On some sense of representing an endofunctor of the category of modules over polynomial rings

Question

If $R$ is commutative ring, $n\in\mathbb{N}$, $\mathsf{M}$ the category of $R[x_1,\dotsc,x_n]$-modules,and $F\colon\mathsf{M}\to\mathsf{M}$ an endofunctor of $\mathsf{M}$ which preserves all finite limits, does it follow in general that there exists an $M\in\mathsf{M}$ and a natural transformation $\tau\colon F\rightarrow \mathrm{Hom}_{\mathsf{M}}(M,-)$ such that for each injective $N\in\mathsf{M}$ the induced morphism $F(N)\rightarrow \mathrm{Hom}_{\textsf{M}}(M,N)$ is an isomorphism of $\mathsf{M}$?

Answer 1

Replacing $R$ by $R[x_1,\ldots,x_n]$ if necessary, we may assume $n = 0$. Note that preserving all finite limits is equivalent to being (additive and) left exact (see e.g. Tag 010N).

Lemma. Let $R$ be a commutative ring, let $F \colon \operatorname{Mod}_R \to \operatorname{Mod}_R$ be a left exact functor, and let $\tau \colon F \to \operatorname{Hom}_R(M,-)$ be a natural transformation that is an isomorphism on injective $R$-modules $I$. Then $\tau$ is a natural isomorphism.

Proof. Let $N$ be an arbitrary $R$-module, and choose a resolution $0 \to N \to I \to J$ with $I$ and $J$ injective. Then we get a commutative diagram $$\begin{array}{ccccccc}0 & \to & F(N) & \to & F(I) & \to & F(J) \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & \operatorname{Hom}_R(M,N) & \to & \operatorname{Hom}_R(M,I) & \to & \operatorname{Hom}_R(M,J) \end{array}$$ with exact rows. Since the right two vertical maps are isomorphisms, so is the left vertical map by the five lemma (three lemma?). $\square$

Thus, we get that $F$ is representable. This is not always the case under your assumptions:

Example. Let $R = \mathbb Z$, and let $F \colon \operatorname{\underline{Ab}} \to \operatorname{\underline{Ab}}$ be the functor $A \mapsto A_{\operatorname{tors}}$, the subgroup of torsion elements of $A$. This is clearly additive and left exact.

To show that $F$ is not representable, note that representable functors preserve all limits (not just finite ones). This is not the case for $(-)_{\operatorname{tors}}$. For example, the natural (injective) map $$\left(\prod_{n \in \mathbb Z_{>0}} \mathbb Z/n\mathbb Z\right)_{\operatorname{tors}} \to \prod_{n \in \mathbb Z_{>0}} (\mathbb Z/n\mathbb Z)_{\operatorname{tors}}$$ is not an isomorphism, since the left hand side does not contain the element $(1,1,\ldots)$. $\square$

Remark. If one really insists on using polynomial rings, one can replace $\mathbb Z$ by $k[x]$ where $k$ is a field, and use a similar argument.

Remark. For $F$ to be representable, it is necessary and sufficient that $F$ preserves all limits and satisfies some solution set condition. See for example Thm. V.6.3 of Mac Lane.