## 108.5 A nonintegral connected scheme whose local rings are domains

We give an example of an affine scheme $X = \mathop{\mathrm{Spec}}(A)$ 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 28.3.1).

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 blowup 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 blowups 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 $X_ k = \mathop{\mathrm{Spec}}(A_ k)$. 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 $X = \mathop{\mathrm{Spec}}(A)$. 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.

Comment #103 by on

In 11 the word neighbourhood is introduced, in this tag (and some others) you are using American spelling.

Comment #4720 by Antin on

What is the definition of the $p_i$?

Comment #4810 by on

@#4720: Not sure what you mean to ask?

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