 This lecture is part of an online course on algebraic geometry and will be about the Zariski tangent space. So, in previous lecture, we defined the tangent space of a variety embedded in, say, affine space a to the m. And we use this to define singular points, where a singular point was one where the tangent space of the variety was in some sense of the wrong dimension. The problem with this is that we can embed the same variety into affine space in many different ways. And it's not at all obvious that the tangent space of a point p in the variety will have the same dimension independent of which space we embed the variety into. So what we want to do is to find a definition of the tangent space that only depends on the variety v and the point p and doesn't depend on how we embed v into affine space. So we have the following definition. The tangent space at point p is the dual of the cotangent space at p. And if you've done differential geometry, you probably think this is a completely stupid definition because the cotangent space in differential geometry is defined to be the dual of the tangent space. So this definition appears to be circular. However, in differential geometry, the tangent space is basic and you define the cotangent space to be its dual. But in algebraic geometry, it turns out the cotangent space is basic and you define the tangent space to be its dual. So how do we define the cotangent space? Well, the cotangent space is defined to be m over m squared. Okay, well, what's m? Well, m is the maximal ideal of the local ring at the point p. So you remember the local ring at a point p can be thought of as the set of, as the ring of functions of the form f over g where g is not equal to zero at the point p. So informally, we can think of the local ring as being rational functions defined at the point p. And the maximal ideal m can then be thought of as functions vanishing at the point p. And the key point of this is that this depends only on the local ring and the local ring of a variety at a point doesn't depend on the embedding of the variety. It only depends on the variety itself. So this gives us a definition of the tangent space that doesn't depend on the embedding, but we still have the problem. Is it the same as the earlier definition we gave? So we have to check that this is equivalent to the earlier definition. So let's look at the earlier definition. If we take a point p, we may as well take it to be the origin for simplicity. And we have a variety contained in a to the n with coordinates x1 up to xn. The maximal ideal of the local ring at p is just generated by the functions x1 up to xn, obviously. So v is going to be the set of zeros of some functions f1, f2 and so on. So these will be the polynomials defining the variety v. And now let's calculate m over m squared. Well, m will be the ideal generated by x1 over up to xn. I'm not quite sure whether we're working in the local ring or the full polynomial ring, but it doesn't really matter very much at this point. And we quotient out by the ideal generated by f1 up to fn in order to get the variety v. And we also have to quotient out by all monomials of degree greater than or equal to 2. So the monomials of degree greater than or equal to 2 are sort of giving you m squared. And this is giving you v. And of course, this is giving you m. And now we can see that this is the same as the vector space spanned by x1 up to xn, because that's just this ideal divided by all monomials of degree at least 2 quotient out by the linear parts of f1 up to fn. Because of course, if you're if you're quotient out by all monomials of degree at least 2, then you can ignore all the nonlinear parts of these polynomials. And we can see that this is just the dual of the tangent space defined earlier. So this confirms that our new definition of the tangent space really is compatible with the old definition of the tangent space. And I should say this vector space can be defined to be r over m and which is equal to k. And why do we bother doing that? Well, it turns out this definition actually makes sense for r is any local ring. We can take the maximal ideal form the quotient m over m squared. And this will be a vector space over the field of quotients of the local ring by its maximal ideal. In fact, this space here is called the Zariski cotangent space for any local ring. And of course, you define the Zariski tangent space of a local ring just be the dual of the cotangent space. So you see that the cotangent space is the more basic object because it's just defined like this. You have to define the tangent space in terms of the cotangent space, which is the opposite way round. There's another way of looking at the local ring or rather the cotangent space of a local ring. So here we take a local ring r and we can form m over m squared for m a maximal ideal, which is a vector space over the field r over m. And the dual of m over m squared can be identified as ring homomorphisms from r to this funny ring k epsilon over epsilon squared. So what's this ring here? Well, it's a two dimensional vector space over k, of course, with a basis consisting of elements one and epsilon. With epsilon squared equals zero. So the ring homomorphism will take m squared to zero and it just takes m each element of m to a multiple of epsilon and as multiples of epsilon are from a one dimensional vector space, we can see this really is an element of the dual of m over m squared. We can also try and picture what does this ring look like. So if we've got any polynomial ring in various generators and we quotient out by an ideal, then this quite often corresponds to a variety or an algebraic set consisting of the points where i is zero. Well, this doesn't really work in this case. The trouble is this only works nicely when the ideal i contains no nil potents and this ideal does actually contain nil potents so it doesn't quite correspond to an algebraic set. Instead it actually corresponds to a scheme called the spectrum of k epsilon over epsilon squared that we'll be looking at a bit later. Well, we can try and imagine what this looks like. Well, first of all, we should take all maximal or prime ideals of this and there is only one prime ideal, which is the ideal generated by epsilon squared. So it's spectrum strictly speaking looks like a point, but it's not really quite a point. So you should think of it as being a sort of something a little bit bigger than a point. Somehow it's really got you can think of it as being something like a point for a little bit sticking out of it where the little bit sticking