§ Stone representation theorem: Proof from Atiyah Macdonald

A boolean ring is one where for every element rRr \in R, we have r2=rr^2 = r. We first study boolean rings abstactly and collect their properties. Secondly, we show an isomorphism between complete lattices and boolean rings. Thirdly, we use the topology from SpecSpec to import a topology on complete lattices, which will be Haussdorf and completely disconnected.

§ In a boolean ring, all prime ideals are maximal.

Let pp be a prime ideal, and let xpx \in p. One is tempted to use jacobson like arguments, and thus one considers the element (1x)(1 - x). Since in this ring, we have x2=xx^2 = x, this means that x(1x)=0x(1 - x) = 0. Hence, we have x(1x)px(1 - x) \in p. This is not so useful, until we notice that we never used the fact that xpx \in p!. So pick some arbitrary element yRy \in R. To show pp a prime ideal is maximal, let y∉py \not \in p. Then we know that 1yp1 - y \in p. Thus, if yy we were added to pp (ie, we try to create a larger ideal m=(y,p)m = (y, p)), we get 1y+y=1p1 - y + y = 1 \in p. Thus, the ideal m=(y,p)=1m = (y, p) = 1 when y∉py \not \in p. Thus, yy is maximal.

§ In a boolean ring, 2=1+1=02 = 1 + 1 = 0. More generally, 2x=02x = 0 for all xx.

Showing 2x=02x = 0 is the same as showing x=xx = -x. We know that (x)=(x)2=x2=x(-x) = (-x)^2 = x^2 = x as in this ring x2=xx^2 = x for all xx. Hence, we have that x=xx = -x or 1=11 = -1 or 2=02 = 0.

§ In the spectrum of a boolean ring, D(f)D(f) for fRf \in R is clopen

That is, the basic open sets of the prime spectrum of the ring are all clopen. The basic open sets D(f)D(f) (where DD stands for doesn't vanish on the prime spectrum) are open by definition. To show that D(f)D(f) is open, we need to find an ideal II such that D(f)=V(I)D(f) = V(I). Consider g1fg \equiv 1 - f. We know that f(1f)=0f(1 - f) = 0 for all ff and we are considering the prime spectrum. Thus, either ff or (1f)(1 - f) must vanish at each point because f(1f)f(1 - f) vanishes at each point: so f(1f)=0pf(1 - f) = 0 \in p implies fpf \in p or (1f)p(1 - f) \in p.
  • To see that V(1f)D(f)V(1 - f) \subset D(f), let pV(1f)p \in V(1 - f). Hence, 1fp1 - f \in p. Thus, we havef∉pf \not \in p: if ff and (1f)(1 - f) are both in pp, then 1p1 \in p which is absurd as pp is a proper ideal.Hence, pD(f)p \in D(f). Thus, V(1f)D(f)V(1 - f) \subset D(f).
  • To see that D(f)V(1f)D(f) \subset V(1 - f), let pD(f)p \in D(f), hence f∉pf \not \in p. But f(1f)=0pf(1 - f) = 0 \in p sincethe ring is boolean. Hence (1f)p(1 - f) \in p by the primality of pp and since f(1f)pf(1 - f) \in p.Thus, pV(1f)p \in V(1 - f). Hence, D(f)V(1f)D(f) \subset V(1 - f).
This shows us that each basic open set is clopen, as each basic open is both open and closed.

§ for all rings, The space Spec(R)Spec(R) is quasi-compact: every open cover of Spec(R)Spec(R) has a finite subcover of Spec(R)Spec(R).

