The Stacks project

Lemma 46.4.5. Let $A$ be a ring. Let $F$ be a module-valued functor such that for any $B \in \mathop{\mathrm{Ob}}\nolimits (\textit{Alg}_ A)$ the functor $TF(B, -)$ on $B$-modules transforms a short exact sequence of $B$-modules into a right exact sequence. Then

  1. $TF(B, N_1 \oplus N_2) = TF(B, N_1) \oplus TF(B, N_2)$,

  2. there is a second functorial $B$-module structure on $TF(B, N)$ defined by setting $x \cdot b = TF(B, b\cdot 1_ N)(x)$ for $x \in TF(B, N)$ and $b \in B$,

  3. the canonical map $N \otimes _ B F(B) \to TF(B, N)$ of Lemma 46.4.3 is $B$-linear also with respect to the second $B$-module structure,

  4. given a finitely presented $B$-module $N$ there is a canonical isomorphism $TF(B, B) \otimes _ B N \to TF(B, N)$ where the tensor product uses the second $B$-module structure on $TF(B, B)$.

Proof. We will use the results of Lemma 46.4.3 without further mention. The maps $N_1 \to N_1 \oplus N_2$ and $N_2 \to N_1 \oplus N_2$ give a map $TF(B, N_1) \oplus TF(B, N_2) \to TF(B, N_1 \oplus N_2)$ which is injective since the maps $N_1 \oplus N_2 \to N_1$ and $N_1 \oplus N_2 \to N_2$ induce an inverse. Since $TF$ is right exact we see that $TF(B, N_1) \to TF(B, N_1 \oplus N_2) \to TF(B, N_2) \to 0$ is exact. Hence $TF(B, N_1) \oplus TF(B, N_2) \to TF(B, N_1 \oplus N_2)$ is an isomorphism. This proves (1).

To see (2) the only thing we need to show is that $x \cdot (b_1 + b_2) = x \cdot b_1 + x \cdot b_2$. (Associativity and additivity are clear.) To see this consider

\[ N \xrightarrow {(b_1, b_2)} N \oplus N \xrightarrow {+} N \]

and apply $TF(B, -)$.

Part (3) follows immediately from the fact that $N \otimes _ B F(B) \to TF(B, N)$ is functorial in the pair $(B, N)$.

Suppose $N$ is a finitely presented $B$-module. Choose a presentation $B^{\oplus m} \to B^{\oplus n} \to N \to 0$. This gives an exact sequence

\[ TF(B, B^{\oplus m}) \to TF(B, B^{\oplus n}) \to TF(B, N) \to 0 \]

by right exactness of $TF(B, -)$. By part (1) we can write $TF(B, B^{\oplus m}) = TF(B, B)^{\oplus m}$ and $TF(B, B^{\oplus n}) = TF(B, B)^{\oplus n}$. Next, suppose that $B^{\oplus m} \to B^{\oplus n}$ is given by the matrix $T = (b_{ij})$. Then the induced map $TF(B, B)^{\oplus m} \to TF(B, B)^{\oplus n}$ is given by the matrix with entries $TF(B, b_{ij} \cdot 1_ B)$. This combined with right exactness of $\otimes $ proves (4). $\square$

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Your email address will not be published.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 06ZB. Beware of the difference between the letter 'O' and the digit '0'.