 Okay so what we are going to do now is continue with our earlier discussion so that is about the noetherian decomposition okay so you see let me again remind you what we are doing is we have if you remember that is the we looked at we took k to be an algebraically closed field and we took the affine space affine n space over k which is just kn with the so called zariski topology and what is the zariski topology the zariski topology is the topology for which the closed sets are given by 0 sets of ideals and the ideals should be taken in the ring of polynomial functions in n variables over k where the n is the same as this n okay so closed sets are also called as algebraic sets and they are just for the form z of i where i inside the polynomial ring in n variables over k is an ideal so this is the zariski topology and then we have seen that there is a you know there is a there is already a dictionary between the closed subsets of an and the ideals of the polynomial ring. In fact if you want a perfect equivalence a bijective correspondence then on this side you should take closed subsets of an on the other side you should take radical ideals in the polynomial ring okay the closed subsets correspond one to one with radical ideals and this correspondence is inclusion reversing okay and the larger the ideal the smaller the 0 set of the ideal okay so the 0 set of the ideal is simply the set of points in kn the intervals of points which at which every function every polynomial function in this ideal vanishes okay and now the point is that we defined we checked that this is a topology on this space okay that is this definition satisfies the axioms for closed sets of a topological space so an becomes so kn becomes an with this topology which is called the zariski topology and then I told you that the beautiful thing is that there is a algebraic geometry side there is a commutative algebra side the algebraic geometry side is the geometric properties of an and it subsets the commutative algebraic side is the commutative algebraic properties of the ideals here okay and as a first result what we showed last the last lecture was that you know the ideal i is prime if and only if or rather the radical of i is prime if and only if z of i is an irreducible closed subset okay so here when we defined irreducibility if you recall irreducibility was a strong form of connectivity okay and whereas for a set to be connected you require that it cannot be written as a disjoint union of two proper non-empty closed subsets the condition for a set to be irreducibly is much more stronger the condition is that it cannot be written as a union not necessarily disjoint but of proper non-empty closed subsets okay so an irreducible set is of course connected but the converse is not true and irreducible sets satisfy all the nice properties that you similar properties as connected sets okay and so the point was that we called irreducible closed subsets of an as affine varieties in an okay and so let me write that irreducible closed subsets of an are called affine varieties in an okay and so the point is so of course the advantage of studying an irreducible closed set is that you have the additional property that it is irreducible okay and this irreducibility is a topologically a very nice property it is a property that is really nice in the sense that as I told you last time for example you can define irreducibility for any subset of a topological space first of all it is not necessary that you should define it only for closed subsets and the point is that if you define irreducibility for any subset you will have to make the definition with respect to the induced topology so you should say that the subset cannot be written as a union of proper non-empty closed subsets closed for the induced topology on the subset induced from the bigger topological space in which the subset is sitting okay and the advantage of irreducibility is that so let me mention some of them well if a subset is irreducible then its closure is also irreducible so adding the boundary which is the same as taking the closure is not going to take away the irreducibility then if you take a non-empty irreducible set of course we by definition we always require that an irreducible set is non-empty because we declare the null set not to be irreducible okay. So well an irreducible set if it is non-empty then it has a nice property that every open non-empty open subset is dense okay and every non-empty open subset is also irreducible therefore you know in an irreducible space a non-empty open subset is enough to test all properties of the space which are going to be preserved when you take closures that is when you take limits okay so it is very important to be able to test on a non-empty open set okay and non-empty open sets are irreducible for an irreducible space okay that is one nice thing then the other nice thing is of course that any non-empty any two open sets any two non-empty open sets will intersect okay that will tell you that the topology is not housed off okay when compared to the usual topology for example if you take k to be complex numbers okay then the an, an, c which is c to the n, c cross, c cross, c n dimensional complex space n dimensional thought of as a vector space c n is with the Zariski topology is not housed off okay that is because of this reason because it is irreducible and the whole space is always irreducible please remember that is because the this corresponds to the zero ideal and the zero ideal the zero set of the zero ideal is the whole space and the zero ideal is of course prime because zero ideal is an ideal in an integral domain okay in a competitive ring with unity which is an integral domain the zero I mean if you take a competitive ring with unity the zero ideal is prime if and only if the ring is