Let be a topological space. is said to be irreducible if and only if no two non-empty open subsets of are disjoint.
Some people (topologists) call irreducible spaces hyperconnected.
Theorem 21.2 (characterisation of irreducible spaces):
Let be a topological space. The following are equivalent:
- is irreducible.
- can not be written as the union of two proper closed subsets.
- Every open subset of is dense in .
- The interior of every proper closed subset of is empty.
We prove 1. 2. 3. 4. 1.
1. 2.: Assume that , where , are proper and closed. Define and . Then are open and
by one of deMorgan's rules, contradicting 1.
2. 3.: Assume that is open but not dense. Then is closed and proper in , and so is . Furthermore, , contradicting 2.
3. 4.: Let be closed such that . By definition of the closure, , which is why is a non-dense open set, contradicting 3.
4. 1.: Let be open and non-empty such that . Define . Then is a proper, closed subset of , since . Furthermore, , which is why has non-empty interior.
We prove 1. 4. 3. 2. 1.
1. 4.: Assume we have a proper closed subset of with nonempty interior. Then and are two disjoint nonempty open subsets of .
4. 3.: Let be open. If was not dense in , then would be a proper closed subset of with nonempty interior.
3. 2.: Assume , proper and closed. Set . Then , and hence is not dense within .
2. 1.: Let be open. If they are disjoint, then .
1. 3.: Assume open, not dense. Then is nonempty and disjoint from .
3. 1.: Let be open. If they are disjoint, then and thus is not dense.
2. 4.: Let be proper and closed with nonempty interior. Then .
4. 2.: Let , proper and closed. Then .
We shall go on to prove a couple of properties of irreducible spaces.
Every irreducible space is connected and locally connected.
1. Connectedness: Assume , open, non-empty. This certainly contradicts irreducibility.
2. Local connectedness: Let , where is open. But any open subset of is connected as in 1., which is why we have local connectedness.
Let be an irreducible space. Then is Hausdorff if and only if .
If , then is trivially Hausdorff. Assume that is Hausdorff and contains two distinct points . Then we find open such that , and , contradicting irreducibility.
Let be topological spaces, where is irreducible, and let be a continuous function (i.e. a morphism in the category of topological spaces). Then is irreducible with the subspace topology induced by .
Proof: Let be two disjoint non-empty open subsets of . Since we are working with the subspace topology, we may write , , where are open. We have
- and similarly .
Hence, and are open in by continuity, and since they further are disjoint (since if , then and thus ) and non-empty (since e.g. if , since , for an and hence ), we have a contradiction.
If is irreducible, is Hausdorff and is continuous, then is constant.
Proof: Follows from theorems 21.4 and 21.5.
We may now connect irreducible spaces with Noetherian spaces.
Let be a Noetherian topological space, and let be closed. Then there exists a finite decomposition
where each is irreducible, and no is a subset of (or equals) any of the other . Furthermore, this decomposition is unique up to order.
First we prove existence. Let be closed. Then either is irreducible, and we are done, or can be written as the union of two proper closed subsets . Now again either and are irreducible, or they can be written as the union of two proper closed subsets again. The process of thus splitting up the sets must eventually terminate with all involved subsets being irreducible, since is Noetherian and otherwise we would have an infinite properly descending chain of closed subsets, contradiction. To get the last condition satisfied, we unite any subset contained within another with the greater subset (this can be done successively since there are only finitely many of them). Hence, we have a decomposition of the desired form.
We proceed to proving uniqueness up to order. Let be two such decompositions. For , we may thus write . Assume that there does not exist such that . Then we may define and then successively
for . Then we set and increase until is a decomposition of into two proper closed subsets (such an exists since it equals the first such that ). Thus, our assumption was false; there does exist such that . Thus, each is contained within a , and by symmetry is contained within some . Since by transitivity of this implies , and . For a fixed , we set , where is thus defined ( is unique since otherwise there exist two equals among the -sets). In a symmetric fashion, we may define , where . Then and are inverse to each other, and hence follows (sets with a bijection between them have equal cardinality) and the definition of , for example, implies that both decompositions are equal except for order.
- Exercise 21.1.1: Let be an irreducible topological space, and let be open. Prove that is irreducible.
Algebraic sets and varieties
Let be a field. Then the sets of the form
where is a subset of the ring of polynomials in variables over (that is ), are called algebraic sets. If for a single , we shall occasionally write
The following picture depicts three algebraic sets (apart from the cube lines):
The orange surface is the set , the blue surface is the set , and the green line is the intersection of the two, equal to the set , where
Three immediate lemmata are apparent.
Proof: Being in is the stronger condition.
Lemma 21.10 (formulas for algebraic sets):
Let be a field and set . Then the following rules hold for algebraic sets of :
- ( a set)
- ( ideals)
- ( sets)
1. Let . If follows . This proves . The other direction follows from lemma 21.9.
2. follows from the constant functions being contained within , and gives no condition on the points of to be contained within it.
