 Let me make a couple of remarks, see what we did was we start with X an affine variety it is an irreducibly close subset of affine space okay and we took a point P of X the point given by coordinates lambda 1 to lambda n alright and we had of course the affine coordinating of X is just the affine coordinating of affine space divided by the ideal of X okay and that is just K X1 etc Xn divided by the ideal of X okay and of course if you calculate the so you know the that is the maximal ideal corresponding to the point P in the this is the maximal ideal P in the whole affine space okay the affine coordinating of the whole affine space and then so we defined we defined the essentially the gradient function the function that associates to each polynomial is gradient at the point P which is given by you know K X1 through Xn 2 I think I call the map psi and that takes so this is into Kn which takes any g to you simply take the gradient namely you take the partial derivatives of g with respect to these n variables and you evaluate at the point P. So this is dou g by dou X1 at P and so on dou g by dou Xn at P okay and what we noticed is that psi is K linear psi maps the generators Xi and it is lambda i of mp which is the subspace of the polynomial ring because everything that we are dealing with I mean all rings we are dealing with when we are worried about varieties all the rings of regular functions and affine coordinating rings they are all K algebras okay and therefore their ideals ideals in such rings will automatically become K subspaces okay and this mp is generated by these by these generators and if you take the gradient of these at the point P you will get the standard basis okay. So it means that the image so the image of mp of the subspace mp contains a basis of Kn therefore it is surjective and what we proved is that the kernel is exactly mp squared okay psi restricted to mp from mp to Kn is surjective that is on to also kernel of psi restricted to mp is exactly mp squared the square of this ideal which is also a subspace mind you of mp and so what we got was an isomorphism we got a K linear isomorphism which I think I also called what I call it I also call that as psi probably the K linear isomorphism from mp mod mp squared to Kn psi dash okay so we got this psi dash which is from mp mod mp squared to Kn alright and now you see now the fact was that the so if you calculate so if you take Ix if you take if f1 etc up to fl generate Ix you take a finite set of generators for the Ix and that is true because Ix is a it is an ideal noetherian ring the polynomial ring is noetherian and therefore every ideal is finitely generated so Ix is finitely generated when you pick a set of generators then you know if you then you calculate in the rank of the Jacobian of this set of generators which is doh fi by doh xj at the point p this turns out to be exactly the image under psi prime of the subspace given by Ix plus mp squared by mp squared okay is equal to dimension it is going to be dimension of the psi prime of Ix plus mp squared over mp squared which is subspace of mp over mp squared mind you Ix I of x is contained in mp because p is a point of x okay so so this is the dimension of this and this is well this is this is one fact then the other fact was that is that if you take mp mod mp squared and you divide by this subspace which is Ix plus mp squared mod mp squared the claim is that this is isomorphic is a k linear isomorphism of this vector space with mp mod mp squared okay where mp is the unique is the unique maximal ideal is unique maximal ideal in o xp which is of course given by it is given by the localization of affine coordinate ring of x at the ideal at the maximal ideal that corresponds to p in x which is given by mp sub x okay and where of course m that is mp sub x is just you know it is mp mod Ix it is just the it is the maximal ideal in the affine coordinate ring of x corresponding to the point p okay just like points of just like points of affine space correspond to maximal ideals in this ring which is affine coordinate ring of the affine space which is version of the Null-Strahlen-Satz the same way the points of the points of x are in bijective correspondence with the maximal ideals of Ax okay and that and the corresponding maximal of ideal of Ax is just gotten by taking the maximal ideal of that point in this polynomial ring and going modulo Ix okay and that is why maximal ideal of the point p in x is just the maximal ideal of the point p in affine space modulo the ideal of x okay and because of this K linear isomorphism and because of this what you get is that if you so the dimension of this minus the dimension of this is equal to the dimension of this but the dimension of this is the same as dimension of its image and psi prime because psi prime is an isomorphism and restricted to subspace it is injective right and therefore what you get is so the upshot of these two things is that the rank of the Jacobian of dou fi by dou xj at the point p this is equal to dimension of this which is equal to well is equal to dimension of this dimension of this minus dimension of mp mod mp squared okay so this will be equal to dimension of capital mp mod mp squared minus dimension of small mp mod mp squared of course dimensions are all over as k vector spaces okay and but of course mp mod mp capital mp mod capital mp squared is isomorphic as vector space to k and so its dimension is n so you end up with getting n here minus dimension of mp mod mp squared and well p is non-singular p is non-singular p is a non-singular point of x if and only if the definition is well if you calculate the rank of this Jacobian the rank of the Jacobian of a set of generators for the ideal of x at the point p should give you the co-dimension of x so this is true if and only if this is co-dimension of x so you get co-dimension of