What is the Spectrum of the Ring of Continuous Functions Over Rn?

[mathwonk]

Hurkyl, We have been getting into using schemes to define differential one forms on singular spaces, in the thread "what is a tensor?" if you are interested.

the "elementary" argument above was supposed to be essentially that a meromorphic function f with divisor p-q where p,q are different, defines an isomorphism between a torus and a sphere, which is not even topologically possible, much less analytically so. did that make sense?

i.e. you just send the point x on the torus to the value of f at x. That defines a holomorphic map from the torus to the complex number plane, and it extends to a continuous map from the torus to the one point compactification of the plane sending the "pole" q to infinity.

that gives a map from the torus to the sphere, but since the divisor of f is p-q that says only one point goes to zero (i.e. p), and only one point goes to infinity (namely q), counting multiplicities.

The general theory of complex maps then implies that every point of the sphere has exactly one preimage, so the map is a homeomorphism, and even an isomorphism.

I.e. for complex maps of compact connected complex one manifolds, each point has the same number of preimages, counting multiplicities, if that number is finite.

 

[Hurkyl]

I've been distracted by a side problem.

Algebraically, I'm afraid I might be using what I'm trying to prove, I have to go back and review. I can see why it's a degree 1 map because if div(f) = [P] - [Q], then f = g (X - P)/(X - Q) where div(g) = 0. I know this implies g is a constant, but I don't remember off hand what was used to prove that... and from here it's obvious that f is an isomorphism. (Of course, I also have not seen the algebraic proof that the EC is a torus, though I believe it)

I've seen the p-functions, so I can follow the argument over the complexes. Passing to the complex plane as the covering of the EC, we have that g is bounded and entire, thus constant.

 

[mathwonk]

Let X,Y be compact Riemann surfaces, i.e. compact real 2 dimensional manifolds,with a complex structure.

And let f:X-->Y be a holomorphic map. Then f is a "branched covering". I.e. away from the points of X where the derivative of f is zero, the map is a covering map. In particular, the preimage of every point of Y which does not contain a zero of the derivative, contains the same number of points.

If f is a meromorphic function on X, to say that div(f) = p-q means that the inverse image of 0 is p, counting multiplicites, and the inverse image of infinity is q, also counting multiplicities.

This impies that the dergee of f is one, i.e. that the number of inverse images of every point is one, counting multiplicities.

Thus the map X-->P^1 is one to one everywhere, hence bijective.

Now the structure of holomorphic mappings implies that this map is thus an isomorphism, hence also a homeomorphism from X to P^1, where P^1 is a sphere.

Hence X is homeomorphic to a sphere.

Now assume that X is an elliptic curve, e.g. a smooth cubic curve in the plane P^2.

We claim that then X is actually homeomorphic to a torus. The easiest way to see this is to note that if we degenerate X to a union of three lines, i.e. a triangle, then X looks like a union of three spheres joined pairwise at points. This is obviously a degenerate torus.

Another way is to note that a smooth cubic curve can be put in the form
y^2 = x^3 + g1x + g2. where g1 and g2 are complex numbers, by some change of coordinates.

Then there exists a meromorphic function called a P function, such that the map from X to P^2 given by P and P' (the derivative of P), takes X isomorphically onto the given cubic curve. Then the theory of (doubly periodic) P functions shows that X is isomorphic to the quotient of C^2 by a lattice, hence to a torus.

Thus every elliptic curve X is topologically a torus, hence X is not homeomorphic to a sphere, hence X cannot have a meromorphic function f with divisor div(f) = p-q, with p different from q.

 

[mathwonk]

oh you want the algebraic proof, over a field other than the complexes where we cannot use the usual topology and homeomorphism argument.

That is more work. Then we have to use some algebraic way to distinguish betwen P^1 and an EC. There are basically two versions of the "genus" of a curve.

One is the number of handles in the usual topology, which we cannot use, and the other is the number of linearly independent differential forms. Those two numbers are equal in the classical case so we can use either of them as the genus. In the algebraic case then we use the number of linearly independent regular algebraic differentials on the curve, defined of course as you konw, by globalizing the I/I^2 definition of differentials.

Here is one argument that an EC and P^1 cannot be isomorphic. First of all, the quotient of two differential forms is a rational function (analog in arbitrary algebraic setting of a meromorphic functrion). Second a rational function defines a map from a smooth projective curve to P^1, which is a branched covering, and again every point of p^1 has the same number of preimages counted properly. This number, the algebraic degree of the mapping, equals the dergree of the field extension defined by the map, i.e. the degree of the field of rational functions on the curve as an extension of the field of rational functions on P^1. [This is why a function with divisor p-q has degree one, because the divisor exhibits the unique inverse image of zero as p. I.e. since the divisor p has degree one, by the theorem the map has degree one. I did not understand your short argument about the degree being one.]

Hence the divisor of any non constant rational function has degree zero. This is not trivial to prove in the algebraic case and uses the theory of finite extensions of rings and the fact that a torsion free finite module over a principal ideal domain (such as the local ring at a smooth point on a curve) is free, of well determined rank. There is a nice proof in Shafarevich's book Basic Algebraic geometry.

Anyway, from the fact that the quotient of two differentials is a rational function and the fact that the divisor of a rational function is zero, it follws that all differentials on a curve have divisors of the same degree.

