 Ok, so you see last lecture we were looking at the notion of dimension ok and we noticed that there is a notion of dimension from the topological point of view that is the definition of dimension of a topological space and then there is also the notion of cruel dimension of a ring ok and somehow these two are related in the sense that the dimension of for example the dimension of affine n space the topological dimension of affine n space is the same as the cruel dimension of the ring of functions on affine n space ok. So let me recall again to refresh your memory and also so that you feel it is comfortable about things let me recall what was going on so see on the so we have as usual the geometric side and we have on this side the commutative algebraic side and here we take affine space over k affine n space over k this is just k n k cross k cross k n times given the Zariski topology where of course k is an algebraically close to it ok as usual you can always think of it as complex numbers if you want to be very concrete. So in that case this would be just c n n copies of c given the Zariski topology and then the ring of functions so the association that I am interested in is called is given by the symbol a of a variety or an algebraic set ok and it is defined only for varieties properly this is well this should be thought of as giving you the ring of functions. So you have the space here and here you have it is ring of functions so on this side you have the space and on this side you have rings of functions ok. So if you take affine space of course what kind of functions are of course polynomial functions we are only interested in algebraic functions polynomial functions. So if you take this this is going to be this is k x1 etc ok and the fact is that you see what I want to say is that there is a there is also an association that goes from this side to this side ok which makes this again a bijective correspondence ok. So again you know this is another instance of you know the interaction between algebraic geometry and commutative algebra that there is always a dictionary which goes from the geometric side to the commutative algebraic side and goes back and forth. So see initially you know we had such a dictionary where we had taken on this side for example we have seen it in other context for example on this side if you took you know close subsets that is algebraic subsets then what you can get on this side as a bijective correspondence are the radical ideals ok and if you and under this correspondence the irreducible algebraic subsets the irreducible close subsets they will correspond to prime ideals ok which are of course radical and the of course the points the singleton sets consisting of points they will correspond to the largest ideals which are the maximal ideals are just proper ideals. So that is one correspondence we have already seen and in that correspondence what goes from this side to that side is given a set here you associate the ideal of that subset ok and what goes from that here as the inverse map is given an ideal you look at the 0 set of the ideal ok and that gave a bijective you know correspondence. Now rather than going from algebraic sets or varieties to ideals what you do is you go from algebraic sets or varieties on this side to rings of functions ok. So that is the change in the point of view but the fact is that again you get a nice bijective correspondence and so let me explain see what is happening is see if you why this is so important is because if I take dimension topological of An this is of course N ok and that is just a translation of the fact that the dimension the cruel dimension of this polynomial ring is N. So the topological dimension of a space here of course when I say space here actually what I am meaning is a variety here ok a variety means an irreducible closed subset of An in particular An itself is an irreducible closed subset of itself ok. So this is the biggest possible variety in An and if you take any variety here then you can associate to it it is ring of functions ok and the dimension of the variety will be the same as the cruel dimension of it is ring of functions ok. So what I want you to understand is that if I take X variety affine variety ok then you know X is of the form Z of some prime ideal ok because you know if a variety I mean if an algebraic subset is always given a zero set of an ideal and if that algebraic subset is irreducible if and only if the radical of the given ideal is prime. So in particular and of course taking the zero set of an ideal and zero set of it is radical does not create any changes the zero set of an ideal is same as zero set of it is radical. So any irreducible closed subset of affine space is always given by the zero set of a prime ideal ok and of course you know this prime ideal is unique it is unique under that bijective correspondence between irreducible closed sub varieties I mean irreducible closed subsets in An and prime ideals of the polynomial ring. So the fact is if you take a affine variety like this then what is it that you are going to get on this side what is it that you are going to get on this side you are going to get the space of the ring of functions algebraic functions on X ok and the ring of algebraic functions on X will be X is of course Z of P and in fact it is actually A of An by P it is just this is just K X1 etc Xn by P this is what it is and so let me draw a line like this because this is this statement is for An