Lemma 89.11.8. Let $R$ be an $S$-algebra, and let $\mathcal{C}$ be a strictly full subcategory of $S\text{-Alg}/R$ containing $R[M]$ for all $M \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$. Let $F: \mathcal{C} \to \textit{Sets}$ be a functor. Suppose that $F(R)$ is a one element set and that for any $M, N \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$, the induced map

$F(R[M] \times _ R R[N]) \to F(R[M]) \times F(R[N])$

is a bijection. Then $F(R[M])$ has a natural $R$-module structure for any $M \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$.

Proof. Note that $R \cong R[0]$ and $R[M] \times _ R R[N] \cong R[M \times N]$ hence $R$ and $R[M] \times _ R R[N]$ are objects of $\mathcal{C}$ by our assumptions on $\mathcal{C}$. Thus the conditions on $F$ make sense. The functor $\text{Mod}_ R \to S\text{-Alg}/R$ of Lemma 89.11.7 restricts to a functor $\text{Mod}^{fg}_ R \to \mathcal{C}$ by the assumption on $\mathcal{C}$. Let $L$ be the composition $\text{Mod}^{fg}_ R \to \mathcal{C} \to \textit{Sets}$, i.e., $L(M) = F(R[M])$. Then $L$ preserves finite products by Lemma 89.11.7 and the assumption on $F$. Hence Lemma 89.11.4 shows that $L(M) = F(R[M])$ has a natural $R$-module structure for any $M \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$. $\square$

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