact that inverse limits commute with finite direct sums. The last two morphsims $0\to 0$ are trivially isomorphisms. Then, a direct application of the Five Lemma implies that $(\lim_{\leftarrow i}M_i)\otimes_A N\cong\lim_{\leftarrow i}(M_i\otimes_AN)$. QED

Claim 2: Let A be a commutative ring. Let $\{M_i\}$ be a surjective system of flat $A$-modules. Let further $N$ be a finitely presented $A$-module. Then, there is an isomorphism of $A$-modules $$(\lim_{\leftarrow i}M_i)\otimes_A N\cong\lim_{\leftarrow i}(M_i\otimes_AN).$$

Proof: Since $N$ is finitely presented, by definition, there are integers $m, n\geq1$ together with two surjective maps of $A$-modules $$\varphi: A^n\twoheadrightarrow N$$ and $$\psi:A^m\twoheadrightarrow\ker\varphi.$$ Both maps give rise to the short exact sequence of $A$-modules $$0\to A^m/\ker\psi\hookrightarrow A^n\twoheadrightarrow N\to0.~~~(1)$$
Since by definition, every $A$-module $M_i$ is flat, for each $i$, there is a short exact sequence of $A$-modules $$0\to M_i\otimes_AA^m/\ker\psi\to M_i\otimes_AA^n\to M_i\otimes_AN\to0.$$ The right exactness of the functor $(-)\otimes_AA^m/\ker\psi$ implies that $\{M_i\otimes_AA^m/\ker\psi\}$ is a surjective system, too. Hence, by Proposition $10.2$ in Atiyah-MacDonald's book Introduction to Commutative Algebra, the sequence of $A$-modules $$0\to\lim_{\leftarrow i}(M_i\otimes_A\frac{A^m}{\ker\psi})\to\lim_{\leftarrow i}(M_i\otimes_AA^n)\to\lim_{\leftarrow i}(M_i\otimes_AN)\to0~~~(2)$$
is exact, too. For every $i$, there is a canonical isomorphism of quotient $A$-modules: $$\frac{M_i}{\{0\}}\otimes_A\frac{A^m}{\ker\psi}\cong \frac{M_i^{\oplus m}}{M_i\otimes\ker\psi}.$$ As the inverse system $\{M_i\otimes_A\ker\psi\}$ is surjective, Proposition $10.2$ in Atiyah-MacDonald’s book implies the following isomorphism: $$\lim_{\leftarrow i}(\frac{M_i^{\oplus m}}{M_i\otimes\ker\psi})\cong\frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}.$$ Thus, we can write short exact sequence $(2)$ in the form $$0\to\frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}\to\lim_{\leftarrow i}(M_i^{\oplus n})\to\lim_{\leftarrow i}(M_i\otimes_AN)\to0.~~~(3)$$
Applying the right exact functor $\lim_{\leftarrow i}(M_i)\otimes_A(-)$ on the short exact sequence $(1)$ yields the exact sequence $$\frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{(\lim_{\leftarrow i}M_i)\otimes_A\ker\psi} \to(\lim_{\leftarrow i}M_i)^{\oplus n}\to (\lim_{\leftarrow i}M_i)\otimes_AN\to0~~~(4)$$ The canonical map $$(\lim_{\leftarrow i}M_i)^{\oplus m}\to\lim_{\leftarrow i}(M_i^{\oplus m})~~~(5)$$ is an isomorphism by the fact that finite direct sums are limits and limits always commute with other limits. At the same time, that isomorphism maps the subspace $\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi$ to $\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)$. Hence, the canonical map $(5)$ descends to a surjective map $$\frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi}\twoheadrightarrow\frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}~~~(6)$$
Now, both exact sequences $(3)$ and $(4)$ fit in the following commutative diagram $\require{AMScd}$ \begin{CD} \frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi} @>>> (\lim_{\leftarrow i}M_i)^{\oplus n} @>>> (\lim_{\leftarrow i}M_i)\otimes_AN@>>>0@>>>0 \\ @VVV @VVV @VVV @VVV @VVV \\ \frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)} @>>> \lim_{\leftarrow i}(M_i^{\oplus n}) @>>> \lim_{\leftarrow i}(M_i\otimes_AN)@>>>0@>>>0. \end{CD} The top and bottom rows are exact. The vertical arrows are the canonical maps. The first vertical arrow is the surjective map $(6)$. The second one is an isomorphisms, as already explained, by the fact that inverse limits commute with finite direct sums. The last two morphsims $0\to 0$ are trivially isomorphisms. Then, a direct application of the Five Lemma implies that $(\lim_{\leftarrow i}M_i)\otimes_A N\cong\lim_{\leftarrow i}(M_i\otimes_AN)$. QED