an integral domain okay. So this is always irreducible okay so any two open subsets of this will always intersect so it is highly non housed off okay house ofness means that you give me any two points you can find open sets small enough such that they contain those two points in their own intersect so this highly non housed off but still that does not deter us from doing good geometry okay that is the point and well so and of course there are other nice things for example continuous image of an irreducible set is again irreducible this is proved in exactly the same as you prove that the continuous image of a connected set is connected okay. So irreducible it is having a studying irreducible close sets is of course very nice and of course from the geometric point of view it is very nice and from the algebraic point of from the competitive algebraic point of view also it is very nice because you are studying prime ideals okay but then the point is how do you get to an arbitrary close set how do you how do you how do you study an arbitrary algebraic set okay. So the fact is that the answer to that is that these irreducible close subsets which are the affine varieties they are the building blocks for the algebraic sets. So there is something called the Noetherian decomposition theorem which says that any algebraic set any close set in an affine n space over k is writable as a union a finite union of irreducible close subsets namely affine varieties and the union is and this decomposition is unique and if you assume that there are no redundancies that means that there is no component which is containing some other component in the union okay. So now how do you prove that the answer to proving such a theorem is the Noetherian property okay. So last time I defined what is meant by Noetherian so an k is a Noetherian topological space. So let me quickly recall this so and again let me tell you the reason one reason for this Noetherian condition is to be able to say that any algebraic set can be broken down into a finite union of affine varieties and the finite union is this decomposition is unique if you assume that there are no repetitions in the union okay of course unique up to permutation of the pieces that occur okay. So if you want to study any algebraic set all you since you can break it down as a union of varieties it is just enough to study varieties and that is why we study only varieties okay that gives us some justification as to why to study varieties and not just algebraic sets that is part of the reason why at least in a first course in algebraic geometry we study only affine varieties and I mean we study affine varieties to begin with and then probably study projective varieties but we do not study non irreducible closed sets okay we study only irreducible closed sets because general non irreducible closed set can always be broken down like this. So the key to showing that any algebraic set can be broken down into affinate union of affine varieties in unique way is due to the fact that An is a Noetherian topological space. So if you so let me recall what I told you the last lecture see the Noetherianness of a topological space is the condition that the closed subset closed subsets of the topological space satisfy the so called descending chain condition that is if you give me a descending chain of closed subsets each one containing the next one okay so the closed sets are between smaller and smaller and smaller then such a chain has to become stationary at some point that it has to become stable that means beyond a certain point all the sets occurring in that chain in the sequence they all have to be one and the same. Another way of saying it is that if you take a descending chain of closed subsets such that every at every step it is a proper containment okay that means it is a strictly descending chain of closed subsets then that has to be only finite you cannot have an infinitely you cannot have infinitely many you cannot have an infinite chain of closed subsets which are becoming smaller and smaller strictly smaller one after the other okay. So this is the descending chain condition for closed subsets and this condition translates into the ascending chain condition for ideals in this polynomial ring that is because after all the closed subsets in the closed subsets in the affine space are they correspond to ideals in the polynomial ring okay and in fact if you want exact correspondence you have to worry about radical ideals okay but the fact is that if you give me any ascending and because the correspondence between the closed subsets of an and the ideals in the polynomial ring is inclusion reversing correspondence the descending chain condition for closed subsets in the affine space will translate to ascending to an ascending chain condition on the ideals in the polynomial ring and but the polynomial ring does have the ascending chain condition on ideals because it is a noetherian ring. So you see one of the definitions of a noetherian ring is that the standard definition is always that it satisfies the ascending chain condition for ideals that is in other words if you have a sequence of ideals becoming larger and larger and larger every ideal being contained being contained in the next one after it then that has to stop at some stage it has to either become stationary or another which means that beyond a certain stage all the ideals should be one and the same in the sequence the other way of saying it is that if you have strictly increasing chain of ideals then it has to be just finite you cannot find an infinite sequence of ideals which are becoming bigger and bigger and strictly bigger and bigger and with this whole thing never coming to an end this cannot happen in a noetherian ring. So the fact is that this is equivalent to the fact that the