# The Stacks Project

## Tag 0BXM

### 32.4. Varieties and rational maps

Let $k$ be a field. Let $X$ and $Y$ be varieties over $k$. We will use the phrase rational map of varieties from $X$ to $Y$ to mean a $\mathop{\mathrm{Spec}}(k)$-rational map from the scheme $X$ to the scheme $Y$ as defined in Morphisms, Definition 28.46.1. As is customary, the phrase ''rational map of varieties'' does not refer to the (common) base field of the varieties, even though for general schemes we make the distinction between rational maps and rational maps over a given base.

The title of this section refers to the following fundamental theorem.

Theorem 32.4.1. Let $k$ be a field. The category of varieties and dominant rational maps is equivalent to the category of finitely generated field extensions $K/k$.

Proof. Let $X$ and $Y$ be varieties with generic points $x \in X$ and $y \in Y$. Recall that dominant rational maps from $X$ to $Y$ are exactly those rational maps which map $x$ to $y$ (Morphisms, Definition 28.46.10 and discussion following). Thus given a dominant rational map $X \supset U \to Y$ we obtain a map of function fields $$k(Y) = \kappa(y) = \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x} = \kappa(x) = k(X)$$ Conversely, such a $k$-algebra map (which is automatically local as the source and target are fields) determines (uniquely) a dominant rational map by Morphisms, Lemma 28.46.2. In this way we obtain a fully faithful functor. To finish the proof it suffices to show that every finitely generated field extension $K/k$ is in the essential image. Since $K/k$ is finitely generated, there exists a finite type $k$-algebra $A \subset K$ such that $K$ is the fraction field of $A$. Then $X = \mathop{\mathrm{Spec}}(A)$ is a variety whose function field is $K$. $\square$

Let $k$ be a field. Let $X$ and $Y$ be varieties over $k$. We will use the phrase $X$ and $Y$ are birational varieties to mean $X$ and $Y$ are $\mathop{\mathrm{Spec}}(k)$-birational as defined in Morphisms, Definition 28.47.1. As is customary, the phrase ''birational varieties'' does not refer to the (common) base field of the varieties, even though for general irreducible schemes we make the distinction between being birational and being birational over a given base.

Lemma 32.4.2. Let $X$ and $Y$ be varieties over a field $k$. The following are equivalent

1. $X$ and $Y$ are birational varieties,
2. the function fields $k(X)$ and $k(Y)$ are isomorphic,
3. there exist nonempty opens of $X$ and $Y$ which are isomorphic as varieties,
4. there exists an open $U \subset X$ and a birational morphism $U \to Y$ of varieties.

Proof. This is a special case of Morphisms, Lemma 28.47.6. $\square$

The code snippet corresponding to this tag is a part of the file varieties.tex and is located in lines 145–226 (see updates for more information).

\section{Varieties and rational maps}
\label{section-varieties-rational-maps}

\noindent
Let $k$ be a field. Let $X$ and $Y$ be varieties over $k$.
We will use the phrase {\it rational map of varieties from $X$ to $Y$}
to mean a $\Spec(k)$-rational map from the scheme $X$ to the scheme $Y$
as defined in Morphisms, Definition \ref{morphisms-definition-rational-map}.
As is customary, the phrase rational map of varieties''
does not refer to the (common) base field of the varieties,
even though for general schemes we make the distinction between
rational maps and rational maps over a given base.

\medskip\noindent
The title of this section refers to the following fundamental theorem.

\begin{theorem}
\label{theorem-varieties-rational-maps}
Let $k$ be a field. The category of varieties and
dominant rational maps is equivalent to the category of
finitely generated field extensions $K/k$.
\end{theorem}

\begin{proof}
Let $X$ and $Y$ be varieties with generic points $x \in X$ and $y \in Y$.
Recall that dominant rational maps from $X$ to $Y$ are exactly those
rational maps which map $x$ to $y$
(Morphisms, Definition \ref{morphisms-definition-dominant-rational}
and discussion following).
Thus given a dominant rational map $X \supset U \to Y$ we obtain a map of
function fields
$$k(Y) = \kappa(y) = \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x} = \kappa(x) = k(X)$$
Conversely, such a $k$-algebra map (which is automatically local as the
source and target are fields) determines (uniquely) a dominant rational
map by Morphisms, Lemma \ref{morphisms-lemma-rational-map-finite-presentation}.
In this way we obtain a fully faithful functor.
To finish the proof it suffices to show that every finitely generated
field extension $K/k$ is in the essential image.
Since $K/k$ is finitely generated, there exists a finite type
$k$-algebra $A \subset K$ such that $K$ is the fraction field of $A$.
Then $X = \Spec(A)$ is a variety whose function field is $K$.
\end{proof}

\noindent
Let $k$ be a field. Let $X$ and $Y$ be varieties over $k$.
We will use the phrase {\it $X$ and $Y$ are birational varieties}
to mean $X$ and $Y$ are $\Spec(k)$-birational
as defined in Morphisms, Definition \ref{morphisms-definition-birational}.
As is customary, the phrase birational varieties''
does not refer to the (common) base field of the varieties,
even though for general irreducible schemes we make the distinction between
being birational and being birational over a given base.

\begin{lemma}
\label{lemma-birational-varieties}
Let $X$ and $Y$ be varieties over a field $k$.
The following are equivalent
\begin{enumerate}
\item $X$ and $Y$ are birational varieties,
\item the function fields $k(X)$ and $k(Y)$ are isomorphic,
\item there exist nonempty opens of $X$ and $Y$ which are isomorphic
as varieties,
\item there exists an open $U \subset X$ and a birational morphism
$U \to Y$ of varieties.
\end{enumerate}
\end{lemma}

\begin{proof}
This is a special case of
Morphisms, Lemma \ref{morphisms-lemma-criterion-birational-finite-presentation}.
\end{proof}

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