Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. **There will be no changes to other Yahoo properties or services, or your Yahoo account.** You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

## Trending News

# When does the Zariski topology on a commutative ring form a locally compact space?

Definitions (just for completion's sake, mostly):

For the definition of the zariski topology, see http://en.wikipedia.org/wiki/Zariski_topology#The_...

X is locally compact = Every point of X has a compact neighborhood.

I don't even know if the answer is known or, if it is known, whether it is interesting. However, I found myself thinking about it after one of the Ph.D students wrote something on my white-board.

Oh, and if the answer is known, but is fairly complex, feel free to direct me to a website or a text - I have the resources to get almost any text.

### 1 Answer

- anonymous dudeLv 51 decade agoFavorite Answer
I'm really more of an analyst, but I can do this one :)

The answer is: always! The I assume that by "compact" you are referring to what the algebraists sometimes (or at least used to) call "quasi-compact" - every open cover has a finite subcover. Here is the basic argument.

Let R be a commutative ring (with 1 - I will discuss the other case at the end) and let X = Spec(R). I shall first show that X is itself compact. This of course means that it is locally compact, but we can say even more. Consider an open cover {U_i} of X where i runs over some index set. Let V_i be the complement of U_i. A little set stuff:

X = Union(U_i)

= Union(Complement(V_i))

= Complement(Intersection(V_i))

So choosing an open cover {U_i} of X is the same thing as choosing a collection {V_i} of closed sets whose intersection is empty. Without loss of generality we can assume that V_i = V(a_i) where a_i is an element of R; this is because the open sets {Complement(V(a)): a in R} form a base for the topology of X. Now, to say that the intersection of the V_i's is empty is the same thing as saying that that there is no prime ideal which contains every a_i. Consider the ideal generated by all of the a_i; since every proper ideal is contained in a maximal (and hence prime) ideal, this ideal must be the whole ring. In particular, 1 is in this ideal. But recall that the ideal generated by a_i consists of all FINITE R- linear combinations of the a_i, so we can conclude that there is an equation of the form c_1 a_{i1} + ... + c_k a_{ik} = 1. Reverse engineering our reasoning so far, we conclude that U_{i1}, ..., U_{ik} constitutes a finite subcover of X.

So X is compact. But actually there are a lot of smaller compact sets, and this may be more interesting to you. Let's look at the open sets U(a) = Complement(V(a)) considered above. U(a) is precisely the set of all prime ideals which do not contain a. Now, consider the multiplicative set S = {a^n: n a natural number} and localize R at this set to form the ring R' = S^(-1)R. A general fact about rings of fractions says that there is a one-to-one correspondence between prime ideals of S^(-1)R and ideals of R which do not intersect S; you can in fact check that Spec(S^(-1}R) is homeomorphic to U(a) equipped with the subspace topology coming from X. But we have shown above that the spectrum of ANY ring is compact, so we conclude that U(a) is compact. So in fact we have constructed a base of compact open sets for the topology of X. One can even show that a subset of X is compact if and only if it is the finite union of some U(a)'s.

The question becomes a bit more interesting in the case where the ring R does not contain 1. However, I don't know much about general non-unital rings. There is one special case where I do know something, however, and it is rather suggestive. It begins with the following interesting observation. Take a locally compact Hausdorff space Y and consider the ring C_0(Y) of all continuous complex valued functions on Y which vanish at infinity; this is a non-unital ring since the constant function 1 (which wants to be the multiplicative identity) does not vanish at infinity. Now consider the space Spec(C_0(Y)) equipped with the Zarisky topology. Consider the subspace M consisting of only maximal ideals equipped with the subspace topology. The space M turns out to be homeomorphic to the original space Y! You can then flip the question around; given a commutative ring R, when is the maximal ideal space in Spec(R) a locally compact Hausdorff space? It turns out that the Spec and C_0 functors determine an equivalence of categories between locally compact Hausdorff spaces and a special kind of commutative ring arising classically in functional analysis called a C* algebra; this is called the Gelfand-Naimark theorem. Now, if you take a locally compact space Y and tensor the non-unital C* algebra C_0(Y) with the complex numbers to form a unital C* algebra, the maximal ideal space of the result is homeomorphic to the one-point compactification of Y. If instead you extend C_0(Y) to contain all bounded continuous complex valued functions, the maximal ideal space of the result is the Stone-Cech compactification of Y. So unitalizing a ring somehow corresponds to compactifying its spectrum; of course in the above cases we are always compactifying the maximal ideal space, so I'm really not sure what is going on with the actual spectrum. Still, I hope that helps a little!