 Okay, so we were finishing the chapter about dimension. So we were talking about dimension and transcendence degree. So I don't really call the definition of transcendence degree, which is relatively clear. So we had, there was this alternative definition definition of the transcendence degree that, so we wanted to prove the following theorem. So if X is a variety, then the dimension of X is equal to the transcendence degree, I don't even know how I denoted it, of the function field, the field of rational functions over K. So the number of algebraically independent elements here. And we had known that these algebraically independent elements can be taken as a subset of the classes of the coordinates, for instance, if you are an AN. And we, so this we are about, we proved this model of the statement. So we, so prove this, assuming the following. The theorem that every variety is vibrational to a hypersurface, okay? And this is somewhat surprising result, because somehow one would think that varieties are much more general than just hypersurface and they also are. But there's, there are open subsets, which are isomorphic to open subsets of hypersurfaces always, okay? And now we want to prove this theorem. So first, so for simplicity, I will prove the theorem under the assumption that the characteristic of K is equal to zero. The proof that I will give can be modified to work for any characteristic, but it's a bit more complicated. And so I will just, so for simplicity assume. So I mean, this I use because I want to use the theorem of the primitive element. Now, one can prove that in the situation where I am, the theorem of the primitive element can be applied. You need a whatever, a separable field extension and the field extension we're looking at will be separable, but one would have to prove that and I don't want to. And in characteristic zero, we know that all finite extensions are separable. So that would be fine. So, so we have the theorem of the primitive element. So let K be a field of characteristic zero and let L over K be a finite field extension. Then there exists a primitive element. So then there exists an element B in L such that L can be obtained by adding, joining this single element B to K. And I expect you had this in the ultra course. So it's not, so now we want to use this to prove our theorem, this one, whatever. So we have to somehow make this hypersurface, no? We have to find the equation of a hypersurface. And we do this by using what we know about this transcendence basis. So we say, so we have Kx, the function field of X. And so let say X1 to X are a transcendence basis of Kx over K. There is some transcendence basis to choose one. So then we know by the theorem that I cited the other time that if I take Kx over Kx1 to X are is a finite algebraic extension. Now we had this theorem. If I have a finite degenerated field extension of K and we divide by K a joint transcendence basis then the quotient, then this is a finite algebraic extension. So now we apply the theorem of the primitive element. So there exists one element which I have to join to this field so that it becomes Kx. So there exists an element Y in Kx such that Kx, large Kx is equal to the small field Kx1 to X are and I join also Y. And Y is algebraic, Kx1 to X are. So if Y is algebraic over Kx1 to X are that, I mean there's a polynomial of which it is a zero. I can basically take the minimal polynomial of this. So thus, so there exists an irreducible polynomial say F which with coefficients in this smaller field Kx1 to X are with variables in a polynomial in X with coefficients in this field such that if I take F of my element Y, this will be equal to zero. It's irreducible so it's not zero anyway. Okay, now we want to basically use this F to define the hyper surface. So this is supposed to be finally the equation of a variety which is by rational to X. But in order to do this, you know, we cannot have it like this with coefficients in the field. You want to clear the denominators. No, this is, so we know that Kx1 to X are is isomorphic to the field of rational functions. Kx1 to X are is isomorphic. Or I could also say is the field of rational functions in variables X1 to X are. So I can write F, see some whatever was some finite sum of some else GL from X1 to X1 to X are divided by HL times X to the L where the GI or the GL and the HL are some polynomials with coefficients in the small k in X1 to X are. No, because we know the elements of this quotient field of this field of rational functions are just quotients of polynomials. Okay, so now we can clear the denominators. So multiply F by the product of the HL and divide afterwards by the greatest common divisor of the new coefficients. So after multiplying, I get some new coefficients of X to the L, I take the greatest common divisor. So then I, in this way, I obtain a new polynomial. So in this way, you get F, which is some polynomial in X1 to X are times the original F, which is equal to whatever G we'll call it