 This video is part of an online algebraic geometry course and will be about singular points of varieties. So in previous lectures, we've been discussing singular points quite a bit without actually defining them. So it's really time we said what the precise definition of a singular point is. So I suppose P is a point of a variety V. And P is called singular if the tangent space to the variety at P has the wrong dimension. So what do we mean by the wrong dimension? Well, by the wrong dimension, we mean it has dimension greater than the dimension of the variety V. So in order to do this, we need to define the tangent space. So we're going to define the tangent space in two or three steps. So first of all, we will take P to be the origin in a fine space. You can change coordinates by just adding a constant to the coordinates to make any point origin. So just for simplicity, let's make P be this point here. And let's take the variety to be a hyper surface. So it's a set of points where F is equal to zero here, F is some irreducible polynomial in several variables. Then the tangent space at P of this hyper surface is the zeros of the linear part of F. So for example, if F is say X minus two Y plus three X squared plus four X Y equals zero, then we take linear part, which is just this bit here and the tangent space is then given by X minus two Y equals zero. So if you want to draw a picture of what's going on, suppose your variety looks say something like that, then the tangent space at the origin will not be this straight line here. So for plain curves, we can describe what the singular points are. So let's take a plain curve, F X Y equals zero. Then the point naught naught is singular is equivalent to saying that first of all, F naught naught equals naught because it's got a lot on the curve if it's a singular point of the curve. And the linear terms vanish because if F has non-zero linear terms and that's going to give you a one dimensional tangent space which is the same dimension as this. However, if the linear terms vanish, then the tangent space is the whole plane. So if the linear terms vanish, this means the tangent space is the whole of the affine plane which is the wrong dimension. Look at the same dimension as the curve. So let's have a look at a couple of examples. If you take Y squared equals X cubed, then this looks like the following curve here. You see the linear parts are zero and in some sense, so the tangent space is by definition two dimensional. And if we take say Y squared equals X squared plus X cubed then it looks like this. And again, the tangent space is two dimensional. You've got to be a little bit careful here because for this curve, it's fairly obvious the tangent space is two dimensional because the tangent space should contain this line here and it should contain this line here and that's obviously generating a two dimensional space. This example is a little bit trickier because if you look at it, you might think the tangent space is just going to be the X axis. And in fact, there's a sense, there's something called a tangent cone and in this particular case, the tangent cone is just the X axis. However, the tangent space is in fact the whole plane even though it sort of looks as if the only tangent effect is in that direction. In n dimensions, things are kind of similar. If we've got a polynomial F1 up to Xn, then naught, naught, naught is singular, is equivalent to saying that F naught, naught equals naught, and all the linear terms vanish. In other words, delta F over delta Xi vanishes at the origin for all i. So that describes singular points of hyper surfaces. Next, we can ask what about co-dimension greater than one? So suppose V is in A to the N and it's the zero set of some polynomials F1, F2, F3 and so on. And let's take the point P to be the origin again, just for simplicity. So all the Fi vanish, we're assuming all these Fi vanish at P, otherwise P wouldn't be on V. Then the tangent space is the common zeros of all linear parts of F1, F2 and so on. I should maybe it'd be better to say the linear parts of all these functions. And we can work out the dimension of this tangent space because if you've got some linear functions, then the dimension where they vanish is given by the rank of a matrix. And the rank of this matrix will be the Jacobian delta Fi over delta Xj. And in fact, the dimension of the tangent space will be the dimension of N minus this. So this is the dimension of the tangent space at point P. And this is a bit of painful to calculate if there's more than one polynomial because you start having to calculate ranks of matrices, which is a bit tiresome to do by hand. So this is why most of the examples we have, we just look at co-dimension one hyper surfaces. However, you can see from this that the subset of singular points is closed. And the reason for this, that the condition for a matrix having rank less than or equal to some constant is a closed condition on the entries. You remember we discussed this when we did determinant or varieties and the condition for having rank less than some constant is the condition that a whole lot of minors of some size should all vanish. So it's a rather complicated closed subset but it's still closed. So the condition of the tangent space being bigger than what it ought to be is a closed condition. The set of points where this happens is a closed subset of the variety. I should add a sort of warning here that for varieties, the set of singular points is closed. However, for more general schemes, this need not be true. There are plenty of examples of schemes such that every singular point is closed. Now we want to show, sorry, there are schemes where the set of closed, where the set of singular points need not be closed. Sorry, I was getting muddled up with the next slide. So the next thing we want to show is that the non-singular points are dense. And this is only going to hold for varieties as I sort of got a bit muddled about a few minutes ago. This isn't true for general schemes. So this is for varieties. Well, the non-singular points form an open set. So to show their dense, we just need to show that the non-singular points are non-empty, because any non-empty open set is automatically dense. And before proving this, let's give something that seems to be a counter example to this. So we're going to take the following example. Let's take the variety x cubed plus y cubed equals one. So over the real numbers, it looks something like this. And we can find all singular points. Well, to