 Okay, so let me recall from the previous lecture, so you know so algebraic geometry is a study of the geometry of the set of common zeros of set of polynomials, so what you do is that you take a commutative ring R and then you take the polynomial ring over that ring in n variables and you take a subset of polynomials and then you look at the zero set defined by that subset, it is a subset of the n, it is a subset of the Cartesian product of R with itself n times consisting of all those n two pills which are zeros common zeros of every polynomial in this set S, okay and then the problem was that so then you associated to this set S the zero set in this n dimensional space over R and of course we want to ensure that this set is not this set does not turn out to be the empty set, so what we do is that we assume that we always work over an algebraically closed field, so we take R to be k where k is an algebraically closed field and then our situation is like this, we on the commutative algebraic side we have the polynomial ring in n variables over k and we take a subset of that ring which means we take a bunch of polynomials which are the elements of the set capital S and then you look at the set of common zeros in n dimensional space over k and of course the important thing is that you should not worry about this as a vector space, you should not think of it as a vector space, you should think of this as an affine space and which means that you are not worried about the vector properties, the addition properties and so on but you are worried about other properties or the topological properties and these topological properties come from what is called the Zariski topology, so I have to tell you what Zariski topologies, so if you remember I told you that the Zariski topology is declared by declaring certain subsets as closed sets and these subsets are the low psi of zeros of a bunch of polynomials, so let me elaborate on that, so what I am going to do is I am going to do the following Zariski topology on kn of course k algebraically closed field, let me write it here okay, so define so definition a subset let me say f or let me not even give symbols a subset of an is called an algebraic set okay, if it is of the form Z of S where S is a subset of the polynomial ring in n variables over k, so look at the definition what it says is an algebraic subset of kn which is k cross k cross k n times the Cartesian product of k with itself n times is nothing but the locus of common zeros of a certain collection of polynomials a subset of polynomials in the polynomial ring with the same number of variables as the number of copies of k you are starting with okay. Now so immediately you have the following I am going to write down some properties and these properties will tell you that you can always declare I mean because of those properties you can declare sets of this form closed sets and that will give you the Zariski topology, so here is a lemma number 1 Z of so the whole space is Z of null set 2 or rather I should write Z of 0 null set is Z of well maybe I can take the whole ring third one is Z of S1 union and so on Z of Sn is Z of S1 dot dot dot Sn fourth one intersection over alpha Z of S alpha is Z of union over alpha S alpha okay. So you see this is the these are four properties and let us try to prove them yeah I will explain what this S1 etcetera Sn means it is the enfold products okay so I will explain the statements one by one so well they are pretty easy to verify we will do that so you see so the proof the first one is well what is what is Z of 0 it is all those points in kn which gives 0 when you substitute them in the polynomial 0 but the polynomial 0 is a constant polynomial it always has the value 0 no matter what you substitute in the polynomial therefore every point in kn satisfies this okay so one is obvious okay and what you must understand here is that we are thinking of 0 as a 0 polynomial okay you are thinking of 0 as a 0 polynomial the 0 polynomial is also when you substitute the variables in the polynomial the value of that polynomial is the value that you get is also 0 okay so you must distinguish between 0 as a polynomial and 0 as a value of the polynomial okay. So the 0 polynomial lives here whereas the value 0 lives in kn and your so you must understand these two things distinct from each other okay so but in any case this is very obvious and look at the second question look at the second statement if you take the subset S to be well you know if you want to be very particular maybe I can even put I can put this because I write 0 of a subset so I should actually give you a set and if I just write 0 it is just a single element but normally we have this abuse of notation that if you have a single element you are just right for a single polynomial f here you do not write z of f within braces but you simply write z of f okay you avoid the braces so if you avoid the braces this becomes z of 0 that is what I wrote first but it will be very strict to begin with let me write as let me write it properly like this okay now look at the second statement second statement says that the 0 set of the whole ring is the empty set that is obvious because in fact any subset of the ring which contains 1 or for that matter any constant nonzero polynomial will never take the