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

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