We generally only call Haussdorf spaces "compact". The covering property is called "quasi-compact" Consider an open covering CiC_i such that Ci=Spec(R)\cup C_i = Spec(R). Since the base of the topology is the doesn't vanish sets, we can write each CiC_i as Ci=jD(fij)C_i = \cup_j D(f_{ij}). Hence we have that Spec(R)=i,jD(fi,j)Spec(R) = \cup_{i, j} D(f_{i, j}). This is the same as saying:
=Spec(R)c=(i,jD(fi,j))c=i,jD(fi,j)c=i,jV(fi,j)c \begin{aligned} &\emptyset = Spec(R)^c \\ &= (\cup_{i, j} D(f_{i, j}))^c \\ &= \cap_{i, j} D(f_{i, j})^c \\ &= \cap_{i, j} V(f_{i, j})^c \\ \end{aligned}
Recall that intersecting vanishing sets is the same as building an ideal containing all those functions. So we have an ideal I(f11,f12,,f21,,fij)I \equiv (f_{11}, f_{12}, \dots, f_{21}, \dots, f_{ij}). Saying that the intersection of all V(fij)V(f_{ij}) is empty is saying that I=RI = R. This is by strong nullstellensatz, which states that every maximal ideal (and hence, every ideal which is contained in some maximal ideal) must have some solution. The only way to not have a solution (ie, to vanish nowhere) is to generate the entire ring. Thus, we must have that I(fij)=RI \equiv (f_{ij}) = R, and hence 1R1 \in R implies 1I=(fij)1 \in I = (f_{ij}). In an ideal, we only ever take finite sums. So 11 is a finite linear combination of some fijf_{ij}. So we have the equation:
1=g1fi1j1+g2fi2j2++gnf1njn 1 = g_1 f_{i_1 j_1} + g_2 f_{i_2 j_2} + \dots + g_n f_{1_n j_n}
Thus we have that 1(fi1j1,fi2j2,finjn1 \in (f_{i_1 j_1}, f_{i_2 j_2}, \dots f_{i_n j_n}, and hence k=1nV(fikjk)=\cap_{k=1}^n V(f_{i_k j_k}) = \emptyset. Complementing both sides, we get that k=1nD(fikjk)=Spec(R)\cup_{k=1}^n D(f_{i_k j_k}) = Spec(R). We know that D(fikjk)CikD(f_{i_k j_k}) \subseteq C_{i_k}, as the basic open set D(fikjk)D(f_{i_k j_k}) was used to cover CikC_{i_k}. Hence, we can "expand out" the finite covering by basic opens to a finite overing by the covering given to us. So we get Spec(R)=k=1nCikSpec(R) = \cup_{k=1}^n C_{i_k}.

§ for all rings, each D(f)D(f) is quasi-compact

This is a generalization of the fact that Spec(R)Spec(R) is quasi-compact, as Spec(R)=D(1)Spec(R) = D(1). Localize at ff, so build the ring R(f)R_{(f)}. Intuitively, Spec(Rf)=D(f)Spec(R_f) = D(f), as Spec(Rf)Spec(R_f) only has ideals where ff does not vanish. If ff vanishes at a prime pp, then fpf \in p. But we localize at ff, hence ff becomes a unit, so we get 1p(f)1 \in p_{(f)}, and thus the ideal is no longer an ideal.

§ Topology: Closed subset SS of a quasi-compact space TT is quasi-compact

Let STS \subseteq T be closed. We wish to show that SS is quasi-compact; that is, any cover of SS has a finite subcover. Let CiC_i be an arbtirary cover of SS. Create a new cover CiC'_i which is CiC_i with ScS^c added. We add ScS^c so that we can cover TT with CiC'_i, and from this extract a cover for SS. This works since ScS^c covers no element of SS; The subcover we get of CiC'_i will have to create a covering for SS using the sets of CiC_i. Ask for a finite subcover FiF_i of CiC'_i. The finite covering of SS is FiScF_i - S^c.

§ In Spec(B)Spec(B), a boolean ring, the sets D(f)D(f) are closed under finite union

We want to show that for each family D(fi)D(f_i), we have a gg such that D(g)=iD(fi)D(g) = \cup_i D(f_i). We will show it for two functions; recurse in general. The idea is that if we have f,gf, g, we want to build a function that does not vanish when either ff or gg vanish. Let's pretend they are boolean functions. Then we are looking for fgf \lor g. We can realise or in terms of and (multiplication) and xor(addition) as hfgf+g+fgh \equiv f \lor g \equiv f + g + fg. To re-ring-theory this, write h=f+g+fg=f+g(1f)h = f + g + fg = f + g(1 - f). See that (1) if ff vanishes (f=0f = 0) then h=gh = g, (2) if gg vanishes (g=0g = 0) then h=fh = f which is as expected. If neither ff nor gg vanish at pp, then in this case, we must have (1f)(1 - f) vanishes at pp, since f(1f)=ff2=0pf(1 - f) = f - f^2 = 0 \in p. Hence ff or f2f^2 belong to the prime ideal, and hence one of them must vanish. If ff does not vanish, then (1f)(1 - f) vanishes, and hence h=fh = f does not vanish. So, hh does not vanish when either ff or gg do not vanish, which means that D(f)D(g)=D(h)=D(f+g+fg)D(f) \cup D(g) = D(h) = D(f + g + fg). Iterate for nn.

