 Okay, so again let us continue with our discussion on algebraic geometry, so you know let me recall the setup in which we are working, K is an algebraically closed field for example you can think of K to be complex numbers if that is convenient for you and then the whole idea is you look at KN with the Zariski topology on the one side which is called as affine N space over K okay and this is the rather this is a geometric side and then on the other side you have the commutative algebraic side and that is supposed to be the ring of functions on that space and in this case of course it is a ring of polynomial functions on that space, so it is actually K of X1 through XN, so X1 through XN are N indeterminates they are N variables and if you think of this ring as a ring of functions on affine space because you give any polynomial you take any polynomial then you can think of it as a map from affine space to AN to A1 which is just A1 is just K okay, so these are functions with values in K okay and therefore this is the ring of polynomial functions they are functions on the space okay, so the geometric side has got to do with the affine space the commutative algebraic side has got to do with the functions see it is a supposed to be a statement of Felix Klein who said that you know the geometry of a space is supposed to be controlled by the functions on the that you allow on the space and this is so geometry actually kind of comes into play when you have a space and you define what the functions on your space are going to be okay, so in this case the space is affine space and the functions are the polynomial functions okay and I told you that so of course the just to recall what we have seen so far you know the given any subset S here you associate to S the common zero locus of S which is a set of all points in affine space which at which every polynomial in S vanishes okay and then there is also a map that goes in this direction given a subset T of affine space then you associate to T which is an ideal in the polynomial ring and this ideal of T is just all those functions which vanish at every point of T okay and then this correspondence goes on the one side the objects that are important here are the sub-objects which correspond to ideals and the objects that are important here are the so called algebraic sets which are the closed sets which has zero sets of this form okay and of course you should remember that if I changed or if I replaced S with the ideal generated by S you will still get the same zero set okay. So basically what you get is you get on this side if you take the set of radical ideals you get a bijection with the set of all closed subsets and this bijection is a it is an inclusion reversing correspondence that which is quite obvious to see because the larger the ideal is the common zeros of all the functions the ideal will grow smaller okay and the other important thing is that so if you look at radical ideals you get closed subsets on the other hand if you look at maximal ideals then under so you should remember so maybe I think let me write this below let me leave some space in between so there are maximal ideals so the maximal ideals are also they are also radical ideals because actually you should perhaps check as an exercise probably you have already done it in competitive a course in competitive algebra or algebra that if an ideal is prime then it is already radical and since a maximal ideal is prime a maximal ideal is also radical but of course there are radical ideals are not even prime okay but in any case maximal ideals the collection of maximal ideals is a subset of this okay and this under this bijective correspondence goes to the smallest possible closed subsets which are actually the points of course when I say points I am thinking of a point here as a singleton subset of an okay so in other words you can actually write an here and of course the notation for this is the maximal spectrum of the ring of functions okay so max spec of a ring competitive ring with one means the collection the set of all maximal ideals in the competitive ring so what you must understand is a point here is a maximal ideal of the ring okay and the fact is that given a maximal ideal you get a point and conversely okay so in particular for example if you take a maximal ideal of the maximal ideal will always look in this form it will be generated by Xi-lambda i for a tuple of a for an n tuple lambda 1 etc lambda n which will be a point of an and this is the correspondence because 0 set of this is this the ideal of this will be that okay so you should remember that in this direction the map is taking the ideal it is the I map and in this direction the map is a Z map which associates the 0 set the common 0 locals okay and in fact I told you that this is that this is this is this is this also uses in a way it is an avatar of the Nullson and Sats probably weaker or stronger probably weaker but let us look at that in the exercises but the point is this is a non-trivial statement okay what is trivial is if you give me an ideal like this it is that it is maximal is reasonably trivial to check but to conversely say that every maximal ideal is of this form which is true only when k is algebraically closed or at least when k is algebraically closed it is