Lemma 46.4.8. Let $A$ be a ring. Let $M$, $P$ be $A$-modules with $P$ of finite presentation. Then $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(\underline{P}, \underline{M}) = 0$ for $i > 0$ where $\mathcal{P}$ is the category of module-valued functors on $\textit{Alg}_ A$.

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

$E : 0 \to \underline{M} \to I^0 \to \ldots \to I^{i - 1} \times _{\mathop{\mathrm{Ker}}(d^ i)} \underline{P} \to \underline{P} \to 0$

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

$E : 0 \to \underline{M} \to F_{i - 1} \to \ldots \to F_0 \to \underline{P} \to 0$

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

$E' : 0 \to \underline{M} \to G_{i - 1} \to \ldots \to G_0 \to \underline{P} \to 0$

Moreover, the canonical maps

$F_ j(B) = B \otimes _ B F_ j(B) \longrightarrow TF_ j(B, B) = G_ j(B)$

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

$\xymatrix{ 0 \ar[r] & \underline{M} \ar[r] \ar[d]^1 & F_{i - 1} \ar[r] \ar[d] & \ldots \ar[r] & F_0 \ar[r] \ar[d] & \underline{P} \ar[r] \ar[d]^1 & 0 \\ 0 \ar[r] & \underline{M} \ar[r] & G_{i - 1} \ar[r] & \ldots \ar[r] & G_0 \ar[r] & \underline{P} \ar[r] & 0 }$

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

$0 \to T\underline{M}(A, N) \to TF_{i - 1}(A, N) \to \ldots \to TF_0(A, N) \to T\underline{P}(A, N) \to 0$

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

$0 \to M \otimes _ A N \to G_{i - 1}(A) \otimes _ A N \to \ldots \to G_0(A) \otimes _ A N \to P \otimes _ A N \to 0$

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

$0 \to K \to G_0(A) \to P \to 0 \quad \text{and}\quad G_2(A) \to G_1(A) \to K \to 0$

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).