§ Recovering topology from sheaf of functions: Proof from Atiyah Macdonald
Let be a compact Haussdorff space. Such a space is normal (T4), so we can separate closed subsets
by open neighbourhoods. This also means that the space obeys the Urhyson lemma, so we can construct
continuous functions that take value at some point and at some other point. We will use this
to argue about zero sets of functions.
- We will first show a strong nullstellensatz like theorem, showing that every maximal ideal of the ring of continuous functions is in bijection with the set of functions that vanish at a point, .
- Let be the ring of all continuous real valued functions on . For each , define to be the set of functions that vanish at . This is a maximal ideal, because it is the kernel of the evaluation map .
- Given some maximal ideal , we will show that there is some point such that. To show this, first consider the common zeros of functions in , .We first show that is non-empty, and we then show that contains exactly one point.
- To show is non-empty, suppose for contradiction that is empty. Thus, for each point ,not all functions in vanish at (otherwise ).So, there is a function that does not vanish at , hence . Since is continuous, there is some open neighbourhood where . (A continuous functionthat does not vanish at a point cannot "suddenly" decay to zero. It will be non-zero over an open nbhd).Since the space is compact, we have a finite number of that cover . Hence, we build a function ( for contradiction) that vanishes nowhere. This means is a unit.But we must have as is built out of functions . This is a contradiction as a unitcannot belong to a maximal ideal. Thus, contains at least one point.
- To show that contains exactly one point, suppose that contains a single point .This means that all functions in vanish at . Thus, , since contains all functions(not just ones in ) that vanish at . But is maximal, and hence . This tells us thatevery maximal ideal corresponds to some vanishing set .
- We will next show that every vanishing set is distinct. We already know that it is maximal. This gives us an injection. Let be two vanishing sets for distinct points. Let be the function constructed fromUrhyson's lemma that is zero at at nonzero at . Thus, we have and . Hence,. This shows that the maximal ideals will be distinct.
- We have thus established a bijection / nullstellensatz between zero sets maximal ideals and functions that vanish at a point .
- We will next show that this provides a homeomorphism. It suffices to consider basic open sets. We know that thesets is a basis for the maximal spectrum of the ring under zariski.We will show that is a basis for the topology of . Then the functionthat takes points to maximal ideals of functions that vanish at that point will provide a topological homeomorphism.Thus, we have shown that the maximal spectrum of the ring allows us to recover the topology of the underlying space!
- We wish to show that the open set form a base for the topology on . So consider an open set . Now think of which is closed. We build the function such that and by invoking Urhyson. Therefore, . So the set can becovered with , which means the sets form a base of the topology on .
- We wish to show that the open sets form a base for the topology on . Let be aclosed set in .
- We wish to show that the open set have homeomorphisms . This completes the isomorphism into a homeomorphism, and we have thus completed the proof that we can recover the topology from the spectrum.
- Consider the function sending the point to the kernel of the evaluation map at .Let be a basic open of . Consider .This will contain all those points such that . This means that it will containpoint such that does not vanish at those points, as (1) implies(2) which implies . Clearly, this is an open subset of , as it is thecomplement of the closed set [zero sets are always closed]. Furthermore, the set maps to what we would expect; it trades the algebraic definition of "does not vanish" to the geometric one,while describing the exact same phenomena.