 So let us continue with our study of irreducible subsets of affine space. So we have just seen in the last lecture that the condition that the zero locus of an ideal is reducible is that the radical of the ideal is prime ideal okay and so let us look at some examples and in fact also examples of the Zariski topology how it looks like. So in fact the Zariski topology is very special when you take A1 namely the affine line which is K itself endowed with the Zariski topology okay. So let me make a few statements. So consider so let us look at some examples these are general examples about also about the Zariski topology and also about reducibility. So you first consider A1 this is just K with the Zariski topology okay. Now so what is the zero set of A1 what is the closed subset of A1? A closed subset of A1 of A1 is Z of S where S inside K of X1 is a subset okay this is the definition of closed subset okay. So S is a bunch of polynomials in one variable okay and of course you know if you want this closed subset to be non-empty you should assume that the ideal generated by S is not the whole ring, Z of S is non-empty if and only if the ideal generated by S is not the whole polynomial ring okay this is essentially one of the versions of the Nuhl-Slernsatz okay or rather weak form of the Nuhl-Slernsatz which says that you know the zero set of an ideal is non-empty if the ideal is a proper ideal okay and so in other words the elements of S do not generate a unit okay and in fact you see what are the irreducible closed subsets of A1? The irreducible closed subsets of A1 are going to correspond to prime ideals in KX1 okay and so the irreducible closed subsets of A1 correspond to R of the form R of the form Z of P where P inside KX1 is a prime ideal okay but then you know if K is a field then the polynomial ring in one variable over the field is a PID it is a principal ideal domain that is every ideal is generated by a single element and you also know that in a PID non-zero prime ideal is also maximal okay so the only chances for P are either that it is the zero prime ideal okay and the other chance is that it is a maximal ideal okay so P is either zero or a maximal ideal okay this means that the zero set of P is correspondingly K1 because if you take zero set of zero okay the set of all points which at which the function zero vanishes when you get all the points so when P is zero you get A1 or yes a single point lambda belonging to K because the maximal ideal will look like X minus X1 minus lambda okay so if P is if P is zero then the zero of P will be A1 it include all the points that in particular tells you that A1 is irreducible okay if you use the fact that the zero set of a prime ideal is irreducible and if P is a prime ideal is non-zero then it has to be a maximal ideal and a maximal ideal is of the form X minus lambda X1 minus lambda for some lambda in K I told you this is also another version of the Nulsel and Sats but actually in this case because it is just in one variable you do not need the Nulsel and Sats okay you just have to use the fact of fact that K is algebraically closed so you know the fact is that if you take any ideal okay which is a proper ideal then you see because it is a principal ideal domain this ideal be generated by single element and therefore the zero set of this ideal will be just the zero set of a single polynomial okay so and then the zero set of a single polynomial will be just the union of the zero sets of the linear factors of the single polynomial any single polynomial is can be broken down into a product of linear factors because K is algebraically closed and each linear factor will correspond to a point so what you will get is that the only closed sets are just finite subsets of points okay so that means the open sets are just complements of finitely many points and from this you can see that you take any two open sets they will certainly intersect which is what you should expect because any two non of course I mean any two non-empty open sets because you already know that A1 is irreducible so any two non-empty open sets have to intersect but the fact is any two non-empty open sets are just complements of finitely many points and but there are infinitely many points the reason that there are infinitely many points is because an algebraically closed field is always infinite a finite field cannot be algebraically closed okay so let me write all that down since K of X1 so let me write here here it is for P is equal to ideal generated by X1 minus lambda since K of X1 is a PID principle ideal domain any ideal is generated by a single element any ideal is generated by a single element so Z of S is just Z of ideal generated by S is just Z of F for some element F in the polynomial ring okay but since K is algebraically closed F can be written as a product of linear factors the number of factors will be equal to the degree of F of course I am assuming that F is not a constant polynomial it is not a constant polynomial because if F is of course if F is a constant polynomial 0 then you will get the whole space if F is a non-zero constant then the ideal generated by F will be the unit ideal it will be the whole ring and the 0 set will be empty okay and the empty set is always a closed set by definition okay so since K is algebraically closed the F is some ideal generated by F is just ideal generated by X1 minus lambda 1 times X2 minus X1 minus lambda 2 times X1 minus lambda m okay and which implies that the 0 set of S this is 0 set of F is just the union of all these points lambda 1 etc up to lambda m of course I am putting the satiating notation some of the lambda s could be repeated they would be multiple zeros of the polynomial but the fact is therefore that any closed set is just