# The Stacks Project

## Tag 0568

### 99.5. A nonintegral connected scheme whose local rings are domains

We give an example of an affine scheme $X = \mathop{\rm Spec}(A)$ which is connected, all of whose local rings are domains, but which is not integral. Connectedness of $X$ means $A$ has no nontrivial idempotents, see Algebra, Lemma 10.20.3. The local rings of $X$ are domains if, whenever $fg = 0$ in $A$, every point of $X$ has a neighborhood where either $f$ or $g$ vanishes. As long as $A$ is not a domain, then $X$ is not integral (Properties, Definition 27.3.1).

Roughly speaking, the construction is as follows: let $X_0$ be the cross (the union of coordinate axes) on the affine plane. Then let $X_1$ be the (reduced) full preimage of $X_0$ on the blow-up of the plane ($X_1$ has three rational components forming a chain). Then blow up the resulting surface at the two singularities of $X_1$, and let $X_2$ be the reduced preimage of $X_1$ (which has five rational components), etc. Take $X$ to be the inverse limit. The only problem with this construction is that blow-ups glue in a projective line, so $X_1$ is not affine. Let us correct this by glueing in an affine line instead (so our scheme will be an open subset in what was described above).

Here is a completely algebraic construction: For every $k \ge 0$, let $A_k$ be the following ring: its elements are collections of polynomials $p_i \in \mathbf{C}[x]$ where $i = 0, \ldots, 2^k$ such that $p_i(1) = p_{i + 1}(0)$. Set $X_k = \mathop{\rm Spec}(A_k)$. Observe that $X_k$ is a union of $2^k + 1$ affine lines that meet transversally in a chain. Define a ring homomorphism $A_k \to A_{k + 1}$ by $$(p_0, \ldots, p_{2^k}) \longmapsto (p_0, p_0(1), p_1, p_1(1), \ldots, p_{2^k}),$$ in other words, every other polynomial is constant. This identifies $A_k$ with a subring of $A_{k + 1}$. Let $A$ be the direct limit of $A_k$ (basically, their union). Set $X = \mathop{\rm Spec}(A)$. For every $k$, we have a natural embedding $A_k \to A$, that is, a map $X\to X_k$. Each $A_k$ is connected but not integral; this implies that $A$ is connected but not integral. It remains to show that the local rings of $A$ are domains.

Take $f, g \in A$ with $fg = 0$ and $x \in X$. Let us construct a neighborhood of $x$ on which one of $f$ and $g$ vanishes. Choose $k$ such that $f, g \in A_{k - 1}$ (note the $k - 1$ index). Let $y$ be the image of $x$ in $X_k$. It suffices to prove that $y$ has a neighborhood on which either $f$ or $g$ viewed as sections of $\mathcal{O}_{X_k}$ vanishes. If $y$ is a smooth point of $X_k$, that is, it lies on only one of the $2^k + 1$ lines, this is obvious. We can therefore assume that $y$ is one of the $2^k$ singular points, so two components of $X_k$ pass through $y$. However, on one of these two components (the one with odd index), both $f$ and $g$ are constant, since they are pullbacks of functions on $X_{k - 1}$. Since $fg = 0$ everywhere, either $f$ or $g$ (say, $f$) vanishes on the other component. This implies that $f$ vanishes on both components, as required.

The code snippet corresponding to this tag is a part of the file examples.tex and is located in lines 135–200 (see updates for more information).

\section{A nonintegral connected scheme whose local rings are domains}
\label{section-connected-locally-integral-not-integral}

\noindent
We give an example of an affine scheme $X = \Spec(A)$ which is
connected, all of whose local rings are domains, but which is not integral.
Connectedness of $X$ means $A$ has no nontrivial idempotents, see
Algebra, Lemma \ref{algebra-lemma-disjoint-decomposition}.
The local rings of $X$ are domains if, whenever $fg = 0$ in $A$, every
point of $X$ has a neighborhood where either $f$ or $g$ vanishes.
As long as $A$ is not a domain, then $X$ is not integral
(Properties, Definition \ref{properties-definition-integral}).

\medskip\noindent
Roughly speaking, the construction is as follows: let $X_0$ be the cross
(the union of coordinate axes) on the affine plane. Then let $X_1$ be
the (reduced) full preimage of $X_0$ on the blow-up of the plane ($X_1$
has three rational components forming a chain).  Then blow up the
resulting surface at the two singularities of $X_1$, and let $X_2$ be
the reduced preimage of $X_1$ (which has five rational components), etc.
Take $X$ to be the inverse limit. The only problem with this construction
is that blow-ups glue in a projective line, so $X_1$ is not affine. Let us
correct this by glueing in an affine line instead (so our scheme will be an
open subset in what was described above).

\medskip\noindent
Here is a completely algebraic construction: For every $k \ge 0$, let $A_k$
be the following ring: its elements are collections of
polynomials $p_i \in \mathbf{C}[x]$ where $i = 0, \ldots, 2^k$ such that
$p_i(1) = p_{i + 1}(0)$. Set $X_k = \Spec(A_k)$. Observe that $X_k$ is
a union of $2^k + 1$ affine lines that meet transversally in a chain.
Define a ring homomorphism $A_k \to A_{k + 1}$ by
$$(p_0, \ldots, p_{2^k}) \longmapsto (p_0, p_0(1), p_1, p_1(1), \ldots, p_{2^k}),$$
in other words, every other polynomial is constant. This identifies
$A_k$ with a subring of $A_{k + 1}$. Let $A$ be the direct limit of $A_k$
(basically, their union). Set $X = \Spec(A)$. For every $k$, we have
a natural embedding $A_k \to A$, that is, a map $X\to X_k$.
Each $A_k$ is connected but not integral; this implies that $A$ is
connected but not integral. It remains to show that the local rings of
$A$ are domains.

\medskip\noindent
Take $f, g \in A$ with $fg = 0$ and $x \in X$. Let us construct a
neighborhood of $x$ on which one of $f$ and $g$ vanishes. Choose $k$
such that $f, g \in A_{k - 1}$ (note the $k - 1$ index).
Let $y$ be the image of $x$ in $X_k$. It suffices to prove that $y$ has
a neighborhood on which either $f$ or $g$ viewed as sections of
$\mathcal{O}_{X_k}$ vanishes.
If $y$ is a smooth point of $X_k$, that is, it lies on only one of the
$2^k + 1$ lines, this is obvious. We can therefore assume that $y$ is one
of the $2^k$ singular points, so two components of $X_k$ pass through
$y$. However, on one of these two components (the one with odd index),
both $f$ and $g$ are constant, since they are pullbacks of functions on
$X_{k - 1}$. Since $fg = 0$ everywhere, either $f$ or $g$ (say, $f$)
vanishes on the other component.
This implies that $f$ vanishes on both components, as required.

## Comments (1)

Comment #103 by Pieter Belmans (site) on December 16, 2012 a 7:40 am UTC

In 004J the word neighbourhood is introduced, in this tag (and some others) you are using American spelling.

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