On a special kind of $R$-linear functor from an $R$-linear additive category

Question

Let $R$ be a Commutative Noetherian ring, let $\mod (R)$ be the category of finitely generated $R$-modules. Let $\mathcal C \subseteq \mod(R)$ be an $R$-linear (https://stacks.math.columbia.edu/tag/09MI ) additive category.

Let $F: \mathcal C \to \mod(R)$ be a covariant $R$-linear functor such that $F(M)\cong M, \forall M \in \mathcal C.$ Let $i_M : \mathcal C \to \mod(R)$ be the inclusion functor that takes every module (and every $R$-linear map between modules ) in $\mathcal C$ to itself.

My question is: Is the functor $F$ naturally Isomorphic to $i_M$?

Answer 1

Here is a variant of Stahl's example that shows the answer can be no even with significantly stronger hypotheses that avoid examples that might seem overly degenerate. In particular, the answer can be no even if $\mathcal{C}$ is a full subcategory and $F$ is fully faithful.

Let $k$ be a field and let $R=k[x,y]$. Let $I$ denote the ideal $(x,y)\subset R$. Let $\mathcal{C}$ be the full subcategory of $\mathtt{Mod}_R$ consisting of all modules that are a finite direct sum of copies of $I$ and $R/I$. Note that $\operatorname{Hom}(I,I)=R$, $\operatorname{Hom}(R/I,R/I)=R/I$, $\operatorname{Hom}(R/I,I)=0$, and $\operatorname{Hom}(I,R/I)=(R/I)^2$ (since we can independently map each of $x$ and $y$ to any element of $R/I$). Now define a functor $F:\mathcal{C}\to\mathtt{Mod}_R$ which is the identity on objects as follows. On morphisms among $I$ and $R/I$, $F$ is the identity on $\operatorname{Hom}(I,I)$ and $\operatorname{Hom}(R/I,R/I)$ but swaps the two coordinates of $\operatorname{Hom}(I,R/I)=(R/I)^2$. This preserves composition, since the only nontrivial compositions just correspond to the $R$-module structure on $\operatorname{Hom}(I,R/I)=(R/I)^2$ and the swap map is $R$-linear. Then you extend $F$ to morphisms between direct sums of $I$ and $R/I$ in the obvious way so that it is additive.

To see that $F$ is not naturally isomorphic to the inclusion, just note that the swap map on $\operatorname{Hom}(I,R/I)=(R/I)^2$ cannot be turned into the identity by composing with isomorphisms $I\to I$ and $R/I\to R/I$ (since any such isomorphisms are just multiplication by some element of $R$).

Answer 2

I claim that such a functor $F$ need not be naturally isomorphic to $i_\mathcal{C}.$ (I denote the inclusion functor from $\mathcal{C}\to\operatorname{Mod}(R)$ by $i_\mathcal{C}$ rather than $i_M$ as it depends on $\mathcal{C},$ not any $R$-module $M$.)

Let $M,N$ be two $R$-modules such that there exists a nonzero morphism $f : M\to N$ in $\operatorname{Hom}_{R}(M,N),$ and let $\mathcal{C}$ be the $R$-linear additive category generated by $M$ and $N$ and the $\operatorname{Hom}$-modules $$ \begin{align*} \operatorname{Hom}_\mathcal{C}(N,N) &= \operatorname{Hom}_R(N,N),\\ \operatorname{Hom}_\mathcal{C}(M,M) &= \operatorname{Hom}_R(M,M),\\ \operatorname{Hom}_\mathcal{C}(M,N) &= \operatorname{Hom}_R(M,N),\\ \operatorname{Hom}_\mathcal{C}(N,M) &= \{0\}. \end{align*} $$ More explicitly, the objects of $\mathcal{C}$ are finite direct sums (possibly empty) of $M$ and $N,$ with morphisms given by $$ \operatorname{Hom}_\mathcal{C}(N^{\oplus n}\oplus M^{\oplus m},N^{\oplus n'}\oplus M^{\oplus m'}) = \operatorname{Hom}_R(N,N)^{\oplus nn'}\oplus\operatorname{Hom}_R(M,N)^{\oplus mn'}\oplus \operatorname{Hom}_R(M,M)^{\oplus mm'} $$ (after appropriate isomorphisms).

Define $F : \mathcal{C}\to\operatorname{Mod}(R)$ by $F(X) = X$ for any $X\in\operatorname{ob}(\mathcal{C}),$ and let $F$ be defined on morphisms by $F(f) = f$ for any endomorphism of $M$ or $N,$ and $F(f) = 0$ for any $f\in\operatorname{Hom}_\mathcal{C}(M,N)\cup\operatorname{Hom}_\mathcal{C}(N,M),$ and extend $F$ on direct sums of objects in the "obvious way." As below, we can check that $F$ is an $R$-linear functor. ($F$ is the $R$-linear functor generated by the concrete description of $F$ given on $M,N,$ and morphisms between combinations of these objects.)