value 0 at any point okay for example you take the polynomial 1 think of 1 as a constant polynomial it always has the constant value 1 so it is never going to take the value 0 so there is no good there is not going to be a single point in this Cartesian product which if you substitute it is going to give you the value 0 because the value is always nonzero and it is a constant therefore this set turns out to be empty okay and look at the third one so for the 2 is also obvious as for the third one I should explain what this s1 through sn means so this s1 through sn means mind you all these si's are subsets of the polynomial ring and in the polynomial ring you have multiplication okay so s1 through sn actually means products 1 you take products consisting of n factors where the ith factor comes from si so let me write that s1, sn is actually the set of all f1, fn such that fi is in si this is what it means okay you take one member from each si and then multiply them out okay and then you take the 0 set of that and my claim is that this and this are the same so as usual how do you check 2 sets are the same you take something here so it is there you take something there so it is here. So let us try to do this if you have a point lambda 1 etc lambda n oops I think I do not want this n to be the same as that n so let me change this to m because this could be any number of subsets but it has to be a finite number okay so let me change it here as well if you take an n tuple which is in z of si okay that what it means is that you take any polynomial in si and you plug in for the variables this n tuple then you get 0 therefore any such product will also vanish so it is clear then clearly this point is in set of s1 etc up to sn okay that is obvious because you know to get a 0 of an element of this means you have to get a 0 of an element which is a product like this product of polynomials and further it is enough just one or it is a 0 of just one of them so that when I plug in this this n tuple into each of these even if one of them vanishes then the product vanishes okay therefore what it tells you is that z of si is always in z of s1 sm so this tells me that z of s1 the union is in the 0 set of the product okay this is obvious what is little not so obvious is the other way round but this is also very easy to do by you can prove it by contradiction conversely if lambda 1 etc lambda n is in z of s1 etc sm okay then you have to show that it is in at least one of the z of si for some i okay so how do you prove it you prove it by contradiction you assume it is not there in any of the z of si if lambda 1 lambda n does not belong to z of si for every i okay then what does it mean that it means then for every i there exist a gi in si such that gi of lambda 1 etc lambda n is not 0 the fact that a point is not in the 0 locus of a set of polynomials means that there is at least one polynomial in that set for which this point is not a 0 that is if you evaluate that polynomial at this point you do not get 0 okay so for e and I can do this for every si okay then if you take the product g1 etc up to gm this belongs to s1 etc up to x sm but you know g1 etc gm of if I evaluate it on lambda 1 etc lambda n is not 0 a contradiction okay so what this will tell you is that this n triple has to belong to some z of si okay so that will give you the inclusion of this into this and we are done okay okay so it is a pretty simple argument okay and well then there is only the fourth one left which is pretty easy to see and once we do that we have this Ariskita apology on space so for the fourth one you take a point lambda 1 etc lambda n is in the is in the intersection of over alpha z of s alpha so here of course you are assuming that s sub alpha is a collection of subsets of the polynomial ring alpha is index and of course alpha varies over some indexing set which I do not want to be very explicit about is that is not what we we are not in so read about that indexing set because we do not want we do not there is no restriction on the indexing set it need not be finite it could be infinite okay so if there is a point in this that means you are saying that this point vanishes in every polynomial in each of the s alphas so it means that this polynomial will also vanish I mean this point will also this point will also be a 0 for if I for the set of polynomials gotten by simply taking the union of all the s alphas right this is obvious what is the first what is the top line it means that you take any polynomial in any s alpha for any alpha and you plug in this evaluate it at this point that is plug in this for the variables then you get 0 but that is exactly what this means if you take all the polynomials in all the s alphas then they all vanish at this point okay so this implies this and the other way is also obvious if if you have a point here then certainly that point is going to be a 0 of every polynomial in s alpha for every alpha so it is going to belong to each zs alpha and therefore it is going to belong to the intersection therefore if this is this is just obvious when provided you think about it for for some time it is just basic it is very simple logic okay so this is this is quite clear okay it is set here so so that is the well that is the end of proof it is a pretty simple exercise but what