polynomial ring in n variables is a noetherian ring and why is the polynomial ring in n variables over k and noetherian ring and that is just because of Hilbert's basis theorem that is just Emean Noetherian Noetherian theorem which says that if you start of the ring if you start of the commutative ring with one if that commutative ring is with one is noetherian that is for example if it satisfies the ascending chain condition of ideals then so does the polynomial ring in n variables over that ring okay. Now a field is always noetherian because a field has only two ideals one is a zero idea one is a the other one is a full field which is unit ideal and therefore a field is always noetherian and therefore if you use Hilbert's basis theorem if you take a polynomial ring in finitely many variables over a field then that will also be noetherian and that is the reason why this is noetherian. So you see the topological side on the topological side the space An is noetherian and on the commutative that is on the algebraic geometric side at the level of topology the affine space is noetherian for the Zariski topology and in the commutative algebra side the ring of functions on affine space namely the polynomial ring that is noetherian ring and these just correspond to each other okay. Now I was trying to so I stated a theorem so this is a theorem that will let us to prove that any algebraic set can be decomposed into a union of affine varieties okay so let me say please recall that theorem and let me try to prove it so here is a theorem if X is a noetherian topological space then any close subset any non-empty close subset Y of X can be written uniquely I should say can be written let me explain what uniquely means rather let me can be written as Y equal to Y1 union Ys okay with YI irreducible closed non-empty this decomposition is unique up to a permutation YIs if no YI is a subset of some Yj for j not equal to I okay. So any non-empty close subset can be written as a finite union of irreducible closed non-empty subsets okay so each of these is irreducible it is closed it is non-empty and this decomposition is called you can make it unique if you make sure that you do not have redundancies okay. So for example if Y1 is if you have Y1 union Y2 and so on and if Y1 is contained in Y2 then Y put Y1 you can throw out Y1 because it is already there in Y2 so once you can write a decomposition like this you can always throw out some of them which are contained in others and finally you can arrive at a decomposition in which none of the sets is contained in the others in any of the others such a decomposition is unique okay such a decomposition is unique and that decomposition is what is called as a noetherian decomposition okay now and the YIs are called the irreducible components of Y okay so each YI is called an irreducible component of Y okay. So the decomposition this decomposition such a decomposition is called a noetherian decomposition and the YI are called the irreducible components they are called irreducible components of Y it is much similar to what you do in topology you take any in topology you define connectedness of a subset okay and then you prove that any topological space can be written as a union of its connected components okay in the same way here you are saying that any topological space if you want any topological space to be written as a union of its irreducible components and you want that to be a finite union the condition you have to put on the topological space then it should be noetherian okay that is a nice condition and that condition is there for us for the affine space as we have already seen. So if you believe this theorem you will immediately get that any algebraic set can be uniquely decomposed into affine varieties in this sense okay so that proves the statement that we want alright so you have to prove we want to prove this it is pretty easy to prove. So the point I want to say is that here we are going to use the various other definitions of noetherianness of a topological space so you know the original definition for example if you take a starting point the definition of a noetherian topological space that there is DCC for closed subsets that is there is you cannot have a strictly decreasing sequence of closed subsets one containing the next which is infinite okay then this is equivalent also saying that any non-empty collection of closed subsets has a minimal element okay and that is also equivalent to saying that any non-empty collection of open subsets has a maximal element because open subsets is just you know complements of closed subsets in any topological space. So you know so DCC for closed sets is the same as ACC for open subsets and DCC for closed subsets is the same as saying that any family of any non-empty family of closed subsets has a minimal element and that is equivalent to saying that any non-empty family of open subsets has a maximal element so these are all various avatars of the definition of the noetherianness of a topological space okay and it is not surprising because if you have seen this in competitive algebra or in algebra you know that there are also several ways of defining a ring to be noetherian one of course a standard way is to say that you know it has ascending chain condition for ideals okay but then there are other equivalent conditions the other useful equivalent conditions are that every ideal is finitely generated okay which is also very very important condition and so and yet another condition is that given any non-empty collection of ideals there is always a maximal element maximal with respect to inclusion okay. So whenever we say maximal it is always nothing is mentioned maximality is always with respect to inclusion subsets okay and of course you know each one has its own each of these axioms has its