$Spec(R) = \cup_{i, j} D(f_{i, j})$. This is the same as saying:
$\begin{aligned}
&\emptyset = Spec(R)^c \\
&= (\cup_{i, j} D(f_{i, j}))^c \\
&= \cap_{i, j} D(f_{i, j})^c \\
&= \cap_{i, j} V(f_{i, j})^c \\
\end{aligned}$

Recall that intersecting vanishing sets is the same as building an ideal containing all those functions.
So we have an ideal $I \equiv (f_{11}, f_{12}, \dots, f_{21}, \dots, f_{ij})$.
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:
$1 = g_1 f_{i_1 j_1} + g_2 f_{i_2 j_2} + \dots + g_n f_{1_n j_n}$

Thus we have that $1 \in (f_{i_1 j_1}, f_{i_2 j_2}, \dots f_{i_n j_n}$, 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 $F_2$