H tilde times F, where H tilde is a polynomial in K X1 to X are and also F is a polynomial K X1 to X are and X. We just have cleared the denominators and I have, and in fact, F is a primitive polynomial. So let's see. I write it like this, I can, this is the same thing by definition of the polynomial ring, but I know the polynomial ring and R plus one variables is the same as the polynomial ring and R variables as polynomials in X with coefficients, polynomials in one variable less. But just, if I viewed now as a polynomial in X, it's a primitive polynomial because I've just said that I wanted the coefficients to be relatively prime or divided by the greatest common divisor. And we know that if I take the original F, so the original F is irreducible in this field therefore also F itself, the small F is irreducible in K as element in K X1 to X are comma X because it's after all just the big F multiplied by a unit in this, in this ring. No, it's just the constant from the viewpoint of this. And so by the Gauss lemma, F is irreducible as a polynomial in this ring. Okay, so we have this irreducible polynomial, I don't know whether. So, and now I want to say, as I said, I want to take this polynomial which is just a polynomial in R plus one variables. I want to take this as the polynomial which determines a hypersurface in AR plus one. So let Y be the zero set of F in AR plus one. This is a hypersurface, it's an irreducible hypersurface that's the zero set of an irreducible polynomial. And if I look at the function field of Y, what is it? It's the quotient field of this ring. The field of rational functions. So this is, so we know if we take the, how is it called? The coordinate ring of Y, this will be K X one to XR X divided by the ideal generated by F. And so the function field is the quotient field of that. Yeah, indeed. And now, yeah, here I'm maybe a little bit fast, but so you could, I claim that this is the same as the function field of K X one to XR and then like this. So you just have to see that. So if you look the way this equation works, you get, you have that this is somehow a finite extension and therefore if you take the quotient field, it doesn't, I mean, you just get this. Anyway, you can try to think about it's quite easy to see. And this is just equal to what we had here. I have unfortunately wiped it out, but we had seen that this was precisely, if this is nothing else as Q S K X one to XR, a joint X divided by F. And this is precisely what we had gotten as K X. K X. Well, if you remember, we had that K X was isomorphic to K X one to XR, a joint some element Y, which was the zero of a polynomial. So if I were here, we have the F, but I can replace over this thing, the difference between the small F and the big F is just a unit. So I can also write F here. And then it's precisely how it was given. And so we see that K X is equal to K Y, so they are birational. So it's a somewhat unexplicit proof because we don't actually see any open set, which is isomorphic to any other open set. We just prove, abstractly, that the function fields are isomorphic. And we also have not really explicitly done anything at all. We take this transcendence basis, which exists by some general existence result. So somehow how you find this hypersurface, which is birational to our given variety is completely mysterious from the proof. But anyway, we have proven that is one. Okay. It's, yeah, yeah, X is birational to Y. That's, K X is equal to K Y. K Y is, and therefore X is birational to Y, sorry. That's, you know that two varieties are birational, if and only if the function fields are isomorphic. Okay, so this proves this result. We will also use it again later in this particular theorem. So either at the end of today's lecture, but that's slightly unlikely, more likely in the next lecture. So with this, we actually finish the story of dimension. So I have this other definition of dimension. So one was with this change of irreducible closed subsets. The other one was with a transcendence degree. And we find that they are both the same. And so as I said, the transcendence degree, you really can think of it as somehow in some sense, at least if you view this XI somehow as coordinates. So the transcendence basis, you can see how many coordinates you have in some sense locally on your variety, so that it has something to do with that. And the other definition is this other thing that each time you cut down by some creation, it gets, the dimension gets smaller by one, and then this way you see the dimension. So there's two quite different views on dimension, which are, which turn out to give the same result. Okay. And so now we want to talk about tangent space, singular and non-singular points. So, you know, if you have a manifold, then the manifold has a tangent space at every point. An algebraic, and this tangent space has always the same dimension as the manifold. If you have an algebraic variety, it will have a tangent space, and this tangent space will not always