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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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