Lemma 24.10.2. In the situation above, let $\mathcal{M}$ be a right graded $\mathcal{A}_ U$-module and let $\mathcal{N}$ be a left graded $\mathcal{A}$-module. Then

as graded $\mathcal{O}$-modules functorially in $\mathcal{M}$ and $\mathcal{N}$.

Lemma 24.10.2. In the situation above, let $\mathcal{M}$ be a right graded $\mathcal{A}_ U$-module and let $\mathcal{N}$ be a left graded $\mathcal{A}$-module. Then

\[ j_!\mathcal{M} \otimes _\mathcal {A} \mathcal{N} = j_!(\mathcal{M} \otimes _{\mathcal{A}_ U} \mathcal{N}|_ U) \]

as graded $\mathcal{O}$-modules functorially in $\mathcal{M}$ and $\mathcal{N}$.

**Proof.**
Recall that the degree $n$ component of $j_!\mathcal{M} \otimes _\mathcal {A} \mathcal{N}$ is the cokernel of the canonical map

\[ \bigoplus \nolimits _{r + s + t = n} j_!\mathcal{M}^ r \otimes _\mathcal {O} \mathcal{A}^ s \otimes _\mathcal {O} \mathcal{N}^ t \longrightarrow \bigoplus \nolimits _{p + q = n} j_!\mathcal{M}^ p \otimes _\mathcal {O} \mathcal{N}^ q \]

See Section 24.6. By Modules on Sites, Lemma 18.27.9 this is the same thing as the cokernel of

\[ \bigoplus \nolimits _{r + s + t = n} j_!(\mathcal{M}^ r \otimes _{\mathcal{O}_ U} \mathcal{A}^ s|_ U \otimes _{\mathcal{O}_ U} \mathcal{N}^ t|_ U) \longrightarrow \bigoplus \nolimits _{p + q = n} j_!(\mathcal{M}^ p \otimes _{\mathcal{O}_ U} \mathcal{N}^ q|_ U) \]

and we win. An alternative proof would be to redo the Yoneda argument given in the proof of the lemma cited above. $\square$

Your email address will not be published.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)