x in an so co-dimension of x in an is equal to n minus dimension over k of mp mod mp squared and that leads but co-dimension of x in an is just dimension of an minus dimension of x so that condition translates to just saying that the dimension of x is equal to the dimension of mp mod mp squared and mind you that dimension of x is always given also by the dimension of the local ring of x at any of its points okay therefore and of course here when I say dimension of x it is topological dimension when I say dimension of the ring here it is the cruel dimension alright this is the topological dimension this is the cruel dimension this is dimension of vector space okay. So the moral of the story is that is that this the point p of x is a smooth point a non-singular point if and only if the dimension of m mod m squared m mod m squared at that point okay where m is the maximal ideal unique maximal ideal of the local ring at that point is equal to the dimension of x which is the same as dimension of the local ring and of course this tells us that really you know finally you see this n which is dimension of fine space in which x was embedded that is finally gone out of the final statement okay so if I had I could have this n did not come in the final picture okay I could have this instead of this n it could have been some other m okay but that would not have affected this final statement alright. So you see one it really did not depend on the embedding of x in an affine space and you also see that after you fixed one embedding of x in affine space as an invisible closed subset it also did not depend on the number of generators this l also could have been arbitrary I could have taken any other set of generators but still I will get the same conclusion so this fact tells you that the definition of non-singularity which is given by this by choosing an embedding in affine space and then choosing a set of generators and calculating the rank of the Jacobian at that point for those generators that though it seems to depend on the embedding in affine space and on the choice of generators for the ideal actually does not depend on those things the definition is therefore in this the definition that the point p is smooth point if and only if this is this rank turns out to be the co-dimension of x okay that is independent of all these choices and finally the condition that the point p is a smooth point or a non-singular point is very nice condition it does not depend on any of these things it is very intrinsic it is the condition is just that the dimension of mp mod mp squared over k as a vector space over k should be the same as a topological dimension of x which is the same as the dimension of the local ring at that point okay. So but perhaps so this isomorphism is something that needs to be this needs to be checked okay this isomorphism needs to be checked but it is kind of pretty easy because you know so as a remark let me let me tell you that as a remark yeah you see you have ix plus mp squared sitting inside this is sitting inside mp as a subspace alright and you well on the other hand you have you know if you take if you take if you take ix ix it is inside mp and it also sits inside this okay so you know if you take the quotient that is you go mod mp mod ix what you will get is the short exact sequence okay namely yes this modulo this is isomorphic this okay so you will get mp mod ix but mp mod ix is actually mp sub x okay and well if you take the quotient here what you will get is ix plus mp squared by ix is just simply going to be mp mod ix the whole squared you will get. So here you will get mp mod ix the whole squared okay which is just mp sub x squared this is what you will get alright and well you see so if you calculate so this is again so when I divide by ix that is the same as dividing mp squared by ix but I claim that there is dividing the same that is the same as dividing mp by ix and then taking squared okay this is something that you can easily check alright and you see you know if you take this inside this and now take the quotient here I will get here I will get mp mod ix plus mp ix plus mp squared okay and here you know here what I am going to get here if I again do this I am simply going to get mp mxp mod mxp squared okay I will get this and the fact is that this is this is an isomorphism alright this will be an isomorphism you can check that alright and further okay so this will be an isomorphism and mind you what I have written on left the left side is exactly mp mod mp squared by x plus mp squared mod mp squared is also mp mod ix plus mp the whole squared alright so and I want to show that that is so statement is that this is linearly isomorphic to this alright now what I want to tell you is once you once you check once you check this isomorphism of course is diagram commutes okay and this is identity map alright if you want you know I can even put 0 here because ix mod ix is 0 right and in fact now what you must understand is that you know you have now you have the short exact sequence 0 I mean you have mxp squared sitting inside mxp and then you have the quotient mpx mod m sub xp squared okay I have this the short exact sequence alright now what you do is you up you just tensor this with the local ring ax localized at mxp over ax this is the short exact sequence of ax modules okay because these are all ideals these two this is an ideal in ax okay this is the maximal ideal of the point p in ax and this is square the square of the ideal and the quotient the quotient will not be an ideal but it will be a module because ideals of a ring are just sub modules okay so this quotient will not be an ideal but it will be a module it will be an ax module so this is the short exact sequence of ax modules and then if you tensor it with ax over tensor it over ax with ax localized at mxp okay you know that if you