have the same dimension. It depends on the point where you are, which dimension it has. But there will be some open subset where the dimension will always be the same. And where the dimension is actually, of the tangent space is actually equal to the dimension of the manifold or of the variety. And these are the non-singular points, so where the thing looks smooth. And where the dimension is different, of the tangent space is different from the dimension of the variety, these will be the singular points. And it somehow also coincides, you know, when you see, if when you make a picture and you see something which looks singular, you will see that this actually is a singular point according to this definition. Okay, so let's see. I will have to do this a bit slowly. First, I do it for hypersurfaces in AN, and then I do it for a fine varieties, and then I do it in general. So we start with the case of hypersurfaces in fine space. So singular and so case of hypersurfaces in AN. So first I talk about the differential of a polynomial, and I will use that to define what I want. So, or rather partial derivatives. So let, so if f is a polynomial in x1, xn, we can consider, so we have, we can consider the partial derivatives, df by dxi, this would be also some polynomial. And you know, these are just, you compute them formally, you know, it's not an analysis thing, you can do this over any field. So if x is a zero set of f in AN, it's a hypersurface, and we assume that the ideal of x is equal to a deal generated by f. So here I'm not assuming that f is irreducible. So if f is irreducible, for instance, the ideal of x is equal to that, but if it could be that x is reducible, but the fact that the ideal of x equal to ideal generated by f means that if you write it as a product of factors, no, none of the factors has a power in it. So, and so anyway, I assume, so x is allowed to be reducible, but it must be that the ideal of x is just f. So a point P and x is called a singular point, where if all partial derivatives of f vanish at this point, so if and on f, the f by dx i at P is equal to zero for all i equals one to n. Okay, so if I have a point which lies in x, so f is zero there, and all partial derivatives are also zero, it's called a singular point. Otherwise, it's called a non-singular point. And maybe one says that the set of non-singular points at x-rag is defined to be the set of all P in x. P is non-singular. So I want to stress one point. A singular point of x must be point of x, okay? It's not, so it's not enough, so by definition, a singular point of x is a point of x. So the condition is not just that all partial derivatives are zero, but also f must be zero. So when I, in the homework and final exam over the years, there was always somebody who got this wrong. So therefore, I stated it again explicitly. So a singular point of a variety is in particular a point of the variety. Okay, so example, so I just want to make some examples. So first, so this condition that the idea of x is f is some more important for this. Yeah, you can also say regular point, yeah. So that's why it's written x-rag here, yeah. But I, you sometimes also say smooth point. One can maybe say that x is called smooth if, and only if x is equal to the set of regular points. So all the points of x are regular. So first, we really want that we really need that the ideal of x is equal to f, not just that x is the zero set of f. I mean, for instance, if I take the zero set of y squared in a two, so the variables x and y, this is just zero set of y, it's just a line, and this is non-singular. It's smooth, so it's called non-singular or smooth. So it's non-singular because you could take the derivative d by dy of this is one, and this will not be zero anywhere. But on the other hand, if you do it for y squared, at every point in the zero set of y, so d by the x of y squared is always zero, and d by dy at every point where y is equal to zero will also be zero. So you would, from this definition, would get that every point of this is a singular point, but that's not true. Actually, every point is non-singular, okay? We can look at some simple examples. If we take x equal to c of x minus y, I do it this way, y minus x squared, then you can easily see that this is a conic, so a parabola. If you make the real picture, it somehow looks like that, so it looks kind of smooth, and indeed you find that this is non-singular. So all the points of it, if you just take d by, d by dy, this will always be one, and well, this is two x, so unless x and y are both zero, anyway, you will see that this will work. Okay, this is a non-singular right, what? So let's just look at, for every point you have to check. So let's check at the point, so if I take d by dy, yeah, and okay, and so it's non-zero. No, but we don't need to show for x. We have to show that not all are zero. So it's a single point if all the partial derivatives are zero. And so here the first one is always non-zero, so it's completely