Lemma 29.50.3. Let $f : X \to Y$ be a birational morphism of schemes having finitely many irreducible components. If $y \in Y$ is the generic point of an irreducible component, then the base change $X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ is an isomorphism.

Proof. We may assume $Y = \mathop{\mathrm{Spec}}(B)$ is affine and irreducible. Then $X$ is irreducible too. If we prove the result for any nonempty affine open $U \subset X$, then the result holds for $X$ (small argument omitted). Hence we may assume $X$ is affine too, say $X = \mathop{\mathrm{Spec}}(A)$. Let $y \in Y$ correspond to the minimal prime $\mathfrak q \subset B$. By assumption $A$ has a unique minimal prime $\mathfrak p$ lying over $\mathfrak q$ and $B_\mathfrak q \to A_\mathfrak p$ is an isomorphism. It follows that $A_\mathfrak q \to \kappa (\mathfrak p)$ is surjective, hence $\mathfrak p A_\mathfrak q$ is a maximal ideal. On the other hand $\mathfrak p A_\mathfrak q$ is the unique minimal prime of $A_\mathfrak q$. We conclude that $\mathfrak p A_\mathfrak q$ is the unique prime of $A_\mathfrak q$ and that $A_\mathfrak q = A_\mathfrak p$. Since $A_\mathfrak q = A \otimes _ B B_\mathfrak q$ the lemma follows. $\square$

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).