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The Zariski topology is a topology on the prime spectrum of a commutative ring. It serves as the basis for much of algebraic geometry.
We consider the definition in increasing generality and sophistication:
First we discuss the naive Zariski topology on affine spaces , consider the classical proofs and discover thereby the special role of prime ideals and maximal ideals;
then we turn to the modern definition of the Zariski topology on affine varieties which takes the concept of prime (and maximal) ideals as primary, and again we provide the classical arguments;
finally we discuss the abstract category theoretic perspective on these matters in terms of Galois connections and obtain slick category theoretic proofs of all the previous statements.
Starting with affine space , then the idea of the Zariski topology is to take as the closed subsets those defined by the vanishing of any set of polynomials over in variables, hence the solution sets to equations of the form
for polynomials. The open subsets of the topology are the complements of these vanishing sets.
It is clear that the vanishing set such a set of polynomials depends only on the ideal in the polynomial ring which is generated by them. Under this translation then forming the intersection of closed subsets corresponds to forming the sum of these ideals, and forming the union of closed subsets corresponds to forming the product of the corresponding ideals. This way the Zariski topology establishes a dictionary between topological concepts of the affine space , and algebra inside the polynomial ring.
In particular one finds that the irreducible closed subsets of the Zariski topology correspond to the prime ideals in the polynomial ring (prop. and prop. below), and that the closed points correspond to the maximal ideals among these (prop. ).
This motivates the modern refinement of the concept of the Zariski topology, where one considers any commutative ring and equips its set of prime ideals with a topology, by direct analogy with the previously naive affine space , which is recovered with a polynomial ring and restricting attention to the maximal ideals (example below).
These sets of prime ideals of a ring equipped with the Zariski topology are called the (topological spaces underlying) the prime spectrum of a commutative ring, denoted .
The Zariski topology is in general not Hausdorff (example below) which makes it sometimes be regarded as an “exotic” type of topology. But it is in fact sober (prop. below) and hence as well-behaved in this respect as general locales are.
We consider here, for a field, the vector space equipped with a Zariski topology. This is the original definition of Zariski topology, and serves well to motivate the concept, but eventually it was superceded by a more refined concept of Zariski topologies of prime spectra, discussed in the next subsection below. In example below we reconsider the naive case of interest in this subsection here from that more refined perspective.
(Zariski topology on affine space)
Let be a field, let , and write for the set of polynomials in variables over .
For a subset of polynomials, let the subset of the -fold Cartesian product of the underlying set of (the vanishing set of ) be the subset of points on which all these polynomials jointly vanish:
These subsets are called the Zariski closed subsets.
Write
for the set of complements of the Zariski closed subsets. These are called the Zariski open subsets of .
(Zariski topology is well defined)
Assuming excluded middle, then:
For a field and , then the Zariski open subsets of (def. ) form a topology. The resulting topological space
is also called the -dimensional affine space over .
We need to show for a set of subsets of polynomials that
for some ;
if is finite then for some .
By de Morgan's law for complements (and using excluded middle) this is equivalent to
for some ;
if is finite then for some .
We claim that we may take
.
(In the second line we have the set of all those polynomials which arise as products of polynomials with one factor from each of the .)
Regarding the first point:
Regarding the second point, in one direction we have the immediate implication
For the converse direction we need to show that
hence that
By excluded middle, this is equivalent to its contraposition, which by de Morgan's law is
This now is true by the assumption that is a field: If all factors are non-zero, then their product is non-zero.
For a field and , consider a subset
of the underlying set of the -fold Cartesian product of with itself. Then the topological closure of this subset with respect to the Zariski topology (def. ) is the vanishing set of all those polynomials that vanish on :
We compute as follows:
Here the first equality is the definition of topological closure, the second is the definition of closed subsets in the Zariski topology (def. ), the third is the expression of intersections of these in terms of unions of polynomials as in the proof of prop. , and then the last one is immediate.
In every topological space the irreducible closed subsets play a special role, as being precisely the points in the space as seen in its incarnation as a locale (this prop.). The following shows that in the Zariski topology the irreducible closed subsets all come from prime ideals in the corresponding polynomial ring, and that when the ground field is algebraically closed, then they are in fact in bijection to the prime ideals. See also at schemes are sober.