non-trivial okay and that uses Hilbert's Nullson and Sats so the what lies in between are the prime ideals the prime ideals the collection of prime ideals they correspond to what is called the spectrum of the commutative ring and you see the spectrum of the commutative ring is supposed to be the set of all of its prime ideals and therefore you think of a prime ideal here as a point in the spectrum so a point here is a maximal ideal and a point here is a prime ideal okay and of course this is contained as I told you prime ideals are radical maximal ideals of prime okay and what happens is that so what corresponds to prime ideals on this side are what are called as affine varieties in An so by this I mean the so the definition is these are all algebraic these are all algebraic sets these are all closed sets which are irreducible okay so prime ideals correspond to irreducible subsets which are closed okay and that is the theorem I stated in the previous lecture and I just stopped with that because the last lecture actually stopped with the definition of what an affine variety is okay an affine variety is an irreducible closed subset of affine space okay and why are they important they are important because we will see later that any variety I mean that any closed subset can be broken down into a finite union of affine varieties and the decomposition is unique if you make sure that there are no redundancies that is no affine variety in this is contained in no affine variety in the decomposition is contained in some other affine variety in the decomposition so every closed subset can be broken down into a finitely many affine varieties only in a unique way essentially unique way I say essentially because you can always you will have you can always permute the pieces in the union but that should not affect the union or the decomposition and so that is one important thing about affine varieties because they are like building blocks of algebraic sets every algebraic set is broken down into union of affine varieties and this is very very important because later on for example I told you probably in the first lecture that there is a more advanced or I should say sophisticated language of algebraic geometry which involves what are called as schemes. And schemes are some spaces with functions okay with rings of functions and in fact the what you have is not just rings of functions you have sheaves of rings which means that you have rings of functions on every open subset of the space so it is data not only with the space with a ring of functions on the whole space but it also comes with for every open set in the space you will have a ring of functions. So your this whole data is called a sheaf of rings and a scheme is something to like that which consists of space and sheaf of rings okay where the sheaf is a collection of rings for every open set in the space but the important point is the technical point is this scheme is supposed to be locally modelled in this way it is supposed to be made of affine pieces. So for all of algebraic geometry no matter how general algebraic geometry you do these affine pieces these are the building blocks these are the building blocks and that is the reason why these have to be first studied okay. So you should understand that affine varieties are important because they are the building blocks even in the most sophisticated form of the theory okay and the other important thing is that the other important thing is of course that these sets are topologically irreducible okay and you know irreducibility I have told you is a very strong form of connectivity so they will have nice properties with respect to maps. So for example in topology you will learn that the image of a connected set under a continuous map is again connected okay so if you have a topological space and you have a map a continuous map from the topological space into another topological space then the image of a connected set if in the source topological space will be a subset of the target topological space which will be connected okay and the same thing will happen for irreducible subsets okay. So irreducibility is a very nice thing to have on a subset okay so now let me try to prove this part so let me recall definition so this I am just recalling a topological space a subset y of a topological space so I am just abbreviating topological space to top x is called irreducible if y cannot be written as y1 union y2 with y1 y2 non-empty proper close subsets okay. If a topological space can be written as y1 union y2 where y1 and y2 are non-empty proper close subsets then we say that the topological space is reducible okay and the definition of irreducible is that it should not be reducible okay and what happens in the case of varieties as we will see what happens in the case of algebraic sets here namely close subsets of avian space you will see that you will get it you will be able to break it down into not just a union of 2 you will be able to break it down into union of finitely many subsets which are each which are themselves irreducible okay and they will be called the irreducible components okay but and that will be called the reducible decomposition of your given close set okay so that is where we are heading to fine. So the of course this definition as I