a finite subset okay and you know what this will tell you is that see this will tell you that you do not at least as far as A1 is concerned you really do not need to use the Null-Celensatz to get all these results okay you do not need to use the Null-Celensatz so because the Null-Celensatz is kind of redundant it is superfluous in the case when N is equal to 1 okay when N equal to 1 Null-Celensatz is just the definition of what an algebraically closed field is okay so you really do not need the Null-Celensatz so all this can be directly seen the moment you define that is ketopology even without using Null-Celensatz okay. So in particular for example how do you see that A1 is irreducible one way is of course from the general theory you can say A1 is irreducible because the 0 set because ideal it is the ideal of A1 is 0 okay and 0 is a prime ideal therefore the 0 set of 0 of the 0 ideal is irreducible namely A1 is irreducible that is one high-five way of seeing it but more easier way of seeing it is that you see if A1 were reducible then you should be able to write it as a union of 2 closed sets 2 proper closed sets but you see any closed set is finite so if you are able to write it as a union of 2 proper closed sets that means you are saying A1 is finite but A1 as a set is just the field. So saying that A1 is reducible is the same as saying that the field has finitely many elements but that is not possible for an algebraically closed field an algebraically closed field is infinite therefore the fact that an algebraically closed field is infinite will automatically tell you in this case that A1 is irreducible okay so note that K note that since an algebraically closed field is infinite A1 has to be irreducible okay. So in this case everything is very easy to see just because of the fact that every ideal is just generated by a single element because it is a principal ideal okay and you know I want you to really think you see let us put for K complex numbers okay put for K complex numbers and K complex numbers are usual to policy okay complex numbers is just the R2 the xy plane if you want okay if you can also think of it as a argon plane okay the point is if you take if you think of R2 in the usual topology see the so many closed sets okay the so many closed sets for example you take a disc and then you take its closure that is a closed set you take a rectangle and take the closure that is a closed set you get all kinds of closed sets but you take see the Zariski topology then the closed sets are only finitely many points you see you see there is so the open sets are complements of just finitely many points it is just the open sets in the Zariski topology for complex numbers is just the plane punctured at finitely many points the huge open sets okay and the closed sets are just finitely many points so you see it is a you see there is such a huge difference between the ordinary topology and the Zariski topology and you can from this also see the Zariski topology is highly not highly non-Hausdorff okay because Hausdorffness is a condition that given any two points I can find open sets which contain these two points and which are disjoint from each other okay I can do that in the usual topology you give me two distinct points on the complex plane then you know I can find two small enough open discs surrounding those two points which will separate these two points the two open discs will not intersect but if you try to do that in the Zariski topology will not be successful okay that is because any two non-empty open sets will intersect so you cannot find two disjoint non-empty open sets containing two given distinct given points so you see the Zariski topology is highly non-Hausdorff and you know this leads to thinking in a philosophical kind of way you see the topological spaces which are not Hausdorff are not well behaved okay because the Hausdorffness is if you look at it from the topological point of view it is a fact that you know if it is the uniqueness of limits okay see in topology the existence of limits and the uniqueness of limits put together is what is called as completeness okay the completeness is the property that limits exist okay for example when you take subspaces of the Euclidean space you know completeness is just the fact that you take any Cauchy sequence in that subspace its limit should also belong to the subspace so and of course the limits limit of a sequence is unique so you get the existence not only of the unit of the limit but also that the limit is unique okay. But the truth is that the uniqueness of the limit alone if you take it out as a property and forget the existence of limit it is a uniqueness of the limit that is promised by a Hausdorff space okay. Now so the fact that complex the complex plane with the Zariski topology being non-Hausdorff gives you the feeling that it is somehow there is no uniqueness of limits if they exist. Now the answer to that is that you should not think that way in algebraic geometry the notion of uniqueness of limit is not Hausdorffness it is another property and that property which is the analog of Hausdorffness is called separatedness okay and so the nice thing is though the complex plane is not Hausdorff it is what is called separated okay and separatedness is the right analog of Hausdorffness in the you know in the Zariski topology. So what you should understand is that though at the outset it looks non-Hausdorff but for all practical purposes there is the analogous property of separatedness which I will explain later okay. So the next thing that I should look at is yeah and in fact let me also remind you of a fact from you know field theory if you have a finite field okay then the algebraic closure of the finite field is countably infinite