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
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