different. Okay, so we can also look at x equal to the zero set of y squared minus x squared times x plus one. So this is, we know is a nodal cubic, so you somehow have a picture, so I always have the same examples, no? Because there are not so many examples which one can actually deal with explicitly. So it looks somehow like that, and you can see there is one point which looks a little bit different, which is the point zero zero. Now when x and y are both zero, something happens. And this, if you look at the real picture, you see really that you have two branches which intersect like that. So you would expect that this is a singular point, and you can easily check that zero zero is a singular point. Now if you just take d by dy, this will be two y, and so it's zero one, y is equal to zero, and d by dx, y you can work this out, it's also obviously zero if y is equal, if x is equal to zero. And in the same way for a cuspular cubic, so you can check that all the other points are non-singular if you want. Yeah, it's the only singular point. And this would be the cuspular cubic, you know that the picture is somehow like this, and we have again a special point here, zero zero, and we find again that zero zero is the only singular point. In fact you can obviously see that this is a singular point, you have the one derivatives two y, and the other one is three. What do you mean? No, this has one singular point. So there's only one, so if you take away the point, it's non-singular, but it is singular. We call it something singular if it has at least a singular point. What, no, maybe please ask a question. Yeah, no, it's called non-singular if all the points are non-singular. No, in fact, we will see in a moment that for any hypersurface, most points are non-singular. It's always an open dense subset where it's non-singular. So therefore, you don't call something non-singular because it's almost everywhere non-singular. You call it non-singular if it is everywhere non-singular. No, it's singular if it is somewhere singular. So otherwise, so it's called smooth if and only if x is equal to x-ray. So if all the points are non-singular, it's called singular if there's at least one point where it's singular. Yeah, yeah, but anyway, I didn't want to actually introduce this, but if you ask whether this is non-singular, no, it's not non-singular. It's non-singular only if it's everywhere non-singular, otherwise it would be. And that, yeah, this is non-singular. It's everywhere non-singular. But these two are singular. Okay, so let's, what? I don't know, that was maybe calf. It sounded like something. Okay, so now let me see. Now I want to prove that if I have an irreducible hypersurface, then it has an open dense subset which is non-singular. So then the set of non-singular points is open and dense. So proposition, so let x in an irreducible hypersurface, so zero set of an irreducible polynomial, then x-ray is open and dense, which means the same as non-empty, but I write it to intense in x. So you could say, okay, so we have the, want to see why this is true. So again, I'm a little bit cheating. I will assume for simplicity that the characteristic of our field is zeroed. Okay, so one can again look a little bit more carefully and see that the proof works with minor modifications if k is, if the characteristic is not zero, but I don't want to take the trouble. So I just will assume the characteristic of k is zero. So we take some polynomial f in k, x1, xn, of which is, so the polynomial of which it is zero set. So irreducible such that x is the zero set of f. So then we know that the ideal of x is the ideal generated by f, no? It's a prime ideal. So now we can say, so maybe I call xing to be the set of singular points of x. And now if you remember the definition, I claim that the singular points of x, what are the singular points of x? These are the points of x, where all the partial derivatives of f vanish. The points of f of x are the zero set of f. And then we want also all the partial derivatives to vanish, so we take also, so it's this zero set. These are the singular points. And so obviously this is a closed subset of x. So this is contained in x by definition and it's closed in x, because it's a zero set of some more polynomials. So what we have to show is that the compliment is dense, which is the same as saying that the compliment is non-empty because the compliment is an open subset of an irreducible of a variety. If I have an open subset of an irreducible variety, it is dense, as long as it's not empty. Obviously when it's empty, it's not dense. Okay, you send this a bit too long. Yeah, anyway, so we have to show. So the only thing we are left to show is that x is not equal to the single locals of x. Well, assume it is. So then we want to come to a contradiction. So you know that then I claim, so the ideal of all polynomials that vanish on x is the ideal generated by f. So therefore I claim that, you know, if x is equal to xing, it follows that df by dx i, if I