(vanishing ideal of Zariski closed subset)#
Let be a field, and let . Then for a Zariski closed subset, according to def. , hence for a set of polynomials, write
for the maximal subset of polynomials that still has the same joint vanishing set:
This set is clearly an ideal in the polynomial ring , called the vanishing ideal of .
With excluded middle then:
Let be a field, let , and let be a Zariski closed subset (def. ). Then the following are equivalent:
is an irreducible closed subset;
The vanishing ideal (def. ) is a prime ideal.
In one direction, assume that is irreducible and consider with . We need to show that then already or .
Now since is a field, we have
This implies that
and hence that
But since , and are all closed, by construction, their intersections are closed and hence this is a decomposition of as a union of closed subsets. Therefore now the assumption that is irreducible implies that
Now for the converse, assume that is a prime ideal, and that . We need to show that or that .
Assume on the contrary, that there existed elements
Then in particular the vanishing ideals would not contain each other
and hence there were polynomials
But since a product of polynomials vanishes at some point once one of the factors vanishes at that point, it would follows that
which were in contradiction to the assumption that is a prime ideal. Hence we have a proof by contradiction.
Proposition gives an injection
The following says that for algebraically closed fields then this is in fact a bijection:
Let be an algebraically closed field and let . Then the function
from prop. is a bijection.
The proof uses Hilbert's Nullstellensatz.
(generalization to affine varieties)
Prop suggests to consider the set of prime ideals of a polynomial ring for general as more fundamental, in some sense, than the set . Morover, the set of prime ideals makes sense for every commutative ring , not just , and hence this suggests to consider a Zariski topology on sets of prime ideals. This leads to the more general concept of Zariski topologies for affine varieties, def. below.
If the field is not a finite field, then the Zariski topology on the affine space (def. ) is not Hausdorff.
This is because the solution set to a system of polynomials over an infinite polynomial is always a finite set. This means that in this case all the Zariski closed subsets are finite sets. This in turn implies that the intersection of every pair of non-empty Zariski open subsets is non-empty.
But the Zariski topology is always sober, see prop. below.
(Zariski topology on set of prime ideals)
Let be a commutative ring. Write for its set of prime ideals. For any subset of elements of the ring, consider the subsets of those prime ideals that contain :
These are called the Zariski closed subsets of . Their complements are called the Zariski open subsets.
(Zariski topology well defined)
Assuming excluded middle, then:
Let be a commutative ring. Then the collection of Zariski open subsets (def. ) in its set of prime ideals
satisfies the axioms of a topology, the Zariski topology.
This topological space
is called (the space underlying) the prime spectrum of the commutative ring.
For write for the ideal which is generated by . Evidently the Zariski closed subsets depend only on this ideal
and therefore it is sufficient to consider the for the case that is not just a subset, but an ideal.
So let be a set of ideals in and let be the corresponding set of Zariski closed subsets. We need to show that there exists such that
;
if is finite set then .
We claim that
,
Regarding the first point:
By using the various definitions, we get the following chain of logical equivalences:
Regarding the second point, in one direction we have the immediate implication
For the converse direction we need to show that
hence that
By excluded middle, this is equivalent to its contraposition, which by de Morgan's law is
This holds by the assumption that is a prime ideal.
We discuss some properties of the Zariski topology on prime spectra of commutative rings.
(topological closure of points)
Let be a commutative ring and consider its prime spectrum equipped with the Zariski topology (def. ).
Then the topological closure of a point is (def. ).
By definition the topological closure of is
Hence unwinding the definitions, we have the following sequence of logical equivalences:
Recall:
Assuming the axiom of choice or at least the ultrafilter principle then:
For a commutative ring and a proper ideal, then is contained in some prime ideal.
The axiom of choice even implies that every proper ideal is contained in a maximal ideal (by this prop.).
(maximal ideals are closed points)
Let be a commutative ring, consider the topological space , i.e. its prime spectrum equipped with the Zariski topology from def. .
