Lemma 64.15.6. With assumptions as in Lemma 64.15.5, let $u\in \text{End}_{A[G]}(P)$ and $v\in \text{End}_{\Lambda [G]}(M)$. Then

**Sketch of proof.**
Reduce to the case $P=A[G]$. In that case, $u$ is right multiplication by some element $a = \sum a_ gg$ of $A[G]$, which we write $u = R_ a$. There is an isomorphism of $\Lambda [G]$-modules

where $\left(A[G]\otimes _ A M\right)'$ has the module structure given by the left $G$-action, together with the $\Lambda $-linearity on $M$. This transport of structure changes $u \otimes v$ into $\sum _ ga_ gR_ g \otimes g^{-1}v$. In other words,

Working out explicitly both sides of the equation, we have to show

This is done by showing that

by reducing to $M=\Lambda $. $\square$

## 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Your email address will not be published.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)