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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    $\begingroup$ An example of such a scheme is given in t3suji's great answer to mathoverflow.net/questions/7477/… $\endgroup$ May 31, 2012 at 20:39
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    $\begingroup$ @Philipp: Why not post this as an answer? [together with an explanation why the other question also answers this one] $\endgroup$ May 31, 2012 at 21:03
  • $\begingroup$ 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? $\endgroup$ Jun 1, 2012 at 13:53
  • $\begingroup$ 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. $\endgroup$ Jun 2, 2012 at 18:16

1 Answer 1


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 $\{0\}$.

  • $\begingroup$ 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.) $\endgroup$ Jun 4, 2012 at 6:16

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