Spectra of Rings

mathwonk

does every continuous map from specB to specA arise from a homomorphism a to B? answer: no.

Easiest case: take C[X] = A = B. then the Zariski topology on the set of prime ideals, which equal {0} and the maximal ideals, is such that the closed sets are the finite subsets, excluding {0}, and the whole space.

The closure of {0} is the whole space. so lets see, i guess any function at all which takes {0} to {0} and is other wise at most finite to one, should be continuous, but hardly likely to be obtained from a homomorphism.

2) why is there a 1-1 correspondence betwen closed sets and radical ideals?

key point: show that if A is a radical ideal and f does not belong to A, then there is a prime ideal containing A but not f. Construction: look at the localization of R/A at the element f. This is a ring and contains maximal hence prime ideals by the zorn lemma, hence the pullback of any one such under the map R/A goes to the localization, is a prime ideal of R not containing A but not f. QED. (R is any ring)

mathwonk

On page 29 of gillman and jerison there is a theorem that says that a prime ideal of C(X) is contained in a unique maximal ideal. Thus there is, on a compact (completely regular, i.e. "good") space, only one common zero of all the functions in the prime ideal.

Thus your example of functions with all derivatives zero at 0, is the typical type of example.

Gillman and Jerison is alittle tedious tor ead as they sue their own peculair notation for everything so you cannot begin except at the beginning, but they had some pair of ideals,

associated to each point p of a compact space called say O(p) and M(p), where M(p) is the amximal ideal of all functions avnishing at p.

Then O(p) was something like all functions f whose zero set contained a whole nbhd of p.

Then they proved that O(p) is a radical ideal, hence an interscetion of prime ideals, hence there exist other prime ideals that are not maximal.

Then they also proved I believe that every prime ideal is caught between two such ideals O(p) and M(p) for some unique point p.

But they did not give, or at least I did not notice it, your very beautiful example, which illustrates the spirit of their other results perfectly.

I only spent a few minutes perusing the book.

Hurkyl

Well, I get the impression that the continuous case is somewhat uglier than the infinitely differentiable case. An "explicit" example will likely require the axiom of choice.

This sounds like an attack on the problem I might be able to carry out. There are three major points:

(a) Prove that all maximal ideals are of the form M(p)
(b) Prove that O(p) is a radical ideal.
(c) Prove that a radical ideal is the intersection of prime ideals.
(d) Show that O(p) cannot be the intersection of maximal ideals.

(c) is the only one that seems difficult at first glance, though it's probably a well-known result. (I admit that part of my motivation for looking at Algebraic Geometry is to strengthen my algebra. )

Merci beaucoup!

mathwonk

I sketched the proof of (c) in post 17 but I was in a hurry.

Let J be a radical ideal in a (commutative) ring R (with identity) and f an element of the ring R but f is not in J. Then since J is radical, no non negative power of f is in J. To show J is an intersection of prime ideals, we must show there is a prime ideal containing J but not containing f.

Consider first the "quotient ring" R/J, whose elements are elements of R but considered equal if they differ by an element of J. For example, R is the ring of polynomials on the plane and say J is the ideal of polynomials vanishing on some curve in the plane, and then R/J is the ring of restrictions to that curve of polynomials on the plane. I.e. two polynomials on the plane are considered equal in R/J if their difference is zero on the curve, i.e. if they have the same values on the curve.

OK, now since J is a proper ideal, the ring R/J is not the zero ring, i.e. it the elements 1 and 0 are different in there. Now moreover the element f is not zero in the ring R/J because it does not lie in J. Moreover no positive power of f is zero in the ring R/J because J is radical.

That means we can "localize" the ring R/J at positive powers of f. I.e. we can enlarge the ring R/J = S by introducing denominators which are non negative powers of f.

I.e. the localization of S at the multiplicative set {1,f,f^2,f^3,....} consists of all elements formally of form s/f^n, with n >=0, and s in S. Of course as usual with fractions we equate s/f^n and t/f^m if sf^m = tf^n.

However, because we are not necessarily in a domain, we must also equate them if there is some power of s, which kills the difference, i.e. also s/f^n = t/f^m, if for some k we have f^k( sf^m - tf^n ) = 0 in S.

This is needed since we are dividing by f in the new ring, so f has become an invertible element. So anything a power of f kills must become zero.

Notice however 1 still does not become zero, since for 1 to equal zero, we would have to have f^k (1-0) = 0, in S, and we chose J so that J is radical, i.e. since f is not in J also no positive power of f is in J, so f^k is not zero in S = R/J.


OK. Now all this implies (by Zorn's lemma) that the localization of S does contain maximal ideals.

