 I am able to remind you about a few things from field theory alright so you know so recall from field theory so you know if you have L over K the field extension you take a field extension L over K ok then we say when an element of L over K is algebraic ok when an element of L is an algebraic element over K if it is an element of capital L is said to be algebraic over capital K if it satisfies a polynomial a non-trivial polynomial with K capital K coefficients in one variable ok so and if every element of L satisfies such a polynomial over K ok then we say L over K is an algebraic extension ok that is every element of L is algebraic over K alright and if you have elements of L which do not satisfy any polynomial any such polynomial with coefficients in K we call such elements as transcendental elements ok and well if L over K is finite extension in the sense that if you treat L as a vector space over K if the dimension of L as a vector space over K is finite then a finite extension is always algebraic alright and in fact but of course an algebraic extension need not be finite ok now you see if L over K is not algebraic then it has transcendental elements ok there are elements which are non-algebraic ok and then given a set of transcendental elements you can define when that set is algebraically independent over K ok and this is analogous to the definition of linear independence over K so a set of you know a set of elements of a vector space is said to be linearly independent if they if no finite subset of that set satisfies a non-trivial linear relation with coefficients in K ok so in the same way a set of elements of L which you know is said to be algebraic over is said to be algebraically independent over K if any finite subset of that set does not satisfy any polynomial relation with coefficients in K so you know if you give me finitely many elements of capital L ok then those finitely many elements are said to be algebraically independent over capital K if those finitely many elements are not zeros of some polynomial over K in those finitely many variables in as many finitely many variables ok and on trivial they should not be they should that element that tuple of elements should not be a zero ok of a polynomial in as many variables with K coefficients right that is when you say those elements are algebraically independent and this is when you take a finite set of elements and an infinite set of elements is said to be algebraically independent if every finite subset of that infinite set is algebraically independent ok and then there is this just like in the case of a vector space you define the dimension as the max you know it is the cardinality of a maximal set of linearly independent elements ok in the same way if you have a field extension which is not algebraic then you can define it is so called transcendence degree ok it is degree of transcendence ok to be the cardinality of a maximal algebra maximal algebraically independent set ok. So if there are elements L which are not algebraic over K you can try to a transcendental elements ok and then you try to take try to find a maximal subset of these such elements which are algebraically independent ok and that cardinality will always be the same ok any two just like if you take any two basis of a vector space are bijective ok in the same way any two sets of maximally any two maximal sets of algebraically independent elements will be bijective and that the cardinality of that set is called the transcendence degree of L over K ok and we say that L over K has a you know we say that L over K has a separating transcendence base ok if you can find it you can find a you know set of elements which forms a transcendence basis ok a transcendence basis is just given by a maximal set of algebraically independent elements ok if you can find a transcendent transcendental basis transcendence basis ok such that you know L over the field adjoined to K given by those given by the transcendence basis is a finite algebraic extension ok. So you know there is a you want a picture like this so L so you have K ok and in L you have this family of elements alpha j where you know j is in some indexing set j capital J alright then you have the you have the field generated by these alpha j's so this is the so this is a transcendence basis and transcendence basis you know it is just a it is just the analog of basis alright so it consists of a maximal subset of algebraically independent elements a usual basis will consist of a maximal subset of linearly independent elements a transcendence basis will consist of a maximal set of algebraically independent elements ok. So you so you want a situation where you can find a transcendence basis ok and you want this extension to be finite separable ok. So you want I mean this is the best thing that can happen with a field extension the best thing that can happen with a field extension is that the field extension splits up into two extensions like this the first one is given we said that this extension is a purely transcendental extension ok because it is the extension which contains only it is gotten by simply adjoining all the transcendental elements from it all the elements of a transcendence basis ok and then this over this will only be a finite extension this will be algebraic ok because there would not be already this contains a maximal set of algebraically independent elements ok this over this cannot be transcendental because if there is an element of this is transcendental over this then I can take that element and add it to this to get a bigger transcendence basis and that is not possible. So this over this has to be algebraic ok and the fact is that you can make it would be nice if this is a finite separable extension ok and that is that always happens if the field capital K is algebraically closed ok. So all this happens if K is algebraically closed well in fact it happens even under a more weaker condition it happens if capital K is what is called a perfect field alright and but let us not worry about it alright basically what is the definition of perfect field is that if a field is of characteristic 0 it is called perfect and if it is of characteristic P it is called perfect you can if you can always find P through ok if you take K power P you should get K ok and you should be able to find P th roots