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). $\square$

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