now I am making a statement for any affine variety closed inside An ok then for any variety the ring of functions is defined like this and then you again see that the dimension the topological dimension of the variety these two are equal here also the dimension the topological dimension of the variety will be the same as the Krull dimension of its ring of functions this will happen. So this is so I want that so this is a generalization of this this is generalization of this statement which is true for all affine space and I also want to tell you what is this this arrow that is coming in the other direction ok that arrow is actually max spec that is the arrow that is going in this direction. So let me explain that so let me explain let us keep this diagram as the basic diagram that we want to understand and let me expand let me expand upon this ok. So the first thing that I want to explain is well just to recall how did you get the topological dimension is equal to the Krull dimension the topological dimension of An to be equal to the Krull dimension of the polynomial ring that is because the topological dimension is supposed to be the you take chains of strictly increasing chains of irreducible closed subsets ok and then you take the largest possible such chain and subtract one from that that is the topological dimension if at all it is finite ok you have to take large if it is not finite then the largest possible will be infinite and it will become infinite dimension ok. So how does one define the dimension of a topological space you simply write out and you try to look at existence of a chain of irreducible closed subsets ok irreducible closed subsets which are contained each one contained in the next try to look at such a chain and try to look at the chain being strict namely that there are no repetitions in successive members of the chain and among such chains try to take maximum the maximal length once namely the ones with as many terms as possible and you if this maximum is a finite number ok you take away one from that that will be the topological dimension ok. Now why does that translate to this side into this side to give the Krull dimension that is because any closed subsets here corresponds to a radical ideal any irreducible closed subsets here corresponds to a prime ideal and if you give me strictly increasing chain of irreducible closed subsets that will correspond to a strictly decreasing chain of prime ideals ok. So what will happen is that the fact is that the largest I mean largest in the sense the chain strictly increasing chain of prime ideals of maximal possible length ok is n plus 1 ok and of course the easiest thing is you take 0 the sort of the 0 ideal then take the ideal generated by one variable x1 then that is contained the ideal generated by x1 and x2 and then at the ith stage is generated by x1, x2, etc up to xi and you go on up to n this is a strictly increasing chain of prime ideals its length is n plus 1 and therefore the Krull dimension becomes n. So the Krull dimension of a competitive ring is defined to be the supremum of the heights of the prime ideals ok and what is the height of a prime ideal? The height of a prime ideal is the maximum possible length of the maximal possible chain that you can descending chain strictly descending chain of prime that you can get starting from that prime ideal and then you have to take away one from that ok that is the reason in all these definitions you will see that instead of taking a chain from Z1 to Zm and taking the supremum and calling this and taking the dimension to be one less than that we rather start with 0 we start with the index 0 ok that is the reason you start with 0 ok. Anyway so the fact is and why is the Krull dimension of this equal to n because whenever you have a finitely generated whenever you have an algebra namely you have a quotient of the polynomial ring which is an integral domain then the Krull dimension of that is actually the trans it is actually the transcendence degree over k of its quotient field and the quotient field of this is simply the field of rational functions in n variables ok it just consist of ratios of polynomials in n variables with the denominator polynomial not equal to 0 and that has transcendence degree n it has so transcendence degree is analog is for algebraic independence the analog of dimension for linear independence just like for linear independence dimension is cardinality of a maximal linearly independent subset for algebraic independence transcendence is the cardinality of a maximal algebraically independent subset ok so it is analogous to what happens in linear algebra. So transcendence degree measures the largest it gives you the number of maximum I mean the maximum number of transcendental elements which are algebraically independent from one another ok that is what it gives so just like dimension of vector space gives you the maximal number of linearly independent elements ok linearly independent vectors so and the transcendence degree of this the quotient field of this is n which is very clear intuitively but you need to prove it ok using some something from you have to use some field theory to prove it ok and now you see so the first thing that I want to explain is why do we define the ring of functions on x like this ok why do we define the ring of functions on x to be the ring of functions on the whole space divided by the ideal of ideal of x so in fact this is I can simply write it as a of a of