Commutative Algebra/Irreducibility, algebraic sets and varieties

Irreducibility edit

Definition 21.1:

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:

  1.   is irreducible.
  2.   can not be written as the union of two proper closed subsets.
  3. Every open subset of   is dense in  .
  4. The interior of every proper closed subset of   is empty.

Proof 1: 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. 

Proof 2: 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  . 

Remaining arrows:

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.

Theorem 21.3:

Every irreducible space   is connected and locally connected.

Proof:

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. 

Theorem 21.4:

Let   be an irreducible space. Then   is Hausdorff if and only if  .

Proof:

If  , then   is trivially Hausdorff. Assume that   is Hausdorff and contains two distinct points  . Then we find   open such that  ,   and  , contradicting irreducibility. 

Theorem 21.5:

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. 

Corollary 21.6:

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.

Theorem 21.7:

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.

Proof:

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. 

Exercises edit

  • Exercise 21.1.1: Let   be an irreducible topological space, and let   be open. Prove that   is irreducible.

Algebraic sets and varieties edit

Definition 21.8:

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

  and
 .

Three immediate lemmata are apparent.

Lemma 21.9:

 .

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  :

  1.   (  a set)
  2.   and  
  3.   (  ideals)
  4.   (  sets)

Proof:

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

 .

4.

  

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.

Lemma 21.11:

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.

Lemma 21.12:

Let   be a field and set  . If  , then the ideal

 

is a maximal ideal of  .

Proof:

Set

 .

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. 

Lemma 21.13:

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.

Proof:

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. 

Definition 21.14:

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:

Theorem 21.15:

Let   be a field. The closed sets of the Zariski topology on   are exactly the algebraic sets.

Proof:

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.

Definition 21.16:

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  .

Proof:

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  . 

Theorem 21.18:

Let   be a field and let   be an algebraic set. Then   is an algebraic variety if and only if   for a prime ideal  .

Proof:

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  . 

Theorem 21.19:

 , equipped with the Zariski topology, is a Noetherian space.

Proof:

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. 

Corollary 21.20:

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.

Proof:

Combine theorems 21.19, 21.7 and 21.18. 

Exercises edit

  • Exercise 21.2.1: Let  . Prove that  .