Then the maximal ideals inside the prime ideals constitute closed points.
Assuming the axiom of choice or at least the ultrafilter principle then also the converse is true:
Then the inclusion of maximal ideals into all prime ideals is precisely the inclusion of the subset of closed points into all points of .
and hence we need to show that
precisely if is maximal.
In one direction, assume that is maximal. By definition contains all the prime ideals such that . That is maximal means that it is not contained in a larger proper ideal, in particular not in any larger prime ideal, and hence .
In the other direction, assume that is a prime ideal such that . By definition this means equivalently that the only prime ideal with is itself. We need to show that more generally for any proper ideal implies that .
But the axiom of choice/ultrafilter principle imply the prime ideal theorem (lemma ), which says that there is a prime ideal with , hence a sequence of inclusions . Since this implies , we have , hence .
(irreducible closed subsets correspond to prime ideals)
With excluded middle then:
Let be a commutative ring, and let be an ideal in , hence is a Zariski closed subset in the prime spectrum of . Then the following are equivalent:
is an irreducible closed subset;
is a prime ideal.
In one direction, assume that is irreducible, and that with . We need to show that then already or .
To this end, first observe that
This is because
where the implication in the middle uses that is a prime ideal.
It follows that
This is a decomposition of as a union of closed subsets, hence the assumption that is irreducible implies that
Now for the converse. Assume that is a prime ideal and that . Observe (as in the proof of prop. ) that this means equivalently that . We need to show that then or that .
Suppose on the contrary that neither nor coincided with . This means that there were elements and such that still , in contradiction to the assumption. Hence we have a proof by contradiction.
As a corollary:
(Zariski topology on prime spectra is sober)
With excluded middle and axiom of choice (or at least the ultrafilter principle) then:
Let be a commutative ring. Then (its prime spectr equipped with the Zariski topology of def. ) is a sober topological space.
We need to show that the function
which sends a point to its topological closure, is a bijection.
By lemma this function is given by sending a prime ideal to the Zariski closed subset . That this is a bijection is the statement of prop. .
(affine space as prime spectrum)
Reconsider the case where is a polynomial ring, for a field, as in the discussion of the naive affine space above.
Observe that, by , the closed points in the prime spectrum correspond to the maximal ideals in the polynomial ring. These are of the form
and hence are in bijection with the points of the naive affine space
There is however also prime ideals in which are not maximal. In particular there is the 0-ideal .
(Spec(Z))
Let be the commutative ring of integers. Consider the corresponding Zariski prime spectrum (prop. ) Spec(Z).
The prime ideals of the ring of integers are
the ideals generated by prime numbers (this special case is what motivates the terminology “prime ideal”);
the ideal .
All the prime ideals are maximal ideals. Hence by prop. these are closed points of .
Only the prime ideal is not maximal, hence the point is not closed. Its closure is the entire space
To see this, notice that in fact is the only closed subset containing the point . This is because
and , because
We now discuss how all of the above constructions and statements, and a bit more, follows immediately as a special case of the general theory of what is called Galois connections or adjoint functors between posets.
(Galois connection induced from a relation)
Consider two sets and a relation
Define two functions between their power sets , as follows. (In the following we write to abbreviate the formula .)
Define
by
Define
by
The construction in def. has the following properties:
and are contravariant order-preserving in that
if , then ;
if , then
The adjunction law holds:
which we denote by writing
both as well as take unions to intersections.
Regarding the first point: the larger is, the more conditions that are placed on in order to belong to , and so the smaller will be.
Regarding the second point: This is because both these conditions are equivalent to the condition .
Regarding the third point: Observe that in a poset such as , we have that iff for all , iff (this is the Yoneda lemma applied to posets). It follows that
and we conclude by the Yoneda lemma.
(closure operators from Galois connection)
Given a Galois connection as in def. , consider the composites
and
These satisfy:
For all then .
For all then .
is idempotent and covariant.
and
For all then .
For all then .
is idempotent and covariant.
This is summarized by saying that and are closure operators (idempotent monads).
The first statement is immediate from the adjunction law (prop. ).