is the the import of this lemma the import of this lemma is that you can declare subsets of kn of this form namely the algebraic subsets you can declare them to be closed sets if you declare them to be closed sets the lemma says that the whole space is a closed set the empty set is a closed set the union of finitely many closed sets is again a closed set the intersection of the arbitrary collection of closed sets is again a closed set these are exactly the axioms for the closed sets in a topology so the lemma tells you lemma above tell implies that kn becomes a topological space if we declare algebraic sets to be closed and so this means that therefore the Zariski topology as it is defined here is defined by a closed sets and of course the open sets are the complements of closed sets always so let me mention that this is the Zariski topology this topology the topology this topology is called the Zariski topology and kn along with this topology is denoted a nk and called affine n space over k ok. So you see we stripped off we stripped off the vector space structure but then to make it into a different thing we call the affine space we added the Zariski topology ok and this is the affine n space over k right so things are quite good if you look at this picture there is the so the space on this side has at least become a topological space so the moment it becomes a topological space you can at least start thinking of continuous functions ok once you have a topological space you can at least think of continuous functions and of course the first question that one would ask is these are of course polynomials are of course functions on the space and then you would ask are they continuous and the answer is yes ok it is an exercise but we will also see it in some other way later so it will happen maybe you want if you want you can try it out as an exercise that you take any polynomial here and think of it as a map from kn to k evaluating every point on the polynomial will produce a element of k so every polynomial becomes a function on the affine space and then check that that function is actually continuous for the Zariski topology because the source space is kn with the Zariski topology is ank and the target space is k which is a1k ok so fact or I will even I will put it as exercise and we will see this later in detail any f in k x1 etc xn gives a continuous map from ank to a1k that is essentially enough it is essentially enough to check that it is essentially enough to check that the inverse image of any point is a closed subset ok but it is an exercise there you will have to do some work you will have to do some work ok so you see see there is there is some there is you know there is some cyclicity going on here ok see you know in the usual topology when we study suppose you study real valued functions or complex valued functions then the set of points where the function vanishes is a closed set of course I am only worried about continuous functions the set of points where a continuous function vanishes is a closed set right and well the reason for that is topological because a single point is closed as a subset is a closed set and the set of points where the function is where the function takes the value 0 is the inverse image of the singleton consisting of the point 0 and that is the inverse image of a continuous function under a continuous function of a closed set so it has to be closed ok so set of points where a continuous function takes a fixed value is always a closed set I mean 0 is a special value but it could be any other value and we use that kind of intuition to define the closed sets here to be given by the common loci of you know bear a bunch of functions vanish ok but then you know what this exercise tells as that it indicates our stand I mean it tells you that our original intuition is correct ok if our original intuition to declare 0 sets 0 loci as closed sets is correct then our functions using which we declared these 0 sets as closed sets they should of course be continuous and the fact is yes ok so this is an exercise it is pretty easy to check but I will come back to it later in a more general form ok so the point I wanted to make is the next point I want to make is again I think I will need this diagram again so I will so let me again draw the diagram I will keep needing the diagram again and again so here is the so let me let me do that this is geometric side and this is a commutative algebraic side so on the geometric side you have An, K which is which is actually Kn plus Zariski topology alright and on the commutative algebraic side you have the polynomial ring in n variables same n as this affine space ok and what we did is well we started with the subset here and then we associated to this an algebraic set which is a set of 0's and this is actually closed by definition of algebraic sets are the closed sets ok and how do you get more general closed sets you get more general closed sets by either taking arbitrary intersections of such closed sets or by taking finite unions of such closed sets that gives you all the closed sets. Now well what I am now next going to worry about is this side of the picture see on this side of the picture what we have is just a subset ok and a subset of a ring is not a very interesting object by itself ok usually