# Spectra of Rings

1. Jun 27, 2004

### Hurkyl

Staff Emeritus
Spectrum of a Rings

So I think I'm finally getting the definition of a scheme, but the texts I'm using aren't exactly ripe with examples. So, I wanted to condier some ring spectra to construct examples...

I played around with the simple examples Z and Z[x], and my next thought was to look at manifolds and stuff...

So I wondered what the spectrum is of A, the ring of continuous functions over Rn... in order for a manifold to be a scheme, the spectrum would have to be homeomorphic to Rn... but I'm almost certain now that it is not.

I can discern some of the prime ideals of A; each of the ideals I(P) = {f in A| f(P) = 0} is prime, which is good. However, so is the ideal Z = {f | f is zero on a set with nonempty interior}, which is annoying because it doesn't even correspond to a subset of Rn. Furthermore, I can't prove to myself that, for instance, there is no subideal of I(P) that is not prime, and I'm having similar troubles when you replace P with some closed connected set of measure zero.

Is there a good way, at all, to describe the spectrum of the ring of continuous functions over Rn? What about differentable, twice differentiable, ..., infinitely differentiable functions? And analytic functions (which I suspect might be nicer)?

Last edited: Jun 27, 2004
2. Jun 29, 2004

### matt grime

Let's make it simple and just consider R. There is a paper by Neeman, though i can't seem to find it at the moment, that demonstrates the derived category of sheaves (of abelian groups) over R is not compactly generated. This kind of result implies some pathological misbehaviour of non-compact manifolds.

The ring of continuous functions seems too big (is it even noetharian?). Obviously it isn't even finitely generated over R, so it's not going to be nice.

3. Jul 2, 2004

### styler

Yeah seems too big.
Somebody suggested that the following ideal contains an infinite ascending chain:
Update: found a mention of it on the webpage of somebody's alg course:
http://www.math.gatech.edu/~singh/sol3.pdf [Broken]

Last edited by a moderator: May 1, 2017
4. Jul 11, 2004

### Hurkyl

Staff Emeritus
Well, I have gotten an answer to my question; I feel silly not noticing it earlier. The spectrum of a ring is always compact, thus cannot be isomorphic to Euclidean n-space.

Considering for the moment the case of infinitely differentiable functions on Euclidean n-space, I can produce two types of "counterexamples"; that is, ideals that are not simply the set of functions zero on a point.

(a) The set of functions zero at the origin and whose derivatives are all zero, form a prime ideal.

(b) Given any ultrafilter (right term?) on closed sets, the set of functions whose zero set is in the ultrafilter form a prime ideal. One can generate a "counterexample" by extending the filter of closed sets with bounded complement.

Now, I haven't been able to come up with a variant of (a) for any of the other rings in question (rings of n-th differentiable functions, or simply continuous functions) and if I work in a compact space, (b) doesn't seem to generate counterexamples because the only ultrafilters seem to be collections of sets containing a given point.

So my amended question is that if I have a compact manifold (compact metrizable space? compact T1 space?) and take the ring of continuous functions (n-th differentiable functions?) on the manifold, is the spectrum isomorphic to the manifold? That is, is the collection of prime ideals of this ring simply those ideals consisting of all functions zero at a given point?

5. Jul 12, 2004

### matt grime

I think you need to say what you mean by isomorphic. Spectra arise from the Zariski topology which is very badly separated, and the zariski topology on R for instance is compact, but obviously not hausdorff, indeed every open set is dense.

I still can't see what you're after. Take the unit circle, T, then whilst its set of continuous functions might have a countable analytic basis, algebraically it is disgusting, and there's little chance of its spectrum being P^1.

Its almost as though you wish to take the spectrum of the spectrum of something, which seems to me at least to be the wrong thing to examine.

Last edited: Jul 12, 2004
6. Jul 12, 2004

### Hurkyl

Staff Emeritus
This is the part I'd like to see shown; I can't figure out why it wouldn't be so. The only prime ideals I can find in this ring are those that are the set of all functions zero at a given point. If, indeed, these are all the prime ideals, then we can identify each prime ideal with its point of the circle, and then the closed subsets of the spectrum will be precisely the closed subsets of the circle.

7. Jul 12, 2004

### matt grime

Well, which topology on T do you want? (given the example of the unit circle)

8. Jul 12, 2004

### Hurkyl

Staff Emeritus
Ok, take T with the topology given by its embedding in R^2. Take the ring of continuous functions on T. Is there a prime ideal that, for all points P of T, is not equal to the set of all continuous functions zero at P?

(actually, I had another related question; given a continuous map f from Spec A to Spec B, is there an associated homomorphism from B to A? I know the other way around is true)

9. Jul 13, 2004

### matt grime

This is proving really tricky to figure out since I have no idea if I should be proving it's true or finding a counter example. Does anyone have references for spectra of (uncountably) infinitely generated rings over R? Perhaps even thinking aobut C-star algebras mght help, though we ought to do this algebraically. Of course, even if we've got the set of prime ideals as those vanishing at some point, you would then need some sort of inclusion reversing correspondence withe radical ideals etc.

10. Jul 13, 2004

### Hurkyl

Staff Emeritus
Suppose the prime ideals of A, the ring of continuous functions over a compact space X, has only prime ideals of the form of all functions zero at a given point.

Suppose I is any ideal of A, and consider the zero set of this ideal. Each element of f is zero on a closed set in the Euclidean sense (since f is continuous), and thus Z(I) is closed in the Euclidean sense.

