Lemma 46.4.9. Let $A$ be a ring. Let $M$ be an $A$-module. Let $L$ be a linearly adequate functor on $\textit{Alg}_ A$. Then $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(L, \underline{M}) = 0$ for $i > 0$ where $\mathcal{P}$ is the category of module-valued functors on $\textit{Alg}_ A$.

**Proof.**
Since $L$ is linearly adequate there exists an exact sequence

Here $P = \mathop{\mathrm{Coker}}(A^{\oplus m} \to A^{\oplus n})$ is the cokernel of the map of finite free $A$-modules which is given by the definition of linearly adequate functors. By Lemma 46.4.8 we have the vanishing of $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(\underline{P}, \underline{M})$ and $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(\underline{A}, \underline{M})$ for $i > 0$. Let $K = \mathop{\mathrm{Ker}}(\underline{A^{\oplus n}} \to \underline{P})$. By the long exact sequence of Ext groups associated to the exact sequence $0 \to K \to \underline{A^{\oplus n}} \to \underline{P} \to 0$ we conclude that $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(K, \underline{M}) = 0$ for $i > 0$. Repeating with the sequence $0 \to L \to \underline{A^{\oplus m}} \to K \to 0$ we win. $\square$

## 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Your email address will not be published.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.

## Comments (0)