have seen in a first course in commutative algebra you would have learned that tensoring with the local ring tensoring with the localization is the same as tensoring with localization is a tensoring a sequence of modules with the localization of the ring is the same as applying the localization to the sequence of modules okay in other words localization is an exact function okay therefore this sequence when you tensor it with ax with this local ring okay then I will again get an exact sequence but after you tensor it if you tensor the maximal ideal with this local ring what you will get is you will get the maximal ideal of that local ring okay so when you tensor mx mpx with this what you will get is the maximal ideal of this local ring which is small mp so after tensoring what you will get is this exactly okay and so you know yeah so the point I want to make is that these two have the same dimension okay this has the same dimension as this alright but then whatever its dimension is okay see the same dimension you will get if you tensor it over this because the residue field is still k the residue field of this local ring okay namely this local ring if you go model of that unique maximal ideal you will still get the same residue field k therefore if you whatever this dimension is as a k vector space you will get the same dimension for this after tensoring over this local ring for the simple reason that residue field of that local ring is the same k okay therefore you will still get the dimension of this over k is the same as dimension of this over k but then these two are isomorphic therefore you get the dimension of this over k the same as the dimension of this over k and that is what we want to show ok, so that is something that you have to check I mean this is an outline of what you have to check alright. So this is one thing then the other thing I want to say is in connection with this proof ok several things in connection with this proof, so the first thing I want to say is let sing x be the set of all points of x such that p x is singular at p look at the set of singular points of a variety ok, so x is a variety and look at all those points of x which are not non-singular points which are not non-singular are called singular ok look at those points. Now the fact is that the fact I want to make is that the set of singular points is a closed set and it is a proper closed set, so here is the theorem then so the first corollary is that the set of singular points is a closed subset of x ok. So here from now on x is any variety ok need not be affine x is any variety mind you for any variety a point is called to called a non-singular point if you know you take that point as a point of an affine open sub variety and there it should be non-singular ok because whether a point is smooth or not is a kind of condition that is checked only in a neighbourhood of the point ok it is something that you should check in a local neighbourhood of the point locally at that point you have to check it ok. So if you have any point of any variety how do you check whether it is smooth whether it is non-singular what you are supposed to do is you take that point that point will certainly be contained in an affine open subset because you know any variety is covered by finitely many affine open subsets namely it is covered by finitely many open subsets each of which is isomorphic to an affine variety ok. So if you want to check a point is smooth or not well you can you have two ways now if you want to check it intrinsically you calculate dimension m mod m squared and check whether it is you get the same thing as a dimension of x if you get the same thing as a dimension of x then it is a smooth point but otherwise commutative algebra a little bit more commutative algebra will tell you that you might end up getting more dimension because dimension of m mod m squared gives you the dimension of the tangent space at that point ok and you could in general at a bad point you could have more tangent vectors dimension of the tangent space could be more than the dimension of the object of the variety. So this could be bigger than this if this is strictly bigger than this then it is a bad point it is a singular point if this is equal to this then it is smooth point. So this is one way of checking the other way of checking is of course you take an open as you can you take an open affine sub variety ok and you embed it in affine space then you take a set of generators for the ideal calculate the rank of the Jacobian and check whether you get the co-dimension namely whether you get the dimension of the ambient affine space minus the dimension of the variety I mean the open affine subset that you have embedded ok that is another way of checking but you know which one is good and which one is not good is depends on your situation if you want an abstract theorem if you want to prove an abstract theorem about points some non-singular points then it is better to use this on the other hand if you have a specific situation where you know the equations of the variety ok then it is that means you know a set of generators for the ideal ok then you can use this ok. So what you have to use is depends on the situation alright now but anyway what I want to say is a set of points which are which are bad what kind of a set do they form they form a close set alright and why is that a corollary of this it is a corollary of this is a corollary of this because you see what you are doing is when I define this map psi ok you know I could have you can think of seeing all these things I had fixed this point p I had fixed this point p in x ok but you think of the point p as a variable point think of the point p as a variable point if you think of the point p as a variable point ok then you know what are going to be the points p which are singular they are all those points