find the singular points, we have to take the derivatives. So we have three x squared equals zero. That's the derivative with respect to x. And the derivative with respect to y must also be zero. And over the complex numbers, this implies x equals y equals naught. And this point doesn't lie on this curve. So it is no singular points over the complex numbers. However, over a field of characteristic three, we see that all points are singular. So we seem to have a variety such that every single point is singular over a certain field. So we can ask, what went wrong? Well, it's very easy. If we look at the equation x cubed plus y cubed minus one, this is actually not irreducible over the field F3. It's actually equal to x plus y minus one or cubed over a field of characteristic three. So this looks like an irreducible variety. And it is indeed an irreducible variety of the complex numbers, but over other fields, weird things can happen. So we're going to try and prove that any variety has at least one singular point. And the first step is we reduce to the case of hypersurfaces. As I mentioned earlier, it's rather tiresome dealing with singular points for varieties of co-dimension, bigger than one in affine space. And the key point is V is birational to a hypersurface. And the easiest way to show this is to cheat a little bit and just quote some bits of field theory. So we look at the function field of our variety. So this is going to be some field K, which is generated over K by finite number of variables. And we can choose the first few variables to be algebraically independent. And then it has some other variables which are all algebraically dependent on these. And we can choose X1 up to Xn. So big K is separable over K X1 up to Xn. That's just quoting bits of field theory. So again, quoting more field theory is generated by one element because any finite separable extension is generated by unique element. So big K is equal to K X1 up to Xn, Y for some Y. And Y has to be algebraically dependent on these. So we have some polynomial f X1 up to Xn, Y equals zero. And our variety V is therefore birational to this hypersurface in an n plus one dimensional space. So now that we've sort of reduced the case of a hypersurface, let's see what happens. So we can write our hypersurface as f X1 up to Xn equals naught. I'm just changing notation from the previous slide by dropping Y. And we're going to assume f is irreducible. And we should also assume we're working over a field that is algebraically closed because if we don't run into some rather strange glitches and characteristic P. And the condition for a singular point is that this should vanish and delta f over delta X1 should be equal to naught and so on and delta f over delta Xn equals naught. And now all point singular implies that delta f over delta XI equals naught at all points of the variety V. So these are just the conditions for a singular point. Well, we've got a problem here because the degree of delta f over delta XI is less than the degree of f. And since f is irreducible, f divides delta f over delta XI and if it divides something that has smaller degree this means that delta f over delta XI equals naught for all i. And now you remember from calculus that if a function of several variables as order of it is vanishing, then it's zero. So f is zero, right? Well, no, because this statement may be true in calculus but it's not actually true in algebraic geometry. So this implies f is equal to naught in characteristic zero. In characteristic P greater than naught it implies that f is the sum of P's powers of P's powers, sorry it's a linear combination of P's powers of monomials times various constants. Now if we're working over an algebraically closed field this implies that f is some P's power of some function G because we can just take the P's roots of all the coefficients. If we're not working over an algebraically closed field then we run into some rather weird problems you actually need to, that there's actually this concept of the point being geometrically regular or regular and these are actually different in characteristic P but in any case we're assuming f is irreducible and we're assuming we're working over an algebraically closed field so f can't be a P's power of some other polynomial so f is equal to zero which contradicts the fact that we've got a hypersurface. So this shows that every variety contains one non-singular point and therefore the non-singular points are dense. Finally we should just remark that at non-singular points this is where we're working over the reals or the complex numbers v looks like it looks locally v looks locally like a smooth manifold and this follows easily from differential geometry and this isn't a differential geometry course I'm not gonna bother giving details. You might ask is the converse true? If v looks locally like a manifold is it non-singular? For example, if we look at a singular point like this you can see it certainly doesn't look locally like a manifold there so it must be non-singular. How do the converse, you've got to be a little bit careful about. So if you look at the following example let's just work over r although this works over the complex numbers as well. If we take the manifold y squared equals x cubed it looks like this and it is actually a topological manifold. So the fact that a variety looks locally like a topological manifold doesn't imply that it's smooth. Of course, if you start trying to make this isomorphic a smooth manifold then it's fairly obvious something is going to go wrong so you can make a case for saying if v looks like a smooth manifold then it's non-singular but you have to be a little bit careful. I mean, the underlying space of this can be given the structure of a smooth manifold if you're willing to be a little bit silly about how you define it. Okay, so there's a big problem with the definition of a tangent space and singular points we've given they seem to depend on how you embed a variety into affine space. So you can have two isomorphic varieties with different embeddings into affine space and it's not clear from the definition we've given that the tangent spaces are independent of the embeddings and in particular it's not clear that the singularity of a point of v is independent of the embedding. So the next video we're going to give a different definition of the tangent space that doesn't depend on the embedding into affine space and that will allow us to show that singular points don't depend on the embedding.