own importance see for example the while the ascending chain condition on ideals gives rise to the descending chain condition for closed sets the fact that every ideal is finitely generated is also very important because that is what tells you that if you take any ideal okay any ideal point if you take any ideal look at the zero sets okay if you use the fact that noetherianness means that every ideal is finitely generated it will tell you that whenever you are looking at the zero set of an ideal you are just looking at the common zeroes of bunch of finitely many polynomials. So you are not an ideal is always a non-zero ideal is always going to be infinite okay they are going to be infinitely many elements in it okay so the point is that it looks like when I take the zero set of an ideal namely all the points in the affine space which at which every one of the functions the ideal manage it looks like I am solving too many equations okay but it is not true in fact you are trying to find common zeroes of infinitely many equations okay but that is not really true what is really happening is that you are only finding the set of common zeroes of only finitely many equations why finitely many because those will be those finitely many equations for example which generate this ideal and that is true because the ideal is an ideal of a noetherian ring and it is finitely generated. So you are always looking at only finite common zeroes of only finitely many polynomials and this finitely is very very important because it allows you to do for example calculations of the computer so if you have a if I want to look at the zero set of an ideal then you know I look at zero sets I take the generators of the ideal and I first look at the zero sets of the first generator and then intersected with the zeroes of second generator and go on and I have to do this process only finitely many times okay so that is the reason why you can do computational computational algebra which will happen algebraic geometry okay. So anyway so let me come back so the point is that the property for noetherian is often the noetherian hypothesis on the topological space I am going to use is not the kind it is not ACC on close subsets that I am going to use not DCC on close subsets that I am going to use but what I am going to use is I am going to use the other equivalent definition that any non-empty family of close subsets has a minimal element okay so that is what I am going to use okay. So let us see how to use it so it is a very simple argument so what you do is basically you assume that there are so you try to contradict this statement okay so what you try to say is you assume that there is a subset which cannot be written as a finite union of irreducible close subsets okay then that is a specimen because that is something that contradicts this theorem okay the point is you should not show that you should show that there is not even a single one like that okay. Now the point is what you do is you put all these specimens together in a subset and apply the existence of a minimal element for that subset okay that is the whole point so what you do is so let me write this let if possible let S script S in B the collection of close subsets of X that cannot be written as a finite union of irreducible close subsets okay if possible let script S be the collection of close subsets of X that cannot be written as a finite union of irreducible close subsets and S non-empty so that means you have to show that the this collection script S is empty, empty you should show that there is no close subset that cannot be written as a finite union of irreducible close subsets. So you go by contradiction what you do is you take script S to be a non-empty collection of close subsets which have a property which contradicts the property of the theorem not even the stronger uniqueness of decomposition it contradicts even the existence of decomposition okay so S is non-empty. Now after all S is just a collection of close subsets of the topological space X and X is noetherian but you know noetherian one equivalent definition of the noetherian condition is that given any non-empty collection of close subsets it will always have a minimal element okay. So since X is noetherian S has a minimal element say Y not so there is a so Y not is a kind of smallest kind of subset irreducible close I mean smallest kind of it is a closed of course non-empty close subsets I you will have to I mean you have to be careful about this should be non-empty closed okay because of course this decomposition is only being made for a non-empty closed set okay. So there is a minimal element which means that Y not is a non-empty closed subset Y not cannot be written as a union of proper close subsets and it is the smallest in the collection S smallest with respect to what with respect to inclusion of subsets which means that if there is an element of S which contains Y not I mean if there is an element of S which is contained in Y not then it has to be equal to Y not that is what minimality means minimality means that if there is some other thing which is smaller than this then it has to be this okay fine. Now you watch carefully since Y not is since Y not is in S the first observation is Y not is not irreducible okay see because if Y not is irreducible then Y not can be written as Y not the fact that it cannot be written as a finite union of irreducible closed sets tells you that it cannot itself be irreducible because if it is irreducible then it is itself when I say union this union can be just one okay. So what you must understand is the elements of the script S by definition they are all not irreducible so what happens is Y not is not irreducible okay note that Y not is not irreducible Y not is not irreducible okay now but then what does it mean Y not can