Lemma 46.4.3. Let $A$ be a ring. Let $F$ be a module valued functor. For every $B \in \mathop{\mathrm{Ob}}\nolimits (\textit{Alg}_ A)$ and $B$-module $N$ there is a canonical decomposition

$F(B[N]) = F(B) \oplus TF(B, N)$

characterized by the following properties

1. $TF(B, N) = \mathop{\mathrm{Ker}}(F(B[N]) \to F(B))$,

2. there is a $B$-module structure $TF(B, N)$ compatible with $B[N]$-module structure on $F(B[N])$,

3. $TF$ is a functor from the category of pairs $(B, N)$,

4. there are canonical maps $N \otimes _ B F(B) \to TF(B, N)$ inducing a transformation between functors defined on the category of pairs $(B, N)$,

5. $TF(B, 0) = 0$ and the map $TF(B, N) \to TF(B, N')$ is zero when $N \to N'$ is the zero map.

Proof. Since $B \to B[N] \to B$ is the identity we see that $F(B) \to F(B[N])$ is a direct summand whose complement is $TF(N, B)$ as defined in (1). This construction is functorial in the pair $(B, N)$ simply because given a morphism of pairs $(B, N) \to (B', N')$ we obtain a commutative diagram

$\xymatrix{ B' \ar[r] & B'[N'] \ar[r] & B' \\ B \ar[r] \ar[u] & B[N] \ar[r] \ar[u] & B \ar[u] }$

in $\textit{Alg}_ A$. The $B$-module structure comes from the $B[N]$-module structure and the ring map $B \to B[N]$. The map in (4) is the composition

$N \otimes _ B F(B) \longrightarrow B[N] \otimes _{B[N]} F(B[N]) \longrightarrow F(B[N])$

whose image is contained in $TF(B, N)$. (The first arrow uses the inclusions $N \to B[N]$ and $F(B) \to F(B[N])$ and the second arrow is the multiplication map.) If $N = 0$, then $B = B[N]$ hence $TF(B, 0) = 0$. If $N \to N'$ is zero then it factors as $N \to 0 \to N'$ hence the induced map is zero since $TF(B, 0) = 0$. $\square$

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