To understand where the definitions of schemes even come from, we
must first study classical algebraic geometry. The set-up is as
follows:
Let
be a an algebraically closed field. This means that if
is a polynomial, every root of
is already in
. This is an important condition for technical reasons
(Intuitively, since we are interested in zero sets of polynomials, we do
not want ’missing information’ in the form of missing would-be vanishing
points). Algebraic Geometry originated with the study of polynomials
over the complex numbers.
We consider
, i.e.
-tuples of elements of
. We can think of multivariate polynomials
as functions
by plugging in an
-tuple of elements of
into the variables of
.
We want to study sets which are "carved out" by multivariate
polynomials. For example, the polynomial equation
, or, rewritten
defines a circle over the real numbers (I know I said we
are working with algebraically closed fields, but it is much easier to
picture the real numbers).
More generally, we define an affine algebraic set (Also called an affine
variety), which will be our objects of interest as follows:
Definition: Let
be a set of polynomials, define the set
, the algebraic set defined by
via
Note that for a point
, if
then
, and if
then
. This closure under addition and absorption tells us that
we can pass to ideals, i.e. if
is the ideal generated by
(i.e. the smallest ideal containing it) then
. This will hopefully allow us to draw algebraic
information from the geometric information of the algebraic set. We can
refine this further, but we will wait until later to do so. For now, we
define the Zariski Topology:
Definition: The Zariski Topology on
is defined by declaring the algebraic sets as the closed
sets. Verify for yourself that the following holds:
If are ideals, then so these sets are closed under finite unions.
If is a family of ideals, then , so these sets are closed under aribtrary intersections.
and (Since no point, not even vanishes on constant nonzero polynomials).
So the algebraic sets have the exact properties of closed sets in a
topological space. If you’ve never seen this, it might be weird to think
as a topology as being defined by its closed sets, but since closed and
open sets are just complements of each other, declaring the closed sets
is the same as declaring the open sets (And the properties they have to
satisfy are dual).
Moreover, the Zariski Topology is quite weird. For example, it is almost
never Hausdorff (In fact, a lot open sets are dense in closed sets! If
is closed and irreducible, i.e. not the union of two
smaller closed sets, then every nomepty open subset of
in the subspace topology is dense, so every two nonempty
open sets intersect nontrivially). Nonetheless, it is useful for
reasoning about varieties, and among other things, gives the correct
notion of ’dimension’.
We now give an open basis for the topology:
Let
, and let
i.e. the set of all points on which
is nonzero. Note that this is the complement of
, i.e. the set of all points on which
is
, hence it is open. Now, this is a basis for the Zariski
topology, since if
is an ideal defining a closed set
, then consider the union
. Convnice yourselves that
, since a point isn’t in the zero locus of
iff it doesn’t vanish under one of the polynomials of
, so
are a basis for the Zariski Topology.
Note that we can also go the other way, i.e. we can talk about ideals
that are determined by subsets of
:
Let
, then the ideal defined by
is
This is dual to the previous definition. Verify for
yourselves that this is an ideal. Now the natural question is what is
? It turns out that this is the smallest algebraic set
containing
:
Theorem:
, i.e. the smallest closed set in the Zariski Topology
containing
is
.
Proof. An equivalent characterisation of the closure is the
intersection of all closed sets containing
, hence by definition
. Converesly, let
. We shall prove that
is in
. Concretely, this means that every polynomial that
vanishes on
vanishes on
as well
( cannot be ’separated’ from
by polynomials).
Indeed, let
be a polynomial that vanishes on
, then
, but
, hence
, simple as that we have
. ◻
In particular, if we start with an algebraic set
, then
. It is now natural to ask what is
for an ideal
, in order to be able to characterise the family of ideals
for which
, (i.e. ideals coming from Algebraic subsets), and establish
a correspondence between algebraic sets (Which we restrict to because
they give
) and these ideals
Note that this correspondence is also order reversing, if
then
(It is easier to vanish on less points) and if
then
(It is easier to a vanishing point of less
polynomials).
The previous theorem showed that
is a left inverse to
on algebraic sets. It turns out that we need to restrict
our attention to a more specific type of ideal. In particular, suppose
that
is the ideal of an algebraic set. Suppose that
is a polynomial such that
, then
for all
, but
if and only if
(Here we mean we are taking the polynomial to the
-th power), hence
as well.