So all we have to do is compute the degree of any divisor of a differential on an EC and the degree of any divisor of a differential on P^1 and show they do not agree. On P^1 just write down dz and note it has a pole of order 2 at infinity and no other zeroes aor poles, so has degree -2. (This is 2g-2 where g=0.)

For an EC with equation y^2 = (x-a)(x-b)(x-c), dx/y is a regular differential everywhere with no zeroes or poles (the zeroes of y at the points (a,0), (b,0), (c,0) are canceled by the zeroes of dx at those points). The degree of the divisor of dx/y is thus zero = 2g-2 where g = 1. Hence we have a contradiction to assuming that EC and P^1 are algebraically isomorphic isomorphic.


There is a more elementary proof that does not develop the theory of the genus, just finding a numerical contradiction, but it is harder to motivate.

I.e. assume there is an isomorphism from P^1 to an EC, say the fermat curve with equation (1) x^3 +y^3 = z^3 . I.e. assume we can express each variable x,y,z as a rational function of t, not all constant.

Then differentiate to obtain (2): x^2x' + y^2y' = z^2z', where ' denotes differentiation w.r.t. t. Now we want to eliminate the z terms. multiply (1) by z' and (2) by z, and subtract, to obtain z'x^3 + z'y^3 = x'x^2z + y'y^2z. Collecting terms, and factoring gives x^2( xz'-zx') = y^2 ( y'z - yz'). If either ( y'z - yz')=0, or ( xz'-zx') = 0, then both do, and hence by the quotient rule for derivatives (which make sense algebraically), we would have x/z and y/z constant. Since x,y are rel prime, x^2 divides (yz'-zy'), in k[t], and thus 2degree(x) <= deg(y)+deg(z)-1. Repeating the argument for each of the other two variables, i.e. eliminating the x and y terms, leads to the same inequality with the variables permuted. Adding the 3 inequalities gives 2[deg(x)+deg(y)+deg(z)] <= 2[deg(x)+deg(y)+deg(z)] - 3, a contradiction. QED.

This argument can be generalized to any EC with some work, but I prefer the more general conceptual approach above.

In algebra, the whole (smooth) curve X with all its geometry is determined just by the algebra of the field of rational functions k(X), as a finitely generated field over k the algebraically closed base field. I.e. if two curves have isomorphic fields k(X) isom to k(Y) as k algebras, then X is isomorphic to Y as curves.

One can define the regular differentials just from this field, hence one can show purely algebraically that the field of an EC is not isomorphic to the field of P^1.

Of course that is what our elementarya rgument above did. I.e. it showed that there is no embedding of the field of the fermat curve as a subfield of the field k[t] of rational functions on P^1. In particular they are not isomorphic.

It also shows further than there is no non constant mapping from P^1 to an EC, even of degree higher than one. In topology thisa could be done by showing that such a map of curves cannot raise genus. And that resault two can be proved using the differential form version of genus. I.e. a surjective algebraic mapping induces in the other direction an injection of differential forms. (In the spirit of the threads in differential geometry and tensors, covector fields are distinguished from vector fields, in that covector fields pull back, although vector fields do not even push forward.)

Anyway it follows from this that the vector space of differential forms on the target curve (or variety of any dimension) must be no larger in dimension that that of the source curve. Thus surjective morphisms, cannot raise the algebraic genus.

 

[mathwonk]

Another proof, by Riemann Roch.

the genus is the most important invariant of a smooth projective curve.

It figures in the famous Riemann Roch theorem as well as follows:

let a divisor D = p1+p2+...+pr -q1-q2...-qs be given on a projective smooth curve X over the algebraically closed field k.

Define L(D) as the vector space of rational functions f on X such that div(f)+D >= 0. (i.e. f has poles at most among the p's, and zeroes at least on the q's.)

Then the dimension of L(D) is finite, and if K is the divisor of a differential form on X, then w e have:

dimL(D) - dimL(K-D) = deg(D) 1-g, where g is the algebraic genus of X.

Let's apply this to X = an EC of genus one, and D = q. Then the constants provide one dimension of functions f with div(f)+q = q>=0, so L(q) has dimension at least one.

However a function f with div(f) = p-q, for some p different from q, would be non constant so we would have dimL(q) >= 2.

Now we know a differential form on an EC has divisor equal to zero, se we get from RRT:

dimL(q) -dimL(-q) = 1-g +deg(D) = 1-1+1 = 1.

Now L(-q) is the zero vector space since no function can have a zero on q without also having a pole somewhere, as noted above.

So RRT gives dimL(q) = 1, a contradiction to the existence of a non constant f in there.

This kind of easy numerical argument shows why people love the RRT, and why it is the most powerful, most useful theorem in the theory of curves.

 

[Hurkyl]

One is the number of handles in the usual topology, which we cannot use, and the other is the number of linearly independent differential forms.

All right, I'll see if I can go from here. Hopefully I won't need to peek!

 

[mathwonk]

did we switch from here to "more questions on alg geom"?

 

[Hurkyl]

I tend to start new posts when I have new questions; I hadn't thought about just keeping them all to one thread...

 

[mathwonk]

sorry this is the only way i can have 15 straight most recentposts, a new record for me.

i feel bad if it totally vacuous though, so here is a toss off question for you. can you show me an example of a ring with infinite krull dimension, but each maximal ideals has finite height?

i.e. each prime ideal represents a subvariety of finite dimension but there are subvarieties of arbitrary dimension?