The Stacks Project


Tag 01RO

Chapter 28: Morphisms of Schemes > Section 28.47: Birational morphisms

Definition 28.47.1. Let $X$, $Y$ be schemes. Assume $X$ and $Y$ have finitely many irreducible components. We say a morphism $f : X \to Y$ is birational if

  1. $f$ induces a bijection between the set of generic points of irreducible components of $X$ and the set of generic points of the irreducible components of $Y$, and
  2. for every generic point $\eta \in X$ of an irreducible component of $X$ the local ring map $\mathcal{O}_{Y, f(\eta)} \to \mathcal{O}_{X, \eta}$ is an isomorphism.

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 11627–11644 (see updates for more information).

    \begin{definition}
    \label{definition-birational}
    \begin{reference}
    \cite[(2.2.9)]{EGA1}
    \end{reference}
    Let $X$, $Y$ be schemes. Assume $X$ and $Y$ have finitely many
    irreducible components. We say a morphism $f : X \to Y$ is
    {\it birational} if
    \begin{enumerate}
    \item $f$ induces a bijection between the set of generic points
    of irreducible components of $X$ and the set of generic points
    of the irreducible components of $Y$, and
    \item for every generic point $\eta \in X$ of an irreducible component
    of $X$ the local ring map
    $\mathcal{O}_{Y, f(\eta)} \to \mathcal{O}_{X, \eta}$
    is an isomorphism.
    \end{enumerate}
    \end{definition}

    References

    [EGA1, (2.2.9)]

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 01RO

    Your email address will not be published.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.




    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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?