## 109.5 The structure sheaf on the fibre product

Let $X, Y, S, a, b, p, q, f$ be as in the introduction to Derived Categories of Schemes, Section 36.23. Picture:

$\xymatrix{ & X \times _ S Y \ar[ld]^ p \ar[rd]_ q \ar[dd]^ f \\ X \ar[rd]_ a & & Y \ar[ld]^ b \\ & S }$

Then we have a canonical map

$can : p^{-1}\mathcal{O}_ X \otimes _{f^{-1}\mathcal{O}_ S} q^{-1}\mathcal{O}_ Y \longrightarrow \mathcal{O}_{X \times _ S Y}$

which is not an isomorphism in general.

For example, let $S = \mathop{\mathrm{Spec}}(\mathbf{R})$, $X = \mathop{\mathrm{Spec}}(\mathbf{C})$, and $Y = \mathop{\mathrm{Spec}}(\mathbf{C})$. Then $X \times _ S Y = \mathop{\mathrm{Spec}}(\mathbf{C}) \amalg \mathop{\mathrm{Spec}}(\mathbf{C})$ is a discrete space with two points and the sheaves $p^{-1}\mathcal{O}_ X$, $q^{-1}\mathcal{O}_ Y$ and $f^{-1}\mathcal{O}_ S$ are the constant sheaves with values $\mathbf{C}$, $\mathbf{C}$, and $\mathbf{R}$. Hence the source of $can$ is the constant sheaf with value $\mathbf{C} \otimes _\mathbf {R} \mathbf{C}$ on the discrete space with two points. Thus its global sections have dimension $8$ as an $\mathbf{R}$-vector space whereas taking global sections of the target of $can$ we obtain $\mathbf{C} \times \mathbf{C}$ which has dimension $4$ as an $\mathbf{R}$-vector space.

Another example is the following. Let $k$ be an algebraically closed field. Consider $S = \mathop{\mathrm{Spec}}(k)$, $X = \mathbf{A}^1_ k$, and $Y = \mathbf{A}^1_ k$. Then for $U \subset X \times _ S Y = \mathbf{A}^2_ k$ nonempty open the images $p(U) \subset X = \mathbf{A}^1_ k$ and $q(U) \subset \mathbf{A}^1_ k$ are open and the reader can show that

$\left( p^{-1}\mathcal{O}_ X \otimes _{f^{-1}\mathcal{O}_ S} q^{-1}\mathcal{O}_ Y \right)(U) = \mathcal{O}_ X(p(U)) \otimes _ k \mathcal{O}_ Y(q(U))$

This is not equal to $\mathcal{O}_{X \times _ S Y}(U)$ if $U$ is the complement of an irreducible curve $C$ in $X \times _ S Y = \mathbf{A}^2_ k$ such that both $p|_ C$ and $q|_ C$ are nonconstant.

Returning to the general case, let $z = (x, y, s, \mathfrak p)$ be a point of $X \times _ S Y$ as in Schemes, Lemma 26.17.5. Then on stalks at $z$ the map $can$ gives the map

$can_ z : \mathcal{O}_{X, x} \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X \times _ S Y, z}$

This is a flat ring homomorphism as the target is a localization of the source (details omitted; hint reduce to the case that $X$, $Y$, and $S$ are affine). Observe that the source is in general not a local ring, and this gives another way to see that $can$ is not an isomorphism in general.

More generally, suppose we have an $\mathcal{O}_ X$-module $\mathcal{F}$ and an $\mathcal{O}_ Y$-module $\mathcal{G}$. Then there is a canonical map

\begin{align*} & p^{-1}\mathcal{F} \otimes _{f^{-1}\mathcal{O}_ S} q^{-1}\mathcal{G} \\ & = p^{-1}(\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{O}_ X) \otimes _{f^{-1}\mathcal{O}_ S} q^{-1}(\mathcal{O}_ Y \otimes _{\mathcal{O}_ Y} \mathcal{G}) \\ & = p^{-1}\mathcal{F} \otimes _{p^{-1}\mathcal{O}_ X} p^{-1}\mathcal{O}_ X \otimes _{f^{-1}\mathcal{O}_ S} q^{-1}\mathcal{O}_ Y \otimes _{q^{-1}\mathcal{O}_ Y} q^{-1}\mathcal{G} \\ & \xrightarrow {can} p^{-1}\mathcal{F} \otimes _{q^{-1}\mathcal{O}_ X} \mathcal{O}_{X \times _ S Y} \otimes _{q^{-1}\mathcal{O}_ Y} q^{-1}\mathcal{G} \\ & = p^{-1}\mathcal{F} \otimes _{q^{-1}\mathcal{O}_ X} \mathcal{O}_{X \times _ S Y} \otimes _{\mathcal{O}_{X \times _ S Y}} \mathcal{O}_{X \times _ S Y} \otimes _{q^{-1}\mathcal{O}_ Y} q^{-1}\mathcal{G} \\ & = p^*\mathcal{F} \otimes _{\mathcal{O}_{X \times _ S Y}} q^*\mathcal{G} \end{align*}

which is rarely an isomorphism.

## Comments (3)

Comment #5574 by Julian on

In lines 4 and 5 it should read $p^{-1}\mathcal{O}_X$ instead of $q^{-1}\mathcal{O}_X$.

Comment #5577 by on

Unfortunately, besides this typo this comment is wrong! Argh! I will fix this ASAP.

## 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 0FLS. Beware of the difference between the letter 'O' and the digit '0'.