okay this is one fact so it is infinite and if you in general if you have any field then the algebraic closure if the algebraic closure will have the same cardinality as that of the original field if it is if it is infinite okay then the algebraic closure of the field will have the same cardinality as the field itself okay this is a fact it is just a counting argument because you get the algebraic closure by putting in zeros of all polynomials in the field after all algebraic you go to the algebraic closure of a field because you want to solve all equations you want to get zeros of all polynomials with coefficients in that field and but then the polynomials when you look at zeros of polynomials polynomials can be counted by degree okay. So you can see that somehow you will get a countable union of subsets which are just as the same cardinality of the field and then you will see that you will again get the union will again be the same cardinality as that of the field okay so this is a fact that you can look up from any you know standard algebra book okay so that is so let me repeat the field is finite then it is algebraic closure will be infinite and countable if the field is infinite then it is algebraic closure will be infinite and have the same cardinality as that of the original field and why I am saying this is also since I brought in the idea of algebraic closure I should tell you why this is important this is important because you see we somehow in this course we are only worried about algebraically closed but you know you can write equations over fields which are not algebraically closed okay for example real numbers then what is a theory the theory is that first of all whatever equations you write you can always go to the algebraic closure of that field and work there and for this you need the existence of an algebraic closure and this is a theorem in algebra from field theory which says that given any field you can find an extension field a larger field which is algebraically closed and which is algebraic over the given field namely that it is gotten from the given field by just adjoining roots of polynomials it coefficients in the in the given field okay so this existence of algebraic closure tells you that no matter what bunch of equations you are trying to look at zeros of over a field that all that will always make sense over the algebraic closure of that field and when you come to the algebraic closure of the field you are in this situation and you are doing variety theory okay so this gives you an idea how to proceed if you are having equations over a field which is not algebraically closed okay that is one thing and the other thing is that many of these results that I have written down are not true if you remove the algebraic closure property okay that should be obvious to you because you know this algebraic closure property that gave rise to that is needed for the null still and such. So and you know the null still and such is somehow you know indispensable in all the main statements that we are making so if you remove algebraic closeness that is if you take a field which is not algebraically closed you will be in lot of one could be in lot of trouble so the simple example is you know if you for example you know if you take an equation like x1 squared plus 1 equal to 0 which makes sense over the real field you know that first of all the zero set is empty if you consider the affine space over the real field because there is no zero the zeros are they are complex roots so the zero set of a polynomial might become empty okay might turn out to be empty and that demonstrates the failure of the null still and such if you work with the field which is not algebraically closed. The other thing is this fact that if you start with the prime ideal and the zero set of the ideal is reducible that also kind of goes wrong for example if you take the if you take the polynomial ring in two variables over R the reals the field of real numbers let us call the variables as x and y okay then if you look at the polynomial xy-1 whose zeros are the is that is the locus of the rectangular hyperbola then you can see that of course it is in two pieces okay it is certainly not connected right but the ideal generated by xy-1 will certainly be a prime ideal because xy-1 is an irreducible polynomial okay and you know in a unique factorization domain an irreducible polynomial an irreducible element will always generate a prime ideal therefore the irreducible polynomial xy-1 will generate a prime ideal in R of xy but the zero set in A2 of R which is R cross R will be the rectangular hyperbola and that is not connected okay. So here is so the fact that the zero set of a prime ideal is irreducible is contradicted as just because you are working over real numbers which is not algebraically closed. So you should always remember that when you are all these results strongly depend on the fact that the field is algebraically closed okay so let me write that down note that if k is not algebraically closed many of the important results obtained so far may not hold examples is for k equal to R real numbers okay x squared plus 1 ideal generated by x squared plus 1 so let me write it as in Rx is proper but z of this in A1 R is empty okay similarly xy ideal generated by xy-1 in Rxy is prime z of xy-1 in A2 R is not connected okay so and if a subset is not connected it cannot be irreducible because irreducible is very very strong. So if the field is not algebraically closed you will not get many of the important results here okay then the third thing is to look at I mean just to look at the whole affine space if you look at the whole affine space then you will of course get that the whole affine space is irreducible for n greater