told you implies that if y is irreducible then it is connected okay because connected is for it to be connected for it to be connected you should not be able to write it as a disjoint union of proper close subsets non-empty close subsets okay and that is certainly not possible if you cannot even write it as a union of non-empty proper close subsets okay. So irreducibility is a very strong form of connectedness I told you that irreducibility has not lots of nice properties one thing that comes is that if space is irreducible then if a subset is irreducible then its closure is also irreducible so its irreducibility is not going to be affected if you add the boundary which is what you do when you take the closure of a set and this is also true for connectedness if a set is connected then its closure is also connected and then but the other more important thing is that you see the more important thing about an irreducible space is that every open every non-empty open subset is dense and is itself irreducible that is another very important property okay which I hope you would have tried as an exercise otherwise you should try it it is a pretty easy exercise. So what it tells you is that if you take an irreducible space and take a non-empty subset then you can test on that subset all those properties which will be preserved when you take a closure okay that subset will because the closure of that subset will be the whole space and that is so you can test on any non-empty open subset and any non-empty open subset will be dense and that also tells you that if you take any two non-empty open subsets they will intersect okay they cannot be disjoint from each other. So these are the nice properties of irreducibility and later on it will come we will again look at it probably it is not so hard you can even check it off hand I think that the image of an irreducible set continues to be irreducible under a continuous map okay which is the same kind of statement that you get for a connected set okay. Now I go to this theorem which I stated last time so the theorem is the following if i in the polynomial ring in n variables over k is an ideal then z of i the zero set of i the points in an which are common zeros of all the polynomials in i is irreducible this is a subset of this topological space you see this topological space is just kn given the Zariski topology okay. So since it is a subset of a topological space this definition applies and you can put the condition that this subset is irreducible and the theorem says that this is irreducible if and only if the radical of i is a prime ideal. So what this means is that if I already started with the radical ideal I mean if I already started with the prime ideal I already started with the prime ideal then z of i will be irreducible and conversely if I already started with the radical ideal then saying z of i is irreducible is same as saying that the ideal itself is prime okay and that is essentially what gives you this correspondence in the middle okay that the affine varieties in an they correspond to prime ideals okay. So well so the proof proof is quite straight forward so let us do both ways so let us begin with let us assume radical of i is prime suppose radical of i is prime what do I have to prove I have to prove z of i is irreducible to show z of i is irreducible okay but actually you see z of i is the same as z of rad i okay this is something that I told you last time two ideals j1 and j2 have the same set of common zeros if and only if the radical of j1 is equal to the radical of j2 so it is the since i and rad i have the same radical namely which is rad i they both have the same zero set okay and but even otherwise this is quite trivial to see directly okay because rad i mind you is defined to be all those polynomials some positive integral power of which lies in i so it is like taking the radical of ideals is like expanding that ideal to include n throats of its members positive n throats of its members okay that is n throats for positive n right and so anyway so how do I check a set is irreducible so the logic is I have to check that it is not reducible I have to check that it is not reducible so I have to check that if it can be written in this form with y1 and y2 as close subsets and if I assume that y1 and y2 are both non-empty and also that y1 and y2 are both proper that should not happen that is what I have to check so what I will do is I will assume that it can be written as in this form with y1 and y2 non-empty but I will assume that I will assume further that y1 is proper close subset and I will try to prove that y2 is not proper namely that y2 is everything if I do that then I am done okay so that proves that it is not reducible in other words that it is irreducible so what I will do is suppose that z of i is y1 union y2 where y1 y2 are non-empty closed so close subsets so here I have to go back to the definition and stress on something which I have not written there with y1, y2 non-empty proper close subsets mind you of y okay that is something that I had not written but I did say that in my last lecture so let me stress it when I say what do you so I told you y is just a subset of a topological space what is the meaning of saying that a subset of y is closed in y that you have closed