 Okay, so what I am going to now do is to tell you about another important property that comes because of the noetherianness okay, so what I just explained the previous lecture was that the noetherianness of a topological space allows every non-empty closed subset to have noetherian decomposition that is it can be broken down into a finite union of irreducible closed subsets non-empty irreducible closed subsets and if you assume that this union is not redundant that is no subset in the union is containing some other subset in the union then this union this decomposition is unique that is the members of the union are unique okay, so and that is the reason why any algebraic set any closed subset of affine space for the zariski topology can be written as a finite union of affine varieties. So this is one of the reasons why we call affine varieties as building blocks of algebraic sets and it is a general philosophy that affines are building blocks of algebraic of structures in algebraic geometry and now I am going to give you another reason for the for the importance of noetherianness okay, so that is got to do with dimension okay, so you know the aim is somehow to try to tell you the obvious thing that you will expect that you know the dimension of affine n space is n okay that is what I will try to explain. But so what I but what I want you to begin with not get confused with is the following see if you take affine space which is an it is just as a set it is just kn okay it is n copies of k where k is an algebraic closed field okay and if you take kn as a vector space over k then it is very clear that it is n dimensional because it is a finite dimensional vector space and you know n copies of k as a of a field k will be dimension n as a vector space over that field okay because you can you for example you can always write down the standard basis okay which consists of 1 in the ith place and 0 is elsewhere right, so but therefore you know if you think that I am trying to say that the dimension of kn over k as a vector space is n then you are mistaken because that is not what I am trying to say. I am trying to define dimension a completely different way I do not want to think of kn especially an affine n space I do not want to think of it as a vector space okay we are not here worried about the vector space properties okay we are worried about the topological properties. So the therefore what this calls for is how to define dimension of a topological space how to define dimension of topological space, so the answer to that is the dimension can be defined by taking the largest possible namely the supremum of lengths of strictly you know decreasing chains of closed subsets okay and in an analogy you can compare it to the vector space situation okay so you know you take any finite dimensional vector space over a field there if you look at subspaces of the vector space then you know the if the vector space of dimension n then the largest strictly increasing or decreasing whichever way you want to see it chain of proper subspaces will be of length n plus 1 because it will start from 0 if it is increasing it will start from the 0 subspace and it will end with the full vector space and at each stage you will get a bigger subspace with dimension 1 more okay so you start with 0 and you end with n so you will exactly get a chain of length n plus 1 of strictly increasing subspaces and that is the maximum possible okay so in the same way the analogy is that for a topological space you can define the dimension to be the supremum of the lengths of you know a strictly increasing chain of irreducible closed subsets okay and this leads to a very good definition because it has got to do in the case of affine space it has got to do with commutative algebra namely with the polynomial ring so in what way I will explain now so let me make this the definition so what is the so the aim is to so my so the aim of this lecture at least the beginning is to show that the noetherian condition is very helpful to define and show that the dimension topological dimension of affine n space is actually n okay and it will involve commutative algebra as you will see so here is my definition for a topological space X we define dimension of X okay so this is called the topological dimension okay some people write dimension with a subscript TOP okay and sometimes we might just omit it okay so whenever I write dimension of a topological space it is always topological dimension what is it this is to be this is defined as a supremum of all n set of all n such that there exists a chain of irreducible closed subsets z0 properly contained in z1 properly contained in z2 and so on up to zn so here is the so here is the definition of what dimension of a topological space okay so you and mind you I am starting with I am starting with z0 alright and I assume that just to check that I am on the right track I need to also say that all of and perhaps I have to put the condition I do not have to okay so z0 is already non-empty because see I was just worrying whether I have to put the condition z0 is non-empty but then I am saying that there are irreducible closed sets and therefore they are already non-empty but mind you I am starting with 0 this is very very important starting with 0 okay. So now you see so you know you see what I want you to understand is that this is let us examine this situation when x is you know affine space okay so put take x equal to An Ank k algebraically closed field so you take k to be an algebraically closed field and look at affine space over k okay then you know that for