The Stacks project

Example 48.15.2. The base change map ( is not an isomorphism if $f$ is perfect proper and $g$ is perfect. Let $k$ be a field. Let $Y = \mathbf{A}^2_ k$ and let $f : X \to Y$ be the blowup of $Y$ in the origin. Denote $E \subset X$ the exceptional divisor. Then we can factor $f$ as

\[ X \xrightarrow {i} \mathbf{P}^1_ Y \xrightarrow {p} Y \]

This gives a factorization $a = c \circ b$ where $a$, $b$, and $c$ are the right adjoints of Lemma 48.3.1 of $Rf_*$, $Rp_*$, and $Ri_*$. Denote $\mathcal{O}(n)$ the Serre twist of the structure sheaf on $\mathbf{P}^1_ Y$ and denote $\mathcal{O}_ X(n)$ its restriction to $X$. Note that $X \subset \mathbf{P}^1_ Y$ is cut out by a degree one equation, hence $\mathcal{O}(X) = \mathcal{O}(1)$. By Lemma 48.15.1 we have $b(\mathcal{O}_ Y) = \mathcal{O}(-2)[1]$. By Lemma 48.9.7 we have

\[ a(\mathcal{O}_ Y) = c(b(\mathcal{O}_ Y)) = c(\mathcal{O}(-2)[1]) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ X, \mathcal{O}(-2)[1]) = \mathcal{O}_ X(-1) \]

Last equality by Lemma 48.14.2. Let $Y' = \mathop{\mathrm{Spec}}(k)$ be the origin in $Y$. The restriction of $a(\mathcal{O}_ Y)$ to $X' = E = \mathbf{P}^1_ k$ is an invertible sheaf of degree $-1$ placed in cohomological degree $0$. But on the other hand, $a'(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}) = \mathcal{O}_ E(-2)[1]$ which is an invertible sheaf of degree $-2$ placed in cohomological degree $-1$, so different. In this example the hypothesis of Tor independence in Lemma 48.6.2 is violated.

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Your email address will not be published.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 0AAC. Beware of the difference between the letter 'O' and the digit '0'.