The Stacks project

Lemma 90.11.4. Let $L: \text{Mod}^{fg}_ R \to \textit{Sets}$, resp. $L: \text{Mod}_ R \to \textit{Sets}$ be a functor. Suppose $L(0)$ is a one element set and $L$ preserves finite products. Then there exists a unique $R$-linear functor $\widetilde{L} : \text{Mod}^{fg}_ R \to \text{Mod}_ R$, resp. $\widetilde{L} : \text{Mod}^{fg}_ R \to \text{Mod}_ R$, such that

\[ \vcenter { \xymatrix{ & \text{Mod}_ R \ar[dr]^{\text{forget}} & \\ \text{Mod}^{fg}_ R \ar[ur]^{\widetilde{L}} \ar[rr]^{L} & & \textit{Sets} } } \quad \text{resp.}\quad \vcenter { \xymatrix{ & \text{Mod}_ R \ar[dr]^{\text{forget}} & \\ \text{Mod}_ R \ar[ur]^{\widetilde{L}} \ar[rr]^{L} & & \textit{Sets} } } \]


Proof. We only prove this in case $L: \text{Mod}^{fg}_ R \to \textit{Sets}$. Let $M$ be a finitely generated $R$-module. We define $\widetilde{L}(M)$ to be the set $L(M)$ with the following $R$-module structure.

Multiplication: If $r \in R$, multiplication by $r$ on $L(M)$ is defined to be the map $L(M) \to L(M)$ induced by the multiplication map $r \cdot : M \to M$.

Addition: The sum map $M \times M \to M: (m_1, m_2) \mapsto m_1 + m_2$ induces a map $L(M \times M) \to L(M)$. By assumption $L(M \times M)$ is canonically isomorphic to $L(M) \times L(M)$. Addition on $L(M)$ is defined by the map $L(M) \times L(M) \cong L(M \times M) \to L(M)$.

Zero: There is a unique map $0 \to M$. The zero element of $L(M)$ is the image of $L(0) \to L(M)$.

We omit the verification that this defines an $R$-module $\widetilde{L}(M)$, the unique such that is $R$-linearly functorial in $M$. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06I6. Beware of the difference between the letter 'O' and the digit '0'.