take the zero set of that, this contains x, no? Because if I take, the zero set should not become smaller by intersecting with it for all i, all i. So in other words, we have that df by dx i is an element in the ideal generated by f, for all i. But now if you think of it, this is more or less impossible because df by dx i has certainly a smaller degree than f. So how is it possible that something of a smaller degree is contained in ideal generated by something? There's only one possibility, namely that this thing is zero. It follows that df by dx i equal to zero for all i. But so now we use the characteristic zero. If the characteristic is non-zero, then there are some ways how this could actually be true, which one then has to exclude. But if the characteristic is zero, then this implies that f is constant. And this contradicts, I mean, f is supposed to be hyper-surface, so it defines a hyper-surface, so it must be non-constant polynomial. So this was this? Yeah, I expect it can also be generated that, but I now just, I mean, certainly we, yeah, I think with a bit more care, one can also do it if it's irreusable, but okay, maybe that you can think of as an exercise, but I think it shouldn't be very difficult to generalize. I just was a bit lazy. Okay, so this was, for the moment, everything I wanted to say about hyper-surface is, I, now I want to talk about tangent space for defined varieties. So the tangent space and singularity and non-singularity. Yeah, I haven't talked about tangent space, we'll see in a moment that what the tangent space was in this case. So one actually finds that according to the definition we are going to give, for a singular point, the tangent space will be the whole of A n, okay, for hyper-surface like that. But let's do it, okay, so let me see. So, I mean, we are not going to get so far, but it is so that there's a big difference in difficulty between treating varieties which are non-singular and varieties which are singular, okay? So many, so one of the reasons is that over the complex numbers, a non-singular variety will be a complex manifold. So you can also, and in fact, so one can then also use differential geometry to study them for once. But at any rate, many other methods of also algebraic geometric methods depend, I mean, or become at least much simpler if the variety is non-singular. So therefore one wants very much to, you know, often wants to restrict the tension to non-singular varieties. So I, in my work, I very seldom encounter singular varieties because I don't, because I like to also use topological and differential geometric tools. And so therefore it is important to, you know, to have to be able to talk about this. So also, for instance, if one studies homology and homology of these varieties, it's much easier to describe if they are non-singular. So it's related to, and so we, if one goes on in the study, one would most of the time want to study non-singular varieties, develop special tools for that. And we will actually, at the end of this lecture, we'll have maybe one or two lectures or one and a half about how it looks like for non-singular curves. That you somehow have a different way of describing the local ring of a non-singular curve. And this can be used to study, you know, many things about non-singular curves, which would be the last chapter of these notes, which we are not going to talk about, I think. But if we did one finds that, for instance, morphisms of non-singular curves have a much nicer description because of the fact that they are non-singular, and therefore the local rings of the non-singular curves have very special properties. So there's really some, I mean, it is really quite useful. And obviously if one is fearless, later when one will also deal with singular things, there is any way one cannot always avoid singularities, one also has to study them, and then it's maybe important in order to understand things, then to understand what the singularities precisely are. And so it's really, I mean, it's really quite important for many things. I mean, algebraic geometry is not only about classification. I mean, to say when two things are isomorphic, that's a very coarse way of looking at it. Obviously it's also true if you have two varieties, and you have an isomorphism between them, then it must map non-singular points to non-singular points and singular points to singular points. So that also you can find something about whether things can be isomorphic from looking at singularities and tangents based. And this is, as you can see until now, as compared to what we did about dimension, somehow talking about tangent space and singular and non-singular is in some sense more elementary. You know, you have actually concrete computations. The only thing why it comes afterwards is that the general definition of non-singular means that the dimension of the tangent space is equal to the dimension of the variety. So obviously you can only make sense of that if you know what dimension is. But by