Regarding the second statement: This holds because applied to sets of the form , we see . But applying the contravariant map to the inclusion , we also have .
This directly implies that the function . is idempotent, hence the third statement.
The argument for is directly analogous.
(closed elements)
Given a Galois connection as in def. , then
is called closed if ;
the closure of is
and similarly
is called closed if ;
the closure of is .
It follows from the properties of closure operators, hence form prop. :
(fixed points of a Galois connection)
Given a Galois connection as in def. , then
the closed elements of are precisely those in the image of ;
the closed elements of are precisely those in the image of .
We says these are the fixed points of the Galois connection. Therefore the restriction of the Galois connection
to these fixed points yields an equivalence
now called a Galois correspondence.
Given a Galois connection as in def. , then the sets of closed elements according to def. are closed under forming intersections.
If is a collection of elements closed under the operator , then by the first item in prop. it is automatic that , so it suffices to prove the reverse inclusion. But since for all and is covariant and is closed, we have for all , and follows.
We now redo the discussion of the Zariski topology on the affine space from above as a special case of the general considerations of Galois connections.
(Zariski closed subsets in affine space via Galois connection)
Let be a field and let , and write for the polynomial ring over in variables. Define a relation
by
By def. and prop. we obtain the corresponding Galois connection of the form
(where now and denote their underlying sets).
sends a set of polynomials to its corresponding variety,
These are just the Zariski closed subsets from def. .
In the other direction,
sends a set of points to its corresponding vanishing ideal
We may now use the abstract theory of Galois connections to verify that Zariski closed subsets form a topology:
(Zariski topology is well defined)
Using excluded middle, then:
The set of Zariski closed subsets of from example constitutes a topology in that it is closed under
arbitrary intersections;
finite untions.
Regarding the first point: From prop. we know that takes unions to intersections, hence that
Regarding the second point, we exploit the commutative ring structure of . It is sufficient to show that the set of Zariski closed sets is closed under the empty union and under binary unions.
The empty union is the entire space , which is (the variety associated with the constant polynomial ),
Hence it only remains to see closure under binary unions.
To this end, recall from prop. that we may replace with the corresponding ideal
without changing the variety:
With this it is sufficient to show that
where is the ideal consisting of finite sums of elements of the form with and .
We conclude by proving this statement:
Applying the contravariant operator to the inclusions and (which are clear since are ideals), we derive and , so the inclusion is automatic.
In the other direction, to prove , suppose and that doesn’t belong to . Then for some . For every , we have since and . Now divide by to get for every , so that .
Let be a field, let and write for the polynomial ring over in variables, and for the set of maximal ideals in this ring.
Define then a relation
by
For a subset we calculate
which is an ideal, since the intersection of any collection of ideals is again an ideal. (However, not all ideals are given as intersections of maximal ideals, a point to which we will return in a moment.)
This is a slight generalization of example since each point induces a maximal ideal
i.e. the kernel of the function
which evaluates polynomials at the point , where we have iff .
Of course it need not be the case that all maximal ideals are given by points in this way; for example, the ideal is maximal in but is not given by evaluation at a point because does not vanish at any real point. However, if the ground field is algebraically closed, then every maximal ideal of is given by evaluation at a point . This result is not completely obvious; it is sometimes called the weak Nullstellensatz.
The set that are closed under the operator in example form a topology.
The proof is virtually the same as in the proof of prop. : they are closed under arbitrary intersections by our earlier generalities, and they are closed under finite unions by the similar reasoning: where is an ideal, so there is no loss of generality in considering for ideals , and . If (meaning ) but doesn’t belong to , i.e., for some , then for every we have . Taking the quotient map to the field , we have , and since we have for every , hence .
Thus the fixed elements of on one side of the Galois correspondence are the closed sets of a topology. The fixed elements of on the other side are a matter of interest; in the case where is algebraically closed, they are the radical ideals of according to the “strong” Nullstellensatz.
We now redo the discussion of the Zariski topology on the prime spectrum of a commutative ring from above as a special case of the general considerations of Galois connections.
Lecture notes include
See also
Last revised on January 4, 2024 at 22:03:59. See the history of this page for a list of all contributions to it.