so the way I am asking you to think is whenever you look at an object you try to think of sub-objects in the suitable sense. So for example you know if you are thinking of groups you should think of sub-groups ok if you are thinking of for example if you are thinking of vector spaces the interesting subsets are subspaces ok. So the same way if you are thinking of subsets of a ring what are the interesting subsets the interesting subsets are well one could be ideals the other could be sub-rings ok but there is no part there is no point in taking a sub-ring here the problem is if you take a sub-ring then it contains 1 and the moment it contains 1 z of 1 will become empty because you know z of s will always become empty the moment s contains a unit because a unit is a constant polynomial it is a non-zero constant polynomial which never vanishes. So if you are subset s contains a unit which in this case is a non-zero constant then the zero set is going to be empty so there is no point in taking a sub-ring here. So the fact is you do not want any units to begin with because if you take units here this set is empty there is nothing to study. So what are the other sub-objects you can think of the ideals and you know an ideal is not reveal if and only if it contains if it does not contain a unit the moment the ideal contains a unit which is an invertible element in the ring then the ideal has to be the whole ring ok. So what this tells you is that it is the interesting things here to look at from the commutative algebra point of view are the ideals ok but what we started out is just a subset now how do you go from the subset to the ideal there is always a philosophy for this whenever you have an object of certain type and you have a subset and you want a sub-object what you do is you take the so called sub-object generated by the subset ok. So if you have a group and you have a subset a subset of a group is not so interesting but you can always take the sub-group generated by the subset which is the smallest sub-group which contains a subset similarly if you have a ring and you have a subset if ideals are what you are interested in as sub-objects and not sub- rings then you can look at the ideal generated by the subset ok. So what you can do is you can replace you can think of replacing this set S by the ideal that is generated by S so what we will do is let us do the following thing let us write S let me use the following notation I will put S with a round bracket and S with a round bracket means the ideal generated by S ok so this is the ideal generated by S in this polynomial ring ok. Now what is this ideal generated by a subset in the polynomial ring of course there are two ways of describing it one is this is also this is equal to the smallest ideal of ideal containing S this is one definition ok and that is and what is the smallest ideal containing S how will you define the smallest ideal containing S it should well it is it first of all it has to be an ideal which contains S and it should be smallest among these which means that if there is a any other ideal which contains S that should contain this as well which tells you that this is the intersection of all the ideals which contain S. So there is another way of stating this set theoretically this is the intersection of all ideals J over J where J contains S ok and of course you know when you want a smallest sub-object containing a subset this is what you always do you take the intersection of all the sub-objects which contain that subset and usually in good situation the intersection of all the sub-objects will again give a sub-object ok. And if you take the intersection of a family of ideals that is again an ideal the intersection of a family of sub-groups is again a sub-group ok. So if you want a subset generated by a sub-group generated by a subset of a group all you have to do is simply take the intersection of all the sub-groups which contain that subset similarly if you want the ideal generated by a subset of a ring all you have to do is just take the intersection of all the ideals which contain that subset okay. So that is the smallest ideal containing S and there is another way of writing it out, the other way of writing it out is just take S linear just take the ring linear combinations of finitely many elements of S just take that okay. So let me write that down this is equal to the set of R so you know k x1 etc xn linear combinations of finitely many elements of S that is what it is that this is the same as this is an exercise I mean if it is a simple exercise which you should have come across in a first course in competitive algebra but if you have not done it would not take you probably more than a couple of minutes to do it and of course what does this mean this means that you know this is set of all the set of all elements of the form sigma fi gi i equal to 1 to l where fi are in S and gi are in the k x1 etc xn this is so you are taking finitely many fi's from the set S and then you are multiplying them with coefficients which come from the ring and then you are taking a linear combination okay and you can check that this is the same as this and that this is the same as this that is a very simple exercise okay. So well the fact is that the picture here becomes better because this is an ideal and that