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I can show that if $X$ is a scheme such that all local rings $\mathcal{O}_{X,x}$ are integral and such that the underlying topological space is connected and Noetherian, then $X$ is itself integral.

This doesn't seem to work without the "Noetherian" condition. But can anyone think about a nice counterexample to illustrate this? So I am looking for a non-integral scheme - with connected underlying topological space - having integral local rings.

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perhaps add "connected" to the question title, to better attract people to this page? –  Ravi Vakil Jul 22 '11 at 18:38
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up vote 28 down vote accepted

Let me try to give a counterexample. (I don't know whether it is 'nice'). First, let us rewrite your properties for an affine scheme $X=Spec(A)$.

Connectedness for $A$ means $A$ has no nontrivial idempotents;

Integrality for $A$ is the usual one ($A$ is a domain);

Local integrality means that whenever $fg=0$ in $A$, every point of $X$ has a neighborhood where either $f$ or $g$ vanishes.

Let us construct a connected locally integral ring that is not integral.

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 gluing in an affine line instead (so our scheme will be an open subset in what was described above).

Here's an algebraic description:

For every $k\ge 0$, let $A_k$ be the following ring: its elements are collections of polynomials $p_i\in{\mathbb C}[x]$ where $i=0,\dots,2^k$ such that $p_i(1)=p_{i+1}(0)$. Set $X_k=Spec(A_k)$. $X$ is a union of $2^k+1$ affine lines that meet transversally in a chain. (It may be better to index polynomials by $i/2^k$, but the notation gets confusing.)

Define a morphism $A_k\to A_{k+1}$ by $$(p_0,\dots,p_{2^k})\mapsto(p_0,p_0(1),p_1,p_1(1),\dots,p_{2^k})$$ (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 $A$ is locally integral.

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 $x$ on $X_k$. It suffices to prove that $y$ has a neighborhood on which either $f$ or $g$ (viewed as functions on $X_k$) 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.

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I think this definitely qualifies as being "nice". Nicely explained too! –  Bjorn Poonen Dec 30 '09 at 3:14
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This is fantastic. I also like how this answer to an essentially algebraic question is motivated by geometry. –  Ravi Vakil Jul 25 '11 at 16:57
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Hochster has an elegant construction which associates a commutative ring to each infinite totally ordered set with the property that strictly between two distinct elements there is a third one.

The spectrum of such a ring is a connected affine scheme of dimension one, all the local rings of which are domains. The ring itself, however is NOT a domain. So, every ordered set with the property mentioned above yields a scheme with the required property.

Here is the link to Hochster's (one-page) construction

http://www.math.lsa.umich.edu/~hochster/614F08/ECdom.sol.pdf

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Thanks for pointing this out. This is actually an algebraic way of stating the geometric construction in my answer. Indeed, I claim that the ring $A$ above is exactly Hochster's construction if the ordered set is the set of all rationals between 0 and 1 of the form $p/2^k$. For index $i=p/2^k$ let $x_i$ be the coordinate on p-th component of $X_k$, which we extend to be constant on other components --- it is zero on components with smaller index, and one on those with higher. Then $A$ is generated by $x_i$ subject to Hochster's relation $x_ix_j=x_j$ if $j>i$ (OK, the order is reversed) –  t3suji Dec 30 '09 at 2:48
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Dear t3suji, this is a very interesting comparison of two aspects of an ingenious example. I had a vague feeling that there was a resemblance between Hochster's and your description , but certainly nothing as precise as your explanation. Thank you for spelling it out and congratulations on your beautiful geometric construction. –  Georges Elencwajg Dec 30 '09 at 9:43
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