the Zariski topology the closed sets here they correspond to the irreducible closed subsets here they correspond to prime ideals in the polynomial ring in n variables which is thought of as a ring of functions on the affine space. So what will this definition translate to here it will translate to the following dimension of An Ank is equal to the topological dimension is a supremum of all the n such that there exists a chain of prime ideals p0 so let me use script p0 properly containing p1 I think I will have to rather I have to number it the other way round p0, p1, pn okay. So by this definition this is what you get okay you have a chain of prime ideals okay and it starts with p0 and goes up to pn and you take the supremum over all n now the fact is that so since you are looking at chains of prime ideals strictly increasing chains of prime ideals what this relates to in commutative algebra is called as height of a prime ideal okay so let me recall what that is okay so what is the commutative algebra involved let me explain that to you so what we do is let R be here commutative ring with one okay let p in R be a prime ideal let this be a prime ideal then what you do is then we define height of p okay height of p written as htp to be the supremum of all n such that there exists a chain of prime ideals p0 properly containing p0 contained in p1 properly contained in so on to pn we define the height of a prime ideal like this alright and with of course with pn equal to p. So in other words you look at a chain of prime ideals which ends with p and you look at the largest possible such chain of course they need not be finite at all okay you may always find a chain like this for every n it may it might happen so but if it but if it happens it does not happen like that then you take the supremum and that will be a finite number and call that the height of the prime ideal okay and see here comes the here comes the following fact suppose R equal to k of so let me write this f of x1 etc up to xm modulo j where f is a field and j is a prime ideal suppose R is of this form alright your question is whether this is an ascending chain or whether this is a descending chain actually the truth is that if I put it either way anyway I am going to get the same number okay whether I put it as if I start with z0 and go up to zn and it is an ascending chain or if I start with z0 and zn it is a descending chain anyway it is that n that I am worried about supremum of that n that I am worried about so it really would not matter if I put you know the inclusion this way or the other way okay but there is a there is an issue when you come to the ring and I will explain that now okay so you see you are assuming R to be of this form that it is a polynomial ring over a field modulo of prime ideal okay that means that R is an integral domain this implies that R is an integral domain okay then so now what you can do is you can look at Q of R this is the this is the quotient field of R or it is otherwise called as field of fractions of the integral domain R this is just the field of fractions of the integral domain R okay mind you this is a ring and you are going modulo of prime ideal ring modulo of prime ideal is a domain okay and if you have an integral domain you can form the field of fractions just like you form rational numbers the field of rational numbers from the ring of integers which is an integral domain. So you take the field of fractions and then what you can do is that you can look at you can see that this will contain F okay clearly Q R contains F because you say F is any way contained in the polynomial ring as constant polynomials okay and you are you are going modulo of prime ideal okay so in particular you are not going modulo of everything okay so the fact is that you are certainly not since this is the proper ideal you are not certainly going modulo the elements of the field okay so the elements of the field still remain invertible in the field of fractions of R okay you can see this for example in commutative algebra either by looking at the universal property of a quotient field or you can use the universal property of localization okay for Q of R is actually the localization of R at the 0 prime ideal which means you invert everything outside 0 okay you localize with respect to the multiplicative set which is the complement of 0 okay so now the point is when you so in other words what you have now is you have an extension of this field you have a field F and you have this field extension now once you have an extension of a field in field theory you can talk about things about talk about many things about the extension first of all you can ask whether it is algebraic if it is not algebraic you can check if it is transcendental and if it is transcendental then you can define what is meant by transcendence degree okay so let me quickly recall if you have a smaller field and you have a bigger field then we say that the bigger field is algebraic or the smaller field if every element in the bigger field is obtained as a 0 of a polynomial coefficients in the smaller field okay and if there is an element which is not the 0 of any polynomial in the smaller field then that element is called a transcendental element for example if you take real numbers over rational numbers okay then the number E which is used in defining the exponential function or the number pi which is used in trigonometry okay they are all transcendental though of course the proofs of these facts are not so easy E and pi are transcendental numbers and they are transcendental numbers because you