Now consider the map from S to the localization of S, taking s to s/1, or to (fs)/f, if you prefer. The inverse image of a prime ideal is always a prime ideal, and any maximal ideal is prime. So pullback any maximal ideal from the localization of S, to get a prime ideal in S. Now a prime ideal in S pulls back further to a prime ideal in R that contains J.

On the other hand, since we have forced f to become an invertible element of the localization of S, f cannot belong to any proper ideal there. Hence the maximal ideal we pulled back did not contain f, and so neither did any of its pullbacks.

So the pullback to R of any maximal proper ideal of the localization of S at powers of f, is a prime ideal of R containing J but not f. QED.

To someone familiar with localizations, this is actually the easier proof, but a direct argument which in essence is exactly the same, would be to check directly that in R any ideal maximal wrt the property that it contains J but not f, is prime.

Actually the construction above is a standard general way of checking that.

mathwonk

to actually prove the pullback of a maximal ideal M does not contain f, proceed as follows: if f became equal to an element s/f^n of M in the localization of S, then there would be a power f^k such that f^k ( f^n+1 - s) = 0 in S, where s is in the pullback of M. But this says that f^(n+k+1) = s, which says that since s belongs to M, so does that power of f. But f is invertible in the localization, so no power of it can belong to a proper ideal.

Hurkyl

I did it without peeking. I have roughly the same proof you gave, but I didn't bother modding by J first... I simply considered the ideal J' = {j / f^n | j in J, n in N} and extended that to a maximal ideal.

I actually started out this way, but this afternoon, this use of localization finally clicked in the sesne that I could recognize that it could be applied usefully.

mathwonk

sorry. I felt kind of foolish giving the proof in detail, since you clearly didn't need it, but I was trying to show off. It seems to have done no harm, except to confirm my reputation as a pedant.

Hurkyl

No, detail is good! I'm not yet an expert on the subject, so seeing full detail can be helpful, bringing to light things I've overlooked, or didn't even know.

For instance, I had missed this fact:

Even worse, until the ride home today, I had never even considered the possibility of localizing at zero divisors; if it hadn't struck me then, this quote would have been fairly enlightening.


Anyways...


BLARG

I finally discovered my explicit example of a nonmaximal prime ideal... and it's very annoying since I had been dancing around the idea for quite a while and is so annoyingly similar to the C^inf example.

Let I be the set of all functions with a zero of infinite degree at P. More precisely,



*sigh* At least I learned something new today, so I don't feel so silly about missing this!

mathwonk

Well I didn't know those examples, and they certainly enlightened me on the topic of primes in rings of functions more general than polynomials or analytic functions. I like them too.

mathwonk

In the world of algebraic geometry, my little project for the last few weeks was learning how to prove the hirzebruch riemann roch theorem for curves and for smooth surfaces in three space, and at least what it says for arbitrary dimensional smooth projective varieties, if you are ever interested in that type of stuff.

also what riemann's approach to the classical result was for compact riemann surfaces. this is all very clasical stuff of course between 50 and 150 years old.

Hurkyl

Well, I'm usually interested in just about anything I can kinda understand! I've seen the theorem in the guise that div f = [p] - [q] --> p = q in the case of elliptic curve functions. It was suggested that it's a fairly difficult theorem, though.

mathwonk

I'm a little tanked right now, but RR is not so hard especially for elliptic functions, it just needs the theory of the weierstrass p function.

mathwonk

oh yes, the fact you stated is very elementary, i.e. if div(f) = p-q, where p is different from q, it means there is a holomorphic map from the elliptic curve E to the complex projective line P^1, such that the inverse image of 0 is p and the inverse image of infinity is q. That means the map is of degree one from E onto P^1. But E is a torus, i.e. topologically a compact surface of genus one, and since P^1 is a sphere that is a contradiction.

mathwonk

the more general theorem that follows from the theory of the P function, is that if D = p1+...pn is any effective divisor, then the space of meromorphic functions f such that div(f) + D >= 0, has dimension equal to n.

In your case above, is p is different from q, then the space of meromorphic functions f such that div(f) +q >= 0, has dimension one, hence equals the constants so f is constant so div(f) = 0, so p = q, contradiction.


the general theorem is that if D is any divisor, and K is the divisor of a differential form, then the dimension of the space of meromorphic functions f such that div(f) + D >= 0, equals 1-g + deg(D) + dim L(K-D),

where L(K-D) is the space of meromorphic functions f such that
div(f) +K-D >= 0.

The general HRR theorem gives a computation of the alternating sum chi(D), (analogous to the euler characteristic V-E+F of a polyhedron), of dimensions of cohomology groups associated to a divisor D, such that chi(D) = Todd(D), where Todd(D) is some number computed from the topological invariants of the divisor D such as its self - intersection numbers etc....

I'm fading...

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.