itself, dimension is in some sense more complicated to handle concept than this. Okay, so now we want to study the tangent space and often a fine algebraic set. So I also do it for a fine algebraic set. I don't necessarily assume that things are irreducible because I don't need it. So let x in an in a fine algebraic set and I want to define the tangent space. So first I define the, so definition. First I define the differential of a polynomial. So let f in k x1 xn be some polynomial and I take p a point, do I need to call it? Yeah, maybe, no. p a point in an, then I can look at the differential at p of f. So the differential of f at p is equal to the sum i equals one to n. I take the partial derivative at p and then I multiply this. So this is just a number, no? I take the partial derivative, which is a polynomial I evaluate at p, this is a number and I multiply this by xi. So this is a polynomial of degree one. A linear polynomial. And therefore it's, I can also view it as a linear map. And now the tangent space to x at p. So if p is a point in x, again the tangent space only defined for points in x. You don't define the tangent space to x at some point which does not lie in x. And is a tpx, which is defined to be the zero set of all the differentials for f in ex. So you just take, so you take all the polynomials in the ideal of x and you take the differential at p. So you see therefore the tpx is actually a vector space. So it's a zero set of linear polynomials in an. So it's a defined subspace of an, which goes through the origin. So it's just a vector space. And now, so p in x is called a non-singular point. If the dimension of this tangent space, you can either take the dimension as a vector space or also the dimension of sub-variety. This is the same because it's an defined space. The dimension of tpx is equal to the dimension of x at p. So what do I mean by that? I think I haven't introduced this. So the dimension of x of p is equal to the maximum of the dimensions of components, of irreducible components of x passing through p. And otherwise p is called a singular point. So in particular if x is irreducible, then x is called a non-singular point, if and only if the dimension of the tangent space at x is equal to the dimension of x. And I should also say this actually is a bit misleading, namely if I have a non-singular point, then this will always lie only on one component. So one can't show, but we cannot in the moment, and I mean it's slightly more, it takes a bit of algebra which we haven't developed. So can show that if p in x is a non-singular point, then p lies only on one component of x. So if the point p lies on the intersection of two components, it's always a singular point. But it's not so... For hypersurfaces you... it would be an exercise to prove that, but in general it takes a bit of algebra. Okay. And again so we have x, equal to the set of p non-singular points of x, and by x-ing is x, set of singular points. And x is called smooth if x is equal to x as before. So there's... Here it's a little bit annoying that we have... So we have to find tpx to be zero set of the differences of all elements in the ideal of x. So that's very many polynomials. So that might not be very easy to check. But so therefore we want to maybe have a way to use less. Just take a set of generators of the ideal. And this is a simple remark. So if the ideal of x is equal to the ideal generated by some polynomial f1 to fr, then the tangent space of x is equal to zero set of dpf1. Okay. So we need to only take the generators. And this is a rather straightforward exercise. So maybe as we have the definition here, I can leave it there for a moment. So by definition obviously the tangent space of x is contained in this common zero set. Because by the definition here, the tangent set is the common zero set of all elements in the ideal of x. And these are certain elements in the ideal of x. So you have to prove the converse inclusion. So if h is an element in the ideal of x, we can write it h equal to some hi fi. We can write it as a linear combination of the fi. Well, then if you look at the definition here of the differential, you can see from the definition that it will fulfill the product rule. I mean the Leibniz rule. Just as a derivative always does, and this is done by derivatives, you find that you have a Leibniz rule, dp of f times g will be equal to dpf times g of t plus f of p times dpg. I mean it's an exercise for a very simple exercise to do this computation. So we apply this here. So if we have dph, this will be equal to some i equals 1 to r, dphi times fi of p plus hi of p times dpfi. Now the fi were supposed to be in the ideal of x and p is a point in x. So p is in x and fi is in the ideal of x. So fi of p is equal to 0. So this term is just not there. And then so you see that this is a combination of the dpfi. So that means that if we instead of taking just the dpfi, we take the dp of all these, this generates the same ideal and so the 0 set is