cannot find them as roots of an equation in one variable polynomial equation in one variable with rational coefficients which is the same as looking at with integer coefficients okay because you can always clear denominators. So the moral of the story is that you do have fields which have transcendental elements so R the field of real numbers is transcendental as a field extension over the field of rational numbers and once you have transcendental elements what you can do is you can actually define you know what is called as a transcendental version of dimension okay so what you can do is you can mimic what you do for a vector space situation see in a vector space situation what you do is how do you define the dimension of vector space the dimension of vector space is defined as the maximal number of linearly independent vectors okay so in other words what you do is you take the maximal subset of vectors which are linearly independent and take its cardinality and call that cardinality as the dimension of your vector space. But the dimension of vector space is just the cardinality of a maximal set of linearly independent vectors okay now you just mimic this in algebra and what you do is instead of linear independence which is used in the situation of vector spaces you now use algebraic independence which is the analog that you use in algebra in ring theory okay so what you do is if you have a bigger field containing a smaller field okay and suppose the bigger field has some elements which are transcendental over the smaller field namely if it has elements which are not zeros of polynomials with coefficients of any polynomials with coefficients polynomials in one variable with coefficients in the smaller field then you can start looking at you can start looking at a collection of transcendental elements okay but put the condition that also put the condition that this collection of transcendental elements is algebraically independent okay so you know a collection of elements finitely many elements in a bigger field is said to be algebraically independent over the smaller field if this if these elements okay they do not satisfy a polynomial in several in as many variables with coefficients in the smaller field okay. So please try to understand when you do it for a vector space you will say that a bunch of vectors finitely many vectors are linearly independent if they do not satisfy a linear polynomial in as many variables with coefficients in the base field now what you are doing is instead of requiring a linear polynomial in so many variables okay as many variables as the number of elements you are looking at you are only saying that now you also assume that you cannot find a polynomial relation you are only saying so let me repeat that if you have finitely many elements of a bigger field we say the finitely many elements of the bigger field is they are algebraically independent if they cannot if they are not they do not have any polynomial relation between them with coefficients in the smaller field in other words they are not 0 of a polynomial in as many variables with coefficients in the smaller field such a subset of elements is called an algebraically independent subset of elements okay now what you do is just like in a vector space situation you took a maximal linearly independent set and took its cardinality and called it the dimension you do the same thing here what you do you do the analogous thing here what you do is you take a maximal set of transcendental elements which are algebraically independent okay take a maximal set of algebraically independent elements and take its cardinality and call that as the transcendent transcendence degree of the bigger field over the smaller field okay so transcendence degree of the bigger field over the smaller field is the is the cardinality of the large maximal number is the cardinality of a maximal set of algebraically independent elements just like in the case of a vector space the dimension of a vector space over the base field is the cardinality of a maximal linearly independent set of vectors the same way the dimension the transcendence degree of a bigger field over a smaller field is the cardinality of an algebra a maximal algebraically independent subset of elements of the bigger field which are algebraically independent over the smaller field okay that is called transcendence degree just mimics what we did for dimension in the linear case okay the vector space case so the beautiful theorem is that if you calculate the transcendence degree of q of r over f that turns out to be what is the cruel dimension of r okay and cruel dimension of r is supposed to be the supremum of the heights of its primitives okay so so let me so let me write that so here so so maybe okay so let me let me let me do so quite a few things that I have mentioned but you can kind of try to at least understand them heuristically now and then do further reading so define for any commutative ring r with one cruel dimension of r to be equal to the supremum of height of p but p in r is prime call this the call this the cruel dimension and the the the notation for that is dimension cruel okay then here is a theorem so it is a theorem from commutative algebra and field theory which says the following if r is equal to f of x1 etc up to xm modulo j where j is a prime ideal then a transcendence degree of quotient field of r over f is equal to the cruel dimension so please understand this theorem say it is it is it is a very basic and of course a very important result what it does is it tells you what the cruel dimension of an