The Stacks project

Proof. This holds because for any $A$-module $M$ we have $M \otimes _ A B = \mathop{\mathrm{colim}}\nolimits M \otimes _ A B_ i$ (see Algebra, Lemma 10.12.9) and because filtered colimits commute with exact sequences, see Algebra, Lemma 10.8.8. $\square$

Proof. Choose an exact sequence $0 \to F \to \underline{M} \to \underline{N}$. This gives the diagram

where the rows are exact (the top one because $B \to B'$ is flat). Since the right two vertical arrows are isomorphisms, so is the left one. $\square$

Proof. Choose an exact sequence $0 \to F \to \underline{M} \to \underline{N}$ where $\underline{M} \to \underline{N}$ is given by $\varphi : M \to N$. By Lemma 46.3.3 it suffices to construct $L \to F$ such that $L(B) \to F(B)$ is surjective for every finitely presented $A$-algebra $B$. Hence it suffices to construct, given a finitely presented $A$-algebra $B$ and an element $\xi \in F(B)$ a map $L \to F$ with $L$ linearly adequate such that $\xi $ is in the image of $L(B) \to F(B)$. (Because there is a set worth of such pairs $(B, \xi )$ up to isomorphism.)

To do this write $\sum _{i = 1, \ldots , n} m_ i \otimes b_ i$ the image of $\xi $ in $\underline{M}(B) = M \otimes _ A B$. We know that $\sum \varphi (m_ i) \otimes b_ i = 0$ in $N \otimes _ A B$. As $N$ is a filtered colimit of finitely presented $A$-modules, we can find a finitely presented $A$-module $N'$, a commutative diagram of $A$-modules

such that $(b_1, \ldots , b_ n)$ maps to zero in $N' \otimes _ A B$. Choose a presentation $A^{\oplus l} \to A^{\oplus k} \to N' \to 0$. Choose a lift $A^{\oplus n} \to A^{\oplus k}$ of the map $A^{\oplus n} \to N'$ of the diagram. Then we see that there exist $(c_1, \ldots , c_ l) \in B^{\oplus l}$ such that $(b_1, \ldots , b_ n, c_1, \ldots , c_ l)$ maps to zero in $B^{\oplus k}$ under the map $B^{\oplus n} \oplus B^{\oplus l} \to B^{\oplus k}$. Consider the commutative diagram

where the left vertical arrow is zero on the summand $A^{\oplus l}$. Then we see that $L$ equal to the kernel of $\underline{A^{\oplus n + l}} \to \underline{A^{\oplus k}}$ works because the element $(b_1, \ldots , b_ n, c_1, \ldots , c_ l) \in L(B)$ maps to $\xi $. $\square$

Proof. Let $x \in F(B)$. The map $p : B \to B[t, t^{-1}]$ is free hence we know that

as indicated we drop the $F(p)$ in the rest of the proof. Write $F(a)(x) = \sum t^ k x_ k$ for some $x_ k \in F(B)$. Denote $\epsilon : B[t, t^{-1}] \to B$ the $B$-algebra map $t \mapsto 1$. Note that the compositions $\epsilon \circ p, \epsilon \circ a : B \to B[t, t^{-1}] \to B$ are the identity. Hence we see that

On the other hand, we claim that $x_ k \in F(B)^{(k)}$. Namely, consider the commutative diagram

where $a'(b) = s^{\deg (b)}b$, $f(b) = b$, $f(t) = st$ and $g(b) = t^{\deg (b)}b$ and $g(s) = s$. Then

and going the other way we see

Since $B \to B[s, t, s^{-1}, t^{-1}]$ is free we see that $F(B[t, s, t^{-1}, s^{-1}]) = \bigoplus _{k, l \in \mathbf{Z}} F(B) \cdot t^ ks^ l$ and comparing coefficients in the expressions above we find $F(a)(x_ k) = t^ k x_ k$ as desired.

Finally, the image of $F(B_0) \to F(B)$ is contained in $F(B)^{(0)}$ because $B_0 \to B \xrightarrow {a} B[t, t^{-1}]$ is equal to $B_0 \to B \xrightarrow {p} B[t, t^{-1}]$. $\square$

Proof. Suppose that the map $A^{\oplus n} \to A^{\oplus m}$ is given by the $m \times n$-matrix $(a_{ij})$. Consider the ring $B = A[x_1, \ldots , x_ n]/(\sum a_{ij}x_ j)$. 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 $\xi \in L(B)$. Note that $\xi $ has the following universal property: for any $A$-algebra $C$ and any $\xi ' \in L(C)$ there exists an $A$-algebra map $B \to C$ such that $\xi $ maps to $\xi '$ via the map $L(B) \to L(C)$.

