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I have some questions about algebraic geometry which might be elementary and boring (sorry).

Let $R$ be a ring - say, an integral domain which is Noetherian. Let $X$ and $Y$ be $R$-varieties - that is, integral separated schemes of finite type over $R$. Let $f: X \to Y$ be a dominant morphism of $R$-varieties. Let $\beta: Y' \to Y$ be a proper, birational morphism.

  • I believe that the fibre product $X \times_Y Y'$ can be reducible but does anyone know a simple example of this phenomenon?
  • Is it true that the fibre product $X \times_Y Y'$ has a unique irreducible component $X'$ which dominates $Y'$?
  • If the answer to the previous question is yes, is it true that the composition $X' \to X \times_Y Y' \to X$ is birational?
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1 Answer 1

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  1. First notice that there exists a dense open subset $V$ of $Y$ such that $\beta^{-1}(V)\to V$ is an isomorphism. Above this $V$, $X\times_Y Y'$ is then canonically isomorphic to $f^{-1}(V)$.

  2. Let $R$ be an algebraically closed field, $Y$ a curve with a node $y_0$ and $Y'\to Y$ the normalization map. Let $X$ be any normal variety with a surjective morphism to $Y$. Then $X\times_Y Y'\to X$ is finite (because $Y'\to Y$ is finite). If $X\times_Y Y'$ were irreducible, then $(X\times_Y Y')_{\mathrm{red}}$ would be integral, finite and birational (by (1) above) to $X$. As $X$ is normal, $(X\times_Y Y')_{\mathrm{red}}\to X$ would be an isomorphism. Let $x_0\in f^{-1}(y_0)$, then the pre-image of $x_0$ by this isomorphism is $\{x_0\}\times \beta^{-1}(y_0)$. But the latter has two points because $y_0$ is a node. Contradiction.

  3. By (1), the pre-image by $X\times_Y Y'\to Y'$ of $\beta^{-1}(V)$ is irreducible and isomorphic to $f^{-1}(V)$. So there exists a unique irreducible component $X'$ in $X\times_Y Y'$ dominating $Y'$. It is equal to the Zariski closure of $f^{-1}(V)\times_V \beta^{-1}(V)$ in $X\times_Y Y'$.

  4. $X'$ is birational to $X$ by (3).

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