## Spec(Z) and three-manifolds

The following question comes from an old friend, John Baez. John is a renowned mathematical physicist and internet guru. You should look at his website and his blog for an enormously informative and entertaining survey of mathematics and physics.

MK

Hi! Why is Spec(Z) 3-dimensional?

I think I understand what the number theorist James wrote in response to “week257” of This Week’s Finds:

“Instead of considering Z, let’s consider F[x], where F is a finite field. They are both principal ideal domains with finite residue fields, and this makes them behave very similarly, even on a deep level. I’ll explain why F[x] is three-dimensional, and then by analogy we can hope Z is, too. Now F[x] is an F-algebra. In other words, X = Spec(F[x]) is a space mapping to S = Spec(F). I already explained why S is a circle from the point of view of the etale topology. So, if X is supposed to,be three-dimensional, the fibers of this map better be two-dimensional. What are the fibers of this map? Well, what are the points of S? A point in the etale topology is Spec of some field with a trivial absolute Galois group, or in other words, an algebraically closed field (even better, a separably closed one). Therefore a etale point of S is the same thing as Spec of an algebraic closure Fbar of F. What then is the fiber of X over this point? It’s Spec of the ring Fbar[x]. Now, *this* is just the affine line over an algebraically closed field, so we can figure out its cohomological dimension. The affine line over the complex numbers, another algebraically closed field, is a plane and therefore has cohomological dimension 2. Since etale cohomology is kind of the same as usual singular cohomology, the etale cohomological dimension of Spec(Fbar[x]) ought to be 2.

Therefore X looks like a 3-manifold fibered in 2-manifolds over Spec(F), which looks like a circle. Back to Spec(Z), we analogously expect it to look like a 3-manifold, but absent a (non-formal) theory of the field with one element, Z is not an algebra over anything. Therefore we expect Spec(Z) to be a 3-manifold, but not fibered over anything.”

But, it would be nice to go a bit deeper than mere “reasoning by analogy with Fbar[x]”, without yet diving into a full-blown calculation of the etale cohomology of Spec(Z). Can you say anything that’s helpful and fun? related question is this. Nowadays various people are making serious attempts to define the field with one element and do algebraic geometry over this field. With these, Z *is* an algebra over the field with one element. How much does this help?