locally noetherian scheme with integral local rings is disjoint union of integral schemes.

Question

I have read in some places that a if you have a locally noetherian connected scheme $X$ such that $\forall x \in X$, $\mathcal{O}_{X, x}$ is a integral domain, then $X$ is the disjoint union of integral schemes. However I haven't been able to find a proof. Could anybody give a proof or a reference to one?

Answer 1

A connected scheme cannot be the disjoint union of integral schemes. So you misread. What is true is that a locally Noetherian scheme $X$ is integral if and only if $X$ is connected and all its stalks $\mathcal{O}_{X,x}$ are integral domains.

Here is a sketch of the main part of the argument, you could google the details. For any scheme $X$ and $x\in X$, the irreducible components of $\text{Spec}(\mathcal{O}_{X,x})$ are in bijective correspondence with the irreducible components of $X$ passing through $x$. In our case $\mathcal{O}_{X,x}$ is an integral domain, so $\text{Spec}(\mathcal{O}_{X,x})$ is irreducible and therefore every point $x$ has a unique irreducible component passing through it. So the irreducible components of $X$ are mutually disjoint. In a locally Noetherian topological space, this is equivalent to saying that the irreducible components of $X$ are the same as its connected components. But since $X$ is connected, there is only one such component.

Caution: The Noetherian assumption is important. There exists a nonintegral connected scheme whose local rings are domains