is a nicer object than just a subset because on the left side you are looking at a competitive ring you better the interesting sub-objects are ideals and then of course the question is you take Z of this take because that is after all it is an ideal there but ideal is also a subset so you can look at the set of common zeros and the fact is that you do not get anything new you get the same thing that is the beautiful point. The beautiful point is see if there is a point here then it is common zero of every polynomial in S therefore if you take any combination like this it will also be a zero of that because every term contains a factor from S and there are only finitely many terms and every term looks like this so something here is always here okay and because this is a subset of that okay something here is also here okay because you know when I take ring linear combinations I can set all the g i's to be whatever I want so I could set all but one g i to be zero and that one g i I can set it to be one okay then I will get that a point here will be a zero of each element of S so this will also be there I mean that will also be here. So the Malo story is the same so what is significance if significance is that instead of looking at zero sets of a bunch of polynomials because a bunch of polynomials a subset of polynomials is very not so very attractive it is not so interesting from the ring theoretic point of view if you look at the zero set of an ideal of polynomials things look better because an ideal is something that is a ring theoretic object and passing from a bunch of polynomials to the ideal does not do you any harm because the zero set is not affected okay. So this tells you that somehow you know now the story is getting better on this side we have close subsets and on that side you just do not have subsets you have ideals okay so you see slowly you are getting some kind of a correspondence here you have close subsets which are the algebraic sets and they are supposed to be zero sets of bunch of polynomials but you can as well take them to be zero sets of ideals so on this side you have ideals okay here you have subsets of here you have the close subsets of affine space and there you have the ideals in the polynomial. So you already have the picture on both sides becoming clear on one side you are studying ideals of the polynomial ring in n variables on the other side you are studying the close subsets of affine space okay. So this is how ideal theory enters into the picture okay but it does not enter without its share of apprehensions and they are as follows see the fact is that you know the very comes from very simple observation see even if my set S was a single element suppose I was just suppose S was a single element that means S is only one polynomial okay then Z of S is just the zero locus of that polynomial there is a set of points in where that polynomial vanishes okay but if you take the ideal generated by that polynomial it is a principal ideal it consists of all multiples of that polynomial and that ideal and the number of polynomials in that ideal is huge okay you have so many polynomials okay and this set is a huge set okay if K is infinite I mean for that matter even if K is finite but the fact is that since K is assumed to be an algebraically closed field some field theory will tell you that it has always been infinite set okay an algebraically closed field cannot be a finite set okay. So the fact is that this even if S is a single polynomial this is an infinite set so it seems to you as if you are looking at zeros of infinitely many polynomials okay now intuition will tell you that if you put too many equations if you have too many equations to solve then you may not get any solutions I mean for example in linear algebra if you have an if you are solving n so many equations in say n variables if there are more equations it is very likely that you do not get any solutions okay and so you can imagine that if I am if I change from one polynomial to say infinitely many polynomials I may not get any solutions so therefore this set may turn out to be empty but that is not the case the case is the fact is because of the Hilbert Null-Stelenzart's you know that because you know these two are the same and since Null-Stelenzart says that the 0 set is of a single non-constant polynomial is non-empty such a thing is not going to happen but there is something more that is happening and that is the following fact the fact is for that matter if you take not just a singleton set if you take any set and take the ideal generated by that that ideal will be generated by only finitely many elements. So what it means is that you are actually always looking at the common zeros of only finitely many equations even if you start with infinitely many equations if at all you get 0 okay then which you will okay so long as the ideal generated by those by that subset is not the whole ring then you are actually studying only the common zeros of finitely many polynomials. So this is so you know this somehow removes this apprehension that you know even though you may be trying to solve a large set of equations I mean you are trying to look trying to get common zeros of a infinite set of polynomials actually in principle you are actually solving only finitely many polynomials