be broken down into 2 as a union of 2 proper non-empty closed subsets okay so Y not so let me continue here so Y not is equal to Y 0 1 union Y 0 2 where Y 0 1 Y 0 2 are proper non-empty closed subsets of Y not okay Y not is not irreducible so it is reducible so it can be broken down as a union of 2 proper non-empty closed subsets but now watch Y 0 1 is a closed set which is smaller than Y not and Y 0 2 is also smaller than Y not so Y 0 1 and Y 0 2 cannot belong to script S see Y 0 1 is a proper subset of Y not and it is a non-empty closed set and Y 0 2 is also similarly a non-empty proper closed subset of Y not so these 2 closed subsets cannot belong to script S because if they belong to script S they will contradict the minimality of Y not Y not is supposed to be the smallest okay so the fact that Y 0 1 and Y 0 2 do not belong to script S means that they can be written as a finite union of irreducible closed sets and if that is true then Y not also can be written as a finite union of irreducible closed sets namely you just take the union of the irreducible closed sets the finite union of irreducible closed sets that gives Y 0 1 and take also the finite union of irreducible closed sets that gives Y 0 2 put them together that will tell you that Y not becomes writable as a finite union of irreducible closed subsets and but that is not possible because Y not is supposed to be in script S so this contradiction tells you that script S has to be empty and this argument finally tells you that every irreducible every closed subset non-empty closed subset is necessarily writable it can necessarily be broken down into finitely many irreducible closed subsets okay so this gives you the existence of a decomposition not without the uniqueness we will get the uniqueness next okay so let me write that down since Y 0 1, Y 0 2 are proper closed subsets of Y not Y 0 1, Y 0 2 do not belong to the collection but then Y 0 1 and Y 0 2 and hence Y 0 is writable as a union of non-empty of finitely many non-empty irreducible closed subsets which contradicts Y 0 belongs to the family thus the family is empty in other words every non-empty closed subset is certainly writable as a union like this of non-empty proper of non-empty irreducible closed subsets okay. So thus any non-empty irreducible sorry non-empty closed subset Y can be written as Y equal to Y 1, Y s with every Y i non-empty closed irreducible because whenever I say irreducible it is automatically it is automatic that is non-empty because we have barred the null set from being taken as irreducible okay. So now of course if some Y i is contained in some other Y j you can throw the you can throw the Y i out so you can assume without loss of generality that no Y i is contained in any other Y j okay and then what the theorem says is that with that assumption this decomposition is unique that is what we are going to prove next okay. So if some Y i is contained in another Y j for j not equal to i we can exclude that Y i from the union thus we may assume that Y i is not contained in Y j for i not equal to i that is the union is not redundant okay this you can assume is obvious the point is that once you assume that this decomposition becomes unique okay that is what we are going to prove next. So let me do that so let me write that down uniqueness of decomposition so let me write a lemma I use a small lemma if z is contained in z1 union z2 union zr where z is irreducible and ez is contained in zi for some i just a moment you also need that all the zi is are let me assume that z is let me assume z is closed and all the zi is are also closed ez i are closed and ez is irreducible then ez is contained in zi for some i that is correct yeah. So if you have a union of closed sets finite union of closed sets and if there is a set which is contained in irreducible set which is contained in finite union of closed sets okay then if it is irreducible it has to go into one of them okay. So proof is well you see I mean you just have to use the see the condition the condition given means that that ez is actually you know z intersection z1 union z intersection z2 union z intersection zi zr okay because if you intersect z is contained in this union so z intersect that union is z but ez intersect that union is just ez intersect each member of the union and then you take the union because the intersection distributes over the union okay. So that is what it means and notice that each z intersection see since zi are closed in ambient topological space the big topological space therefore z intersection zi is closed in z okay so the each z intersection zi is closed in z okay that is very clear because that is what induced topology means induced topology means induced topology in a subset is that a subset of that subset is closed if that subset which you are claiming to be closed is supposed is gotten by intersecting with a closed set in the bigger space okay. So z1 any zi is closed in the big topological space therefore it is intersection with a subset is closed in that subset okay fine then the other thing is that course you know I can in this union I can simply forget if some z intersection zi is empty okay. So if some z intersection zi is empty I do not write it at all so I can assume no z intersection zi is empty or rather is non-empty for every i okay I mean if it is empty just forget that forget that index and re-index you will get a lesser number of indices okay. So after this assumption maybe this r will come down it might come down to a smaller number okay but let us not it is not really one is not really worried about what that smaller number is one is only worried about that is the fact that it is finite okay. Then the other thing that you may understand is if some z intersection so you see if some z intersection zi is equal to z then we are