Such ideals are called Radical Ideals, and the wonderful
Hilbert’s Nullstellensatz tells us that this is the exact property that
characterises the ideals of variaties. It is also important in giving us
the motivation for schemes:
Theorem: (Hilbert’s Nullstellensatz): Let
be an ideal, then
Although the proof is quite nice, we skip it here for
brevity, we are only skimming this story after all, and even so there is
still so much to do and define! As a corollary, we get that there is a
bijective correspondence given by
and
between algebraic sets and Radical ideals, since
for a radical ideal. Note that if
, then
by the Nullstellensatz, since
is radical (the proof given for
for
an algebraic set works for a general subset
as well), so ideals do not distinguish between algebraic
sets and their closures, just like algebraic sets don’t distinguish
between subsets and the (radical) ideals generated by them.
Moreover, it gives us the following important corollary:
Corollary: The maximal ideals of
are in bijective correspondence to points
, in particular
, and
Proof. First, let
. Note that
, hence
. Conversely, if
, then note that
, but this means that in the quotient
we have
, so
. It is possible to see this in more concrete ways as well:
we can define
. Now we have
Now if we just show that
, then every other term will be in
, hence we will have
. Indeed, we have
as all other terms vanish and
by assumption, hence
.
Now, suppose that we have
with
, then for every
, we have
, hence
, so
for all
, hence
. This shows injectivity of the correspondence
Finally, we show surjectivity: Let
be a maximal ideal, then
is nonempty (By the Nullstellensatz, otherwise we would
have
). Let
, then
(Here we are using the order-reversing property + the
Nullstellensatz), hence by maximality
. This shows that the ideal
is indeed maximal, completing the proof. ◻
This lays the ground for classical algebraic geometry, but keep in
mind there is a whole world of theory that we have not discussed and
will not be getting into. Just keep in mind that there are a lot of
powerful theorems and techniques.
This correspondence between maximal ideals (purely algebraic) and points
(geometric) gives a natural idea of how to generalise the powerful
techniques of classical algebraic geometry and use them to study
arbitrary commutative rings: Just substitute maximal ideals for points,
and follow the geometry (Define a Zariski Topology on the sets of
maximal ideals so that it acts as in the classical case, in particular
it should make
homeomorphic to
via the identification of maximal ideals and points). We
shall explore this method and see if it holds up next time. In
particular, we give some more motivation through coordinate rings.
Comments
Was quite surprised at this order of exposition!
For context, I’ve learned Nullstellensatz from Hulek, which states it as follows:
, and proceeds to prove these in order: maximal-ideals part via Zariski’s lemma, existence-of-zeros trivially from there, and via Rabinowitch’s trick. So deriving maximal-ideals as a corollary kind of reads circular: didn’t we need it to prove NSS?
But, as I understand, these three are all equivalent (separately from being true), and there’re other proofs of NSS (not routing through maximal-ideals?), so this way is also valid. (Or for that matter, if we trust NSS now and derive maximal-ideals from it, but re-prove maximal-ideals later in the course of proving NSS, that’s also valid.)
It is very common to first prove Zariski’s Lemma and that use that to prove the Nullstellensatz, but there are many paths to this proof. Indeed, the proof I was first exposed to in my Commutative Algebra course first proved a much more general Nullstellensatz related to Jacobson Rings, from which all the claims easily follow.
For this write-up I wanted to emphasize the correspondence between radical ideals and algebraic sets, because I think it’s important to notice it subsumes the points to maximal ideals correspondence. I think it is also pedagogically worthwhile to see more than one path to the proof. In particular, there are constructive proofs of the Weak Nullstellensatz, which one can upgrade using Rabinowitch’s Trick to Hilbert’s Nullstellensatz or even using Gröbner Bases.
While pedagogically I think it’s correct to see Zariski’s lemma first (At least the first time you prove the theorem) I allowed myself to call this a corollary because, well, it’s true, and i’m expositing about the gener structure of varieties and algebraic geometry, and I think this order is more correct for the general flow I was going for (In particular I like noticing that points already give maximal ideals, but Hilbert’s Nullstellensatz is exactly the tool we need to go back, so it completes this Galois Connection of I and V).