§ Stone representation theorem: Proof from Atiyah Macdonald
A boolean ring is one where for every element , we have . 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 to import a topology on complete lattices,
which will be Haussdorf and completely disconnected.
§ In a boolean ring, all prime ideals are maximal.
Let be a prime ideal, and let . One is tempted to use jacobson like arguments,
and thus one considers the element . Since in this ring, we have ,
this means that . Hence, we have . This is not so
useful, until we notice that we never used the fact that !. So pick some arbitrary
element . To show a prime ideal is maximal, let . Then we know
that . Thus, if we were added to (ie, we try to create a larger ideal ),
we get . Thus, the ideal when . Thus,
§ In a boolean ring, . More generally, for all .
Showing is the same as showing . We know that
as in this ring for all . Hence, we have that or or .
§ In the spectrum of a boolean ring, for is clopen
That is, the basic open sets of the prime spectrum of the ring are all clopen. The basic
open sets (where stands for doesn't vanish on the prime spectrum)
are open by definition. To show that is open, we need to find
an ideal such that . Consider . We know
that for all and we are considering the prime spectrum.
Thus, either or must vanish at each point because vanishes
at each point: so implies or .
This shows us that each basic open set is clopen, as each basic open is both open and closed.
- To see that , let . Hence, . Thus, we have : if and are both in , then which is absurd as is a proper ideal. Hence, . Thus, .
- To see that , let , hence . But since the ring is boolean. Hence by the primality of and since . Thus, . Hence, .
§ for all rings, The space is quasi-compact: every open cover of has a finite subcover of .
We generally only call Haussdorf spaces "compact". The covering property is called "quasi-compact"
Consider an open covering such that . Since the base of the topology
is the doesn't vanish sets, we can write each as . Hence we have that
. 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 .
Saying that the intersection of all is empty is saying that .
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 , and hence implies .
In an ideal, we only ever take finite sums . So is a finite linear combination of some . So we have
Thus we have that , and hence .
Complementing both sides, we get that . We know that , as
the basic open set was used to cover . Hence, we can "expand out" the finite covering by basic opens
to a finite overing by the covering given to us. So we get .
§ for all rings, each is quasi-compact
This is a generalization of the fact that is quasi-compact, as . Localize at , so
build the ring . Intuitively, , as only has ideals where does not vanish.
If vanishes at a prime , then . But we localize at , hence becomes a unit, so we get ,
and thus the ideal is no longer an ideal.
§ Topology: Closed subset of a quasi-compact space is quasi-compact
Let be closed. We wish to show that is quasi-compact; that is,
any cover of has a finite subcover. Let be an arbtirary cover of .
Create a new cover which is with added. We add so that
we can cover with , and from this extract a cover for . This works
since covers no element of ; The subcover we get of will have
to create a covering for using the sets of . Ask for a finite subcover
of . The finite covering of is .
§ In , a boolean ring, the sets are closed under finite union
We want to show that for each family , we have a such that .
We will show it for two functions; recurse in general. The idea is that if we have ,
we want to build a function that does not vanish when either or vanish. Let's
pretend they are boolean functions. Then we are looking for . We can realise
or in terms of and (multiplication) and xor(addition) as .
To re-ring-theory this, write . See that (1) if vanishes ( )
then , (2) if vanishes ( ) then which is as expected. If neither
nor vanish at , then in this case, we must have vanishes at , since .
Hence or belong to the prime ideal, and hence one of them must vanish. If
does not vanish, then vanishes, and hence does not vanish. So,
does not vanish when either or do not vanish, which means that .
Iterate for .
§ In , for a boolean ring, the sets are the only subsets that are clopen.
We know that all the are clopen. We need to show that these are the only ones.
So pick some clopen set (for "ajar", a pun on clopen). Since is open, we must that
is a (possibly infinite) union of basic opens .
Since is closed and is quasi-compact, is also quasi-compact.
Thus, we can extract a finite subcover of to write .
The sets are closed under finite union. So there exists some such that .
Thus, any clopen set can be written as for some .
§ , for a boolean ring, is Haussdorf
Intuitively, since every prime ideal is maximal, given two distint prime ideals , we can find
functions such that vanishes only on and vanishes only on . Since the
basic opens are clopen, we can then build opens that separate from by complementing
the vanishing sets of .
Pick two points , . These are maximal ideals (all prime ideals in are maximal).
Thus, neither contain the other; So we must have elements , and . So
we have that and . Since is clopen, we know that
and are also open. So we get neighbourhoods
and such that and , and .
Thus we are able to separate the space.
§ , 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, for a boolean ring is compact.
§ A boolean lattice can be converted into a boolean ring.
Take a boolean lattice define the zero of the ring to be bottom, so ,
and the one of the ring to be the top, so . The addition operation
is XOR, and the multiplication is intersection; So we define ,
and multiplication as . It's easy to check that this does obey the
axioms of a commutative ring, and is boolean because .
§ Boolean rings are boolean lattices of the clopen sets of the spectra
Take a boolean ring , build its spectra . Take the set of all clopens. We have
seen that this is exactly the sets . Let us show that and
where is the exclusive or of the sets. This induces
a map from the ring operations to the lattice operations.
§ Boolean lattices are the clopen sets of spectra of boolean rings .
Take a lattice , treat it as a ring, and consider the clopens generated from the ring.
We know that for two elements we have that . From the previous
argument, we know that . This gies ,
a lattie homomorphism. we get by complementing; Since
every set is clopen, we can complement a clopen set to get some clopen set .
But every clopen set can be written as for some .
§ Bonus: quotient ring for prime ideal is