done because if some z intersection zi is z then you are just saying that z is contained in zi alright so we are done okay z is contained in zi alright. So if some z intersection zi is equal to z then you are done okay. If not if suppose z intersection zi is properly contained in z for all i this is the other possibility you have to show that this possibility does not occur okay you have to show that this possibility does not occur and the only possibility that occurs is z intersection zi is z for some i and therefore z is contained in zi for some i that is essentially the lma okay. Now suppose z intersection zi is proper subset of z so what it will tell you is that you have broken down z into a finite union of non-empty proper closed subsets now that is not possible because z is that is because z is irreducible okay the irreducibility says that you cannot break it down into a union of two proper non-empty closed subsets but you can extend it to finitely many the reason is because you see suppose I have this condition what I can assume is that I can also assume that z is not contained in any union of a sub collection of the zi okay. So in other words I am saying that you can also assume without loss of generality that you know z is not in the union of say for example you throw out z1 it is not in this union because if it is in this union I would work with this. So try to make sure that z is not contained in any smaller union a union of a subset proper subset of the collection of all the zi's you can assume that okay assume that z we can in fact we can assume so let me rewrite this we may assume that z is not contained in zi1 union zi l for a subset a proper subset i1 etc i l of 1 2 etc r you assume that okay. So you can assume that otherwise if there is something like this then work with this call this subset as 1 through replace this collection with that subset and what you will do is you will keep on reducing to a stage until you will come to a point where you get a collection like this okay that is z is contained in the union okay z intersection zi is proper subset of z for every i and z is not contained in any smaller union you will come to a stage like that okay. Now you have to say that is not possible you see then you see you will have z is equal to z union z1 union z union z2 z union z r I will break it down like this see this is this is non-empty closed proper okay and this guy oh sorry these are all intersections I have been care I am a little careless yeah so this is if you look at it both of these guys here are non-empty closed proper. So what I have done is I have broken an irreducible set into two pieces which each of which is non-empty closed proper that is a contradiction to the irreducibility of z. So this can never happen therefore the only thing that can happen is that z has to be in zi for some time okay. So this implies zi z not irreducible and that implies that implies a contradiction which proves the lemma so a lemma is proved. So the model of the lemma is that if an irreducible set is contained in a union of closed finite unit of closed sets it has to go into one of them in other words it cannot fall into two pieces in a union it has to go exactly to one of them okay. Now let us apply that to the uniqueness of the conversation so let me go back here okay so let me go back here. Now assume now let y equal to y1 union etc ys and let that also y equal to y1 prime union ys, y prime s prime with every yi, y prime j irreducible, irreducible closed of course non-empty. What do you have to prove? You have to prove that s is equal to s prime and you have to prove that every yi is some y prime j for unique j okay. So of course you know and of course you are assuming that there are no redundancies yi is not contained in yj for i not equal to j y prime y prime oops y prime i is not contained in y prime j for i not equal to j where of course here the ij's are indices from 1 to s prime and here the ij's are indices from 1 to s okay that is to show s equal to s prime and each yi is precisely some yj y prime j okay that is what you have to show. So and how do you do that? It is very very simple you see so let me first take it in words I am going to use that lemma take y1 now y1 is a sub of this union so it is a sub of this union okay now the lemma says that they said y1 I do not need the fact that y1 is closed I only need the fact that y1 is irreducible y1 is contained in a finite union of closed sets therefore it has to be in exactly one of them it is contained in at least one of them not exactly it is contained in at least one of them so if I apply the lemma I will get that y1 is contained in some y prime j okay now again apply the lemma to y prime j that y prime j is contained in all the union of the yi's and that y prime j is irreducible therefore y prime j is contained in some yj prime okay so if you write this down you will get that y1 is equal to y prime j you will get y1 is equal to y prime j. So by since so let me write that since y1 is contained in the union of yi's we have by the lemma just using the fact that y1 is irreducible that y1 is contained in y prime j for some j okay similarly since y prime j is contained in the union of yi's we have since y prime j is irreducible that y prime j is contained in some y let me I do not want to use k but let me use l if you want okay so you will get y1 is contained in y prime j which is contained in yl okay so finally you get y1 is contained in yl but you are not supposed to have that because the yi's are all they are non redundant none of them is contained in the others. So what this will tell you is that l equal to 1 l has to be 1 and that will tell you that y1 is equal to y prime j this implies l equal to 1 that is y1 is equal to y prime j okay. Now the point is you renumber all these okay you renumber