an ok divided by I of x, I of x is p script I is the ideal of the given subset ok so I of x is p right so here also I can write that p is equal to I of x ok and so in a way you know previously we were associating to x I of x which is a prime ideal now what you are doing is you are not associating x to the prime ideal but you are associating it to the quotient by that prime ideal instead of looking at the ideals on this side you are looking at quotients by those ideals that is the only difference but the looking the advantage of looking at quotients by those ideals is the fact that you get rings of functions and why is that that is what I am trying to explain it is very easy see what you do is you see you take you look at a of x so this is you define it to be functions from so functions from x to k which are given by polynomials by restriction by polynomials so look at this definition so a of x are just functions they are maps from x to k and the point is that this map is it should be an algebraic function ok which means it should be a polynomial ok and of course both polynomial polynomial in n variables because x is a subset of an and any polynomial in n variables is a function on an you can evaluate it at each point of an ok so actually what is happening is when you go from here to here there is something that is already happening so if you apply this condition you see if you if you if you apply this definition so if I write a of an this will be functions from an to k which are given by polynomials and you see like if I take the ring of all polynomial functions in n variables then I have a map like this that map is you start of the polynomial in n variables and you use that f to define a function think of c f is here now a polynomial define that as a define use that f to define a function let me if I want to be very strict to distinguish the fact that here I am thinking of f as a polynomial and there instead of the polynomial I am looking at the function that it defines what I will do is I will put evf this is evaluation of f evaluation of f from an to k is a function this is what f goes to what does this do give me any point in an you simply evaluate that point f at that point evaluate the polynomial at that point this is the map so you are just associating f to evaluation of f ok and what you must understand is you see if you take functions like this which are given by polynomials then it is very clear that if you take two such functions their sum is also given by corresponding sums of polynomials right so if a function is given by a polynomial then a sum of two such functions is also given by sum of the corresponding polynomials product of the two functions will be a function that is given by product of the corresponding polynomials and of course there are constant functions here which correspond to the constant polynomials in other words what I am saying is that this map is actually a ring home of sum I am saying this is a ring home of sum this fellow on this side is actually ring that is what you must understand take it here also if you when I say functions are given by polynomials if I have two functions are given by polynomials then their sum is given by the sum of polynomials ok and the product is given by product of polynomials so these are all rings these are all rings ok they are commutative rings and they contain constants the constant functions are always there because constant functions come from evaluation of constant polynomials so constant functions are there so this is these are k algebras a k algebra is nothing but a ring a commutative ring with one with k which is also vector space over k ok and in such a way that the multiplication in the ring is compatible with the scalar multiplication lambda times f g is the same as f into lambda times g ok so the vector space structure and the ring structure they are compatible that is what an algebra is ok so these are these two are both k algebras ok so their rings and the fact is here also I can start with take a polynomial in n variables I can have this map what is this map this is just f going to evaluation of f restricted to x from x to k it is the same map but you restrict it to the subset x after all because x is a subset of an you can always restrict the map that is what I mean to say that these are functions which are given by polynomials when you say they are functions given by polynomials they are actually given by restrictions of polynomials that is what this map is ok and what you must again understand in this case also is that this is a ring homomorphism it is also ring homomorphism because f plus g evaluation of f plus g will be evaluation of f plus evaluation of g evaluation of f into g will be evaluation of f into evaluation of g and evaluation of lambda will be just lambda if lambda is a constant polynomial alright so you can see that this is a this is not just a ring homomorphism it also preserves the vector space structure mind you what is there on the left side is also k algebra this is a commutative ring with one which is also vector space over k and the scalar multiplication is compatible with the ring multiplication if you take a scalar lambda and if you take two ring elements to polynomials f and g then lambda times f g is the same as lambda f times g that is the same as f times lambda g it really does not matter to which factor of a product you multiply a constant you are going to get the same polynomial right in a product of polynomials. So