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.

  • 3
    $\begingroup$ 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? $\endgroup$
    – Todd Trimble
    Jan 29, 2021 at 19:37

1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$
    – MAS
    Jan 30, 2021 at 6:07
  • $\begingroup$ 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 ? $\endgroup$
    – MAS
    Jan 30, 2021 at 15:55
  • 1
    $\begingroup$ (1) No, I don't think right exactness can be dropped. (2) Yes, always. $\endgroup$
    – Todd Trimble
    Jan 30, 2021 at 16:01
  • $\begingroup$ Thank you very much. You are very helpful $\endgroup$
    – MAS
    Jan 30, 2021 at 16:04
  • 1
    $\begingroup$ Yes, that's right. $\endgroup$
    – Todd Trimble
    Jan 30, 2021 at 19:44

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