## Tag `0BAJ`

Chapter 28: Morphisms of Schemes > Section 28.48: Generically finite morphisms

Lemma 28.48.6. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be a birational morphism between schemes which have finitely many irreducible components. Assume

- either $f$ is quasi-compact or $f$ is separated, and
- either $f$ is locally of finite type and $Y$ is reduced or $f$ is locally of finite presentation.
Then there exists a dense open $V \subset Y$ such that $f^{-1}(V) \to V$ is an isomorphism.

Proof.By Lemma 28.48.5 we may assume that $f$ is finite. Since $Y$ has finitely many irreducible components, we can find a dense open which is a disjoint union of its irreducible components. Thus we may assume $Y$ is irreducible. By Lemma 28.47.5 we find a nonempty open $U \subset X$ such that $f|_U : U \to Y$ is an open immersion. After removing the closed (as $f$ finite) subset $f(X \setminus U)$ from $Y$ we see that $f$ is an isomorphism. $\square$

The code snippet corresponding to this tag is a part of the file `morphisms.tex` and is located in lines 12031–12043 (see updates for more information).

```
\begin{lemma}
\label{lemma-birational-isomorphism-over-dense-open}
Let $X$, $Y$ be schemes. Let $f : X \to Y$ be a birational morphism
between schemes which have finitely many irreducible components.
Assume
\begin{enumerate}
\item either $f$ is quasi-compact or $f$ is separated, and
\item either $f$ is locally of finite type and $Y$ is reduced or
$f$ is locally of finite presentation.
\end{enumerate}
Then there exists a dense open $V \subset Y$
such that $f^{-1}(V) \to V$ is an isomorphism.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-finite-over-dense-open} we may assume that $f$ is finite.
Since $Y$ has finitely many irreducible components, we can find a dense
open which is a disjoint union of its irreducible components. Thus we may
assume $Y$ is irreducible. By Lemma \ref{lemma-birational-birational} we find
a nonempty open $U \subset X$ such that $f|_U : U \to Y$ is an open immersion.
After removing the closed (as $f$ finite)
subset $f(X \setminus U)$ from $Y$ we see that $f$ is an isomorphism.
\end{proof}
```

## Comments (0)

## Add a comment on tag `0BAJ`

Your email address will not be published.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 lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

There are no comments yet for this tag.