Note that $B$ is a graded $A$-algebra, hence we can use Lemmas 46.3.7 and 46.3.5 to decompose the values of our functors on $B$ into graded pieces. Note that $\xi \in L(B)^{(1)}$ as $(x_1, \ldots , x_ n)$ is an element of degree one in $\underline{A^{\oplus n}}(B)$. Hence we see that $\varphi (\xi ) \in \underline{M}(B)^{(1)} = M \otimes _ A B_1$. Since $B_1$ is generated by $x_1, \ldots , x_ n$ as an $A$-module we can write $\varphi (\xi ) = \sum m_ i \otimes x_ i$. Consider the map $A^{\oplus n} \to M$ which maps the $i$th basis vector to $m_ i$. By construction the associated map $\underline{A^{\oplus n}} \to \underline{M}$ maps the element $\xi $ to $\varphi (\xi )$. It follows from the universal property mentioned above that the diagram commutes. $\square$

Proof. By Lemma 46.3.6 we may assume that $F = \bigoplus L_ i$ is a direct sum of linearly adequate functors. Choose exact sequences $0 \to L_ i \to \underline{A^{\oplus n_ i}} \to \underline{A^{\oplus m_ i}}$. For each $i$ choose a map $A^{\oplus n_ i} \to M$ as in Lemma 46.3.8. Consider the diagram

Consider the $A$-modules

Then we see that $\mathop{\mathrm{Coker}}(\varphi )$ is isomorphic to the kernel of $\underline{Q} \to \underline{P}$. $\square$

Proof. Choose an injection $G \to \underline{M}$. Then we have an injection $G/F \to \underline{M}/F$. By Lemma 46.3.9 we see that $\underline{M}/F$ is adequate, hence we can find an injection $\underline{M}/F \to \underline{N}$. Composing we obtain an injection $G/F \to \underline{N}$. By Lemma 46.3.9 the cokernel of the induced map $G \to \underline{N}$ is adequate hence we can find an injection $\underline{N}/G \to \underline{K}$. Then $0 \to G/F \to \underline{N} \to \underline{K}$ is exact and we win. $\square$

Proof. Choose an injection $F \to \underline{M}$ and an injection $G \to \underline{N}$. Denote $F \to \underline{M \oplus N}$ the diagonal map so that

commutes. By Lemma 46.3.10 we can find a module map $M \oplus N \to K$ such that $F$ is the kernel of $\underline{M \oplus N} \to \underline{K}$. Then $\mathop{\mathrm{Ker}}(\varphi )$ is the kernel of $\underline{M \oplus N} \to \underline{K \oplus N}$. $\square$

Proof. The statement on direct sums is immediate. A general colimit can be written as a kernel of a map between direct sums, see Categories, Lemma 4.14.12. Hence this follows from Lemma 46.3.11. $\square$

Proof. Let $B$ be an $A$-algebra, $\xi \in F(B)$, and $b \in B$. We have to show that $\varphi (b \xi ) = b \varphi (\xi )$. Consider the ring map

This ring map is faithfully flat, hence $G(B) \subset G(B')$. On the other hand

because $x, y$ are units in $B'$. Hence we win. $\square$

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

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

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

We will show that the element

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

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

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. $\square$

Proof. Choose an exact sequence $0 \to F \to \underline{M} \to \underline{N}$. If we can show that $(\underline{M} \oplus G)/F$ is adequate, then $G$ is the kernel of the map of adequate functors $(\underline{M} \oplus G)/F \to \underline{N}$, hence adequate by Lemma 46.3.11. Thus we may assume $F = \underline{M}$.

We can choose a surjection $L \to H$ where $L$ is a direct sum of linearly adequate functors, see Lemma 46.3.6. If we can show that the pullback $G \times _ H L$ is adequate, then $G$ is the cokernel of the map $\mathop{\mathrm{Ker}}(L \to H) \to G \times _ H L$ hence adequate by Lemma 46.3.10. Thus we may assume that $H = \bigoplus L_ i$ is a direct sum of linearly adequate functors. By Lemma 46.3.14 each of the pullbacks $G \times _ H L_ i$ is adequate. By Lemma 46.3.12 we see that $\bigoplus G \times _ H L_ i$ is adequate. Then $G$ is the cokernel of

where $\xi $ in the summand $(i, i')$ maps to $(0, \ldots , 0, \xi , 0, \ldots , 0, -\xi , 0, \ldots , 0)$ with nonzero entries in the summands $i$ and $i'$. Thus $G$ is adequate by Lemma 46.3.10. $\square$

Proof. Choose an exact sequence $0 \to F \to \underline{M} \to \underline{N}$. Then $F'(B') = F(B') = \mathop{\mathrm{Ker}}(M \otimes _ A B' \to N \otimes _ A B')$. Since $M \otimes _ A B' = M \otimes _ A A' \otimes _{A'} B'$ and similarly for $N$ we see that $F'$ is the kernel of $\underline{M \otimes _ A A'} \to \underline{N \otimes _ A A'}$. $\square$

Proof. Choose an exact sequence $0 \to F' \to \underline{M'} \to \underline{N'}$. Then

Thus $F$ is the kernel of $\underline{M} \to \underline{N}$ where $M = M'$ and $N = N'$ viewed as $A$-modules. $\square$

Proof. This is true because an $A$-algebra $B$ is canonically a product $B_1 \times \ldots \times B_ n$ and the same thing holds for $A$-modules. Setting $F(B) = \coprod F_ i(B_ i)$ gives the correspondence. Details omitted. $\square$

Proof. The functors $B \to F(B \otimes _ A A')$ and $B \mapsto F(B \otimes _ A A' \otimes _ A A')$ are adequate, see Lemmas 46.3.18 and 46.3.17. Hence $F$ as a kernel of a map of adequate functors is adequate, see Lemma 46.3.11. $\square$