The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06ZG. Beware of the difference between the letter 'O' and the digit '0'.