§ In Spec(B)Spec(B), for a boolean ring, the sets D(f)D(f) are the only subsets that are clopen.

We know that all the D(f)D(f) are clopen. We need to show that these are the only ones. So pick some clopen set AA (for "ajar", a pun on clopen). Since AA is open, we must that AA is a (possibly infinite) union of basic opens D(fi)D(f_i). Since AA is closed and Spec(B)Spec(B) is quasi-compact, AA is also quasi-compact. Thus, we can extract a finite subcover of AA to write A=k=1nD(fik)A = \cup_{k=1}^n D(f_{i_k}). The sets D(f)D(f) are closed under finite union. So there exists some gg such that A=D(g)A = D(g). Thus, any clopen set AA can be written as D(g)D(g) for some gg.

§ Spec(B)Spec(B), for a boolean ring, is Haussdorf

Intuitively, since every prime ideal is maximal, given two distint prime ideals p,qp, q, we can find functions f,gf, g such that ff vanishes only on pp and gg vanishes only on qq. Since the basic opens are clopen, we can then build opens that separate pp from qq by complementing the vanishing sets of f,gf, g. Pick two points p,qSpec(B)p, q \in Spec(B), pqp \neq q. These are maximal ideals (all prime ideals in BB are maximal). Thus, neither contain the other; So we must have elements fpqf \in p - q, and gqpg \in q - p. So we have that V(f)={p}V(f) = \{ p \} and V(g)={q}V(g) = \{ q \}. Since Spec(B)Spec(B) is clopen, we know that V(f)cV(f)^c and V(g)cV(g)^c are also open. So we get neighbourhoods NpV(f)V(g)cN_p \equiv V(f) \cap V(g)^c and NqV(g)V(f)cN_q \equiv V(g) \cap V(f)^c such that NpNq=N_p \cap N_q = \emptyset and pNpp \in N_p, and qNqq \in N_q. Thus we are able to separate the space.

§ Spec(B)Spec(B), for a boolean ring, is compact

Compact is just a definition that asks for (1) Haussdorf, and (2) quasi-compact, both of which we have shown above. Thus, Spec(B)Spec(B) for a boolean ring is compact.

§ A boolean lattice LL can be converted into a boolean ring.

Take a boolean lattice LL define the zero of the ring to be bottom, so 00 \equiv \bot, and the one of the ring to be the top, so 11 \equiv \top. The addition operation is XOR, and the multiplication is intersection; So we define a+b(a¬b)(¬ab)a + b \equiv (a \land \lnot b) \lor (\lnot a \land b), and multiplication as ababa \cdot b \equiv a \land b. It's easy to check that this does obey the axioms of a commutative ring, and is boolean because a2=aa=aa^2 = a \land a = a.

§ Boolean rings BB are boolean lattices of the clopen sets of the spectra BRing(Clopen(Spec(B)))BRing(Clopen(Spec(B)))

Take a boolean ring BB, build its spectra Spec(B)Spec(B). Take the set of all clopens. We have seen that this is exactly the sets D(f)D(f). Let us show that D(fg)=D(f)D(g)D(fg) = D(f) \cap D(g) and D(f+g)=D(f)D(g)D(f + g) = D(f) \oplus D(g) where \oplus is the exclusive or of the sets. This induces a map from the ring operations to the lattice operations.

§ Boolean lattices LL are the clopen sets of spectra of boolean rings Clopen(Spec(R(L)))Clopen(Spec(R(L))).

Take a lattice LL, treat it as a ring, and consider the clopens generated from the ring. We know that for two elements l,ml, m we have that lm=lmlm = l \land m. From the previous argument, we know that D(lm)=D(l)D(m)D(lm) = D(l) \cap D(m). This gies D(lm)=D(lm)=D(l)D(m)D(l \land m) = D(lm) = D(l) \cap D(m), a lattie homomorphism. we get D(lm)=D(l)D(m)D(l \lor m) = D(l) \cup D(m) by complementing; Since every set is clopen, we can complement a clopen set D(l)D(l) to get some clopen set D(l)cD(l)^c. But every clopen set can be written as D(l)D(l') for some ll'.

§ Bonus: quotient ring R/pR/p for prime ideal pp is F2F_2