integral domain which is a finitely generated algebra over a field measures it actually measures the number of algebraically independent elements in the fraction field of the integral domain over the base field okay so this is a so you know in some sense this side is analogous to what you do in linear algebra when you have vector space over a field then the dimension of the vector space over the field is the cardinality of a maximal linearly independent subset of vectors which are linearly independent over the base field in the same way when you have a field extension of a field then the transcendence degree of the field extension over the smaller field is the cardinality of a maximal set of linear of algebraically independent elements here over which are algebraically independent over the base field okay and in the case of linear independent the condition is that those elements those finitely many elements do not satisfy a linear polynomial in as many variables with quotients in the base field and in the case of algebraic independent the condition is that those finitely many elements do not satisfy a polynomial of higher degree in as many variables with quotients in the base field that is the analogy okay and that these two are equal is the theorem is the theorem from computational theorem okay I need also another I need also another theorem and I think this stating this theorem will in retrospect check whether I have muddled with the inclusions in this definition or in this definition okay so what is the theorem so what this theorem says is that this theorem actually you know connects the height of the ideal with the dimension of the quotient okay so the theorem is so with r as above that is r is a polynomial ring in finitely many variables over a field model of prime ideal okay height of j plus dimension cruel dimension of r is equal to the cruel dimension of this yeah so what I want to basically say is that if you are looking at if you are looking at the if you are looking at j equal to 0 okay if you are looking at j equal to 0 and you look at height of the 0 prime ideal then the height of the 0 prime ideal is just 0 okay because I can start with 0 and I have to and that is it I cannot make it any larger so the height of the 0 of the 0 prime is just 0 okay plus the cruel dimension of r will just give me again the cruel dimension of r because r in this case if I put j equal to 0 then r is actually f of x1 etc to xm and the cruel dimension so of course if I put j equal to 0 I will get I do not get anything but the point is that the cruel dimension of this is actually m because the cruel dimension of a ring is actually the transcendence degree of the quotient field of that ring over the quotient field of that integral domain over the base field. So if you take a polynomial ring in m variables and look at this quotient field you will get the field of quotients in n variables okay so the quotient field of this will be f round bracket x1 through xm this is the set of all quotients of polynomials in m variables with of course the denominator being on 0 and as you can easily see you have to check that there are this if you take the quotient field of this the number of linearly I mean algebraically independent variables will be m it will be these x1 through xm okay you cannot have more than m algebraically independent elements over f okay. So what this will tell you is probably I do not need this now maybe I will need it later when I look at the general case of an affine variety but for the moment what you get immediately from all this is that the dimension of the your affine space okay the dimension of your affine space will be by definition it will be the supremum of all these things okay and you can see that this is the same as the Krull dimension of the ring of polynomial functions on the affine space which is equal to m okay. So what I wanted to understand probably I do not need this now this so maybe I will do the following thing for the moment let me so let me so let me not let us not worry about this okay let us not worry about this this is not immediately relevant for the discussion but we are this is what is important put R equal to k x1 etc out of xn we get dimension topological dimension of an is equal to dimension Krull of k x1 etc xn which is by definition equal to I mean which is by this theorem equal to the transcendence degree over k of the quotient field of k x1 etc xn and that is of course equal to transcendence degree over k of q of x1 etc xn this is the notation for the quotient field of sorry k round bracket x1 through xn is the quotient field of k square bracket x1 through xn and k round bracket x1 through xn consists of ratios of polynomials from this ring polynomials in these n variables with of course the denominator polynomial being non-zero okay and that is this is equal to n transcendence degree of this is n okay that is again a fact that we will we will accept from field theory after from competitive algebra that polynomial ring what you have done to the polynomial ring is what you have done in constructing this polynomial ring is that you started with the field k and you added n in determinants and these n elements are algebraically independent by definition because any try to understand that if you look at x1 through xn they are elements of this ring and this ring sits inside its quotient field so they are also elements of the of this quotient field you can think of each xi as xi by 1 divided by 1 just as you think of an integer as a rational number given by the integer divided by 1 okay and therefore these xi's are