How to use $5$-lemma to prove that $F(M) \otimes_RM' \overset{\simeq}{\longrightarrow} F(M \otimes_R M')$ is a (natural) isomorphism?

I am describing the question details, though the main question is short as below.

Let $$O$$ be the ring of integers of the finite extension $$K$$ of the $$p$$-adic field $$\mathbb{Q}_p$$. Let $$R$$ be a finite $$\mathbb{Z}_p$$-algebra. Let $$\bar K$$ be the algebraic closure of $$K$$ and $$G_K:=\text{Gal}(\bar K/K)$$ Then consider a right exact functor $$F: \mathscr{C} \to \mathscr{D}$$ between the abelian categories $$\mathscr{C}$$ and $$\mathscr{D}$$, where $$\mathscr{C}$$ is the category of finite $$O \otimes_{\mathbb{Z}_p}R$$-modules and $$\mathscr{D}$$ is the category of continuous representations on finite $$R$$-module (I think any abelian categories is ok for the below question).

Question: For a finite $$O \otimes_{\mathbb{Z}_p}R$$-module $$M \in \mathscr{C}$$ and any finite $$R$$-module $$M'$$, there is a natural isomorphism $$F(M) \otimes_RM' \overset{\simeq}{\longrightarrow} F(M \otimes_R M').$$ Some hints is given. It is asked to take a presentation of the module $$M'$$ and then to use the $$5$$-lemma to prove the above isomorphism. So I was trying the following way: $$------------------------------------$$ My efforts:

For natural numbesr $$m,n$$, Take the finite presentation $$R^{\otimes n} \to R^{\otimes m} \to M' \to 0$$ of the finite $$R$$-module $$M'$$ and then do the following two things:

$$(i)$$ First take with tensor product $$(O\otimes_{\mathbb{Z}_p} R)$$-module $$M$$ and then apply the given right exact-functor $$F$$ to this finite presentation to obtain the exact sequence $$F(M \otimes R^{\otimes n}) \to F(M \otimes R^{\otimes m}) \to F(M \otimes M') \to 0$$ $$(ii)$$ Take tensor product of the finite presentation $$R^{\otimes n} \to R^{\otimes m} \to M' \to 0$$ by $$F(M)$$ to obtain the exact sequence $$F(M) \otimes_R R^{\otimes n} \to F(M) \otimes_R R^{\otimes m} \to F(M) \otimes_R M' \to 0.$$ I think now from $$(i)$$ and $$(ii)$$, we have the following commutative diagram to apply $$5$$-lemma:

\begin{align} \matrix{ F(M \otimes_R R^{\otimes n}) &\to&F(M \otimes_R R^{\otimes m}) &\to&F(M \otimes_R M')&\to&0 \cr \downarrow f_1&&\downarrow f_2&&\downarrow {\color{red}{f_3}}&&\downarrow f_4&& \cr F(M) \otimes_R R^{\otimes n}&\to&F(M) \otimes_R R^{\otimes m}&\to& F(M) \otimes_R M' &\to& 0} \end{align} I think using $$5$$-lemma, we need to show $${\color{red}{f_3}}$$ is an isomorphism to answer our question. I think we need to define $$f_1, f_2$$ in order to satisfy $$5$$-lemma criteria. But here we have $$4$$ columns and we can use $$4$$-lemma at best instead of $$5$$-lemma. How do get the $$5$$-th column according to the given hints.

Am I doing correct so far? Any guidance and help lease.

• What you have looks good so far. Couldn't you just use a fifth column that looks like $0 \to 0$, to the right of the last column you have? Jan 29, 2021 at 19:37

Okay, I might as well answer. Whoever gave the hint "5-lemma" might have been using that term as a shorthand for "apply some standard homological algebra result", knowing that some application or other of the 5-lemma would get the job done. But that hint doesn't seem optimized.

Let $$R$$ be a commutative ring. For any category $$\mathcal{C}$$ enriched in the category of $$R$$-modules and with finite colimits, there's a sensible notion of tensoring an object $$C \in \mathcal{C}$$ by a finitely presented $$R$$-module $$M$$, and this is implicit in what you write: for a natural number $$n$$, define $$C \otimes_R R^n$$ to be $$C^n$$, the coproduct of $$n$$ copies of $$C$$, guided by the intuition

$$C \otimes_R R^n \cong C \otimes_R (R \oplus \ldots \oplus R) \cong (C \otimes_R R) \oplus \ldots \oplus (C \otimes_R R) \cong C \oplus \ldots \oplus C = C^n.$$

This definition is easily made functorial: if we have an $$R$$-module map $$f: R^m \to R^n$$, i.e., a matrix, then we get an induced action $$C^m \to C^n$$ of this matrix by utilizing the $$R$$-module enrichment. Then, if we have a finite presentation $$R^n \to R^m \to M \to 0$$ of an $$R$$-module $$M$$, we may define $$C \otimes_R M$$ by taking the cokernel of the induced map $$C^n \to C^m$$. It may be checked that this definition of $$C \otimes_R M$$ doesn't depend essentially on the chosen presentation. (Side comment is that if $$R$$ is Noetherian, then finite presentability is equivalent to being finitely generated.)

Now suppose given an right exact additive functor $$F: \mathcal{C} \to \mathcal{D}$$ between finitely cocomplete categories enriched in $$R$$-Mod. This means $$F$$ preserves finite direct sums and cokernels. In particular, for the $$f_1$$ in the diagram above, just transcribe the canonical isomorphism

$$F(C \otimes_R R^n) \cong F(C^n) \cong F(C)^n \cong F(C) \otimes_R R^n$$

where the middle isomorphism just results from the fact that an enriched functor must automatically preserve finite direct sums. Same for $$f_2$$.

Now apply $$F$$ to the cokernel of $$C^n \to C^m$$, i.e., to the exact sequence $$C^n \to C^m \to C \otimes_R M \to 0$$. That gives essentially your top horizontal sequence. But $$F$$, being right exact, preserves this exact sequence, so $$F(C \otimes_R M)$$ is isomorphic to the cokernel of the map $$F(C^n) \to F(C^m)$$, which we have isomorphically identified with the map $$F(C)^n \to F(C)^m$$. But the cokernel of the latter is $$F(C) \otimes_R M$$ by our definitions.

• Many many thanks for your beautiful answer. I got stuck to prove a theorem due to this. I now got it. May be I can ask few comments later if necessary but for the time being lots of gratefulness
– MAS
Jan 30, 2021 at 6:07
• Just few questions. $(1)$ Do you think this exactness condition of the functor $F$ in our question can be dropped ? $(2)$ Is the category of modules over commutative ring both cocomplete and complete ?
– MAS
Jan 30, 2021 at 15:55
• (1) No, I don't think right exactness can be dropped. (2) Yes, always. Jan 30, 2021 at 16:01
• Thank you very much. You are very helpful
– MAS
Jan 30, 2021 at 16:04
• Yes, that's right. Jan 30, 2021 at 19:44