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 $\mathsf{M}$?



On some sense of representing an endofunctor of the category of modules over polynomial rings
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](http://stacks.math.columbia.edu/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.


