A boolean ring is one where for every element $r \in R$, we have $r^2 = r$. 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 $Spec$ to import a topology on complete lattices, which will be Haussdorf and completely disconnected.

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

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

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

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

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

That is, the basic open sets of the prime spectrum of the ring are all clopen. The basic open sets $D(f)$ (where $D$ stands for doesn't vanish on the prime spectrum) are open by definition. To show that $D(f)$ is open, we need to find an ideal $I$ such that $D(f) = V(I)$. Consider $g \equiv 1 - f$. We know that $f(1 - f) = 0$ for all $f$ and we are considering the prime spectrum. Thus, either $f$ or $(1 - f)$ must vanish at each point because $f(1 - f)$ vanishes at each point: so $f(1 - f) = 0 \in p$ implies $f \in p$ or $(1 - f) \in p$.

• To see that $V(1 - f) \subset D(f)$, let $p \in V(1 - f)$. Hence, $1 - f \in p$. Thus, we have $f \not \in p$: if $f$ and $(1 - f)$ are both in $p$, then $1 \in p$ which is absurd as $p$ is a proper ideal. Hence, $p \in D(f)$. Thus, $V(1 - f) \subset D(f)$.
• To see that $D(f) \subset V(1 - f)$, let $p \in D(f)$, hence $f \not \in p$. But $f(1 - f) = 0 \in p$ since the ring is boolean. Hence $(1 - f) \in p$ by the primality of $p$ and since $f(1 - f) \in p$. Thus, $p \in V(1 - f)$. Hence, $D(f) \subset V(1 - f)$.

This shows us that each basic open set is clopen, as each basic open is both open and closed.

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

We generally only call Haussdorf spaces "compact". The covering property is called "quasi-compact"

Consider an open covering $C_i$ such that $\cup C_i = Spec(R)$. Since the base of the topology is the doesn't vanish sets, we can write each $C_i$ as $C_i = \cup_j D(f_{ij})$. Hence we have that $Spec(R) = \cup_{i, j} D(f_{i, j})$. This is the same as saying:

Recall that intersecting vanishing sets is the same as building an ideal containing all those functions. So we have an ideal $I \equiv (f_{11}, f_{12}, \dots, f_{21}, \dots, f_{ij})$. Saying that the intersection of all $V(f_{ij})$ is empty is saying that $I = R$. 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 $I \equiv (f_{ij}) = R$, and hence $1 \in R$ implies $1 \in I = (f_{ij})$. In an ideal, we only ever take finite sums . So $1$ is a finite linear combination of some $f_{ij}$. So we have the equation:

Thus we have that $1 \in (f_{i_1 j_1}, f_{i_2 j_2}, \dots f_{i_n j_n}$, and hence $\cap_{k=1}^n V(f_{i_k j_k}) = \emptyset$. Complementing both sides, we get that $\cup_{k=1}^n D(f_{i_k j_k}) = Spec(R)$. We know that $D(f_{i_k j_k}) \subseteq C_{i_k}$, as the basic open set $D(f_{i_k j_k})$ was used to cover $C_{i_k}$. Hence, we can "expand out" the finite covering by basic opens to a finite overing by the covering given to us. So we get $Spec(R) = \cup_{k=1}^n C_{i_k}$.

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

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

§ Topology: Closed subset $S$ of a quasi-compact space $T$ is quasi-compact

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

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

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

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

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

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

Intuitively, since every prime ideal is maximal, given two distint prime ideals $p, q$, we can find functions $f, g$ such that $f$ vanishes only on $p$ and $g$ vanishes only on $q$. Since the basic opens are clopen, we can then build opens that separate $p$ from $q$ by complementing the vanishing sets of $f, g$.

Pick two points $p, q \in Spec(B)$, $p \neq q$. These are maximal ideals (all prime ideals in $B$ are maximal). Thus, neither contain the other; So we must have elements $f \in p - q$, and $g \in q - p$. So we have that $V(f) = \{ p \}$ and $V(g) = \{ q \}$. Since $Spec(B)$ is clopen, we know that $V(f)^c$ and $V(g)^c$ are also open. So we get neighbourhoods $N_p \equiv V(f) \cap V(g)^c$ and $N_q \equiv V(g) \cap V(f)^c$ such that $N_p \cap N_q = \emptyset$ and $p \in N_p$, and $q \in N_q$. Thus we are able to separate the space.

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

§ A boolean lattice $L$ can be converted into a boolean ring.

Take a boolean lattice $L$ define the zero of the ring to be bottom, so $0 \equiv \bot$, and the one of the ring to be the top, so $1 \equiv \top$. The addition operation is XOR, and the multiplication is intersection; So we define $a + b \equiv (a \land \lnot b) \lor (\lnot a \land b)$, and multiplication as $a \cdot b \equiv a \land b$. It's easy to check that this does obey the axioms of a commutative ring, and is boolean because $a^2 = a \land a = a$.

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

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

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

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

§ Bonus: quotient ring $R/p$ for prime ideal $p$ is $F_2$