Flavius Aetius
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    Proposition 10.2 in Atiyah-Macdonald applies to short exact sequences of inverse systems, whereas here we have only a right exact sequence. Maybe one can replace the modules on the left with their image in the middle module to have something left exact but you would have to check that the rest of the argument goes through... – marlu Nov 17 '22 at 01:41
  • I do not see why for the proof of Prop. 10.2 in Aliyah-Macdonald, we need the sequence to be also left exact. However, if this turns out to be a problem, then we can overcome it by assuming that $N$ is flat. Then we have a short exact sequence of $0\to\ker(A^n\to N)\to A^n\to N\to0$ and ${M_i\otimes_A\ker(A^n\to N)}$ remains a surjective system. – Flavius Aetius Nov 17 '22 at 19:22
  • I will improve the proof later as soon as I understand why Prop. 10.2 cannot be applied directly despite the fact that the sequence is only right exact. I do not see the problem clearly yet. – Flavius Aetius Nov 17 '22 at 20:20
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    The proof of Prop. 10.2 in Atiyah-Macdonald uses the snake lemma (Prop. 2.10). In the snake lemma it is enough to assume the first row is right exact and the second row is left exact. To have the left exactness of the second row I think it is necessary to assume the given sequences are left exact. – marlu Nov 18 '22 at 00:58
  • Ok, I edited my proof. Perhaps with the current hypotheses, the proof of the claim works. I would be interested in your opinion when you have time. – Flavius Aetius Nov 18 '22 at 23:14
  • This looks correct, except the flatness of $\mathrm{ker}(\varphi)$ is not needed to have that the inverse system ${M_i \otimes_A \mathrm{ker}(\varphi)}$ is surjective. Also, the original question was for the case $N = A/I$ which is not flat as an $A$-module – marlu Nov 19 '22 at 22:44
  • Ok, I added another claim and a proof of it which states that if ${M_i}$ is a surjective system of flat $A$-modules and $N$ is finitely generated over $A$, then the functor $(-)\otimes_AN$ commutes with the inverse limit. – Flavius Aetius Nov 19 '22 at 23:56
  • By the way, I am not deluding myself that the map $(\lim_{\leftarrow i}M_i)\otimes_A\ker\varphi\to\lim_{\leftarrow i}(M_i\otimes_A\ker\varphi)$ is surjective, do I? – Flavius Aetius Nov 20 '22 at 00:03
  • I don't see that surjectivity. Maybe it's best to assume finitely presented instead of finitely generated? Then $\mathrm{ker}(\varphi) \cong A^m$, and the surjectivity of $(\varprojlim_i M_i) \otimes_A A^m \to \varprojlim_i (M_i \otimes_A A^m)$ is clear. – marlu Nov 21 '22 at 00:58
  • Where does the surjectivity fail? Since the system is surjective, for each $m_i\otimes b$ in $(m_i\otimes b)i\in \lim{\leftarrow i}(M_i\otimes\ker\varphi)$, there is an element $m_{i+1}\otimes b$ which is mapped on $m_i\otimes b$. Then, we take the element $(m_i)i\otimes b$ as a preimage in $\lim{\leftarrow i}(M_i)\otimes\ker\varphi$. I do not see where that fails. – Flavius Aetius Nov 21 '22 at 11:22
  • Also, if the module $N$ is finitely presented, then $\ker\varphi$ is not isomorphioc to $A^n$ but to a quotient of $A^n$, right. – Flavius Aetius Nov 21 '22 at 11:27
  • So, Claim $1$ and Claim $2$ are both incorrect the way they stand, as the map $\lim_{\leftarrow i}(M_i)\otimes\ker\varphi\to\lim_{\leftarrow i}(M_i\otimes\ker\varphi)$ is not surjective. Or is the map in Claim $1$ surjective? I do not follow. – Flavius Aetius Nov 21 '22 at 12:24
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    Your proof of surjectivity fails because a general element in $\varprojlim_i (M_i \otimes_A \mathrm{ker}, \varphi)$ is not of the form you write. Firstly, instead of $m_i \otimes b$ it should be $m_i \otimes b_i$ (not necessarily the same $b_i$ for all $i$), and then it should also be finite sums of those rather than simple tensor products of two elements. That's where the proofs of the two claims currently fail. You are right that assuming $N$ being finitely presented might actually not help because the kernel of $\varphi$ will be only a quotient of $A^m$... – marlu Nov 22 '22 at 19:52
  • In fact, finite presentation does the job! The quotient does not violate the desired surjectivity. I edited my proofs by replacing finitely generated with finitely presented. I think everything should work fine. – Flavius Aetius Nov 22 '22 at 22:28
  • But the connecting morphisms in ${M_i\otimes\ker\varphi}$ are of the form $f_{ij}\otimes1: M_i\otimes\ker\varphi\to M_{j}\otimes\ker\varphi$ for $i\geq j$ , correct? – Flavius Aetius Nov 22 '22 at 23:23
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    Yes, it looks correct now. Maybe the proof of Claim 2 could be abberviated using a phrase like "...and from here we can proceed as in the proof of Claim 1". Regarding the question in the comment: yes, the connecting morphisms are those. – marlu Nov 24 '22 at 21:25
  • Thanks, marlu! That means that the surjectivity of the first vertical map in the commutative diagram in Claim $1$ and $2$ has been established now, right? – Flavius Aetius Nov 25 '22 at 01:39
  • Yes, the surjectivity has been established. – marlu Nov 25 '22 at 02:01