# Why do we need Noetherianness to conclude that a connected scheme is integral iff its local rings are?

I'm trying to proof/understand the following Statement:

If $$X$$ is a Noetherian and connected scheme, then $$X$$ is integral if and only if $$X$$ has integral stalks.

It can quickly by verified that if $$X$$ is integral, all stalks have to be integral. It is the reverse implication that troubles me.

Of course for a scheme being integral is the same as being irreducible and reduced. If $$X$$ has integral stalks, it has reduced stalks, and reducedness is a stalk-local property. Hence, if $$X$$ has integral stalks, it is reduced.

It remains to show that $$X$$ is irreducible. If $$Y\neq Z$$ are two irreducible components of $$X$$ which meet at a point $$p$$, and if $$p$$ is contained in some affine open $$\operatorname{Spec} A$$, then $$Y \cap \operatorname{Spec} A$$ is a irreducible component of $$\operatorname{Spec} A$$, hence corresponds to some minimal prime $$\mathfrak p_Y$$ of $$A$$. But the same applies to $$Z$$, so we have two different minimal primes $$\mathfrak p_Y$$ and $$\mathfrak p_Z$$ of $$A$$, so $$(0)$$ isn't prime in $$A$$, and $$A$$ is not integral. If $$\mathfrak p$$ is the prime corresponding to the point $$p$$ of $$\operatorname{Spec} A$$, both $$\mathfrak p_Y$$ and $$\mathfrak p_Z$$ are contained in $$\mathfrak p$$, therefore there are two different minimal prime ideals in $$A_{\mathfrak p} = \mathcal O_{X,p}$$.

Hence we have shown: If two different irreducible components meet at a point $$p$$, $$\mathcal O_{X,p}$$ is not integral. But since all stalks are integral by assumption, every pair of irreducible components of $$X$$ has empty intersection, and since $$X$$ is connected there can only one irreducible component.

My problem is that I don't know where we used Noetheriannes, but I guess it was in the last step. If $$X$$ is Noetherian this step will work for sure, as we can write $$X$$ as a finite union of irreducible components. But why would this fail if we had an infinite union of irreducible components?

• I'm not sure about this statement so I won't take any risks, but generally speaking there are a lot of statements out there needlessly assuming noetherian, because it's a comfortable keyword that makes you feel safe about what you are saying. It can be a bit scary to state things without this hypothesis sometimes because you can always have a doubt that somewhere along the proof you used something that is actually only true for noetherian schemes, and thus a lot of authors will just not take any risk and add that hypothesis, since anyway usually it's not very restrictive in practice. – Captain Lama May 9 at 16:34
• I found this; stacks.math.columbia.edu/tag/0568 - but I haven't yet deduced where exactly your proof fails. – Jesko Hüttenhain May 9 at 17:32
• PS: this came from MO: mathoverflow.net/a/9967 – Jesko Hüttenhain May 9 at 17:34

## 1 Answer

The irreducible components of $$X$$ are closed subsets of $$X$$. If there are finitely many of them and they are disjoint, this means each one is also open, since it is the complement of the union of the other irreducible components (and that is a finite union so it is still closed). So, the irreducible components are clopen, and this contradicts connectedness if there are more than one of them. But if there are infinitely many irreducible components, they are only closed, not open, so there is no way to contradict connectedness.

The following example may be illustrative: consider $$\mathbb{R}$$ with its usual topology. The irreducible components of $$\mathbb{R}$$ are just singletons, since $$\mathbb{R}$$ is Hausdorff. These irreducible components are pairwise disjoint, but $$\mathbb{R}$$ is still connected, since they are only closed, not open.

• This is a great answer. I guess the fact that irreducible components can be disjoint but still form a connected topological space is what contradicted my intuition. – LurchiDerLurch May 9 at 21:08