than 1 a nk is also irreducible since it corresponds to the to the 0 ideal in polynomial ring which is prime okay and you know of course the 0 ideal in a ring is a competitive ring with 1 is prime if and only if the ring is an integral domain and you know that the ring of polynomials in n variables for a field is an integral domain that is a product of 2 polynomials cannot be 0 without either of the polynomials being 0 you cannot have 2 non constant polynomials whose product is 0 that can never happen okay and more generally the fact from competitive algebra is that you know if you have a competitive ring and you look at the polynomial ring over that competitive ring then the if the original ring is an integral domain then the polynomial ring in n variables over that integral domain is again an integral domain okay this is the fact that you can check. So more generally instead of k if I put an any integral domain then the polynomial ring over that will continue to be an integral domain okay so this is this is one fact and okay so maybe I will continue with my discussion and for that so I will go to what is called as noetherian decomposition so I will go into what is called as noetherian decomposition so this is another important feature of irreducible cross search okay it is the in some sense it is the analog of the noetherianness of a competitive ring okay. So you know so let me make a definition a topological space X is called noetherian if it satisfies DCC for closed sets so DCC is abbreviation for descending chain condition okay and what does what does that mean it means the following thing if you have a sequence of closed subsets which is becoming smaller and smaller okay a descending sequence of closed sets then that sequence has to stabilise beyond a certain stage okay it has so what it means is that if you have such a sequence then beyond a certain stage all the sets occurring in that sequence have to be the same. Another way of saying it is that if you if you are going to look at only strict sequences namely a sequence in which every next set is a strictly smaller subset of the previous one then it has to be only finite length you cannot have infinite length sequence like that and the actually I should tell you at the outset what this what is the importance of this topologically the importance of this topologically is the this gives you a handle on being able to define the dimension of a space okay you can define the whole notion of dimension of a space can be defined if you have this notion okay so you know so it is if you think of linear algebra okay then suppose you have an n dimensional vector space okay then if you take a sequence of subspaces then you see that the sequence of subspaces has to stabilise in fact if you take a sequence of strict subspaces then you know it has to stop because as you take a stricter subspace the dimension falls and the dimension cannot go below 0 and the 0 dimensional space will be just the 0 vector single point okay and it cannot go below beyond that so you if you have a vector space of dimension n and you take a sequence of strictly decreasing subspaces then you know you can get only something of length n plus 1 starting with the whole space which is n dimensional then a hyper plane which is n minus 1 dimensional then a hyper plane that which is n minus 2 dimensional and you can go on up to 0 so that will give you n plus 1 terms from 0 to n okay and it cannot be any longer than that so in the same way what you are doing for a general topological space is that you are of course in the case of vector space you have subspaces okay but in a general topological space what you do is you do not use subspaces but you use closed subsets okay and that is the that is the difference but the fact is that this DCC allows you to define dimension okay so let me write that down given sequence of closed subsets f1 or let me use z1 containing z2 containing z3 okay there exists or there exists m greater than or equal to 1 such that zm equal to zm plus 1 and so on. So the sequence of a descending sequence of closed subsets at some point has to become constant okay of course the other way is that if you have strictly descending sequence of closed subsets that has to be of only finite only it cannot be infinite okay that is another way of saying it. So let me write that also yes I am continuing here another way of saying this is that any strictly descending sequence of closed subsets has to be finite that is given z1 properly containing z2 properly containing z3 and so on the sequence has finite length finitely many terms okay. So this is another way of saying it okay a strictly decreasing sequence of closed subsets has to terminate does stop okay and so you of course the main purpose of this is I told you is that it allows you to define dimension okay and but there is something more it gives rise to what is called noetherian decomposition okay. So let me write that probably I will erase this part of the board probably before that let me do the following thing let me convince you that the affine space is noetherian okay. So the affine space An with the Zariski topology is noetherian and the reason is because this noetherian condition on the affine space actually translates if you apply the if you apply the this if you apply the I function to the ascending chain condition on ideals of the polynomial ring which holds because the polynomial ring is noetherian okay. So that is in a Jiffy saying that the affine space is noetherian okay. So let me write it like this geometric so there is again a geometric and there is a commutative algebraic and the geometric thing is so I will use the following or equivalent okay for a topological for a topological space X number 1 X is noetherian okay