 This lecture is part of an online commutative algebra course and will be about regular local rings. So just to set up notation r will be a notarian local ring and m is its maximal ideal. And there are several properties of local rings. The one we're going to talk about today are regular is regular rings. And there's a sort of nested family of properties, sort of go as follows. So a slightly more general class of local rings are local complete intersection. And local complete intersection rings are themselves a special case of Goronstein rings. Goronstein rings are a special case of Cohen-McCauley rings. And Cohen-McCauley rings are a special case of general local rings. So what we're going to do over the next few lectures is give some examples of each of these sorts of ring and explain why they might be interesting. So very informally regular local rings are the ones that correspond to non-singular points of manifolds or varieties. So you remember local ring tells you something about an algebraic variety near a point. So these correspond to saying the point is non-singular. Local complete intersection corresponds roughly to saying that your variety is defined by the minimum possible number of equations. Goronstein rings are those that have nice duality properties. Cohen-McCauley rings are something to do with not mixing bits of different dimension. For instance, a typical example of ring that isn't Cohen-McCauley is a sort of a union of a line and a plane where you're mixing things of two different dimensions. And finally we have general local rings. So this lecture is going to be about the best possible ones which are regular ones. So let's first define regular local rings. So we recall that if you've got a local ring with maximal ideal m, then the dimension of m over m squared is greater than or equal to the dimension of the local ring r. Here this is the dimension as a vector space over the field r over m. And this is its dimension in the sense of local ring theory. That's crawl dimension or whatever. And that's because the reason this inequality holds is that a set of generators for m is a system of parameters. And any system of parameters has cardinality at least the dimension of the ring r. And the ring is regular if equality holds. So the dimension of m over m squared is equal to the dimension of r. So let's have a few examples of this. So the first example, let's just take the ring kxy over y squared minus x cubed. So this is the coordinate ring of a cusp which is y squared equals x squared cubed. And we're going to localize at the origin in other words at the ideal xy. So at this point here you can see the variety is obviously singular at this point in some sense. There's no longer a smooth manifold at the origin. And if we look at the local ring, so if we localize this at the ideal xy, we see that the maximal ideal is the ideal xy. And m modulo m squared has dimension 2 because neither x nor y are killed off by this polynomial or by the ideal. On the other hand, if r is this local ring, the dimension of r is just equal to 1. And you see that so the dimension of the cotangent space is bigger than the dimension of the ring. And this means the ring is not regular and this corresponds to the fact this point here is kind of singular. So that's a local ring that isn't regular. Now let's see some examples of local rings that are regular. First of all, let's take r to be just say a power series ring in n variables. So that's a local ring. The maximal ideal is x1 up to xn. And we can see that m over m squared has dimension n because it has a basis x1 up to xn. Well, now we need to figure out what is the dimension of the ring r. And the easy way to figure this out is to remember that if we quotient out by an element xn which is not a unit or a zero divisor, then the dimension of r over x, the ideal xn, is the dimension of r minus 1. Now the ring r over xn is just equal to the power series ring in x1 up to xn minus 1. So every time you add a variable, you just increase the dimension by 1. So it's pretty obvious now that the dimension of r is just equal to n. So we see that in this case the dimension of the ring r is equal to the dimension of its cotangent space. So r is regular. You can also see that the dimension of the localization of the polynomial ring, so suppose we localize this at the ideal x1 up to xn, in other words we sort of look at the local ring at a point, is also regular. That's because its completion is the power series ring and a ring has the same dimension as its completion and it has the same cotangent space as its completion. So if the completion of a local ring is regular, then the ring is regular. Now we say a ring r, which is not necessarily a local ring, so let's take any ring s, it's called regular, if all the localizations of s at prime ideals p are regular. In fact it's sufficient to do all maximal ideals p, because Seher proved that the localization of any regular local ring is also regular. So in localization of s at some non-maximal ideal, you can get by first of all taking the localization of the maximal ideal and then localizing that. So by Seher's theorem this is just regular. We see that the ring of polynomials in n variables is a regular ring and that's because if you take the localization of this at a maximal ideal it's isomorphic to this ring here, which is regular. And that's just as well because this is the coordinate ring of n-dimensional affine space and n-dimensional affine space obviously ought to be non-singular under any reasonable definition of non-singular. So this shows that our definition of regular ring does at least seem to be behaving in a reasonable way. It is giving you the non-singular points of an algebraic variety. Now we can ask what do regular rings look like? Well it can't really say what a general regular ring looks like, but we can say quite a lot about complete regular rings or rather