such that when you calculate the rank of this Jacobian you see the rank of this the rank of this matrix you know this should not be equal to the co-dimension alright. So it will be it will be lesser than that ok so as I told you you see at a bad point this is going to give you the dimension of the tangent space ok at a bad point you will have more tangent vector so the tangent space this is this is the dimension of the tangent space this is going to go up if this goes up this is going to go down but this is exactly the but this is exactly the rank of the Jacobian matrix at that point. So what are the bad points p the bad points p are those points for which the rank goes down ok and what are the points p for which the rank goes down therefore they are the defined by the they are those points which satisfy the polynomial equations given by taking all the you know minors ok you take you take the minus minor determinants of this matrix ok but you take not the biggest but you know or you take the biggest you take the largest possible minors and they all should vanish ok. If when does the rank of a matrix when is a matrix of full rank when the largest possible when at least one of the largest possible minors minor determinants does not vanish ok. So when is it going to be of lesser rank it will be of lesser rank if you take all possible maximal minor determinants they should all vanish so the condition that the rank falls is a vanishing condition it is given by vanishing of some minor determinants and therefore you know the points p which are singular they correspond to the points which they correspond to the points of x at which if you take the maximal minors determinants of this which are again by the way polynomials ok those polynomials given by the maximal minor determinants they should also vanish ok and that is of course a closed set ok. So it is so I because of that observation what happens is that if x is an affine variety in an then you get that the singular points of x is a closed set ok and but the topologically a set being closed is something that you can check on a cover ok and therefore the same is true for any variety because any variety admits a cover by open sets which are affine varieties ok. So the moral of the story is that the set of singular points is a closed set ok. So what is an ideal for this I mean if x is affine an ideal for this is just given by you take the ideal of x ok and you also take you add in also these polynomials ok without the p you take all the maximal minor determinants they will also be some more polynomials finitely many polynomials you add them also to the ideal of to take them together with the ideal of x and the 0 set of this is going to be exactly the set of singular points of x and that is therefore the 0 set of it is an algebraic set therefore it is a closed subset ok therefore this is always a closed subset alright. Now what is more serious what is more serious is that and this is very very important what is more serious is that the bad situation that all points are singular does not occur in other words this closed set is a smaller closed set it is not the whole space ok what it means is the compliment is a non-empty open set and you know non-empty open set is irreducible and dense ok so which means that there is a huge open subset of points which are all good points they are all non-singular. The set of bad points is only a smaller closed set ok it is a small closed set so here is a more serious theorem so here is a theorem the theorem is that the singular points of x is a proper closed set subset ok this is an important theorem and what it implies is that I mean what it says you know from the viewpoint of the if you think of it with respect to you know usual analysis where we study where we think of you know smooth geometric objects as being given by manifolds. So what it says is well every variety has a huge open set which looks like a manifold ok and it is the compliment of that open set which is the boundary of that open set that is the one which contains the non-manifold points ok the manifold points are the smooth points the non-singular points and that is a huge open set ok and the non-manifold points are the non-smooth points they are the singular points those which are not non-singular and that forms a proper closed set ok. So you if you want to think of it if you want to think in terms of analysis you must think of a variety as a nice manifold with a boundary consisting of points which are not manifold points and of course the singular points may be empty also ok we have already seen last time that you know if you take any affine space then that is smooth which means it is non-singular every point of affine space is a smooth point it is a non-singular point that is because that is basically because of this isomorphism ok that is basically given by this isomorphism. So this isomorphism will tell you that the dimension of if you take the local ring of the point p in the affine space then m mod m squared is just kn ok and therefore the dimension of the local ring is equal to n which is equal to dimension of affine space. So this calculation itself tells you after you accept this theorem at calculation that first calculation itself tells you that every point of affine space is smooth. So affine space if you take x equal to affine space then this is empty the singular point of x is empty. So the singular points can be empty ok in which case we say the variety is a smooth variety or non-singular variety if the singular points are not empty well still life is not so bad they where the set of singular points is only a small set ok you have a huge set of good points right. So now to prove this theorem one needs to go into a small detour of which involves a little bit of field theory and you know which come up because of concepts of issues of birationality.