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Is the empty set considered a variety in affine and projective space? By variety, i mean a closed irreducible set in the Zariski topology.

On one hand it seems that the empty set satisfies the definition of e.g. an affine variety: it is an algebraic set and it is irreducible (to show that it is irreducible, we may argue that it can not be written as the union of two proper closed subsets simply because it does not have any proper subsets).

On the other hand, if the empty set is an affine variety, then the points of $\mathbb{A}^n$ are not minimal algebraic sets, since they contain the empty set.

Finally if we view the empty set as a projective variety in projective space, then we have an ambiguity since the empty set is both the zero set of the entire ring and the zero set of the maximal homogeneous ideal.

Any comments/flaws on my arguments above?

Manos
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    I include non-empty in the definition of irreducible. I believe the standard sources do as well. To appeal to the ultimate authority: (2.1) in EGA${}_0$. – Hoot May 28 '14 at 17:21
  • Well, I thought an ideal answer would explain why this is the convention. I think there are good reasons for it, but I would have to collect my thoughts. – Hoot May 28 '14 at 17:24
  • @Hoot: Ok this answers my question. – Manos May 28 '14 at 17:25
  • @Hoot: I can think of a reason: if the empty set was a projective variety then we get an ambiguity as to what its vanishing ideal is: is it the entire ring or is it the homogeneous maximal ideal? – Manos May 28 '14 at 17:28
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    I find myself always confused on this point. Hartshorne e.g. defines a scheme as integral if every open set has integral coordinate ring, but the empty set is open and apparently has zero coordinate ring and he says the zero ring is not integral. In Zariski Samuel moreover the unit ideal is considered prime, hence so presumably is the integer 1. It is apparently difficult to be wholly consistent, and although we tend now to read mostly the same sources, hence to agree on these matters, agreement is not universal. Similarly, people differ on whether the empty set is considered connected. – roy smith Apr 28 '18 at 19:48
  • notice also that in Mumford's redbook, irreducibility is defined so that the empty set qualifies, and it is stated that irreducible sets correspond to prime ideals, so (1) is prime. Also the sheaf on an affine variety in section I.4, defn 4, assigns the whole fraction field to the empty set! (rather than the zero ring) and as Ravi Vakil says: "you don't argue with Mumford." As Ravi also observes, here the category of subrings of the fraction field has that field as final object, whereas for general rings, (0) is the final object. – roy smith Apr 29 '18 at 02:37

1 Answers1

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The empty set is a variety (it's the spectrum of the zero ring), but it's not an irreducible variety.

The usual definition of an irreducible variety does not do the right thing here, and one must use a slightly different one. An irreducible variety is a variety with exactly one irreducible component, and the empty variety has zero irreducible components.

This is the same kind of thing that causes $1$ to not be prime; the analogous definition of prime that does not work at $1$ is "a prime number is a positive integer which cannot be written as the product of two smaller positive integers." $1$ satisfies this definition but it still shouldn't be prime because that would make unique factorization false. See too simple to be simple for more details.

Qiaochu Yuan
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