Let $X$ be a connected scheme. Then $X$ irreducible iff $\forall x, \text{ Spec}(\mathcal{O}_{X,x})$ is

I could use some help for the second part of exercise I-34 from Eisenbud and Harris' Geometry of Schemes:

Let $X$ be a connected scheme. Show that $X$ is irreducible if and only if for all $x \in X$, the stalk local ring has a unique minimal prime ideal.

I guess it must be related to the first part of the exercise, that states:

An arbitrary scheme is irreducible iff every open affine subset is irreducible.

My ideas

I know that an affine scheme $\text{Spec }R$ is irreducible precisely when $R$ has a unique minimal prime ideal, so that leaves us with the to be proven statement: $$X \text{ is irreducible} \quad\iff \quad \forall x \in X, \text{ Spec}(\mathcal{O}_{X,x}) \text{ is irreducible}.$$ but the RHS seems even harder to prove: $\text{ Spec}(\mathcal{O}_{X,x})$ doesn't really seem to be "accessible" in some way. For example, this result cannot be applied since $\text{ Spec}(\mathcal{O}_{X,x})$ cannot be identified with affine patches as one would like to do. Do you have ideas?

1 Answer

This is false. Indeed, there exists a ring $A$ which has no nontrivial idempotent, which is not a domain, and such that the localization of $A$ at any prime ideal is a domain. See http://stacks.math.columbia.edu/tag/0568 for details of the construction. Note that such a ring automatically is reduced, since all of its localizations at prime ideals are reduced.

Since $A$ has no nontrivial idempotent elements, $\operatorname{Spec} A$ is connected, and since every localization of $A$ at a prime ideal is a domain, every stalk has a unique minimal prime. But $\operatorname{Spec}A$ is reducible, since any nonzero $f,g\in A$ with $fg=0$ give proper closed subsets whose union is all of $\operatorname{Spec} A$ (the vanishing sets of $f$ and $g$ are proper subsets since $f$ and $g$ are not nilpotent).

On the other hand, here are some positive results. First, the forward direction is always true. Indeed, if $X$ is irreducible and $x\in X$, let $U=\operatorname{Spec} A$ be an affine open subset containing $x$. Then $U$ is irreducible, so $A$ has a unique minimal prime. The same is then true of any (nonzero) localization of $A$, in particular the local ring $\mathcal{O}_{X,x}$.

Second, the reverse direction is true if $X$ is Noetherian. Indeed, in that case $X$ can be written as a finite union of irreducible components. If $X$ is not irreducible, it has more than one irreducible component, and if $X$ is connected, there must be two distinct irreducible components $A$ and $B$ of $X$ which intersect (otherwise, each irreducible component would be clopen). Let $x\in A\cap B$. Then the generic point of $A$ and the generic point of $B$ give distinct minimal primes in the local ring $\mathcal{O}_{X,x}$.

• Dear @Eric Wofsey, thank you very much for your answer. By now I read the first half of it, and I have the following question. Given $f,g \neq 0$ with $fg=0$, how do you know that $\mathcal{V}(f) \not \subseteq \mathcal{V}(g)$ (and vice versa)? It looks like (although I am not sure)it comes down to $g \notin \sqrt{(f)}$ and vice versa, but I don't see why that couldn't be true – Koenraad van Duin Jul 23 '17 at 6:09
• Well since $fg=0$, you know $V(f)\cup V(g)$ is all of $\operatorname{Spec} A$, so if $V(f)\subseteq V(g)$ that means $V(g)=\operatorname{Spec} A$. That can only happen if $g$ is nilpotent, but $g$ is nonzero and $A$ is reduced. – Eric Wofsey Jul 23 '17 at 6:14
• o of course, thank you. – Koenraad van Duin Jul 23 '17 at 6:42
• your answer is very useful. Although, with your answer a contradiction shows up. Do you have any idea who could be wrong? The most likely would be that I misinterpreted Eisenbud and Harris. If you have access to a copy of that book, could you read and point out what is wrong? – Koenraad van Duin Jul 24 '17 at 13:15
• I think it's just an error in the book. Probably they had the argument for Noetherian schemes in mind and didn't realize it doesn't work in general. – Eric Wofsey Jul 24 '17 at 15:30