subset of y this is the language of induced topology a subset of y is said to be closed subset of y if it is gotten by intersecting y with a closed subset of the ambient space the larger space x in which y sits okay so when I write z of i is y1 union y2 where y1 y2 are non-empty closed subsets of z of i what you must understand is that since z of i is already closed in x it follows that y1 y2 are not just closed subset of z subsets of z of i but they are actually closed subset of x subsets of x itself because a closed subset of a closed subset will continue to be a closed subset okay see in other words if when I say y1 is a closed subset of z of i it means y1 is z of i intersected with a closed subset of x okay but then you see z of i itself is closed in x and so if I intersect another closed subset of x the intersection of finitely many closed subsets is again a closed subset for in the topology because the axiom of topology for axioms for a topology if you take the axioms for closed sets tells you that you take any finite number of closed sets and you take the intersection the result is again a closed set in the whole space. So what this tells you is that y1 which is supposed to be the intersection of z of i with a closed subset of x is itself a closed subset of x so let me stress that suppose z of i is y1 union y2 where y1 y2 are non-empty closed subsets of z of i with y1 a proper subset of z of i of course I will then try to prove that y2 is equal to z of i that in other words that y2 is not proper okay then note that y1 y2 are closed in the larger space x which is actually in our case an because by definition as I just told you y1 has to be z of i intersection with a closed subset of an but z of i is already closed in an and the intersection of two closed subsets of a topological space is again a closed subset so the reason is since z of i is already closed in a by definition because you know that is how the Zariski topology was defined the Zariski topology was defined just by taking for the closed sets subsets of the form z of i okay so but what does that mean saying that y1 y2 are closed in an a n means that y1 y2 are the 0 sets of some ideals okay so this implies so y1 is z of i1 y2 is z of i2 for i1 i2 ideals in kx1 etc etc okay this is what you get but then what is now y1 union so if you take z of rad i which is z of i which is y1 union y2 is actually z of i1 union z of i2 okay and you know but this is the same as z of i1 i2 because this is something that we this is how we proved that sets of the form z of i they form a topology by declaring such sets as such subsets as closed subsets in fact what we proved is if you took z of s1 union z of s2 union etc up to z of sm where si's are subsets not even ideals then the union is just z of the product s1 times s2 times etc sm we proved that okay. So this is z of i1 i2 okay and you see it is at this point that you know I will use so I will now use the null cell insert okay so what will happen is you see apply i if you apply i the null cell insert tells you that since i of z of j is rad j okay mind you this is this is a statement that involves the null cell insert and that is valid for any ideal j actually i of z of j always contains rad j it is very easy to see the non-trivial thing is to say that i of z of j is contained in rad j namely it is a statement that if f if a polynomial is in i of z of j namely if a polynomial vanishes on z of j then some power of the polynomial is in j you cannot have a polynomial some power of which is not in j to vanish on all the zeros of j that cannot happen okay. So this is this statement uses in null cell insert and if I apply this on both sides what I will get is I will get radical of rad i is equal to radical of i1 i2 okay I just want to say that this contains i1 i2 okay this is the way I have to go okay so you see radical of an ideal always contains the ideal okay because the radical is supposed to be all those elements some positive power of which is in the ideal and the first positive power of every element of the ideal is in the ideal so the ideal itself is contained in its radical always and see the fact is that this is rad i okay because taking rad more than once is not going to change anything and this is prime this is prime. So what you are getting is you are getting a prime ideal contains the product of ideals now you see we use this following fact from commutative algebra it is very simple fact that if a prime ideal contains a finite product of ideals then it has to contain at least one of them okay. So here is a lemma very simple lemma from commutative algebra if a prime ideal contains a finite product of ideals then it has to contain at least one of those ideals one of the ideals in the product of course you know of course all this is in this statement the background I am assuming is that you are working in a commutative ring with one and you are having finitely many ideals and if you have finitely many ideals J1, J2 through Jm then their product is J1 dot J2 dot etc Jm which consists of just finite sums of products of M tuples taken from the Cartesian product of all the Js okay and if a prime ideal contains the product J1, J2, Jm then it has to contain some Ji and this is very easy to see because it is just definition of prime ideal that if a