all elements here okay and the fact that they are all algebraically independent is the fact that if you write a polynomial in the xi's with coefficients from k and if it is equated to 0 then all the coefficients have to be 0 that is what it means to say that the xi's are you know indeterminates the fact that xi's are indeterminates says that they are transcendental over k and any polynomial relation amongst them is 0 if and only if all the coefficients are 0 okay that is in other words they are all algebraically independent so it is very clear that x1 through xn are algebraically independent and therefore the transcendence degree has to be at least n and then if you do some field theory you can check that the transcendence degree is exactly n. So by this definition you will get that the dimension of an the topological dimension of an is the same as the Krull dimension of this ring and that is equal to n okay now let me come back to this statement here and that has got to do with trying to do it to all this dimension count even for an affine variety okay so let me do this for any affine variety and use this. So you see suppose y inside an is an affine variety suppose y inside an is an affine variety okay so y is an irreducible closed subset okay and of course y is well 0 set of the ideal of y with of course ideal of y a prime ideal in the polynomial okay. Now you see so what I want to say is that a statement similar to this also can be made for affine variety so what is happening here is the topological dimension of affine space is the Krull dimension of the ring of functions on affine space see that is what the first statement says the topological dimension of affine space with respect to Zariski topology is the Krull dimension of the ring of functions on the affine space and the Krull dimension of the ring of functions on the affine space is the transcendence degree of its quotient field over the base field okay which is essentially this theorem okay. A similar statement holds for any irreducible closed subset so what do you expect the theorem to be the theorem the fact will be that if you take the topological dimension of y okay notice that y is a subset of affine space and affine space has a Zariski topology so y has also the induced topology and in fact y is itself a closed subset so therefore any closed subset the closed subsets of y are precisely the closed subsets of affine space which are contained in y there is no difference okay and so y if you take the Zariski topology induced on y and you look at the topological dimension of that that will turn out to be equal to the dimension the Krull dimension of the ring of functions on y see look at the statement the topological dimension of the space is the Krull dimension of the ring of functions on the space and the ring of functions on the space is all the polynomials so here I should write ring of functions on y okay but what is the ring of functions on y the ring of functions on y to get the ring of functions on y you will of course all polynomials which are functions on the whole affine space are also going to be functions on y because after all you take a polynomial I can evaluate it on affine space I can also evaluate it on a subset of the affine space so I can take all those polynomials and restrict it to y but the point is two such polynomials two different polynomials there may still define the same function on y that is because their difference may be a function which vanishes on y what this tells you is that the functions on y are the same as the functions here on affine space up to translation by elements of the ideal I y in other words what you are doing is you are looking at cosets of I y in the polynomial ring which means you are actually looking at the quotient ring so the moral of the story is that the ring of functions on y has to be the ring of functions on the bigger space which is affine space modulo the ideal of y okay so this is the cruel dimension of the ring of functions on the affine space which is the bigger space k x1 etc xn modulo the ideal of y okay notice that the ideal of y is prime okay and the ideal of y is prime and therefore this quotient is an integral domain okay and we are in this we are in the situation of this theorem you are having a polynomial ring over a field you are going modulo prime ideal and then the cruel dimension is actually the transcendence degree of the quotient field of the integral domain over to the base field so if you use that theorem you will get that this is transcendence degree so this is by definition transcendence degree yeah but before I do that let me use this theorem okay let me use this theorem alright so what this theorem will tell you is that it is the cruel dimension of r is the cruel dimension of the bigger ring minus the height of the ideal by which you are going to get r so this will tell you that this will be just n minus height of the ideal of y and of course this will be equal to transcendence degree of the over k of the quotient field of this quotient ring kx1 which is an integral domain okay so the moral of the story is that you can calculate dimensions for you have a formula for dimensions for affine close affine sub varieties of affine space okay so as an example you can look at y equal to say z of x1 okay or more generally suppose I take the ideal generated by x1 through xr okay where inside affine space where r is less than or equal to n okay then you can see that so then dimension of y as a topological space is by definition going to be n minus height ideal of y okay and the fact is that the ideal see the