The Stacks project

Lemma 46.3.6. Let $A$ be a ring. Let $F$ be an adequate functor on $\textit{Alg}_ A$. Then there exists a surjection $L \to F$ with $L$ a direct sum of linearly adequate functors.

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

\[ \xymatrix{ A^{\oplus n} \ar[r] \ar[d]_{m_1, \ldots , m_ n} & N' \ar[d] \\ M \ar[r] & N } \]

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

\[ \xymatrix{ A^{\oplus n} \oplus A^{\oplus l} \ar[r] \ar[d] & A^{\oplus k} \ar[d] \\ M \ar[r] & N } \]

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 $. $\square$

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