prime ideal contains a product then it has to contain one of the finite product then it has to contain one of the factors of the product it is just a restatement of that if you try to work it out. So what this lemma will tell you is that rad i has to contain I1 or rad i has to contain I2 but then this now you apply Z okay you apply Z to take the zero locus and remember that when you apply Z the inclusion is reversed. So what you will get is Z of rad of I is contained in Z of I1 or Z of rad of I is contained in Z of I2 this is what you get and mind you but Z of rad of I is mind you is just same as Z of I and Z of I1 is Y1 and Z of rad of I again is Z of I here and this is Y2. So what you are saying is Z of I is contained in Y1 or Z of I is contained in Y2 okay but what we have already started with was that Y1 and Y2 are contained in Z of I so what this means is that either Z of I is equal to Y1 or Z of I is equal to Y2 but then you are assuming that Z of I is not equal to Y1 so what this will tell you is that Z of I has to be equal to Y2 and that tells you that you cannot reduce Z of I okay. So let me write that down this implies that Z of I is equal to Y1 or Z of I is equal to Y2 which implies that Z of I is equal to Y2 since Z of I is supposed to properly contain proper of sorry supposed to properly contain Y1 okay and this implies that Z of I is reduced okay. So we have started with rad I prime okay and we are able to reduce that Z of I is reduced now we will do the other way we will assume Z of I is irreducible and show that rad I is prime okay so conversely assume that Z of I is irreducible assume that Z of I is irreducible we will show rad I is prime okay. So how do you so this is again this is again translation you just have to check the condition for a prime ideal you have to take a product how do you check something is a prime ideal how do you check an ideal is a primary you take a product of two elements of the ring as belonging to the prime ideal and demonstrate that one of the two factors of the product is that ideal okay and so what we will do is so let F times G belong to rad I let the product be in rad I okay. So what this will tell you this will tell you that the ideal generated by FG F into G is a subset of rad I okay because if an element belongs to an ideal and the ideal generated by that element is also in that ideal because the ideal generated by an element is just simply multiples of that element by ring elements okay. So if I now you apply Z if you apply Z I will get Z of FG contains Z of rad I okay and but you know Z of FG is just Z of F union Z of G okay this is exactly the same statement that Z of I1 union Z of I2 Z of I1 I2 okay. So Z of FG is Z of F union Z of G okay and of course when I write Z of F for a single element F by that I mean Z of a single element F is the same as Z of the subset consisting of the single element F and this is also the same as Z of the ideal generated by F they are all one and the same okay. So what happens is that so you know so now you can so what is what this tells you is that you see Z of rad I has been written as ZF intersection Z of rad I union ZG intersection Z of rad I you you see Z of F is a close Z F this union contains this so you intersect this with the smaller subset you will get back the smaller subset. So if I intersect this with Z of rad I I should get Z of rad I and that is and intersection you as you know distributes over the union by simple set theory okay so you get this but what you must realize is that this is if I call this as Y1 and if I call this as Y2 what you will notice is that Y1 is a closed set it is a closed set because it is the intersection of two closed sets so it is a closed set similarly Y2 is a closed set closed set and you have written Z of rad I as union of two closed sets but mind you Z of rad I is the same as Z of I but what is the assumption on Z of I is the assumption on Z of I is that it is irreducible. So the moral of the story is that either one of these is empty okay and if both are non-empty then one of them has to be then one of them has to be the whole Z I itself okay so let us let us write that out let me write that out here since Z of I equal to Z of rad I is irreducible we have and Y1, Y2 are closed we have the following possibilities so I want to say Y1 is non-empty Y2 is non-empty okay so let us write down Y1 is if Y1 is empty this will tell you that that means so that will imply that Z of F intersection Z of rad I is empty well this intersection is supposed to be Z of F union Z of the ideal generated by F union rad I right this is what it is supposed to be by definition because what is how do you show that the how do you show that the closed sets form a topology how do you show algebraic sets form a topology. So what you if you recall you can recall that Z of if you take intersection over alpha or lambda in capital lambda some indexing set of Z of S lambda this is just Z of the ideal generated by the union of S lambdas okay if S lambdas are subsets of the polynomial ring okay and you take the Z S lambdas these are a collection of closed sets how do you show that the intersection of see how do you show that the intersection of an arbitrary collection of closed sets is closed it follows from this calculation okay so Z of F intersection