The Stacks project

Proof. Choose a set $\{ L_ i\} _{i \in I}$ of linearly adequate functors such that every linearly adequate functor is isomorphic to one of the $L_ i$. This is possible. Suppose that we can find $Q(F) \to F$ with (1) and (2)' or every $i \in I$ the map $\mathop{\mathrm{Hom}}\nolimits (L_ i, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (L_ i, F)$ is a bijection. Then (2) holds. Namely, combining Lemmas 46.3.6 and 46.3.11 we see that every adequate functor $G$ sits in an exact sequence

with $K$ and $L$ direct sums of linearly adequate functors. Hence (2)' implies that $\mathop{\mathrm{Hom}}\nolimits (L, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (L, F)$ and $\mathop{\mathrm{Hom}}\nolimits (K, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (K, F)$ are bijections, whence the same thing for $G$.

Consider the category $\mathcal{I}$ whose objects are pairs $(i, \varphi )$ where $i \in I$ and $\varphi : L_ i \to F$ is a morphism. A morphism $(i, \varphi ) \to (i', \varphi ')$ is a map $\psi : L_ i \to L_{i'}$ such that $\varphi ' \circ \psi = \varphi $. Set

There is a natural map $Q(F) \to F$, by Lemma 46.3.12 it is adequate, and by construction it has property (2)'. $\square$

Proof. This lemma just collects some facts we have already seen so far. Part (1) is clear from the definitions, the characterization of weak Serre subcategories (see Homology, Lemma 12.10.3), and Lemmas 46.3.10, 46.3.11, and 46.3.16. Recall that $\mathcal{P}$ is equivalent to the category $\textit{PMod}((\textit{Aff}/\mathop{\mathrm{Spec}}(A))_\tau , \mathcal{O})$. Hence (2) by Injectives, Proposition 19.8.5. Part (3) follows from Lemma 46.4.1 and Categories, Lemma 4.24.5. Parts (4) and (5) follow from Homology, Lemmas 12.29.1 and 12.29.3. $\square$

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

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

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 $TF(B, 0) = 0$. $\square$

Proof. We will use the results of Lemma 46.4.3 without further mention. Denote $h : \textit{Alg}_ A \to \textit{Sets}$ the functor given by $h(C) = \mathop{\mathrm{Mor}}\nolimits _ A(B[N], C)$. Similarly for $h_1$ and $h_2$. The map $B[N] \to B[N_2]$ corresponding to the surjection $N \to N_2$ is surjective. It corresponds to a map $h_2 \to h$ such that $h_2(C) \to h(C)$ is injective for all $A$-algebras $C$. On the other hand, there are two maps $p, q : h \to h_1$, corresponding to the zero map $N_1 \to N$ and the injection $N_1 \to N$. Note that

is an equalizer diagram. Denote $\mathcal{O}_ h$ the module-valued functor $C \mapsto \bigoplus _{h(C)} C$. Similarly for $\mathcal{O}_{h_1}$ and $\mathcal{O}_{h_2}$. Note that

where $\mathcal{P}$ is the category of module-valued functors on $\textit{Alg}_ A$. We claim there is an equalizer diagram

in $\mathcal{P}$. Namely, suppose that $C \in \mathop{\mathrm{Ob}}\nolimits (\textit{Alg}_ A)$ and $\xi = \sum _{i = 1, \ldots , n} c_ i \cdot f_ i$ where $c_ i \in C$ and $f_ i : B[N] \to C$ is an element of $\mathcal{O}_ h(C)$. If $p(\xi ) = q(\xi )$, then we see that

where $z, y : B[N_1] \to B[N]$ are the maps $z : (b, m_1) \mapsto (b, 0)$ and $y : (b, m_1) \mapsto (b, m_1)$. This means that for every $i$ there exists a $j$ such that $f_ j \circ z = f_ i \circ y$. Clearly, this implies that $f_ i(N_1) = 0$, i.e., $f_ i$ factors through a unique map $\overline{f}_ i : B[N_2] \to C$. Hence $\xi $ is the image of $\overline{\xi } = \sum c_ i \cdot \overline{f}_ i$. Since $I$ is injective, it transforms this equalizer diagram into a coequalizer diagram

