# The Stacks Project

## Tag 01JD

Example 25.14.3 (Affine space with zero doubled). Let $k$ be a field. Let $n \geq 1$. Let $X_1 = \mathop{\rm Spec}(k[x_1, \ldots, x_n])$, let $X_2 = \mathop{\rm Spec}(k[y_1, \ldots, y_n])$. Let $0_1 \in X_1$ be the point corresponding to the maximal ideal $(x_1, \ldots, x_n) \subset k[x_1, \ldots, x_n]$. Let $0_2 \in X_2$ be the point corresponding to the maximal ideal $(y_1, \ldots, y_n) \subset k[y_1, \ldots, y_n]$. Let $U_{12} = X_1 \setminus \{0_1\}$ and let $U_{21} = X_2 \setminus \{0_2\}$. Let $\varphi_{12} : U_{12} \to U_{21}$ be the isomorphism coming from the isomorphism of $k$-algebras $k[y_1, \ldots, y_n] \to k[x_1, \ldots, x_n]$ mapping $y_i$ to $x_i$ (which induces $X_1 \cong X_2$ mapping $0_1$ to $0_2$). Let $X$ be the scheme obtained from the glueing data $(X_1, X_2, U_{12}, U_{21}, \varphi_{12}, \varphi_{21} = \varphi_{12}^{-1})$. Via the slight abuse of notation introduced above the example we think of $X_i \subset X$ as open subschemes. There is a morphism $f : X \to \mathop{\rm Spec}(k[t_1, \ldots, t_n])$ which on $X_i$ corresponds to $k$ algebra map $k[t_1, \ldots, t_n] \to k[x_1, \ldots, x_n]$ (resp. $k[t_1, \ldots, t_n] \to k[y_1, \ldots, y_n]$) mapping $t_i$ to $x_i$ (resp.  $t_i$ to $y_i$). It is easy to see that this morphism identifies $k[t_1, \ldots, t_n]$ with $\Gamma(X, \mathcal{O}_X)$. Since $f(0_1) = f(0_2)$ we see that $X$ is not affine.

Note that $X_1$ and $X_2$ are affine opens of $X$. But, if $n = 2$, then $X_1 \cap X_2$ is the scheme described in Example 25.9.3 and hence not affine. Thus in general the intersection of affine opens of a scheme is not affine. (This fact holds more generally for any $n > 1$.)

Another curious feature of this example is the following. If $n > 1$ there are many irreducible closed subsets $T \subset X$ (take the closure of any non closed point in $X_1$ for example). But unless $T = \{0_1\}$, or $T = \{0_2\}$ we have $0_1 \in T \Leftrightarrow 0_2 \in T$. Proof omitted.

The code snippet corresponding to this tag is a part of the file schemes.tex and is located in lines 2513–2556 (see updates for more information).

\begin{example}[Affine space with zero doubled]
\label{example-affine-space-zero-doubled}
Let $k$ be a field. Let $n \geq 1$.
Let $X_1 = \Spec(k[x_1, \ldots, x_n])$,
let $X_2 = \Spec(k[y_1, \ldots, y_n])$.
Let $0_1 \in X_1$ be the point corresponding to the maximal ideal
$(x_1, \ldots, x_n) \subset k[x_1, \ldots, x_n]$.
Let $0_2 \in X_2$ be the point corresponding to the maximal ideal
$(y_1, \ldots, y_n) \subset k[y_1, \ldots, y_n]$.
Let $U_{12} = X_1 \setminus \{0_1\}$ and
let $U_{21} = X_2 \setminus \{0_2\}$. Let
$\varphi_{12} : U_{12} \to U_{21}$ be the isomorphism
coming from the isomorphism of $k$-algebras
$k[y_1, \ldots, y_n] \to k[x_1, \ldots, x_n]$
mapping $y_i$ to $x_i$ (which induces $X_1 \cong X_2$ mapping
$0_1$ to $0_2$).
Let $X$ be the scheme obtained from the glueing data
$(X_1, X_2, U_{12}, U_{21}, \varphi_{12}, \varphi_{21} = \varphi_{12}^{-1})$. Via the slight abuse
of notation introduced above the example we think of
$X_i \subset X$ as open subschemes.
There is a morphism $f : X \to \Spec(k[t_1, \ldots, t_n])$
which on $X_i$ corresponds to $k$ algebra map
$k[t_1, \ldots, t_n] \to k[x_1, \ldots, x_n]$
(resp.\ $k[t_1, \ldots, t_n] \to k[y_1, \ldots, y_n]$)
mapping $t_i$ to $x_i$ (resp.\  $t_i$ to $y_i$).
It is easy to see that this morphism identifies
$k[t_1, \ldots, t_n]$ with $\Gamma(X, \mathcal{O}_X)$. Since
$f(0_1) = f(0_2)$ we see that $X$ is not affine.

\medskip\noindent
Note that $X_1$ and $X_2$ are affine opens of $X$.
But, if $n = 2$, then $X_1 \cap X_2$ is the scheme
described in Example \ref{example-not-affine} and hence not affine.
Thus in general the intersection of affine opens of a scheme
is not affine. (This fact holds more generally for any $n > 1$.)

\medskip\noindent
Another curious feature of this example is the following.
If $n > 1$ there are many irreducible closed subsets $T \subset X$
(take the closure of any non closed point in $X_1$ for example).
But unless $T = \{0_1\}$, or $T = \{0_2\}$ we have
$0_1 \in T \Leftrightarrow 0_2 \in T$. Proof omitted.
\end{example}

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