3. follows by
since clearly .
We will first prove for the case . Indeed, let , that is, neither nor . Hence, we find a polynomial such that and a polynomial such that . The polynomial is contained within and , since every field is an integral domain. Thus, .
Assume holds for many sets. Then we have
From this lemma we see that the algebraic sets form the closed sets of a topology, much like the Zariski-closed sets we got to know in chapter 14. We shall soon find a name for that topology, but we shall first define it in a different way to justify the name we will give.
Let be a field and . Then
we recall that is the radical of .
Proof: "" follows from lemma 21.9. Let on the other hand and . Then for a suitable . Thus, . Assume . Then , contradiction. Hence, .
From calculus, we all know that there is a natural topology on , namely the one induced by the Euclidean norm. However, there exists also a different topology on , and in fact, on for any field . This topology is called the Zariski topology on . Now the Zariski topology actually is a topology on , for a ring, isn't it? Yes, and if , then is in bijective correspondence with a subset of . Through this correspondence we will define the Zariski topology. So let's establish this correspondence by beginning with the following lemma.
Let be a field and set . If , then the ideal
is a maximal ideal of .
This is a surjective ring homomorphism. We claim that its kernel is given by . This is actually not trivial and requires explanation. The relation is trivial. We shall now prove the other direction, which isn't. For a given , we define ; hence,
Furthermore, if and only if . The latter condition is satisfied if and only if has no constant, and this happens if and only if is contained within the ideal . This means we can write as an -linear combination of , and inserting for gives the desired statement.
Hence, by the first isomorphism theorem for rings,
Thus, is a field and hence is maximal.
Let be a field. Define
(according to the previous lemma this is a subset of , as maximal ideals are prime). Then the function
is a bijection.
The function is certainly surjective. Let , and assume for a certain . Then , and thus
Thus, contains a unit and therefore equals , contradicting its maximality that was established in the last lemma.
Let be a field. Then the Zariski topology on is defined to consist of the open sets
- , open
where and are given as in lemma 21.13 (that is, the Zariski topology on is defined to be the initial topology with respect to ).
It is easy to check that the sets , really do form a topology.
There is a very simple different way to characterise the Zariski topology:
Let be a field. The closed sets of the Zariski topology on are exactly the algebraic sets.
Unfortunately, for a set , the notation is now ambiguous; it could refer to the algebraic set associated to , or to the set of prime ideals of satisfying . Hence, we shall write the latter as for the remainder of this wikibook.
Let be closed w.r.t. the Zariski topology; that is, , where is the function from lemma 21.13 and . We claim that . Indeed, for ,
Let now be an algebraic set. We claim . Indeed, the above equivalences prove also this identity (with replacing ).
In fact, we could have defined the Zariski topology in this way (that is, just defining the closed sets to be the algebraic sets), but then we would have hidden the connection to the Zariski topology we already knew.
We shall now go on to give the next important definition, which also shows why we dealt with irreducible spaces.
Let be a field and let be an algebraic set. If is irreducible w.r.t. the subspace topology induced by the Zariski topology, is called an algebraic variety.
Often, we shall just write variety for algebraic variety.
We have an easy characterisation of algebraic varieties. But in order to prove it, we need a definition with theorem first.
Theorem and definition 21.17:
Let be an algebraic set. We define
and call the ideal associated to or the ideal of vanishing of . We have
and any set such that is contained within .
Let first be any set such that . Then for all and , and hence . Thus .
Therefore, , and hence by lemma 21.9. On the other hand, if , then for all by definition. Hence . This proves .
Let be a field and let be an algebraic set. Then is an algebraic variety if and only if for a prime ideal .
Let first be a prime ideal. Assume that , where are two proper closed subsets of (according to lemma 21.10, all subsets of closed w.r.t. the subspace topology of have this form). Then there exist and . Hence, there is such that and such that . Furthermore, since for all either or , but neither nor .
Let now be an algebraic set, and assume that is not prime. Let such that neither nor . Set and . Then and are strictly larger than . According to 21.17, and , since otherwise or respectively. Hence, both and are proper subsets of . But if , then . Hence, either or , and thus either or . Thus, is the union of two proper closed subsets,
and is not irreducible. Hence, if irreducibility is present, then is prime and from 21.17 .
, equipped with the Zariski topology, is a Noetherian space.
Let be an ascending chain of open sets. Let and be given as in lemma 21.13 and definition 21.14. Set for all . Then, since , being a function, preserves inclusion,
Since is a Noetherian ring, so is (by repeated application of Hilbert's basis theorem). Hence, the above ascending chain of the eventually stabilizes at some . Since is a bijection, . Hence, the stabilize at as well.
Every algebraic set has a decomposition
for certain prime ideals such that none of the is a proper subset of the other. This decomposition is unique up to order.
That is, we can decompose algebraic sets into algebraic varieties.
Combine theorems 21.19, 21.7 and 21.18.
- Exercise 21.2.1: Let . Prove that .