This diagram is compatible with the direct sum decompositions $I(B[N]) = I(B) \oplus TI(B, N)$ and $I(B[N_ i]) = I(B) \oplus TI(B, N_ i)$. The zero map $N \to N_1$ induces the zero map $TI(B, N) \to TI(B, N_1)$. Thus we see that the coequalizer property above means we have an exact sequence $TI(B, N_1) \to TI(B, N) \to TI(B, N_2) \to 0$ as desired. $\square$

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

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

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). $\square$

Proof. Let $B$ be given. Let $0 \to N_1 \to N_2 \to N_3 \to 0$ be a short exact sequence of $B$-modules. Part (1) follows from a diagram chase in the diagram

with exact horizontal rows and exact columns involving $TF$ and $TG$. To prove part (2) we do a diagram chase in the diagram

with exact horizontal rows and exact columns involving $TF$ and $TH$. Part (3) follows from part (2) as $G \times _ H H'$ sits in the exact sequence $0 \to F \to G \times _ H H' \to H' \to 0$. $\square$

Proof. Choose an injective resolution $\underline{M} \to I^\bullet $ in $\mathcal{P}$, see Lemma 46.4.2. By Derived Categories, Lemma 13.27.2 any element of $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(\underline{P}, \underline{M})$ comes from a morphism $\varphi : \underline{P} \to I^ i$ with $d^ i \circ \varphi = 0$. We will prove that the Yoneda extension

of $\underline{P}$ by $\underline{M}$ associated to $\varphi $ is trivial, which will prove the lemma by Derived Categories, Lemma 13.27.5.

For $F$ a module-valued functor on $\textit{Alg}_ A$ say $(*)$ holds if for all $B \in \mathop{\mathrm{Ob}}\nolimits (\textit{Alg}_ A)$ the functor $TF(B, -)$ on $B$-modules transforms a short exact sequence of $B$-modules into a right exact sequence. Recall that the module-valued functors $\underline{M}, I^ n, \underline{P}$ each have property $(*)$, see Lemma 46.4.4 and the remarks preceding it. By splitting $0 \to \underline{M} \to I^\bullet $ into short exact sequences we find that each of the functors $\mathop{\mathrm{Im}}(d^{n - 1}) = \mathop{\mathrm{Ker}}(d^ n) \subset I^ n$ has property $(*)$ by Lemma 46.4.7 and also that $I^{i - 1} \times _{\mathop{\mathrm{Ker}}(d^ i)} \underline{P}$ has property $(*)$.

Thus we may assume the Yoneda extension is given as

where each of the module-valued functors $F_ j$ has property $(*)$. Set $G_ j(B) = TF_ j(B, B)$ viewed as a $B$-module via the second $B$-module structure defined in Lemma 46.4.5. Since $TF_ j$ is a functor on pairs we see that $G_ j$ is a module-valued functor on $\textit{Alg}_ A$. Moreover, since $E$ is an exact sequence the sequence $G_{j + 1} \to G_ j \to G_{j - 1}$ is exact (see remark preceding Lemma 46.4.7). Observe that $T\underline{M}(B, B) = M \otimes _ A B = \underline{M}(B)$ and that the two $B$-module structures agree on this. Thus we obtain a Yoneda extension

Moreover, the canonical maps

of Lemma 46.4.3 (4) are $B$-linear by Lemma 46.4.5 (3) and functorial in $B$. Hence a map

of Yoneda extensions. In particular we see that $E$ and $E'$ have the same class in $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(\underline{P}, \underline{M})$ by the lemma on Yoneda Exts mentioned above. Finally, let $N$ be a $A$-module of finite presentation. Then we see that

is exact. By Lemma 46.4.5 (4) with $B = A$ this translates into the exactness of the sequence of $A$-modules

