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In Connes work on the Riemann Hypothesis he talks about constructing $\overline{\text{Spec}\mathbb{Z}}$ as a curve over the field with one element. I just want to know what Spec means. Is the same as spectrum from algebraic geometry?

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Maybe some context. Where are you getting this symbol, so that we can try to give you the definition and why it's interesting? In fact, why do you think it's interesting? Give us a bit of your motivation, maybe, so that we can try to help? –  Charles Siegel Nov 19 '09 at 7:56
    
Seconding Charles' comment. –  Yemon Choi Nov 19 '09 at 9:12
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up vote 13 down vote accepted

If you remove the overline, you have the affine scheme Spec Z. It is the spectrum of a noetherian domain of Krull dimension one. This description also holds for any affine algebraic curve over a field, so we have the basis for an analogy. Z has many structural features in common with the ring of polynomials with coefficients in a field (e.g., basic number-theoretic machinery like a Euclidean algorithm), so there are good reasons to think of Spec Z as a curve.

However, there are several differences. First, Z is the initial object in the category of commutative rings, so Spec Z is the final object in the category of locally ringed spaces. In particular, Spec Z is not a curve over any base field. Second, there are maps from the spectra of fields of many different characteristics into Spec Z. This makes objects like zeta functions and L-functions much more transcendental (in appearance), since the logarithm doesn't behave as nicely as it does over finite fields. You can also find a definition of genus of a number field in Neukirch's Algebraic Number Theory, and while it is zero for Q, it tends to be transcendental in general (essentially due to the presence of logs).

Despite the visible flaws in the analogy, there are good reasons for thinking that Spec Z should have a compactification, in a manner similar to the compactification of Spec F[t] into the projective line. The most basic is that Z has a valuation that is not captured by the points of the topological space, namely the usual Archimedean absolute value. For the ring of polynomials in a field, this translates to the degree. By adding this extra valuation you find that if you take all of the absolute values of a rational number (normalized appropriately) and take their product, you get one. Logarithmically, the sum of valuations is zero just like the residue theorem in the function field case. A more sophisticated reason comes from arithmetic intersection theory. If you take a curve defined over Z, you can compute a notion of intersection between two integral points, in an manner analogous to computing the intersection class of two curves in a surface. In the topological setting, you need some conditions for this to be well-behaved, e.g., the surface should be compact, so you can't push the intersections off the end of the surface. This is achieved over Z using Arakelov theory - the intersection over infinity is computed by base changing the curve to the complex numbers, making some distribution on the resulting Riemann surface using the integral points, and computing the integral of that distribution.

I don't think there is general agreement on how to make a geometric object with all of the properties we want from a compactification of Spec Z. In particular, there doesn't seem to be a satisfactory theory of the "base field" yet. There is a way to approximate the compactification using Berkovich spaces, which are an analytic refinement of schemes. The Berkovich spectrum of Z already comes with an Archimedean branch, and if you weaken the triangle inequality, you can have points corresponding to arbitrary non-negative powers of the Archimedean absolute value. Compactification means adding the point $|-|^\infty_\infty$. The local ring at this point is the closed interval [-1,1] in R, with the multiplication operation (i.e., you have some kind of logarithmic structure, instead of an actual ring), and the global functions on this object are given by the multiplicative monoid {-1,0,1}, which is a rather boring looking candidate for the field with one element.

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Yes, this is the same Spec as in algebraic geometry. As far as I understand, the symbol comes from the often cited analogy of Spec Z with an open algebraic curve. The overline then is supposed to suggest a "compactification", i.e. one wants to also consider the valuation given by the absolute value geometrically. I think this can be understood as an algebraization of Arakelov theory.

As far as I understand, since such a compactification does not exist in the category of usual schemes, one passes to a bigger category with the "scheme associated to the field with one element" as terminal object, and the hope is that in this category a suitable compactification of $\text{Spec } \mathbb{Z}$ exists. There are different approaches to $\mathbb{F}_1$-schemes, see for example here or here.

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Why is Spec Z analagous to an open curve? –  Abtan Massini Nov 19 '09 at 8:26
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Not only $\text{Spec} \mathbb{Z}$ but all rings of integers in number fields. This has been discussed on MO before: mathoverflow.net/questions/1367/… Check out the table in section 2.6 of www-math.mit.edu/~poonen/papers/curves.pdf in particular. –  Lars Nov 19 '09 at 8:51
    
@Abtan: Spec of a ring is an "affine" scheme, these don't have "points at infinity" and so are considered "open". Examples include the affine line. An example of a closed curve is the projective line. –  Rob Harron Nov 19 '09 at 19:06
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