these are actually k algebras so what is happening is that you are getting ring homomorphisms which are actually homomorphisms of k algebras these homomorphisms also satisfy they also satisfy they also they are linear they also respect the vector space structure so evaluation of f plus g is evaluation of f plus evaluation of g this respects addition it respects multiplication and it respects scalar multiplication ok and evaluation of lambda is lambda this all this is true and this is for f g in polynomial ring and every lambda in the every constant polynomial ok. So this is so in fact instead of writing ring homomorphism let me write k algebra ring homomorphism k algebra means it is a ring homomorphism which is also a k vector space map and of course the algebra condition is on the rings that the scalar multiplication and the ring multiplication are compatible that is you know you also have lambda times f g is equal to lambda f times g is f times lambda g this is the algebra condition that multiplication by a scalar is well behaved with respect to multiplication between two ring elements not of a vectors ok fine so now what I want you to understand is that this map is an isomorphism this map is actually isomorphism as you see if first of all if you have two polynomials if so you know it is a ring homomorphism to show that it is an isomorphism I have to show it is surjective I have to show it is injective to show it is injective I have to show it is kernel is 0 if you want because it is a ring homomorphism it is surjective by definition because my definition of ring of functions is that they are given by polynomials so it is automatically surjective both of these maps are surjective because by definition a function is supposed to be coming from a polynomial so it is so both maps are surjective might be the only thing that you will have to worry about is injectivity ok so if a function f is such that evaluation at f is 0 ok for every point of an then that function has to be the 0 polynomial that is that is because if it is not the 0 polynomial ok then you should be able to find some point where if you evaluate it you will get a non-zero element of k ok so that tells you that this is injective it is already surjective so it is an isomorphism and this is what tells you that this definition is correct that to think of the ring of functions on an as the polynomial ring is correct because of this because of this argument now in the same way if you look at this argument what you will get is you will get you will get this actually ok what will happen is that this ring homomorphism is surjective what is the kernel of this ring homomorphism this ring homomorphism is not injective it is it has a kernel what is a kernel it is all those functions whose evaluations when restricted to x become 0 but that is precisely all those functions which vanish on x but what are the functions which vanish on x they are precisely the functions in this they are precisely the polynomial in the ideal of x therefore the kernel of this map is actually the kernel of this map is actually ix and is surjective therefore source modulo the kernel is isomorphic to the target so you get that k x1 etc xn modulo ix is isomorphic to ax by this map ok so these two statements justify that this these definitions are correct ok so they give you the naturalness behind these two definitions ok so let me write that down let me write so let me write here kernel of evaluation is 0 this is this is so you know this evaluation this ring homomorphism is just ev because it is f going to ev of f and this ring homomorphism is ev restricted to x you evaluate and then restrict to x or you do the evaluation on x. So kernel of evaluation is 0 kernel of evaluation restricted to x is ix and so it tells you so this tells you that a of an is isomorphic to k x1 etc xn and a of x is isomorphic to k x1 etc xn modulo ideal of x ok. So what we are what you are doing is that see here by brute force we write like this that is we are able to think of the functions as think of the polynomials as functions ok but if you want to be more strict you should do it like this this is a stricter way to do it ok so in which you distinguish between the polynomial and the function that it defines by evaluation ok. So when you make this distinction you get these isomorphisms not equal to you get isomorphisms in a very nice way and why is that nice because this can also be written mind you this is also equal to k of x1 etc xn modulo ideal of an because ideal of an is 0 right. So what it tells you is that this formula ax is k x1 etc xn by ix works even for x equal to an ok. So you get this you know you get this kind of you get a philosophy that if x is embedded in as an irreducible closed subset of a bigger space ok this is irreducible closed in a bigger space then a of x has to be a of y divided by the ideal of x in y in a of y you get something like this. So what this tells you is that it tells you that you know the notion of so the point that you must appreciate here is that this quotient ring homomorphism this is a quotient homomorphism because it is surjective homomorphism always a quotient ok because the target is a quotient of the source by the kernel so it is a quotient and the quotient is actually thought of as restriction of functions ok. So the moral of the story is that the whenever you see a quotient in competitive algebra algebraic geometrically it means that you are reciting functions to a subset closed subset