ideal of y is just ideal of z of x1 etc through xr okay and you know if you take i of z of some ideal you get its radical which is you know one of the important consequence of the null seven sets so what you will get here is the radical of x1 through xr and you can check that it is its own radical that is because its prime why is it prime because if you take the polynomial ring modulo this ideal you will get the polynomial ring in the other variables that is a very easy check and since when you go modulo this you get a polynomial ring in some variables which is an integral domain this has to be prime and since this is prime it is already radical okay so you will get this and therefore and you know if you take height of iy you will get r okay because you are going to look at maximal chain like this which starts with something smaller and goes up to the ideal and you know you can see that since the maximal the maximal chain maximal chain of primes will be for example like 0 properly contained in x1 properly contained in x1, x2 properly contained in and so on x1 through xr okay. So you will see that the height is r okay and but I am saying for example because one has to prove it okay the fact that you have a chain like this tells you that the height of this ideal is at least r okay because height is supposed to be supreme the fact is that you cannot get a chain of bigger length that is the fact that needs to be proved okay but if you assume that you can if you believe that then height of iy is r and you will get dimension of the topological dimension of y is equal to n-r okay and this is so in other words what you are saying is the topological dimension of the 0 set of x1 through xr is n-r of course you know if I put n equal to r then x1 through xn will be a maximal ideal that will correspond to the origin so the 0 set will be a single point and the dimension will become 0 n-n the point will have 0 dimension okay. So this is a so what this demonstrates to you is that you get the most natural thing namely if you go if you take if you cut down by r equations then your dimension also cuts down by that many equations roughly this is what is happening but then it is technical to check that you know this the height of this is exactly r okay but what I am trying to demonstrate to you is so of course here I have used this result I mean this allows you to do dimension calculations for sub varieties of affine space close sub varieties of affine space and here is a standard example. So the when you take z of x1 through xr you are taking the locus given by the intersection of all the hyper surfaces when you take 0 set of any xi you are looking at the equation xi equal to 0 that is called a hyper surface because it is cutting by one equation and now what you are doing is now you are successively cutting by r of these equations and obviously you should expect the dimensions should also go down by r and that is exactly what this computation says and the fact is that this is exactly what you would expect this is exactly what happens but the commutative algebra that intervenes is locked in these two theorems is captured by these two theorems and that involves the definition of transcendence degree it involves definition of Krull dimension and which in turn depends on dimension of I mean the definition of height of a prime ideal ok. So these are the I mean this is the commutative algebra that comes in so the moral of the story is the following the moral of the story is that if I draw a diagram with you know so maybe I can do that if I draw a diagram with if I draw a diagram with the geometric side here and the commutative algebraic side here side here if I start with an ok then what you go to is k of x1 etc to xn which is the ring of functions on an. So now you see now this dictionary what I am writing down is not what I is different from what I wrote down earlier earlier I was looking at subsets of an say closed subsets of an here and I was looking at ideals here ok but I am not doing that now I what I am doing is I am defining a relation a function which to every set gives its set of functions. So this is this is a of so this is the symbol standard symbol so a of an k is the set of functions on an if you give me z affine variety z of p affine variety affine sub variety which is a closed subset irreducible closed subset of an ok then and if you apply this a you will get a of z of t is just k of x1 through xn mod p the affine the ring of functions on an affine variety is just the quotient of ring of function the ambient space model of the prime ideal whose zeros define that particular irreducible closed subset. So this can also be written as a of an by I of z of t ok so you see now you get a picture is here on this side you have the affine variety an mind you an itself is an affine variety because it is an irreducible closed subset ok and you have its irreducible closed subsets which are proper affine varieties and there are the corresponding rings of functions. So the model of the story is that we have this correspondence between space the geometric spaces on this side and the rings of functions so in all of algebraic geometry the point is that everything that is geometric here gets translated into commutative algebra and vice versa ok. So here is so for example the topological dimension on this side is the cruel dimension on the other side that is what the theorem says ok. So the notion of topological dimension here corresponds to the notion of cruel dimension there that is the that is the translation ok that is what you must understand ok so I will stop here.