Hence the sequence of $A$-modules $0 \to M \to G_{i - 1}(A) \to \ldots \to G_0(A) \to P \to 0$ is universally exact, in the sense that it remains exact on tensoring with any finitely presented $A$-module $N$. Let $K = \mathop{\mathrm{Ker}}(G_0(A) \to P)$ so that we have exact sequences

Tensoring the second sequence with $N$ we obtain that $K \otimes _ A N = \mathop{\mathrm{Coker}}(G_2(A) \otimes _ A N \to G_1(A) \otimes _ A N)$. Exactness of $G_2(A) \otimes _ A N \to G_1(A) \otimes _ A N \to G_0(A) \otimes _ A N$ then implies that $K \otimes _ A N \to G_0(A) \otimes _ A N$ is injective. By Algebra, Theorem 10.82.3 this means that the $A$-module extension $0 \to K \to G_0(A) \to P \to 0$ is exact, and because $P$ is assumed of finite presentation this means the sequence is split, see Algebra, Lemma 10.82.4. Any splitting $P \to G_0(A)$ defines a map $\underline{P} \to G_0$ which splits the surjection $G_0 \to \underline{P}$. Thus the Yoneda extension $E'$ is equivalent to the trivial Yoneda extension and we win. $\square$

Proof. Since $L$ is linearly adequate there exists an exact sequence

Here $P = \mathop{\mathrm{Coker}}(A^{\oplus m} \to A^{\oplus n})$ is the cokernel of the map of finite free $A$-modules which is given by the definition of linearly adequate functors. By Lemma 46.4.8 we have the vanishing of $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(\underline{P}, \underline{M})$ and $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(\underline{A}, \underline{M})$ for $i > 0$. Let $K = \mathop{\mathrm{Ker}}(\underline{A^{\oplus n}} \to \underline{P})$. By the long exact sequence of Ext groups associated to the exact sequence $0 \to K \to \underline{A^{\oplus n}} \to \underline{P} \to 0$ we conclude that $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(K, \underline{M}) = 0$ for $i > 0$. Repeating with the sequence $0 \to L \to \underline{A^{\oplus m}} \to K \to 0$ we win. $\square$

Proof. Choose an exact sequence $0 \to F \to \underline{M^0} \to \underline{M^1}$. Set $M^2 = \mathop{\mathrm{Coker}}(M^0 \to M^1)$ so that $0 \to F \to \underline{M^0} \to \underline{M^1} \to \underline{M^2} \to 0$ is a resolution. By Derived Categories, Lemma 13.21.3 we obtain a spectral sequence

Since $Q(\underline{M^ q}) = \underline{M^ q}$ it suffices to prove $R^ pQ(\underline{M}) = 0$, $p > 0$ for any $A$-module $M$.

Choose an injective resolution $\underline{M} \to I^\bullet $ in the category $\mathcal{P}$. Suppose that $R^ iQ(\underline{M})$ is nonzero. Then $\mathop{\mathrm{Ker}}(Q(I^ i) \to Q(I^{i + 1}))$ is strictly bigger than the image of $Q(I^{i - 1}) \to Q(I^ i)$. Hence by Lemma 46.3.6 there exists a linearly adequate functor $L$ and a map $\varphi : L \to Q(I^ i)$ mapping into the kernel of $Q(I^ i) \to Q(I^{i + 1})$ which does not factor through the image of $Q(I^{i - 1}) \to Q(I^ i)$. Because $Q$ is a left adjoint to the inclusion functor the map $\varphi $ corresponds to a map $\varphi ' : L \to I^ i$ with the same properties. Thus $\varphi '$ gives a nonzero element of $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(L, \underline{M})$ contradicting Lemma 46.4.9. $\square$