ok that is the that is the geometric meaning of what a quotient is that is what this is ok. So the quotient homomorphism is just gotten by restricting functions literally right so that tells you why you know it is correct to define the ring of functions for a fine variety like this ok of course we will really not worry about all the time writing isomorphic to isomorphic to ok so by abuse of notation we will simply put equal to ok but what that equality is to be very precise it is an isomorphism because when you simply write the ring they are actually polynomials but you want to think of them as functions you must evaluate those polynomials so there is a difference and it is isomorphism is to signify that difference but then one forgets this and by abuse of notation keeps writing equal to ok fine so this much is this much is clear then I want to explain why this and this are one and the same why are this and this one and the same see that is that is quite that is again quite clear because if you look at the look at the way it goes you see what is see if you take x ok and you in x if you give me a sequence of irreducible closed subsets ok so these are all irreducible closed in x ok then you see what these will correspond to is see mind you x is sitting inside x itself is an irreducible closed inside a n x itself itself is an irreducible closed inside a n ok so if you go to the if you go to the polynomial ring the ring of functions on a n which is k x1 etc xn ok then you see this x as a closed subset of a n on this side will correspond to the ideal of x which is prime ok mind you that there is an inclusion like this because all these are all closed in x and that is further closed ok so what will happen is you will get something like this you will get ideal of ideal of Zm which way so it should go the other way so ideal of Z0 will contain so it will be like this ideal of so let me write it proper containment ideal of Zm-1 and so on proper containment ideal of Z0 so this diagram on this side translates to this diagram on the other side ok and of course you know why this is because we have a on this side to that side if you take the if you apply this I you know that is a that is a bijection because it is inverse is given by Z you know this bijection this is a bijection between radical ideals and closed subsets and if you are only looking at irreducible closed subsets then it is a bijection between irreducible closed subsets and prime ideals ok and of course it is inclusion reversing alright. So see what see what this will tell you is that in the earlier lecture I define what is meant by height of a prime ideal what is the height of a prime ideal you start from that prime ideal and then you take a descending sequence of strictly decrease descending sequence of prime ideals ok and take the maximal possible length that is the height of the prime ideal ok so you know if I start with Ix and if I take a strictly descending sequence of prime ideals starting from Ix and going down all the way to the smallest prime ideal which is 0 mind you 0 is a prime ideal therefore because these are all integral domains 0 is a prime ideal polynomial ring is an integral domain. So then what this will tell you is the height of that plus this m ok and if I take the see here if I take m to be maximum if I take the maximum such m I am going to get dimension of x ok dimension of x is actually supremum false h m ok and suppose as in this case if what this will tell you is that if the dimension of x is say if the dimension of x is say r ok then you have a maximal chain here which is length r ok that will correspond to that will tell you that this has length r ok I mean this will be r ok plus you know if I further continue because I because of maximality this will be if I further continue it to go all the way down to 0 which will give me the height then if I add both I should get the dimension of the polynomial ring ok if I take this ok with m equal to r which is the maximal possible and then if I further go down ok from I of x to 0 that part will give me the height that plus this r will give me the dimension of the polynomial this is just this is just as this is just reflection of this fact that you know height of a prime ideal plus dimension cruel dimension of r mod p is equal to cruel dimension of r ok I wrote down a formula like this last time that is if r is an integral domain which is finitely generated as a k algebra then and p is a prime ideal of r then the height of the prime ideal plus the cruel dimension of the quotient is equal to the cruel dimension of the whole ring that is just a reflection of this fact that I explained just now. So moral of the story is that the dimension of x will be actually if the dimension of x is r then the height of I x will be n minus r so the dimension of x is r here this will correspond to height of I x is equal to n minus r because r plus n minus r adds up to n which is the dimension the cruel dimension of the whole polynomial. So the moral of the story is the but then what is n minus r? n minus r is the dimension of the quotient but what is the quotient? The quotient is a ring of functions that is why the dimension of the topological dimension of x here corresponds to the cruel dimension of the ring of functions ok that explains why these two are equal ok you understand that so I am just trying to give an explanation why these two are one and the same right. So I will stop here and then we will continue in the next talk.