And again, as below, if $\eta : F\Rightarrow i_\mathcal{C}$ is a natural transformation, then we must have a commutative diagram $$ \require{AMScd} \begin{CD} M @>\eta_M>> M \\ @V0VV @VVfV \\ N @>>\eta_{N}> N\\ \end{CD} $$ for any $f : M\to N.$ But if $f$ is nonzero, there is some $m\in M$ such that $f(m)\neq 0,$ and we have $$ \eta_M(f(m)) = 0, $$ which implies that $\eta_M$ cannot be an isomorphism, and thus that $\eta$ cannot be a natural isomorphism.

Edit: Below is my original example, which is an $R$-linear but not additive category satisfying the condition. Thanks to Eric Wofsey for the helpful comments.

Let $R$ be any nonzero ring, and let $\mathcal{C}$ be the subcategory of $\operatorname{Mod}(R)$ whose objects are $\operatorname{ob}(\mathcal{C}) = \{R, R^2\}$, and whose morphisms are given by $$ \begin{align*} \operatorname{Hom}_\mathcal{C}(R,R) &= R,\\ \operatorname{Hom}_\mathcal{C}(R^2,R^2) &= R,\\ \operatorname{Hom}_\mathcal{C}(R,R^2) &= \left\{r\mapsto (\alpha r,0)\mid\alpha\in R\right\}\\ \operatorname{Hom}_\mathcal{C}(R^2,R) &= \{0\}. \end{align*} $$ More precisely, for either element of $\mathcal{C}$, we only consider endomorphisms which come from scaling by elements of $R$, and the only nonzero morphisms between the two objects are scalar multiples of inclusion of $R$ into the first factor of $R^2$.

Let us define $F : \mathcal{C}\to\operatorname{Mod}(R)$ by setting $F(M) = M$ for every $M\in\operatorname{ob}(\mathcal{C})$ and $$ F(f) := \begin{cases}f,&f\in\operatorname{Hom}_\mathcal{C}(R,R)\textrm{ or }\operatorname{Hom}_\mathcal{C}(R^2,R^2),\\ 0, &f\in\operatorname{Hom}_\mathcal{C}(R,R^2)\textrm{ or }\operatorname{Hom}_\mathcal{C}(R^2,R).\end{cases} $$ Clearly $F(\operatorname{id}_M) = \operatorname{id}_{F(M)}$ for any $M\in\operatorname{ob}(\mathcal{C}).$ If $f$ and $g$ are both endomorphisms of the same object, then $F(f\circ g) = f\circ g = F(f)\circ F(g).$

On the other hand, suppose that one of $f$ or $g$ is an element of $\operatorname{Hom}_\mathcal{C}(R^2,R) = \{0\}.$ Then $f\circ g = 0,$ so we have $$F(f\circ g) = F(0) = 0 = F(f)\circ F(g).$$ The only cases left to consider are for compositions of the form $$ R\xrightarrow{g} R\xrightarrow{f} R^2 $$ and $$ R\xrightarrow{g} R^2\xrightarrow{f} R^2. $$ In the first case, $$ 0 = F(f\circ g) = F(f)\circ F(g) $$ holds because $F(f) = 0,$ and in the second it holds because $F(g) = 0.$ So, $F$ is indeed a functor.

To see that $F$ is $R$-linear, observe that $F_{M,N} : \operatorname{Hom}_\mathcal{C}(M,N)\to\operatorname{Hom}_R(F(M),F(N))$ is either the inclusion of a submodule isomorphic to $R$ or the zero map.

Now, let $\eta : F\Rightarrow i_\mathcal{C}$ be a natural transformation, and let $f : R\to R^2$ be the inclusion of $R$ into the first factor of $R^2$ in $\operatorname{Hom}_\mathcal{C}(R,R^2).$ Then we have a commutative diagram $$ \require{AMScd} \begin{CD} F(R) @>\eta_R>> i_\mathcal{C}(R) \\ @VF(f)VV @VVi_\mathcal{C}(f)V \\ F(R^2) @>>\eta_{R^2}> i_\mathcal{C}(R^2).\\ \end{CD} $$ Simplifying, this means that we have a commutative diagram $$ \require{AMScd} \begin{CD} R @>\eta_R>> R \\ @V0VV @VVfV \\ R^2 @>>\eta_{R^2}> R^2.\\ \end{CD} $$ It follows that $$ f(\eta_R(1)) = (\eta_R(1),0) = (0,0), $$ and the injectivity of $f$ implies $\eta_R(1) = 0,$ so that $\eta_R$ (and thus $\eta$) cannot be an isomorphism. Thus, there is no natural isomorphism $\eta : F\to i_\mathcal{C}.$