for all your elements alright. So an algebraically closed field is always a perfect field because finding P th roots is just amounting to solve equations and over an algebraically closed field you can always solve equations because that is the definition of algebraically closed therefore an algebraically closed field is always perfect and for a perfect field you have this very beautiful situation of course it is important that I need a finite separable extension here alright and this part will be a purely transcendental extension ok and this is some field theory alright but the important thing is that we want to apply it when capital K is small k our algebraically closed field and when L is a function field of a variety ok. So applies if capital K is small k for a small k is the algebraically closed field over which we are doing algebraic geometry that is the field over which we are studying varieties and capital L is the function field of X for a variety X over K ok so this is our application right. So in all these things our view point is that this small k this capital K is our small k and this L is a function field of a variety and then the theorem is that the function field of the variety is a finite separable extension of a purely transcendental extension of k and you know that if L is K X then the transcendental degree of K X will over small k will give you the dimension of the variety therefore but you know the transcendental degree will be just the cardinality of this J because the transcendental degree is just the size or the cardinality of a transcendence basis. So you know if the variety X has dimension R then this J will have R elements so this will just be a this will be small k and this will be small k adjoined with R indeterminates or R transcendental elements alright. So you know so the picture will look like this you will get small k and then you will have small k of X1 etc XR ok and then you will have L of L which is actually K X I do not have to use L so let me just and this part will be finite separable ok this is a so this is the picture that we need if you take the function field of any variety X then that function field what kind of an extension is it of K you can break it up into two pieces the first piece is a purely transcendental extension ok it is an extension which is gotten by simply adding as many variables as the dimension of X ok. So this part will correspond we will contribute to the transcendence degree this will give you the transcendence degree SR and this part will not have any transcendence this will because all the transcendence because X1 through XR are already a maximal set of algebraically independent elements. So this over this will be algebraic and not only algebraic it will actually be finite ok and of course of finite extension not always algebraic and not only that it will actually be a separable extension ok and the separability is a technical condition and it is very very important and the reason why it is important for example in field theory is that whenever you have finite separable extension that can always be generated by a single by adjoining a single element and that is called the so called theorem of the primitive element the theorem of the primitive element says that whenever you have a field extension which is finite and separable then the bigger extension can be gotten by adjoining a single element to the smaller field ok. So this finite separable extension actually tells you that this K of X is this small K X1 etc XR comma Y you can you can find a capital Y in K X ok such that you adjoin this capital Y to this field you get this field and that is all of capital K X alright. So you know the so this you get this capital Y because of the fact that this over this is finite separable and you are using the so called theorem of the primitive element which says that a finite separable extension can be gotten by adjoining just one element alright. And in fact the you know you see in the theorem of the primitive element says more it says that if you take a set of finite finitely many generators for this ok then that primitive element is even a linear combination of those generators with coefficients coming from the smaller fields ok. So you know so the moral of the story is that you have very nice description in terms of fields if you are working with a variety alright I mean if you are working with function field of a variety right now you see now what I want to you to understand is that well you see you take now you know you take you take X to be you take X to be a variety over K alright suppose the dimension of X is small r alright and you take K X it is field of rational functions then you can find of course you know X1 through Xr which you can think of as r algebraically independent rational functions on X ok. So capital X1 through capital Xr or r algebraically independent rational functions on X ok and then if you take the subfield generated by these r independent algebraically independent rational functions you get this field alright and then K X over that is a finite separable extension therefore by the theorem of the primitive element you can get this from this by adjoining single element Y which is yet another rational function on X ok and you know the point is that this Y is actually belonging to K X ok and K X over this intermediate field is finite so it is algebraic so in fact Y satisfies a polynomial with coefficients here ok Y satisfies a polynomial with coefficients here and the moral of the story is that Y if you clear denominators Y will satisfy a polynomial with you know coefficients in the polynomial ring in r variables you can have that polynomial to be an irreducible polynomial ok alright and therefore you know that polynomial in an affine space with so many variables it is 0 set will give a hypersurface and that hypersurface its function field will exactly be this ok. So this argument tells you that the function field of X ok is the same as the function field of a hypersurface in r plus 1 dimensional affine space where r is a dimension of X ok and we have already seen last lecture that I think last lecture or maybe couple of lectures go that if two varieties have the same function fields then they are birational. So all this argument tells you together that any variety X of dimension r is birational to a hypersurface