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Without finiteness assumptions, the irreducible and the connected components of a scheme may behave in strange ways. More precisely, let us consider a scheme $X$ and the following properties:

(1) $X$ is the sum of its irreducible components;

(2) The irreducible and the connected components of $X$ coincide;

(3) The irreducible components of $X$ are pairwise disjoint.

It is clear that (1) implies (2), that (2) implies (3), and that if the set of irreducible components of $X$ is locally finite then all three statements are equivalent (see [EGA]). However, (2) does not necessarily imply (1) in general: An affine counterexample is given by the spectrum of the product of infinitely many fields (which is non-discrete and totally disconnected). So, out of pure curiosity we may ask the following:

Is there an (affine) scheme fulfilling (3) but not (2)?

One can note that this is equivalent to the following:

Is there a nonempty, reducible, connected (affine) scheme whose irreducible components are pairwise disjoint?

(One can also note that for topological spaces that are not necessarily underlying spaces of schemes it is easy to find an example that fulfils (3) but not (2) - every connected, separated space with at least two points does so.)

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An example of such a scheme is given in t3suji's great answer to… – Philipp Hartwig May 31 '12 at 20:39
@Philipp: Why not post this as an answer? [together with an explanation why the other question also answers this one] – Martin Brandenburg May 31 '12 at 21:03
Dear @Philipp, thank you for the link (which seems to have slipped my previous search of MO but answers my question in a satisfying way). May I second Martin in asking you to post it as an answer? – Fred Rohrer Jun 1 '12 at 13:53
You may also want to see Prime ideal structure in commutative rings. by Mel Hochster (1969). In it he classifies what topological spaces arise as a $\text{Spec}$ of a commutative ring. – Karl Schwede Jun 2 '12 at 18:16
up vote 7 down vote accepted

In the answers to Non-integral scheme having integral local rings , and in particular in t3suji's beautiful geometric construction, it is shown that there is a (non-empty) affine scheme $X=\mathrm{Spec}(A)$ whose underlying topological space is connected, such that all local rings of $X$ are integral domains and such that $X$ is however not integral. This scheme fulfills the conditions you are requiring: It is reduced and not integral, hence reducible. Assume two irreducible components $Y_1$ and $Y_2$ of $X$ met in a point $x\in X$. The point $x$ corresponds to a prime ideal $\mathfrak{q} \subset A$, and the irreducible components $Y_1$ and $Y_2$ are of the form $Y_1=V(\mathfrak{p}_1)$ and $Y_2=V(\mathfrak{p}_2)$ for minimal prime ideals $\mathfrak{p}_1$ and $\mathfrak{p}_2$ of $A$. As $x\in Y_1,Y_2$, we would have $\mathfrak{p}_1, \mathfrak{p}_2 \subset \mathfrak{q}$.

Then $\mathfrak{p}_1$ and $\mathfrak{p}_2$ would correspond to distinct minimal prime ideals of $A_\mathfrak{q}=\mathcal{O}_{X,x}$, which is impossible, as $\mathcal{O}_{X,x}$ is by construction a domain, so that its only minimal prime ideal is $\{0\}$.

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I'm fighting with LaTeX here, can someone please explain to me why it doesn't work as expected? It compiles fine using for example pdflatex. – Philipp Hartwig Jun 2 '12 at 12:45
You need to escape the math in certain cases, or it gets interpreted as markdown (italics in this case). There's a box to the right under 'How to write math' that explains this, or see the FAQ. – Dylan Thurston Jun 2 '12 at 13:24
I fixed it by doing what Dylan said. – Will Sawin Jun 2 '12 at 13:27
Thank you both. – Philipp Hartwig Jun 2 '12 at 13:57
Dear @Philipp, thank you very much for your answer. That the irreducible components are pairwise disjoint follows also from the general observation that irreducibility of the stalk at a point of a scheme $X$ is equivalent to this point lying in precisely one irreducible component of $X$ (cf. [EGA I.2.1.9]). (In particular, irreducibility of all stalks is equivalent to the irreducible components being pairwise disjoint.) – Fred Rohrer Jun 4 '12 at 6:16

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