Z of rad I will be Z of F union rad I ideal generated by F union rad I and so what that will imply is if I apply I to both sides if you apply by I to both sides and use again use this use this Z of I of Z of J is rad J so what I will get is if I apply I to both sides you will get radical of the ideal generated by F union rad I will be if I apply I to the null set okay then I will get the whole ring okay because what is I of subset it is all those polynomials which vanish on the subset okay I of a subset of affine space is all those polynomials in the polynomial ring which vanish on that subset but if that subset is empty there is nothing to test every polynomial will satisfy this condition therefore I of null set will be just the whole polynomial ring okay and you know if this happens I mean I essentially have to show that if this happens I am done otherwise I have to proceed further okay so what does this mean this means that this means that F the ideal generated by F union rad I is itself the polynomial ring see so this is again a fact I am using that you know if an ideal if the radical of an ideal contains a unit then the ideal itself contains a unit because saying that the radical of saying that the radical of an ideal contains a unit say 1 tells you that there is some power of this which is equal to 1 there is some power of this the ideal generated by this union which is equal to 1 but then that if the some power of an element is equal to 1 then that element itself is unit okay that means that the ideal generated by this itself is the whole polynomial ring okay and what this will tell you is that so you know there is some so what this will tell you is the following that so there exists g1 etc gm in rad I such that sigma fg plus fg plus sigma over i i equal to 1 to m fi gi is 1 this is what it tells you okay I mean an element in the ideal generated by the union like this will look like this you will have to pick actually you have to pick finitely many elements from the subset and then take ring linear combinations of that and such a ring linear combination is equal to 1 because 1 is there on the right side and now what I want to say is that from this we will have to say that so you know if this happens that is if y1 is empty it should more or less follow that if y1 is say that again yes if y1 is empty yes z of z of rad i is equal to the remaining thing on that oh I see you will just get z of rad i is equal to z of i g intersection z of rad i and then you will get therefore z of rad i contain in z of g okay so if you apply i on both sides you will get g belongs to rad i right you will get g belongs to rad i okay okay okay so probably so this is not this is not this is not required you are right okay so let me get rid of this good let me get back to the easier part of the argument but I would strongly encourage you to think about that whatever I wrote down okay so and the fact is that also will lead to something okay that will also lead to what you want but you have to so the point is you have to keep translating back to the geometric side if you are on the geometric side you should translate to the ideal side if you are on the ideal side you should translate to the geometric side by applying this i and z appropriately okay so as you as one of you has rightly pointed out what you should do what one does is that if y1 is empty then I mean z of it is obvious that z of rad i is just z of g intersection z of rad i and so which means that z of rad i is contained in z of g right and because the right side is contained in z of g and now you apply i you will get i of z of rad i which is just rad of rad i which is rad i containing i of z of g will be just radical of the ideal generated by g okay and to which g belongs okay so if y1 is empty you get g is in rad i okay alright so you assume y1 is not empty if y2 similarly if y2 is empty then you will get f in rad i with mind you f times g is in rad i so you have to prove either f is in rad i or you have to prove g is in rad i okay and y1 equal to empty directly gives you g is in rad i similarly y2 is empty implies f is in rad i okay and you have done so assume both are not true okay so assume both y1 and y2 are non-empty okay so you will have to thrash out all the possibilities okay. Suppose suppose y1 is proper then irreducibility of your right z of rad i implies that if this is proper then that cannot be proper if y1 is proper then y2 cannot be proper so y2 has to be everything okay and in that case you see g belongs to rad i alright so you are done essentially so let me write that down suppose y1 is proper then irreducibility of z of rad i implies that z of rad i is y2 okay and this implies that again the same argument literally z of rad i is equal to z of g intersection z of rad i and this will imply that g I guess this will imply g is in rad i okay so if y1 is proper you will get g is in rad i similarly if y2 is proper okay you will get f is in rad i okay similarly y2 proper will imply f is in rad i and that complies with proof okay. So it is very clear that an ideal here is prime if and only if the radical of an ideal here is prime if and only if the corresponding 0 locus here is irreducible okay that proves it okay so I will stop here and then we will continue in the next part with some examples