Lemma 46.3.14. Let $A$ be a ring. Let $0 \to \underline{M} \to G \to L \to 0$ be a short exact sequence of module-valued functors on $\textit{Alg}_ A$ with $L$ linearly adequate. Then $G$ is adequate.

Proof. We first point out that for any flat $A$-algebra map $B \to B'$ the map $G(B) \otimes _ B B' \to G(B')$ is an isomorphism. Namely, this holds for $\underline{M}$ and $L$, see Lemma 46.3.5 and hence follows for $G$ by the five lemma. In particular, by Lemma 46.3.7 we see that $G(B) = \bigoplus _{k \in \mathbf{Z}} G(B)^{(k)}$ for any graded $A$-algebra $B$.

Choose an exact sequence $0 \to L \to \underline{A^{\oplus n}} \to \underline{A^{\oplus m}}$. Suppose that the map $A^{\oplus n} \to A^{\oplus m}$ is given by the $m \times n$-matrix $(a_{ij})$. Consider the graded $A$-algebra $B = A[x_1, \ldots , x_ n]/(\sum a_{ij}x_ j)$. The element $(x_1, \ldots , x_ n) \in \underline{A^{\oplus n}}(B)$ maps to zero in $\underline{A^{\oplus m}}(B)$ hence is the image of a unique element $\xi \in L(B)$. Observe that $\xi \in L(B)^{(1)}$. The map

$\mathop{\mathrm{Hom}}\nolimits _ A(B, C) \longrightarrow L(C), \quad f \longmapsto L(f)(\xi )$

defines an isomorphism of functors. The reason is that $f$ is determined by the images $c_ i = f(x_ i) \in C$ which have to satisfy the relations $\sum a_{ij}c_ j = 0$. And $L(C)$ is the set of $n$-tuples $(c_1, \ldots , c_ n)$ satisfying the relations $\sum a_{ij} c_ j = 0$.

Since the value of each of the functors $\underline{M}$, $G$, $L$ on $B$ is a direct sum of its weight spaces (by the lemma mentioned above) exactness of $0 \to \underline{M} \to G \to L \to 0$ implies the sequence $0 \to \underline{M}(B)^{(1)} \to G(B)^{(1)} \to L(B)^{(1)} \to 0$ is exact. Thus we may choose an element $\theta \in G(B)^{(1)}$ mapping to $\xi$.

Consider the graded $A$-algebra

$C = A[x_1, \ldots , x_ n, y_1, \ldots , y_ n]/ (\sum a_{ij}x_ j, \sum a_{ij}y_ j)$

There are three graded $A$-algebra homomorphisms $p_1, p_2, m : B \to C$ defined by the rules

$p_1(x_ i) = x_ i, \quad p_1(x_ i) = y_ i, \quad m(x_ i) = x_ i + y_ i.$

We will show that the element

$\tau = G(m)(\theta ) - G(p_1)(\theta ) - G(p_2)(\theta ) \in G(C)$

is zero. First, $\tau$ maps to zero in $L(C)$ by a direct calculation. Hence $\tau$ is an element of $\underline{M}(C)$. Moreover, since $m$, $p_1$, $p_2$ are graded algebra maps we see that $\tau \in G(C)^{(1)}$ and since $\underline{M} \subset G$ we conclude

$\tau \in \underline{M}(C)^{(1)} = M \otimes _ A C_1.$

We may write uniquely $\tau = \underline{M}(p_1)(\tau _1) + \underline{M}(p_2)(\tau _2)$ with $\tau _ i \in M \otimes _ A B_1 = \underline{M}(B)^{(1)}$ because $C_1 = p_1(B_1) \oplus p_2(B_1)$. Consider the ring map $q_1 : C \to B$ defined by $x_ i \mapsto x_ i$ and $y_ i \mapsto 0$. Then $\underline{M}(q_1)(\tau ) = \underline{M}(q_1)(\underline{M}(p_1)(\tau _1) + \underline{M}(p_2)(\tau _2)) = \tau _1$. On the other hand, because $q_1 \circ m = q_1 \circ p_1$ we see that $G(q_1)(\tau ) = - G(q_1 \circ p_2)(\tau )$. Since $q_1 \circ p_2$ factors as $B \to A \to B$ we see that $G(q_1 \circ p_2)(\tau )$ is in $G(B)^{(0)}$, see Lemma 46.3.7. Hence $\tau _1 = 0$ because it is in $G(B)^{(0)} \cap \underline{M}(B)^{(1)} \subset G(B)^{(0)} \cap G(B)^{(1)} = 0$. Similarly $\tau _2 = 0$, whence $\tau = 0$.

Since $\theta \in G(B)$ we obtain a transformation of functors

$\psi : L(-) = \mathop{\mathrm{Hom}}\nolimits _ A(B, - ) \longrightarrow G(-)$

by mapping $f : B \to C$ to $G(f)(\theta )$. Since $\theta$ is a lift of $\xi$ the map $\psi$ is a right inverse of $G \to L$. In terms of $\psi$ the statements proved above have the following meaning: $\tau = 0$ means that $\psi$ is additive and $\theta \in G(B)^{(1)}$ implies that for any $A$-algebra $D$ we have $\psi (ul) = u\psi (l)$ in $G(D)$ for $l \in L(D)$ and $u \in D^*$ a unit. This implies that $\psi$ is a morphism of module-valued functors, see Lemma 46.3.13. Clearly this implies that $G \cong \underline{M} \oplus L$ and we win. $\square$

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