Lemma 45.10.9. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $k''/k'/k$ be finite separable algebras and let $X$ be a smooth projective scheme over $k'$. Then

**Proof.**
We will use the results of Lemma 45.10.2 without further mention. Write

for some finite separable $k'$-algebra $l$. Write $F' = H^0(\mathop{\mathrm{Spec}}(k'))$, $F'' = H^0(\mathop{\mathrm{Spec}}(k''))$, and $G = H^0(\mathop{\mathrm{Spec}}(l))$. Since $\mathop{\mathrm{Spec}}(k') \times \mathop{\mathrm{Spec}}(k'') = \mathop{\mathrm{Spec}}(k'') \amalg \mathop{\mathrm{Spec}}(l)$ we deduce from axiom (B)(a) and Lemma 45.9.9 that we have

The map from left to right identifies $F''$ with $F' \otimes _{F'} F''$. By the same token we have

as modules over $F' \otimes _ F F'' = F'' \times G$. This proves the lemma. $\square$

## 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Your email address will not be published.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)