I mean you are looking at zeros of only finitely many polynomials and this fact follows from the so called Hilbert basis theorem or any no ethereum which is one of the fundamental theorems that you study when you study no ethereum rings in commutative algebra. So let me recall that and write that down so recall from commutative algebra a ring commutative with one is called no ethereum every ideal is finitely generated okay a no ethereum ring is a ring in which every ideal is finitely generated if R is no ethereum then so is R x1 etc xn is a polynomial ring in finitely many variables over a no ethereum ring is again a no ethereum ring and this is the so called Hilbert basis theorem or it is also called I mean more generally it is called any no ethereum. So this is the fact from commutative algebra right so what this tells you is that this implies that if z of s is non-empty okay then even if s is an infinite subset the ideal generated n by s is f1 etc up to fm for some fi in k x1 etc xn so you know what I must remind you at this point is that if I take R equal to a field k then a field k is of course no ethereum because a field has you know the only ideals in a field are the 0 ideal and the full ideal okay and the 0 ideal is obviously finitely generated by the element 0 and the whole field as an ideal is generated by 1 so both there are only 2 ideals they are finitely generated therefore a field is always no ethereum and therefore a polynomial ring over a field is also no ethereum by this by this fact therefore what it means is that this polynomial ring in n variables over a field has every ideal finitely generated. So if I start with a subset s of that no matter that subset is probably even an infinite subset if I pass to the ideal generated by that subset because it is an ideal in this no ethereum ring it is finitely generated which means that it is the ideal generated by finitely polynomial okay and so this implies that z of s which is z of becomes just z of f1 etc fn. So the moral of the story is you know when we started defining algebraic sets we were looking at common zeros of bunch of polynomials which could be even infinite but in principle finally we are only studying common zeros of finitely many polynomials and that is very heartening okay. So why it is heartening is because if there are finitely many polynomials this is something that you can compute you can do some computation and you can expect that you know this is not going to be a computation that will never end and this is one of the reasons that lot of algebraic geometry is nowadays done on the computer okay. So you have some special software programs like Macaulay and cocoa and things like that which allow you to do ring theoretic computations on the computer and you can do these computations so you will be doing computations on the computative algebra side and from those computations you can get consequences which means something geometrically. So you can try to prove geometrical statements by doing computations on this side and the fact that you can do the reason why you can do computations is one of the reasons is this you are never dealing with an infinite bunch of equations you are always dealing with only finitely many equations and in fact you may ask when I write here for some fi it seems as though that these fi are very abstract that you cannot catch them but that is not true in fact there is a if you look at a proof of the Hilbert basis theorem then the proof also can be refined to a constructive procedure where you can get all these finitely many polynomials you can really get them and so you can do this on a computer okay I can really do this on a computer and that is what helps me to do computations in computative algebra which when translated to algebraic geometry give nice results okay. So the moral of the story is because of Hilbert's basis theorem this the closed sets you are looking at in affine space in the Zariski topology are just zeros of finitely common zeros of finitely many polynomials so you are simply studying finitely many polynomials okay so the moral of the story is that when you do algebraic geometry you are actually trying to look at the geometry of the set of common zeros of a bunch of polynomials and usually the bunch of polynomials is having co-efficiency in a certain commutative ring okay even if you go to a general commutative ring but make sure that the commutative ring is noetherian then the Hilbert basis theorem will tell you that even algebraic geometry in that very broad sense you are only looking at finitely many equations. So mind you Hilbert basis theorem will work if R is any noetherian ring okay so this tells you that you will always be looking at only finitely many equations if you are working with coefficients from a noetherian ring okay that is the importance of this that is the geometric significance okay and the other important thing is of course you must not forget that we are looking at the case when the ring is an algebraically closed field because of Hilbert's Null's theorem which assures that you really get zeros that is it ensures that the zero set is not the empty set provided this ideal generated by this collection is not the whole ideal that means it does not contain a unit okay. So we will see more about that in the next lecture okay.