which is DCC for close sides okay it satisfies DCC for close sides number 2 the you know but before I come to number 2 let me let me take the case when X is An okay if you take so I go from this picture to this picture on this side I am writing it only for any general topological space but I am going from here to here only when I am looking at An okay and what happens is that you see you will see that again the following are equivalent for number 1 K X1 etc Xn satisfies ACC for ideals okay. So this is I mean this is you know this is one of the definitions for a ring to be noetherian for a commutative ring to be noetherian ideal should satisfy the ascending chain condition okay and this polynomial ring is certainly noetherian because that is because of the Hilbert basis theorem or ME noetherian theorem that if a given ring is noetherian then the polynomial ring infinitely many variables over that ring is also noetherian okay. So you see this corresponds to this but then if you look at the definition of a there are several other equivalent formulations of the definition of a noetherian ring of course the there is one other definition which says that if you give me any non-empty collection of ideals it has a maximal element. So this is another condition so any non-empty collection of ideals in K X1 etc Xn has a maximal element okay and you know usually it is a zone slimmer axiom of choice kind of argument that tells you that you know these two are equal right and you see the point is that there is something corresponding here the corresponding thing here is you see ideals if you take an ideals correspond to closed subsets. So you know the translation of that will be any non-empty collection of closed subsets has a minimal element because you see the when you take an into the picture on the commutative algebraic side you will take you are actually looking at K of X1 etc Xn which is the ring of functions on an. So ideals correspond to closed subsets and you know maximal means with respect to inclusion they will correspond to minimal subsets because the because this correspondence is inclusion to reversing. So in fact it is true that for a general topological space also that is correct that any this condition that X satisfies dcc for closed sets is equivalent to any non-empty collection of closed sets of closed subsets. So of course here I mean closed subsets has a minimal element okay any non-empty collection of closed subsets has a minimal element and you know but you can write something more here and what you can write more is just because in this case you have topology the complements of closed subsets are open subsets you can also say that any non-empty collection of open subsets has a maximal element okay because that is just the complement of this open sets are just complements of closed sets. So in fact I can also write any non-empty collection of open subsets has a maximal element okay so you get that also on this side okay. So it is clear that it is immediately clear from this comparison that An is a noetherian topological space with the Zariski topology. So I will just draw a line here and say clearly An is a noetherian topological space and you know let me tell you again but I will come back to this in a later lecture and that is you see the whole point about this noetherianness is to study finite dimensional noetherian topological spaces and you know by any amount of intuition even by common sense you should expect that the dimension of An must be N okay you should expect that and that is what we have to prove okay. But for that you know there are two things first of all you should know how to define dimension okay and then you have to prove that the dimension is N okay and it is very easy to define what dimension is. What you do is that you imitate what you did for what you see happening for a vector space of dimension N. If you take a vector space of dimension N if you take the longest possible strictly descending sequence it will have N plus 1 terms starting with the full space of N dimension and ending with the 0 subspace which is 0 dimension there will be N plus 1 terms. So what you do is you define the dimension of the topological space to be you take sequences like this strictly descending sequences and take their lengths and take the supremum. We have to use the word supremum because there could be I mean you could have larger and you could have lengths of any I mean you could have sequences strictly descending sequences of any length okay. See the definition of noetherian only tells you that if you have a strictly descending sequence it has to be a finite length but it does not tell you that the lengths cannot exceed a certain finite quantity. So you could have sequences different sequences each of different finite lengths and with the lengths increasing going being unbounded. So you take all possible sequences like this strictly decreasing sequences and take their length okay and in fact you should take length minus 1 okay because you know if you go by the vector space analogy then the length of strictly decreasing sequence max the largest possible strictly decreasing sequence is N plus 1 terms and you have to take away 1 from that to get N which is the dimension of the vector space okay. So that is how you define dimension and then after you define dimension you will have to show that dimension of An is N this is what you want okay. So what you should try to understand is that on the geometry side you see we have the Zariski topology and now we are talking about dimension which is a topological property okay. So you know this is the this is the this is how you start studying the topology of varieties okay and as I told you this is the first step in geometry you look at you look at some