Furthermore, if F is closed in the euclidean sense, then I(F) = {f | f is zero on F} is an ideal with Z(I(F)) = F. So, we have a 1-1 correspondence between the closed sets in the Euclidean sense and in the scheme-theoretic sense.

(Incidentally, another question I had is if, in an arbitrary Spec A, the closed sets really are in 1-1 correspondence with radical ideals)

11. Aug 10, 2004

### mathwonk

Take a closed bounded interval, like [0,1] and the ring of continuous functions on that. For each point p of the interval, there is an evaluation map from the ring of continuous functions to the reals, taking f to f(p). This is obviously an algebra surjection onto a field, so the kernel, or set of functions mapping to zero, i.e. the ideal of f such that f(p) = 0, is a maximal ideal of the ring R.

What about in the opposite direction? take any maximal ideal M in R. I claim it is the set of functions vanishing at some point of the interval. The proof is by contradiction.

Suppose for each point p of the interval there is a function in the ideal not vanishing at p. Then it also does not vanish on some open nbhd of p. Since this gives an open cover of the interval, which is compact, we obtain a finite cover and a corresponding finite number of functions. Then the sum of the squares of these fucntions does not vanish anywhere on the interval, hence is an unit, so the ideal contains a unit and hence is not maximal, but is the unit ideal.

Thus in fact every proper ideal of the ring R of continuous functions on a compact interval has a common zero, hence the ideal is contained in the maximal ideal corresponding to that point, and by maximality must equal it.

The same proof seems to work for almost any compact space like the circle.

I seem to recall from topology class, that in general the space of maximal ideals of the ring of continuous functions, on a space on which the topology can be defined by continuous functions, is a compactification of the original space called the Stone Cech compactification.

In fact I believe I recall (after 35 years), there is a one one correspondence betwen all constant - containing, point - separating, subalgebras of this ring, and all compactifications of the space. So the full set of maximal ideals defines the largest possible compactification.

Thus for a non compact space like R^n, the space of maximal ideals defines a larger compact space, indeed much larger.

for example, the set of functions "vanishing at infinity" is a maximal ideal, where f vanishes at infinity if for every epsilon, there is a compact subset K in R^n such that f is less than epsilon outside K.

If you take as your subalgebra the set of functions having limits at infinity, defined in an analogous way, you get the famous one point compactification of R^n, i.e. you get a sphere S^n.

Note that in a space like R^n, the Zariski topology defined by continuous functions agrees with the usual topology.

12. Aug 11, 2004

### Hurkyl

Staff Emeritus
There are two sorts of ways my hypothesis could fail:

There exists a prime ideal M such that, for each p, there's a function f in M with f(p) != 0

There exists a prime ideal M that is a proper subset of some maximal ideal.

I was fairly convinced the first way couldn't happen on a compact set; I didn't work out all the details, but it's good to see proof that the maximal ideals are precisely what they "should" be. Do you have any idea about how to tackle the other problem?

13. Aug 11, 2004

### mathwonk

I am not sure I am understanding, so let me just blather on.

In general, in small rings like polynomials in two variables over a field, there are a lot of non maximal prime ideals. Indeed the definition of dimension of a spectrum is (one less than) the length of a strictly nested chain of prime ideals, such as (0), (x), (x,y), in K[x,y], where K is a field.

Now in the more flexible ring of continuous functions on an interval like [0,1], there may possibly be no prime ideals that are not maximal, but I do not see a proof immediately.

In general algebraic geometry the prime ideals correspond to "irreducible" algebraic sets, such as a line in the plane. I.e. no product of two polynomials in x,y can vanish on a whole line without one of them doing so.

But in the world of continuous functions, for any subinterval say, we can find a function vanishing on part of it and another vanishing on the rest, so their product is in the ideal of functions vanishing on that set, yet neither factor is. so that ideal is not prime.

So I do not see an analogous family of "irreducible" subsets of an interval that are not just one point sets.

There is a famous book from the 60's called Rings of Continuous functions by Gillman and Jerison, but I do not have a copy, that might discuss this stuff.

14. Aug 11, 2004

### mathwonk

In an arbitrary commutative ring A, there is a 1-1 correspondence between the clsoed sets of the spectrum and radical ideals of the ring.

By definition a set S is closed if for some ideal J , S equals the set of primes which contain J. Then rad(J) determiens the same clsoed set, and conversely, the intersection of all the prime idelas in S equals rad(J).

see a basic introd to schemes such as Mumford's redbook.

15. Aug 11, 2004

### Hurkyl

Staff Emeritus
Here are two examples of the reasons why I still have some doubt:

If, instead of continuous functions, you take the ring of analytic functions, then you get all sorts of irreducible closed sets. (if you're in more than 1 variable)

In the ring of infinitely differentiable functions, an example of a non-maximal prime ideal is {f | f(i)(0) = 0 for n in N}.

16. Aug 11, 2004

### mathwonk

well, analytic functions are not so different from polynomials, but I like the example from C infinity.

I guess I see the proof. If the lowest order derivative of f that is not zero is r, and the lowest order non zero derivative of g has order s, then fg has non zero (r+s)th derivative, by the producr rule.

17. Aug 12, 2004

### 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)

Last edited: Aug 13, 2004
18. Aug 16, 2004

### 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.

19. Aug 16, 2004

### Hurkyl

Staff Emeritus
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!

20. Aug 17, 2004

### 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.

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