in affine r plus 1 space ok so let me write that down see why is gotten from the primitive element theorem and in fact since k of X is finite over k of X1 etc Xr and Y is in kx we have Y algebraic over k of X1 etc Xr ok. So Y satisfies a polynomial an irreducible polynomial it satisfies an irreducible polynomial in one variable over k X1 etc Xr ok but what are the elements of k round bracket X1 etc Xr they are actually quotients of polynomials in those r variables ok when you put square brackets it is the polynomial ring in r variables but when you put round brackets you are going to its quotient field ok. So why satisfies an irreducible polynomial in one variable ok and in fact you know if you take the degree of that polynomial that degree will be the same as the degree of this finite extension ok for any finite extension generated by a single element the degree of the finite extension is the same as the degree of the minimal polynomial of that element and that the minimal polynomial of that element is the unique irreducible polynomial that that element satisfies it is the polynomial of least degree that that element satisfies ok. So so this so here I am looking at the minimal polynomial of Y ok over this extension right and so now but you know now the quotients of this if in fact you can even make that polynomial monic ok if you want but the point is that is because the leading term can always be made one by dividing ok by its coefficient but the fact is that if I but you know if you think of these as quotients of polynomials in r variables and you clear denominators what you will get is that you will get that you know you will get that Y satisfies a polynomial in r plus 1 variables ok if you clear by clearing denominators Y satisfies a polynomial and irreducible polynomial in r plus 1 variables. So you know so what you are doing is Y satisfies an irreducible polynomial in one variable call that one variable as S ok and the coefficients are all here ok and but the coefficients are therefore quotients of polynomials in the X i's alright if you clear denominators finally you know and you rewrite instead of the X i's you put the T's ok you will get a irreducible polynomial in r plus 1 variables ok which the whole tuple X 1 through X r up to Y satisfies alright and now you know see the but you know this is a polynomial it is an irreducible polynomial in n so I should not put n here it should be r sorry this should be r. So this is an irreducible polynomial in r plus 1 variables what does it is 0 what is it is 0 set it is a hyper surface in a r plus 1 ok so the 0 set of f of T 1 etcetera T r, S this is a hyper surface in a r plus 1 we have already seen this a hyper surface in affine space is simply given by the 0 set of a single irreducible polynomial a hyper surface is by definition a co-dimension 1 sub variety and irreducible closed subset of dimension 1 less than the dimension of affine space ok. So this is a hyper surface in a r ok and for this hyper surface ok what is the what is the function field the function field will precisely be K X ok this hyper surface the function field of this hyper surface how do you get it what you have to do is you have to first get its affine co-ordinate ring which is gotten by taking this polynomial ring in r plus 1 variables and you have to divide by f ok you have to divide by f because that is the ideal of the hyper surface ideal of the hyper surface is generated by f. So you have to divide by the ideal generated by f and then you will get affine co-ordinate ring of the hyper surface and then you have to take its quotient field alright but then if you divide by f you the the moment you divide by f and then invert everything you will get back K X because that is how f was gotten. So what you will get is that the the the the function field of this hyper surface is K X by definition the function field of this hyper surface is exactly K X ok because it will be the polynomial ring in these r plus 1 variables you divide by the ideal generated by f ok and then you will get a finitely you will get an integral domain you take its quotient field ok. So essentially what you are doing is your your your inverting your inverting everything except that your inverting it is like taking the quotient field of you know the polynomial ring in these r plus 1 variables but putting the additional condition that this f of this is equal to 0 but putting the condition that f of that is equal to 0 will give you the smaller field ok which is K X right by construction. But so the so the moral of the story is that now the function field of the hyper surface is same as the function field of X now we have already seen that if 2 are at least of the same function field then they are birational so this implies that this implies that X is birational to this hyper surface ok. So X is birational to this hyper surface so what this tells you is that I mean the whole purpose of my recalling all this though I did it very quickly and on your part this will demand some more reading ok is that is just to tell you that if you have a variety of dimension r then it is birational to a hyper surface in a r plus 1. So this is a fact that I need to use I am going to use the fact that given a variety of dimension r it is having an open set which is isomorphic to an open subset of a hyper surface in a r plus 1 dimensional affine space. So you know roughly what I am saying is the following thing I am saying take a variety of dimension r there is an open set where the open set looks like the 0 set even by single equation in an affine space of dimension 1 more the variety has x has dimension r ok you take an affine space of dimension 1 more which is a r plus 1 and there is a hyper surface there and an open subset of the hyper surface which looks like the open subset of your variety. So I am just saying that given a variety of dimension r there is an open set which looks like the 0 set of a single equation in affine space of dimension 1 more ok. So that is this fact and now I use that to prove this theorem ok and I do that in the following way what I say is well you know if I take so this is standard taken algebraic geometry you want to prove something about an arbitrary variety you want to prove some open condition ok something that is happening on an open set ok. So for example here I am trying to