Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
perhaps add "connected" to the question title, to better attract people to this page? –  Ravi Vakil Jul 22 '11 at 18:38

2 Answers 2

up vote 29 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.

share|improve this answer
I think this definitely qualifies as being "nice". Nicely explained too! –  Bjorn Poonen Dec 30 '09 at 3:14
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
I have just noticed that this answer is now in the stacks project (tag 0568). url: stacks.math.columbia.edu/tag/0568 –  Ravi Vakil Jan 7 at 18:27

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


share|improve this answer
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
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
This construction is also now given in Fred Rohrer's paper "Irreducibility and integrity of schemes", arXiv:1411.5901v2, in Expos. Math. –  Ravi Vakil Jan 7 at 18:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.