out is something to do with this this element epsilon. So we should think of the spectrum of k of epsilon over epsilon squared is not quite a point, but a point with a with a tangent vector sticking out of it. And then maps from the spectrum of this to a variety consists of picking a point of the variety and then choosing a sort of tangent vector at this point. So there's yet another way of looking at tangent vectors or there's a risky tangent space. Suppose you look at vector fields on a smooth manifold be over the reals. Then let's look at the ring. C of v, which is just smooth functions. Then if we look at the space of all vector fields, smooth vector fields, of course, then the space of vector fields is just a module. Over the ring of smooth functions on the manifold because of course, if you've got a vector at a point we can just multiply it by the value of that function at that point. And we can define a module w of cotangent vector fields just to be the dual of the module of vector fields. And now you remember from differential geometry that there's actually map D, which goes from functions to cotangent vector fields. So it's going from the ring C of v to its module. This takes F to the one form DF. So here this is a ring. This is a module. You have to be a little bit careful because D is not actually linear over C of v. Instead, it satisfies Leibniz's rule DFG equals FDG plus DF times G. So if it were linear, it would satisfy this rule here, but it's now satisfying Leibniz's rule so it's not linear over C of v. It's linear over the field of real numbers, of course. So more generally in algebraic geometry, instead of the ring of smooth functions, we might look at the ring of polynomials over a variety, which should be some ring like this. And now what we want to do is take an analog of this. We want to find a map from R to a module W, which will be a module over R. And we want to think of W as being the module of cotangent vectors. So we're just going to, the idea is we are going to take this construction of cotangent vector fields over a smooth manifold and try and find an analog of it for algebraic varieties. At least I find one. So how do we construct the module W? Well, W is generated over as a module over R by elements of the form DR for R in R. And it satisfies the relations. First of all, DR is equal to zero for R in the field K we're working over. And secondly, DFG is equal to DF times G plus F times DG. And now we can define W to be the module with these generators and relations. So let's work out an example of this. Suppose we take R to be affine space KX1 up to XN. Then it's fairly easy to check that W is just the free module on elements DX1 up to DXN. And having defined W as the module of cotangent vectors, we can also define the module of tangent vector fields. Just to be the module of all homomorphisms over R from W to R, which is the sort of dual of W. If we want the tangent space of this at a point, we can just localize this module which gives us a module over the local ring and then take the Zariski tangent space of that. Well, it's a bit difficult. I mean, in general, this module of cotangent vector fields is kind of a bit complicated. It's usually not a free module over the ring. In fact, it's quite rare for it to be a free module. It's only a free module if the cotangent bundle happens to be a sum of trivial bundles, which is actually quite rare. So it's useful to have an alternative description of the module of cotangent vector fields. And we can define the module W like this. Suppose we take the homomorphism from R tensor R to R, so all tensor products will be over K. So this just takes A tensor B to AB. And we're going to define I to be the kernel of this map. And then I'm going to claim that we can define W to be I over I squared. So here I is an ideal of R tensor over R. And we need to define a map D from R to W, of course. And I'm going to define D by saying D of R is equal to 1 tensor R minus R tensor 1. And I also need to say how W is an R module. So I'm going to say R acts by multiplication on the left R in R tensor with R. So we could make this into an R module by having R acting on the right or on the left. And I'm just going to fix acting on the left. So we can easily check Leibniz rule DRS equals RDS plus DR times S. The big problem is to check the universal property. So what we mean is that by this is that we want to show that this really is the module that generates this relation. In other words, suppose we've got any module M with a map from D from R to M obey Leibniz's rule. We've also got a map from a module W from R to W. And we want to show there's a unique map from W to M making this diagram commute. So we can do this as follows. We've got our ideal I containing R tensor with R which maps to R. And we've got a map from R to M given by D. So we can define a map F from R tensor R to M by just mapping A tensor B to A D B. So this gives a map from I to M. And what we want is not a map from I to M but a map from I over I squared to M. So we need to check that this map F is not on I squared. And to do this, we just do an explicit calculation. So pick some element sum of Si tensor Ti in I and pick another element sum of Ui tensor Uj tensor Vj in I. So this means that the sum of Si Ti is equal to naught and the sum of Uj Vj is equal to naught because that's the definition of being in I more or less. And now we calculate F of sum of Si tensor Ti times sum of Uj tensor Vj. So this is an element of M squared and we want to show that F of this is equal to zero. So is this equal to zero? Well, let me just calculate it. So this is the sum of Si Uj d Ti Vj. And now we apply Leibniz as normally find this is the sum of Si d Ti times sum of Uj Vj plus sum of Si Ti times sum of Uj d Vj. So now we notice that this sum here is zero because we said so up here. And we notice that this sum here is zero because we said it was zero up there. So this expression here is equal to zero, which is what we wanted to prove. And now it's easier to check from this that this does indeed make W into the universal module so that there's a map from D from R to W satisfying Leibniz's rule. So in general, we say that a local ring is regular if the dimension of the Zariski tangent space is equal to the dimension of the local ring that we may discuss this in more detail later. And we say that a variety is non-singular at a point of its local ring is regular. Okay, well that's all for the Zariski tangent space for this lecture. Next lecture will be about some particular examples of singularities called Duval singularities.