Zariski Topology for Rings
In this article, we generalize earlier results in algebraic geometry to apply to general rings. Recall that points on an affine variety V correspond to maximal ideals . For general rings, we have to switch to taking prime ideals because of the following.
Lemma 1.
If
is a ring homomorphism and
is a prime ideal of B, then
is a prime ideal of A.
Proof
The composition has kernel
. Hence this induces an injective ring homomorphism
. Since
is a domain, so is
so
is prime. ♦
However it is not true that if is maximal in B,
must be maximal in A. For instance consider the inclusion
. The zero ideal is maximal in
but
is not maximal in
.
Now we define our main object of interest.
Definition.
Given a ring A, the spectrum of A is its set of prime ideals
. The Zariski topology on
is defined as follows:
is closed if and only if it is of the following form
for some subset
.
Note that since where
is the ideal generated by S, we lose no generality by taking S to be ideals of A. If S = {f} we will write V(f) instead of V({f}) or V((f)) to reduce the clutter.
Immediately, we prove that the above gives a bona fide topology.
Proposition 1.
For any collection of ideals
of A and ideals
of A, we have
,
,
.
Proof
The first claim is obvious. For the second, contains
for each i if and only if it contains
.
Finally, for the third claim, we have and
so we have
Finally suppose . Then
and
so there are
and
. Then
since
is prime so
. Hence if
then
. ♦
Structure of Spec A
Thanks to Zorn’s lemma, we have the following results on the structure of Spec A.
Proposition 2.
If A is a ring, every proper ideal
is contained in a maximal ideal.
Proof
Let be the collection of all proper ideals
containing
, ordered partially by inclusion. We claim that every chain
has an upper bound. Since
we may assume without loss of generality
.
Take . Let us show that
is an ideal of A.
Clearly . Let
and
. Then
and
for some
. Since
is totally ordered, either
or
. Assuming the former, this gives
so
. Thus
is an ideal of A.
Since none of the contains 1, we have
. Hence
is an upper bound of
.
By Zorn’s lemma has a maximal element, which is precisely a maximal ideal of A containing
♦
We say that minimal if it is a minimal element with respect to inclusion.
Proposition 3.
Any
contains a minimal prime
.
Proof
Let be the collection of all prime ideals of A contained in
, ordered partially by reverse inclusion, i.e.
if
. To apply Zorn’s lemma we need to show that every chain
has a maximal element (i.e. minimal with respect to inclusion). Without loss of generality we may assume
.
Let , an ideal of A. We need to show that it is prime. Suppose
so there exist
such that
and
. Since
is a chain either
or
; assuming the former we have
. Since
is prime we have
so
.
Now apply Zorn’s lemma to obtain a maximal element, which is a minimal prime contained in . ♦
Corollary.
If A is non-trivial,
must contain a maximal ideal and a minimal prime ideal.
Homomorphisms
Suppose is a ring homomorphism. We saw that this induces a map
.
Proposition 4.
The map
is continuous with respect to the Zariski topology; in fact
Note that is not an ideal of B in general, but that is okay since we defined V on arbitrary subsets.
Proof
Let . This lies in the LHS if and only if
contains
, if and only if
contains
. ♦
Clearly the identity ring homomorphism on A induces the identity map on , also for ring homomorphisms
and
we have
Exercise A
Prove that for an ideal , the canonical map
induces
which is a subspace embedding onto the closed subset .
Comparison with Algebraic Geometry
Proposition 5.
Let
be a closed subset of
. Then
,
the radical of
.
Proof
(⊇) Suppose ,
for some n > 0. Now each
contains
and hence contains
. Since
is prime we have
.
(⊆) Suppose ,
does not contain any power of x. We will prove the existence of
not containing x.
Let be the set of all ideals
not containing any power of x; thus
. Order
by inclusion. We will show that any chain
has an upper bound. Indeed, let
. As before, since
is a chain of ideals ordered by inclusion,
is an ideal of A. It does not include any power of x so
.
Hence by Zorn’s lemma, has a maximal element
; by construction
contains
but not any power of x. It remains to show
is prime: suppose not so there are
with
. Now
and
strictly contain
, so by maximality of
, we have
. Hence
a contradiction. Thus is prime. ♦
Exercise B
From proposition 5, prove that we have a bijection
The correspondence reverses inclusion. For and
in Spec A, write down the corresponding operations for radical ideals of A.