§ Recovering topology from sheaf of functions: Proof from Atiyah Macdonald

Let XX 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 00 at some point and 11 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 mm of the ring of continuous functions C(x)C(x) is in bijection with the set of functions that vanish at a point, V(x)V(x).
  • Let C(X)C(X) be the ring of all continuous real valued functions on XX. For each xXx \in X, defineI(x)C(X)I(x) \subseteq C(X) to be the set of functions that vanish at xx. This is a maximal ideal, because it is the kernel of the evaluation map ff(x)f \mapsto f(x).
  • Given some maximal ideal mC(X)m \subseteq C(X), we will show that there is some point pXp \in X such thatm=I(p)m = I(p). To show this, first consider the common zeros of functions in mm, V(m){xX:f(x)=0fm}V(m) \equiv \{ x \in X : f(x) = 0 \forall f \in m \}.We first show that V(m)V(m) is non-empty, and we then show that V(m)V(m) contains exactly one point.
  • To show V(m)V(m) is non-empty, suppose for contradiction that V(m)V(m) is empty. Thus, for each point xXx \in X,not all functions in mm vanish at xx (otherwise xV(m)x \in V(m)).So, there is a function fxmf_x \in m that does not vanish at xx, hence fx(x)0f_x(x) \neq 0. Since fxf_xis continuous, there is some open neighbourhood xUxx \in U_x where f(Ux)0f(U_x) \neq 0. (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 XX is compact, we have a finite number of UxiU_{x_i} that cover XX. Hence, we build a functioncifxi2c \equiv \sum_i f_{x_i}^2 (cc for contradiction) that vanishes nowhere. This means cc is a unit.But we must have cmc \in m as cc is built out of functions fximf_{x_i} \in m. This is a contradiction as a unitcannot belong to a maximal ideal. Thus, V(m)V(m) contains at least one point.
  • To show that V(m)V(m) contains exactly one point, suppose that V(m)V(m) contains a single point xx.This means that all functions in mm vanish at xx. Thus, mI(x)m \subseteq I(x), since I(x)I(x) contains all functions(not just ones in mm) that vanish at xx. But mm is maximal, and hence m=I(x)m = I(x). This tells us thatevery maximal ideal mm corresponds to some vanishing set I(x)I(x).
  • We will next show that every vanishing set I(x)I(x) is distinct. We already know that it is maximal. This gives us an injection. Let I(p),I(q)I(p), I(q) be two vanishing sets for distinct points. Let zpz_p be the function constructed fromUrhyson's lemma that is zero at pp at nonzero at qq. Thus, we have zpI(p)z_p \in I(p) and zp∉I(q)z_p \not \in I(q). Hence,I(p)I(q)I(p) \neq I(q). This shows that the maximal ideals I(p),I(q)I(p), I(q) will be distinct.
  • We have thus established a bijection / nullstellensatz between zero sets maximal ideals V(m)V(m) and functions that vanish at a point I(p)I(p).
  • We will next show that this provides a homeomorphism. It suffices to consider basic open sets. We know that thesets Dspec(f)={mC(X):f∉m}D_{spec}(f) = \{ m \in C(X) : f \not \in m \} is a basis for the maximal spectrum of the ring under zariski.We will show that Dtop(f){xX:f(x)0}D_{top}(f) \equiv \{ x \in X: f(x) \neq 0 \} is a basis for the topology of XX. 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 Dtop(f)D_{top}(f) form a base for the topology on XX. So consider an open set UXU \subseteq X. Now think of UcU^c which is closed. We build the function d(x,U)d(x, U) such that d(x,U)(x)=1d(x, U)(x) = 1and d(x,U)(Uc)=0d(x, U)(U^c) = 0 by invoking Urhyson. Therefore, xDtop(d(x,U))Ux \in D_{top}(d(x, U)) \subseteq U. So the set UU can becovered with {Dtop(d(x,U)):xU}\{ D_{top}(d(x, U)): x \in U \}, which means the sets D(d(x,U))D(d(x, U)) form a base of the topology on XX.
  • We wish to show that the open sets Dspec(f)D_{spec}(f) form a base for the topology on maxSpec(C(X))maxSpec(C(X)). Let UU be aclosed set in maxSpec(C(X))maxSpec(C(X)).
  • We wish to show that the open set Dspec(f)D_{spec}(f) have homeomorphisms Dtop(f)D_{top}(f). 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 zero:XmSpec(C(X))zero: X \rightarrow mSpec(C(X)) sending the point xx to the kernel of the evaluation map at xx.Let Dspec(f)mSpec(C(X))D_{spec}(f) \subseteq mSpec(C(X)) be a basic open of mSpec(C(X))mSpec(C(X)). Consider zero1(Dspec(f))zero^{-1}(D_{spec}(f)).This will contain all those points xXx \in X such that zero(x)Dspec(f)zero(x) \in D_{spec}(f). This means that it will containpoint xXx \in X such that ff does not vanish at those points, as (1) zero(x)Dspec(f)zero(x) \in D_{spec}(f) implies(2) f∉zero(x)f \not \in zero(x) which implies f(x)0f(x) \neq 0. Clearly, this is an open subset of XX, as it is thecomplement of the closed set f(x)=0f(x) = 0 [zero sets are always closed]. Furthermore, the set zero1(Dspec(f))zero^{-1}(D_{spec}(f))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.