Is there a non-noetherian scheme with infinitely many irreducible components passing through a point? (I expect the answer to be yes, but I do not know of an example.)

For extra internet points, I would love it if somehow the local ring of that point was still a domain - that the usual procedure for building zero divisors at a point meeting a finite number of irreducible components fails somehow. For instance, this would require that there was no way to subdivide these infinitely many irreducible components into a finite collection of closed sets.


Let $R$ be the quotient of a polynomial ring $k[x_1,x_2,\dots]$ in infinitely many variables over a field by the ideal generated by all products $x_ix_j$ for $i\neq j$. Note that if $P\subset R$ is a prime ideal, then there can be at most one $i$ such that $x_i\not\in P$. It follows that if $P_i$ is the ideal generated by all the $x_j$ for $j\neq i$, every minimal prime of $R$ is of the form $P_i$ (each $P_i$ is prime since $R/P_i\cong k[x_i]$). All of these minimal primes are contained in the maximal ideal $M$ generated by all the $x_i$. So $\operatorname{Spec} R$ has a point $M$ which is in all of the infinitely many irreducible components of $\operatorname{Spec} R$. (Geometrically, you should think of $\operatorname{Spec} R$ as infinitely many lines meeting at the point $M$. Indeed, there is a natural bijection between $\operatorname{Spec} R$ and infinitely many copies of $\mathbb{A}^1_k$ with the origin in all of them identified.)

On the other hand, it is impossible for the local ring at an intersection point of infinitely many irreducible components to be a domain. To see this, we may assume our scheme is affine; say it is $\operatorname{Spec} R$ for some ring $R$, and the intersection point is some prime $P\subset R$. The irreducible components of $\operatorname{Spec} R$ correspond to the minimal prime ideals of $R$, so there are infinitely many minimal primes of $R$ contained in $P$. But there is an inclusion-preserving bijection between the prime ideals of the localization $R_P$ and the prime ideals of $R$ that are contained in $P$. So every minimal prime of $R$ corresponds to a minimal prime of $R_P$, and $R_P$ must therefore have infinitely many minimal primes. In particular, it cannot be a domain.

  • $\begingroup$ Great example, thanks. As for this ring, this feels like a "limit" of the union of the coordinate rings of coordinate axis in $A^n$, as $n \to \infty$. Can this be made precise? I like this second way of arguing that a point meeting multiple components cannot be a domain. I had always built the zero divisors (very pedestrian), but this idea of producing too many minimal primes is cool. $\endgroup$
    – Elle Najt
    Nov 23 '15 at 1:54
  • 1
    $\begingroup$ Yes, you can make that precise. If you let $R_n=k[x_1,\dots,x_n]/(x_ix_j)_{i\neq j}$, then there are natural inclusions $R_n\to R_{n+1}$ for each $n$, and my ring $R$ is the colimit (aka direct limit) of the sequence $R_1\to R_2\to\dots$. It follows that $\operatorname{Spec} R$ is the inverse limit of the schemes $\operatorname{Spec} R_n$. These schemes are just the unions of the coordinate axes in $\mathbb{A}^n$, and the maps between them are just the maps that collapse the extra axes down to the origin. $\endgroup$ Nov 23 '15 at 2:01

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