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Tag 0BX8

Chapter 28: Morphisms of Schemes > Section 28.46: Rational maps

Lemma 28.46.2. Let $S$ be a scheme. Let $X$ and $Y$ be schemes over $S$. Assume $X$ has finitely many irreducible components with generic points $x_1, \ldots, x_n$. Let $s_i \in S$ be the image of $x_i$. Consider the map $$ \left\{ \begin{matrix} S\text{-rational maps} \\ \text{from }X\text{ to }Y \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} (y_1, \varphi_1, \ldots, y_n, \varphi_n)\text{ where } y_i \in Y\text{ lies over }s_i\text{ and}\\ \varphi_i : \mathcal{O}_{Y, y_i} \to \mathcal{O}_{X, x_i} \text{ is a local }\mathcal{O}_{S, s_i}\text{-algebra map} \end{matrix} \right\} $$ which sends $f : U \to X$ to the $2n$-tuple with $y_i = f(x_i)$ and $\varphi_i = f^\sharp_{x_i}$. Then

  1. If $Y \to S$ is locally of finite type, then the map is injective.
  2. If $Y \to S$ is locally of finite presentation, then the map is bijective.
  3. If $Y \to S$ is locally of finite type and $X$ reduced, then the map is bijective.

Proof. Observe that any dense open of $X$ contains the points $x_i$ so the construction makes sense. To prove (1) or (2) we may replace $X$ by any dense open. Thus if $Z_1, \ldots, Z_n$ are the irreducible components of $X$, then we may replace $X$ by $X \setminus \bigcup_{i \not = j} Z_i \cap Z_j$. After doing this $X$ is the disjoint union of its irreducible components (viewed as open and closed subschemes). Then both the right hand side and the left hand side of the arrow are products over the irreducible components and we reduce to the case where $X$ is irreducible.

Assume $X$ is irreducible with generic point $x$ lying over $s \in S$. Part (1) follows from part (1) of Lemma 28.40.4. Parts (2) and (3) follow from part (2) of the same lemma. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 11270–11301 (see updates for more information).

    \begin{lemma}
    \label{lemma-rational-map-finite-presentation}
    Let $S$ be a scheme. Let $X$ and $Y$ be schemes over $S$. Assume $X$ has
    finitely many irreducible components with generic points
    $x_1, \ldots, x_n$. Let $s_i \in S$ be the image of $x_i$.
    Consider the map
    $$
    \left\{
    \begin{matrix}
    S\text{-rational maps} \\
    \text{from }X\text{ to }Y
    \end{matrix}
    \right\}
    \longrightarrow
    \left\{
    \begin{matrix}
    (y_1, \varphi_1, \ldots, y_n, \varphi_n)\text{ where }
    y_i \in Y\text{ lies over }s_i\text{ and}\\
    \varphi_i : \mathcal{O}_{Y, y_i} \to \mathcal{O}_{X, x_i}
    \text{ is a local }\mathcal{O}_{S, s_i}\text{-algebra map}
    \end{matrix}
    \right\}
    $$
    which sends $f : U \to X$ to the $2n$-tuple with
    $y_i = f(x_i)$ and $\varphi_i = f^\sharp_{x_i}$. Then
    \begin{enumerate}
    \item If $Y \to S$ is locally of finite type, then the map is injective.
    \item If $Y \to S$ is locally of finite presentation, then the map is bijective.
    \item If $Y \to S$ is locally of finite type and $X$ reduced,
    then the map is bijective.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Observe that any dense open of $X$ contains the points $x_i$
    so the construction makes sense. To prove (1) or (2)
    we may replace $X$ by any dense open. Thus if $Z_1, \ldots, Z_n$
    are the irreducible components of $X$, then we may replace
    $X$ by $X \setminus \bigcup_{i \not = j} Z_i \cap Z_j$.
    After doing this $X$ is the disjoint union of its irreducible
    components (viewed as open and closed subschemes). Then both the
    right hand side and the left hand side of the arrow are products
    over the irreducible components and we reduce to the case where
    $X$ is irreducible.
    
    \medskip\noindent
    Assume $X$ is irreducible with generic point $x$ lying over $s \in S$.
    Part (1) follows from part (1) of
    Lemma \ref{lemma-morphism-defined-local-ring}.
    Parts (2) and (3) follow from part (2) of the same lemma.
    \end{proof}

    Comments (1)

    Comment #2994 by Minsik Han on November 8, 2017 a 8:09 pm UTC

    $f:U\to X$ in the 4th line should be $f:U\to Y$ instead.

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