topology look at the topological properties try to talk about dimension connectedness with reducibility and things like that try to understand what are the closed sets what are the open sets and so on and so forth. And then after you have done enough topology then you start worrying about more complicated things like things that are you know connected with analysis like you know tangents spaces and smoothness and singularities and things like that okay which also we will do in this case okay. So what I am trying to say is that at this level we are still looking at the topology okay and so I need to tell you also one more thing. So you might wonder how this notion of dimension is going to translate into on this direction and the topological dimension on that side of An being N is going to translate into what is called as a Krull dimension okay named after the commutative algebraist Krull of this ring okay and the Krull dimension of this ring is N and that will correspond to the topological dimension of the affine space okay. So we have to translate all this to things here and to prove that the Krull dimension of this is N you need some field theory and some commutative algebra which I will recall okay. So that is to indicate to you how this is going to go about alright but apart from that what is this thing I started with I started with this Noetherian decomposition. So if a topological space is Noetherian the nice thing is that every non-empty closed subset has a Noetherian decomposition okay and that is kind of the importance of having Noetherian topological space okay this break you can break up any closed subset into a finite union of irreducible closed subsets such that no which can be made unique up to permutation of the factors in such a way that it can be made unique if you assume that none of one of these is contained in some other okay. So it is Noetherian decomposition that is very important so let me write that down that is a theorem and we will probably look at a proof of that in the next lecture. So theorem if x is a Noetherian topological space then any closed subset any non-empty closed subset y of x can be written can be decomposed as y is equal to y1 union y2 union and so on ym and where dy is irreducible closed subsets and this decomposition is unique and the decomposition is unique up to the permutation of the sets appearing provided no yi is contained in some in another yj this is the theorem on Noetherian decomposition. If you have Noetherian topological space you take a non-empty closed subset you can break it down into irreducible closed subsets and finitely many of them and you can make the decomposition unique if you assume that there are no redundancies that is no yi is contained in some other yj okay and if you believe this theorem what this will tell you immediately is that since An the affine space with the Zariski topology is Noetherian it will tell you that any closed subset of An namely the zero set of an ideal can be broken down into a finite union of affine varieties any irreducible closed subset of An which may not be irreducible it can be broken into a finite union of irreducible closed subsets and you know the irreducible closed subsets are called the affine varieties and therefore you get a decomposition of any closed subset into a finitely many affine varieties and those finitely many affine varieties are called the irreducible components it is just like an topology when you have a space which is not connected you know you can break the space down into union of its connected components okay. In the same way what you can do here is that an irreducible any subset of a Noetherian topological space any closed subset of a Noetherian topological space can be broken down into irreducible components and these irreducible components are closed so these yi's are called the irreducible components of y okay then the yi's are called the irreducible components components okay. So I will end with this and I will just make a final remark to tell you that you see the notion of Noetherianness has is essentially needed on one hand to make sense of dimension okay and on the other hand it is very helpful because it allows you to break down any closed set into irreducible closed sets in a nice way okay and that for us is important and all this for as important is because you can talk about dimensions of affine varieties okay and you can also break down any closed subset of affine space into its irreducible components which will be affine varieties and which will be essentially unique okay so that is the importance of the Noetherianness and so this is the geometric part of it but if you come to the commutative algebraic part of it what is it? It is just the usual Noetherianness of the polynomial ring. So you see this is the beauty of algebraic geometry that some nice property on one side gives rise to properties on the other side which have nice consequences so after all the fact that the polynomial ring is Noetherian okay is a completely commutative algebraic property but when you translate it to geometry you see the advantage is that it allows you to define dimension okay and we are eventually going to prove that the dimension of An is n as you would expect okay and it also allows you to decompose any closed subset into affine varieties okay. So you see this is how you see how a nice commutative algebraic property translates into such nice things in geometry okay see this is essentially the attraction of algebraic geometry to go to take something on one side and try to go and investigate what it means on the other side you get interesting things on the other side and here we are going from this from the commutative algebra side to the algebraic geometry side okay. So I will stop here.