The Stacks Project


Tag 01II

25.9. Schemes

Definition 25.9.1. A scheme is a locally ringed space with the property that every point has an open neighbourhood which is an affine scheme. A morphism of schemes is a morphism of locally ringed spaces. The category of schemes will be denoted $\textit{Sch}$.

Let $X$ be a scheme. We will use the following (very slight) abuse of language. We will say $U \subset X$ is an affine open, or an open affine if the open subspace $U$ is an affine scheme. We will often write $U = \mathop{\rm Spec}(R)$ to indicate that $U$ is isomorphic to $\mathop{\rm Spec}(R)$ and moreover that we will identify (temporarily) $U$ and $\mathop{\rm Spec}(R)$.

Lemma 25.9.2. Let $X$ be a scheme. Let $j : U \to X$ be an open immersion of locally ringed spaces. Then $U$ is a scheme. In particular, any open subspace of $X$ is a scheme.

Proof. Let $U \subset X$. Let $u \in U$. Pick an affine open neighbourhood $u \in V \subset X$. Because standard opens of $V$ form a basis of the topology on $V$ we see that there exists a $f\in \mathcal{O}_V(V)$ such that $u \in D(f) \subset U$. And $D(f)$ is an affine scheme by Lemma 25.6.6. This proves that every point of $U$ has an open neighbourhood which is affine. $\square$

Clearly the lemma (or its proof) shows that any scheme $X$ has a basis (see Topology, Section 5.5) for the topology consisting of affine opens.

Example 25.9.3. Let $k$ be a field. An example of a scheme which is not affine is given by the open subspace $U = \mathop{\rm Spec}(k[x, y]) \setminus \{ (x, y)\}$ of the affine scheme $X =\mathop{\rm Spec}(k[x, y])$. It is covered by two affines, namely $D(x) = \mathop{\rm Spec}(k[x, y, 1/x])$ and $D(y) = \mathop{\rm Spec}(k[x, y, 1/y])$ whose intersection is $D(xy) = \mathop{\rm Spec}(k[x, y, 1/xy])$. By the sheaf property for $\mathcal{O}_U$ there is an exact sequence $$ 0 \to \Gamma(U, \mathcal{O}_U) \to k[x, y, 1/x] \times k[x, y, 1/y] \to k[x, y, 1/xy] $$ We conclude that the map $k[x, y] \to \Gamma(U, \mathcal{O}_U)$ (coming from the morphism $U \to X$) is an isomorphism. Therefore $U$ cannot be affine since if it was then by Lemma 25.6.5 we would have $U \cong X$.

    The code snippet corresponding to this tag is a part of the file schemes.tex and is located in lines 1598–1681 (see updates for more information).

    \section{Schemes}
    \label{section-schemes}
    
    \begin{definition}
    \label{definition-scheme}
    \begin{history}
    In \cite{EGA1} what we call a scheme was called a ``pre-sch\'ema'' and the
    name ``sch\'ema'' was reserved for what is a separated scheme in the
    Stacks project. In the second edition \cite{EGA1-second} the terminology
    was changed to the terminology that is now standard. However, one may
    occasionally encounter the terminology ``prescheme'', for example in
    \cite{Murre-lectures}.
    \end{history}
    A {\it scheme} is a locally ringed space with the property that
    every point has an open neighbourhood which is an affine scheme.
    A {\it morphism of schemes} is a morphism of locally
    ringed spaces. The category of schemes will be denoted
    $\Sch$.
    \end{definition}
    
    \noindent
    Let $X$ be a scheme.
    We will use the following (very slight) abuse of language.
    We will say $U \subset X$ is an {\it affine open}, or an {\it open affine}
    if the open subspace $U$ is an affine scheme. We will often
    write $U = \Spec(R)$ to indicate that $U$ is isomorphic
    to $\Spec(R)$ and moreover that we will identify (temporarily)
    $U$ and $\Spec(R)$.
    
    \begin{lemma}
    \label{lemma-open-subspace-scheme}
    Let $X$ be a scheme. Let $j : U \to X$ be an open immersion
    of locally ringed spaces. Then $U$ is a scheme. In particular,
    any open subspace of $X$ is a scheme.
    \end{lemma}
    
    \begin{proof}
    Let $U \subset X$. Let $u \in U$.
    Pick an affine open neighbourhood $u \in V \subset X$.
    Because standard opens of $V$ form a basis of the topology
    on $V$ we see that there exists a $f\in \mathcal{O}_V(V)$
    such that $u \in D(f) \subset U$. And $D(f)$ is an affine scheme
    by Lemma \ref{lemma-standard-open-affine}. This proves that every point
    of $U$ has an open neighbourhood which is affine.
    \end{proof}
    
    \noindent
    Clearly the lemma (or its proof) shows that any scheme
    $X$ has a basis (see Topology, Section \ref{topology-section-bases})
    for the topology consisting of affine opens.
    
    \begin{example}
    \label{example-not-affine}
    Let $k$ be a field.
    An example of a scheme which is not affine is
    given by the open subspace
    $U = \Spec(k[x, y]) \setminus \{ (x, y)\}$
    of the affine scheme $X =\Spec(k[x, y])$.
    It is covered by two affines, namely $D(x) = \Spec(k[x, y, 1/x])$
    and $D(y) = \Spec(k[x, y, 1/y])$ whose intersection is
    $D(xy) = \Spec(k[x, y, 1/xy])$. By the sheaf property
    for $\mathcal{O}_U$ there is an exact sequence
    $$
    0 \to
    \Gamma(U, \mathcal{O}_U) \to
    k[x, y, 1/x] \times k[x, y, 1/y] \to
    k[x, y, 1/xy]
    $$
    We conclude that the map $k[x, y] \to \Gamma(U, \mathcal{O}_U)$
    (coming from the morphism $U \to X$) is an isomorphism.
    Therefore $U$ cannot be affine since if it was then by
    Lemma \ref{lemma-category-affine-schemes} we would have $U \cong X$.
    \end{example}

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