regular local rings. So suppose r is a complete regular local ring. Now we're going to suppose that r contains a field mapping isomorphically to the field r over m. For example, if we take a ring of power series then this obviously contains a field mapping to r over m which is k. And you might ask well how can it not contain such a field? Well here's an example when it doesn't. If we take the ring of p-addict integers which is the completion of z at the prime p then this is a complete regular local ring. But it doesn't actually contain a field mapping onto the quotient field which is the finite field of order p. So some complete regular local rings contain a field isomorphic to the field of quotients and some don't. Well now we're going to say suppose it does. Then r is isomorphic to a power series ring kx1 up to xn for some n. And the proof of this is fairly easy. What we do is we pick x1 up to xn to be a basis for m over m squared. And then we get a map, we get a homomorphism from the ring of power series x1 up to xn onto our ring r. Here this is any complete notary and local ring. So so far we haven't used the fact that r is regular and always done is we've got a map from the power series ring onto r. But now we suppose that r is regular. What we want to do is show that this is injective. So if r is regular this implies this map is injective. Well suppose a is in the kernel. Then a is not a unit or a zero divisor. So kx1 up to xn over a has dimension n minus one. So the ring r has dimension at most n minus one. But this gives you a contradiction because the cotangent space of r has dimension n and r is regular. So m over m squared has dimension n and r is regular. So these two numbers would have to be equal which isn't possible if there's something in the kernel. So this map here is actually an isomorphism. So we found all the complete regular local rings with a field mapping onto their quotient field. This is actually a special case of the Cohen structure theorem for complete regular local rings which describes more generally what happens if the ring r doesn't contain a field. And then as you can see by looking at the piatic integers things get kind of a bit more complicated. So you need a discrete valuation ring appearing in the conclusion. So let's have a look at some more examples. What about suppose you take a hypersurface fx1 xn in n dimensional affine space. So here we're just looking at some sort of surface given by the space f equals zero. And we can ask which points have regular local rings. So if you think of the zeros as being some sort of space we're asking which points of this space are non-singular. Whatever non-singular means. Well what non-singular means is that the local ring is regular because that's the definition of non-singular. So what we've got we may as well look at the point zero zero zero just to simplify notation because we can change to that point just by changing coordinates. So first of all we've got to ask what is the dimension of the local ring where we take kx1 up to xn. We're quotient out by f and we, sorry, we localize at zero which means we localize the ideal x1 up to xn. Well we know the dimension of the local ring of kx1 up to xn is n because we worked this out earlier. Now we know that f is not a unit or a zero divisor. Well it's not a zero divisor because this ring is an integral domain and it's not a unit because we should have said it's not a unit. So f is not a unit. If it is a unit then this space is empty and it's a bit of a pointless working out as dimension. So the dimension of kx1 up to xn, let's take the local ring and then quotient out by f is n-1. It's not terribly surprising this is just saying a hypersurface in n dimensional space has dimension n-1 as it ought to. So next we need to work out what is the dimension of the local ring, sorry the dimension of the cotangent space m over m squared. So let's work out m over m squared for the kx1 up to xn localized at the origin. Well this is obviously just, we're going to take space the ideal generated by x1 up to xn and quotient out by all products of the form xi xj which gives you m squared. And this is just an n dimensional vector space spanned by x1 up to xn. So that's the cotangent space of affine space. Now we need to look at the cotangent space of the hypersurface. And for this we just take x1 the ideal x1 up to xn and now we have to quotient out by all these numbers xi xj and we also have to quotient out by f. So what we're doing is we're taking the vector space generated by x1 up to xn and we're quotient out by the linear part of f. And now we can see that this is dimension is given by n if f has no linear part and it's given by n-1 if f has a linear part. And f has a linear part if and only if one of its first partial derivatives at zero is none zero. So we see that the dimension of a local ring of the hypersurface f equals zero at a point is, sorry the dimension of not the local ring, the cotangent space is either n or n-1 depending on whether some partial derivative of f is none zero. So it's n-1 if some partial derivative is none zero and n otherwise. So this tells us what the singular points of a curve are. They're exactly the points where all partial derivatives of f and f also vanish. For instance if f is y squared minus x cubed the singular points are those satisfying the following equations. First we must have y squared equals minus x cubed equals zero because obviously the singular points of a curve must actually lie on the curve. And then we must have two y equals zero and three x squared equals zero. And at least in characteristic zero you see this implies x equals y equals zero. So the singular point is exactly the origin which is what our geometric intuition would tell us because it's the only point that looks kind of a bit funny. So what we'll do next lecture is give some more examples of local rings that are or aren't regular.