also the same. OK, do I need this? Yeah. OK, so I want to also, OK, so this is the definition. I also want to talk about the, ah, so maybe I should, as an example, we have to see that our previous definition of non-singularity somehow coincides with this one, no? So if x is the 0 set of f in a n plus 1 and the ideal of f is equal to, no, the ideal of x is equal to the ideal generated by f, then so we have, there's only one f. So the tangent space of x will be equal to the 0 set of dpf, no? And dpf is equal to the sum i equals 1 to n df by dxi at p times xi. So basically, so if, so this is a linear polynomial, so the 0 set has co-dimension 1, except of course, if this polynomial is 0. So we have either, we have that df by dxi at p is equal to 0 for all i equals 1 to n, then it follows that the tangent space of x at p is equal to the whole of a n. And this means that the dimension of it is not equal, the dimension of the hypersurface is by definition n minus, is by our theorem n minus 1. So, and thus p is a singular point. Or otherwise, we have that at least one of these derivative is non-zero, then the dimension of the, then we have indeed the hypersurface, and so the, I mean the tangent space is a hypersurface, and so it also has dimension n minus 1. And so we see that the definitions do indeed coincide, and also what tangent space is in this case. So now I want to say just very briefly, if you have a morphism of a fine varieties, then it also induces a map between the tangent spaces at the corresponding points, and this is given in terms of somehow the Jacobian of this map, which also I can write in terms of differentials. So, definition, so this is a bit like, you know, the second tier of analysis, what? Oh, yeah, yeah, I don't know why I wrote, I mean, it could have been n plus 1, but at least afterwards I had decided it was n. So the Jacobian of f1 to fr in kx1 is r, is the matrix n times n matrix, what is it n times r times n matrix, so j of f1, fr, which is the matrix, df1 by dx1 until dfr by dx if 1 by dxn, and here dfr by dx1 dfr by dxn. So this is a polynomial, so it's a matrix in this. So matrix is coefficient in this polynomial, and maybe I should note just that you, if I, I mean, I can view the differential at p as linear map, so as a matrix. And so the matrix, so if I take the differential of, if I take the differential at p of f1 as a matrix, this is df1 by dx1, and so on, df1 by dxn. So therefore I see that I could also say that dp f1, so for instance if I take the 0 set of dp f1 to dp fr, this will therefore be equal to the kernel of the Jacobian evaluated at p, because it's just a different way of writing df1. I can also write a corresponding column here, it's just, okay, so now I want to define the tangent map definition, so let x in n, y in m, close sub-varieties, or you don't need, maybe I don't actually need them to be a reducible node, because I have defined what the morphism or polynomial map between, anyway, so let p be a point in x, and q a point in y, and let p be a morphism from x to y, so I can write it as f1 to fm to y, where the fi are some polynomial in the variables on the source, and we assume that p of p is equal to q, so then I want to just say what the map on, induced map on tangent factors is, so the differential of phi at p, I can write it in two ways, this is equal, is dp phi, which is equal to dp f1 to dp fr, and I claim this is a map from tpx to tqy, so by definition dp phi is just, one can see that the coordinates on a n are in some way mapped to the coordinates on a m, so it is a map from a n to a m, and it's an easy exercise to show that it maps the tangent space of x to the tangent space of y, I mean, just if you go through the definitions, it's basically obvious, so but check, it maps tpx to tqy, and as you can see here, as I just said in words before, you can also write this, if you want instead of writing it in components like this, you can also write it as a matrix, and the matrix is just a Jacobian, so you can also write dp phi, okay, this is this map, and you can see that, I mean, just the obvious things that if I take the, do this for the identity, then this is the identity, if you take the identity from x to x, it's easy to check that if you take the dp psi composed with phi, this will be d phi of p psi composed with dp phi, like, you know, you learn also in analysis, and so thus, if phi is an isomorphism, then dp phi from tpx to p phi of p, y is an isomorphism of vector spaces or whatever, for all p in x, so this actually can be quite useful if you want to prove that something is not an isomorphism, it's enough to find one point where this tangent map, which is very easy to compute, is not an isomorphism. So this can be useful. So for instance, if we take our standard example, so if, again, we take c to be the cascaded cubic, so then we knew we had this nice morphism from a1 to c, which was given by sending a point t in a1 to t, t squared, t to the 3, no? Now, then it follows that d, so if I take a