Schemes in algebraic geometry 2 : prime spectra and generic points

10 March 2009 at 10:07 pm 2 comments

I just explained how the affine plane could be described by the ring \mathbb Z[x,y]. A point M of the affine plane whose coordinate ring is R is a morphism \mathbb Z[x,y] \to R defined by the assignment P \mapsto P(a,b) \in R, where (a,b) are the coordinates of M. In the case of points corresponding to morphisms \mathbb Z[x,y] \to \mathbb Z, there is a natural way of recovering the point from the ring morphism by looking at his equations, which are elements of the kernel of the morphism. If M satisfies the equations x=a and y=b, then M has the form (a,b). This motivates the abstract definition of point of the affine plane as a morphism \mathbb Z[x,y] \to R to some ring.

Conversely, the set of equations of M defines a canonically associated point p_M, which is the morphism \mathbb Z[x,y] \to \mathbb Z[x,y]/I, where I is the ideal generated by the equations. But this morphism has no reason to totally recover M if it wasn’t a point with integral coordinates. For example, the point (2,3) is a special point, satisfying a lot of equations, which characterize it. But (\log 2, \pi) do not satisfy any polynomial equations with integral coefficients, so the set of its equations is empty, and cannot be used to recover it. Moreover, the point (e, \log 3) does not satisfy equations either: their algebraic properties are exactly the same. These points are called generic.

The prime spectrum of a ring is a convenient way of describing equivalence classes of points of a given ring.
Definition. The prime spectrum of A = \mathbb Z[x,y] is the set of points p_I: A \to A/I for prime ideals I. If M: A \to R is any point of the affine plane with coordinates in an integral domain R, then Mis canonically associated to some p_M := p_I, where I is the kernel of the map A \to R.

Usually non-integral domains can be (almost) expressed as products of integral domains, which explains the restriction in the definition, which in turn explains why we restrict to prime ideals (which are those defining integral quotient rings). Working on a field simplifies things a bit : the rational affine plane \mathbb A^2_{\mathbb Q}, represented by the prime spectrum \mathrm{Spec} \mathbb Q[x,y] contains:

  • the usual points with rational coordinates corresponding to the ideal (x-a,y-b).
  • algebraic points of higher degree: these are not rational, but have equations describing precisely their coordinates: the point (\sqrt 2, 1) is such a point, having equations x^2 - 2 = 0 and y -1=0, but it cannot be distinguished from (-\sqrt 2, 1). Both points are represented by the same point of the prime spectrum.
  • generic points of dimension 1: one of their coordinates does not satisfy any algebraic equation, for example (\pi, \pi^2-2). However, it satisfies at least one algebraic equation, here y=x^2-2. We say this point is a generic point of the curve y=x^2-2. All such points are represented by a single point of the prime spectrum, which is the point for the principal ideal I = (y-x^2+2), which has coordinate ring \mathbb Q[x,y]/(y=x^2-2) \simeq \mathbb Q[x]. The coordinate rings of the generic point are not rings of usual numbers, but rather polynomial rings, where indeterminates are abstract symbols for the generic coordinate
  • “the” generic point : both coordinates are algebraically independant and do not satisfy any polynomial equation. Then (\log 2, \pi) is such a point, and all generic point are represented in the prime spectrum by the same point, whose coordinate ring is \mathbb Q[x,y] itself.

Of course, many points which you would consider as different tend to become equal in the prime spectrum, but the interesting thing is that replacing the whole functor of points (the collection of all points in all coordinate rings) by the prime spectrum does not change how point belong or not to subvarieties defined by polynomial equations (elements of \mathbb Q[x,y]). For example, the points (\pi, \pi^2-2) and (e, e^2-2) satisfy the same equations in \mathbb Q[x,y]. Evaluation at (\pi, \pi^2-2) is a ring morphism \mathbb Q[x,y] \to \mathbb Q[\pi] whose image is isomorphic to the quotient ring \mathbb Q[x,y]/(y-x^2+2) : it makes sense to say that this is the coordinate ring of the point. The residue field is the fraction field of this coordinate ring: it is the field where rational functions (fractions of functions) would take their values.

The prime spectrum is also equipped with the Zariski topology: remember that any point in the plane with complex coordinates defines a canonical point of \mathbb A^2_{\mathbb Q} which describes the equations it satisfies. One of the properties used in Serre’s GAGA theorem is that this morphism \mathbb C^2 \to \mathbb A^2 is continuous. To define Zariski topology correctly, remark that any element of \mathbb Q[x,y] defines a function \mathbb C^2 \to \mathbb C, but also a function \mathbb A^2 \to \mathbb A^1 (check this…). Then the Zariski topology on \mathbb C^2 is the finest topology making polynomial functions continuous, and the Zariski topology on \mathbb A^2 coincides with the quotient topology of the Zariski topology on the usual complex affine plane.

The abstract definition of the Zariski topology says that in order to make polynomial functions continuous, we only have to say that points of the prime spectrum satisfying an equation form closed sets (these are prime ideals containing the function), and define the Zariski topology to be generated by these closed sets (principal hypersurfaces).

An unfortunate property of prime spectra and Zariski topology is that the prime spectrum \mathbb A^2 is not the cartesian product of sets \mathbb A^1 \times \mathbb A^1, which only contains points looking like (\sqrt 2, -1) or (x,y), but no points of the form (x, x^2) having relations between coordinates. The language of functor of points and categories is thus necessary to recover a more traditional definition of product of schemes.

Next post will be entitled glued schemes and sheaves.

Entry filed under: algebraic geometry, commutative algebra, english, schemes in algebraic geometry. Tags: , , , .

Schemes in algebraic geometry I : the affine plane Why can we have a Fast Fourier Transform ?

2 Comments Add your own

  • 1. Matt DeLand  |  11 March 2009 at 7:10 pm

    Another “unfortunate” property is that knowing the points of a ring A does not allow you to recover the ring. The problem is that the intersection of the prime ideals of A define the Nilradical of A, so that the points of A are the same of the points of A/Nil(A). This is why we need to associate to the topological space associate to the ring A a certain sheaf of rings….

    • 2. remyoudompheng  |  11 March 2009 at 8:19 pm

      Aside from non-reducedness, another issue is probably the fact that all Spec of fields look set-theoretically like a point, but the fact that spectra of fields are different can be seen on fiber products (field extensions).


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