all these guys so that you call y that y prime j call that as y prime 1 and the fact is that you can now strike off from the union the y1 and the y prime 1 the fact that you can strike off uses this topological fact that any non empty subset of irreducible set is dense okay so I mean the fact is that if you take this union okay and you throw out y1 okay you will get an open set because what you have thrown out is a closed subset okay because you are taking a compliment if you are removing y1 from this union is like taking the compliment of y1 in y that is a compliment of a closed set okay so it is open but it is not if it is non empty then it is dense so if I take y1 out and take the closure I will get the union of the other pieces because what I have taken out when I take y1 out from y I am taking out also y1 from y2 through ys okay so what I will get is essentially subset of y through to ys okay and what I have thrown out from each of the y2 to ys is a closed subset it is a proper closed subset okay therefore whatever that is left out if I close it up I will get back y2 through ys so in other words from this union if I take out y1 and if I close it up what I will get is y through to ys so this is you can think of this as a cancellation property okay so well so let me write that or rather that leads to a cancellation property so let me write that out it is just a couple of more lines then we can wind up so let me write that down a re-number re-label or re-index the y prime j the y prime l so that y prime j is y prime 1 okay that y prime j which we got to be equal to y1 call that as y prime 1 re-number it okay then y is equal to y1 union y2 union ys that is also equal to y1 prime union y2 prime union y y prime s prime okay and you know the point is that this is the same as y1 is equal to y1 this is y1 okay now what I want to say is that if you I want to say that this implies that y2 union ys is actually equal to y1 I mean y2 prime union y prime s prime which is like you cancel off the y1 that is on both sides of the union and this cancellation property is actually because of the irreducibility property okay that any non-empty open subset of an irreducible space is dense and in fact actually this is equal to y-y1 closure you take this union from that union you remove y1 okay what you get will be an open subset of y through to ys and it is a non-empty open subset of y through to ys therefore its closure will be y through to ys because in each piece the elements for example in y2 that you have removed which are common with y1 you will get an open subset of y2 if I close that up I will get my y2 back that is because this open subset that I have removed I mean this closed subset that I have removed is the common elements between y1 and y2 and that is not the whole of y2 because no yi is contained in some other yj. So if I remove y1 from y2 and close it up I will get back y2 so the same way from this whole thing if I remove y1 which is equivalent to removing the intersection of y1 with each of these pieces and then close it up I will just get the union of these other pieces and now what you can do is you can continue the induction you can next strike of y2 and y2 prime okay I mean you can strike of y2 and some other yprime j okay because they will be equal by the same argument and that yprime j can be re-indexed so that you get that yprime j becomes yprime 2 okay then you can strike that off and you can keep on doing this and this process has to end finitely and when it ends you cannot this s has to be equal to sprime because if it is not if s is less than sprime then at some point you will get a null set here equal to a set there which is not empty and if s is greater than sprime at some point you will get a null set on the right side which is equal to a non-empty set on the left side. So this argument can go on you can go on only finitely many times and the fact that you cannot you cannot have a non-empty set equal to an empty set it will force that s has to be equal to sprime and every yi is unique yj, yprime j okay so continuing as before we find that y2 is equal to yprime j for some j re-number re-index yprime j as yprime 2 and we get after cancelling of yprime 2 that you will get y3 union ys is equal to yprime 3 union yprime s, yprime sprime and by induction we will end with s equal to sprime and ys equal to ysprime okay that will be the end of the proof okay. So finally you get this uniqueness of decomposition okay so the moral of the story is that uniqueness of decomposition and the noetherian decomposition itself holds for a noetherian topological space and the nice thing is that it holds for affine space and therefore any irreducible any close subset of affine space namely any algebraic set is finite union of affine varieties and affine varieties are unique if you assume that the union is non redundant that is none of them is contained in any of the other and this is called the noetherian decomposition of a close subset okay. So this tells you partly why it is worthwhile to study affine varieties rather than just studying algebraic sets because the affine varieties are building blocks for any algebraic any algebraic set it can be decomposed into affine varieties so affine varieties are the building blocks for our algebraic sets and in fact it is true also in the widest generality of algebraic geometry in the most sophisticated language of algebraic geometry it is the affine pieces that are the building blocks. The most general object in algebraic geometry the most sophisticated object is called a scheme and the definition is that it is built up by the building blocks which are called affine schemes and this is the this is the philosophy the affines are like the bricks that make up the whole building okay they are the building blocks okay so I will stop here.