you know what I am trying to show is that I am trying to say that the condition of non-singularity is open because it is complement is you know it is equivalent to saying that the condition of singularity is closed and I want to say it is non-empty open ok. So what I have to show is that you know the set of points where x is non-singular is actually non-empty open set ok. So you know I am so I am just saying that you know if I prove that you take a hyper surface for a hyper surface if you show that the singular points is a proper closed set then I am done because if I show for any hyper surface the set of singular points is a proper closed set then I am saying that there is an open set which consists of good points. But you know but an opens but any variety has an open set which is isomorphic to an open subset of the hyper surface therefore it will have good points because under isomorphism of varieties local rings are preserved ok since local rings are preserved this smoothness condition will be preserved. So under isomorphism of varieties a smooth point will go only to a smooth point. So if you give me any variety alright then I know that I if I can I know that there is an open subset of that variety which is isomorphic to an open subset of a hyper surface and I know open subset of the hyper surface will contain smooth points if I know that then I will know that x will contain smooth points the moment x contain smooth points this becomes a proper subset. So you know this theorem is just to show that every variety contains at least one smooth point ok and I prove this theorem by reducing to the case of a hyper surface because I know every variety of dimension R is birational to a hyper surface in R plus 1 dimension of a space ok. So because of this fact assume that x is assume that the variety x is a hyper surface right. So now the proof is just probably a few lines. So assume x is a hyper surface in z of f in a R plus 1 ok what do I have to show I have to show the singular points of x is a proper subset ok. So if let us look at this let us go by contradiction the singular points of x is all of x ok. Then what it means is that doh f by doh xi ok vanishes on z of f for all i from 1 to etc up to R plus 1. So here now you know the xi's are all the affine coordinates in R plus 1 dimensional space and if the singular points is the whole of x that means these I mean these components of the gradient of f ok they all need to vanish ok. A point will be a smooth point if at least one of them does not vanish at that point but the only way that every point is not smooth ok is that all of these guys vanish alright. But then you see what does this mean if see after all these are polynomials and if they vanish on the variety then they have then you know that by the Hilbert Nullstall and Schwarz they have to be some power of I mean they have to be in the ideal generated by f ok. So you see so by the Nullstall and Schwarz you see doh f by doh xi they all belong to the ideal generated by f for every i that this is because of Nullstall and Schwarz which says that you know if a polynomial vanishes on a variety then some power of that polynomial should be contained in the ideal of the variety ok. So you know if this vanishes on z of f then some power of this is contained in the ideal of f ideal of z of f but ideal of z of f is just the ideal generated by f ok and if some power of that is contained in f then that itself is has to be contained in f because f is irreducible and therefore the ideal generated by that is prime alright. So you will get this but then what is the degree of f this is more than the degree of this will have degree at least one less right because you have taken a partial derivative. And therefore this condition will tell you that these are all identically 0 so this will imply that doh f by doh xi are identically 0 for all i ok it will tell you that all these partial derivatives are identically 0. Now you see if you are in characteristic 0 this cannot happen ok because if the when will the partial derivatives of all the variables be 0 I will be 0 for a polynomial if those variables do not appear in the polynomial at all ok if the if the if the variable appears in a polynomial ok and if you are in characteristic 0 ok then the that variable come with the coefficient alright. So when you take a partial derivative ok the coefficient will not kill it because you are in characteristic 0 and the partial derivatives cannot vanish ok so the fact that all the partial derivatives vanish so this implies that characteristic of k is not 0 ok because in characteristic 0 this cannot happen and in characteristic so if characteristic is not 0 then the characteristic of k is positive is a prime positive alright and in positive characteristic if a polynomial is in a certain variable partial derivative is 0 it means that the polynomial should be a polynomial in the pth power of that variable. So this implies that f is equal to f is polynomial in xi power p for every i this is a result from characteristic p and this will tell you that you know f is g power p because you know since you are in since your field is algebraically closed you can take pth roots of all the coefficients and you can use the fact that a plus b plus I mean if you want a plus b whole power p is a power p plus b power p ok. So if a polynomial is a polynomial in all these xi power p ok then you take the pth root of all the coefficients then you can write the polynomial itself as g power p and but this will contradict the fact that f is irreducible contradicts irreducibility of f. So this contradiction tells you that if x is a hyper surface then the singular points cannot be the whole of x ok that means there are smooth points but already we have seen as a corollary the singular points is a closed set. So that means the compliment of this closed set namely the set of smooth points is non-empty ok so the set of smooth points of a hyper surfaces irreducible non-empty and dense and since any arbitrary variety is birational to a hyper surface each singular points will also be only a proper closed set ok. So that proves that on any variety you have a open dense irreducible subset consisting of smooth points non-singular points ok and the bad points namely the singular points they only correspond to a proper closed set ok. So I will stop here.