point a in a1, dA phi will be 2A 3A squared, and so we see that this tangent map is actually the 0 map at the point 0, 0. And it happens very often that when you have a morphism, which is not an isomorphism, then either the morphism is not injective or the tangent map is not injective. I mean, I'm not saying it's injective, but if you want to check it in examples, it's something you should check first, instead of looking for more complicated ways. Yeah, yeah, no, I mean, well, then... No, no, it is, we are also just starting, no? I've got my notes, so I have to... No, no, I mean, the beginning is quite easy, maybe because I decide to make it easy in the beginning. I mean, to start slowly. Okay, I don't know why I didn't bring it, but so let's... Now, this does it for the moment for the case of a fine variety. Now I want to, in general, define the tangent space, and I will give a completely different definition, which is much more abstract, but it is intrinsic. So, here we always have defined the tangent space in terms of the equation by t. The thing is somehow embedded in n, and then given by some equation and we get the tangent space somehow as a sub-space of the defined space. Now, we want to define the tangent space in such a way that it is automatically you can automatically see it only is invariant and isomorphism, and in fact you want to define it in terms of the local ring at the point. We have defined the local ring at the point and in terms of that we want to define the tangent space. This is not so great in many cases if you want to do concrete computations, but it is for theoretical things it's much more powerful to have some intrinsic definition. If you want to prove any kind of theorem it's not so good with such things. So now let's make tangent space and it is what generalizes best for general varieties. So let's just let the definition let X be a variety I want to say what the tangent space is and it's kind of strange. So we have the we'll be in terms of the local ring so let OXP so let P in X a point so the tangent space TPX is TPX it's the dual vector space MP we take MP the maximal ideal so P a point and we should have written it OXP the local ring and MP the local ring the maximal ideal so then the tangent space is you take MP model MP squared now MP is a no it's an idea MP model MP squared is a vector space no so it's a certainly a K vector space and I take its dual so if you I don't know whether you have encountered this in differential geometry so one definition in differential geometry of the tangent space is precisely that we look at the functions I mean the differentiable functions in a neighborhood of P modulo the differential functions in a neighborhood of P so which vanish at P modulo those which vanish modulo the square of those and this is one possible definition of the tangent space so it is the same definition as in differential geometry but no here in this algebraic setting but you most like you had another definition I don't know but there is a definition which looks precisely like that so more explicitly I could say this is all the the linear maps so K linear maps from well this by definition if it's a dual vector space it's just from mp modulo mp squared to K but it might be more useful to say it as linear maps from mp to K such that maybe I call them V from mp to K such that V restricted to mp squared is equal to 0 obviously that's the same thing I cannot really identify and then P in X is called non-singular if Tpx the dimension of Tpx now this is a vector space so if the dimension of this vector space is equal to the dimension of X and then we have the usual thing remember it's called singular so we have singular if it's not non-singular and we have X, we have the singular locus and we have regular locus and if X is equal to X we have X is equal to if X so we have set this three times today so if X is equal to Xreg then X is called non-singular or smooth okay this is just the same kind of sermon each time okay so this is this definition now this looks a bit in the moment it is maybe not even completely evident that this dimension has to be finite we have this mp which is a some some subspace of the of the local ring the local ring is certainly infinite dimensional as a vector space divided by mp squared and now we say that this would be regular if the dimension is equal to the dimension of X we don't even know whether it has finite dimensions so that doesn't sound so great but we will have to see that this would be okay and if for instance X is in a fine variety a close sub-variety of a fine space we have now two different definitions of the tangent space so they had better be equal otherwise this is not okay and so this is the the first thing that we will prove next time that these are equal which is a slightly tricky thing not very but you just have to set up the definition and then then we will go on from there and I think the next time also start talking about non-singular curves in the last bit okay thank you