 This lecture is part of an online commutative algebra course and will be about examples of regular local rings. So in the previous lecture we looked at some examples of regular local rings, especially over algebraically closed fields. The problem is if you start looking at non-algebraically closed fields in characteristic p greater than zero, then some rather odd things can happen with regular local rings and the first couple of examples will be to illustrate what might happen. So suppose we take a variety v over some field. So generally then its coordinate ring will be the ring of polynomials over the field in n variables, quoted out by some ideal, and then we can localize at some point. Well, if you're working over an algebraically closed field a point is just going to be point x1 up to xn in k to the n. However, if the field isn't algebraically closed the points can be a bit more general than this, so a point here will correspond to a maximal ideal. And we can take the local ring at this point and we ask the following problem. If the local rings at the points are regular is the same true over the algebraic closure k bar. So in this case instead of taking the ring of polynomials over k, we take the ring of polynomials over the algebraic closure in n variables and again quotient out by the ideal i. So if all these local rings are regular are these ones. Well in characteristic p this isn't actually true. This was discovered I think by Zariski and when Zariski came up with this example everybody got a little bit confused by it at first. So an example might be something like this. Let's take the curve x to the p plus y to the p equals a. Here we're working over a field k of characteristic p greater than zero and a is not a piece power. So in particular the field can't be algebraically closed. Then this has a maximal ideal given by y because if we take k x to the p plus y k x y modulo x to the p plus y to the p minus a and then also quotient out by y then this is just k of x over x to the p minus a. This is irreducible because a is not a piece power so this is a field. So this is actually maximal ideal because the quotient by it is a field. And the local ring has dimension one. So if we localize it y this local ring has dimension one and the maximal ideal y is generated by one element. So the ring is regular. Since the maximal ideal is generated by one element this means the cotangent space is also one dimensional. So over this none algebraically closed field we've got a maximal ideal that gives us a regular local ring. Now let's look at what happens over the algebraic closure k bar where in this case the curve becomes k bar x y over x to the p plus y to the p minus a and now something a bit weird happens because this is just k bar of x y modulo x plus y minus b to the p where b to the p is equal to a. So it not only become this not only factorizes but but this this ring has nil potent elements in it. And this means in fact the ring the the the the well it's not really a variety anymore but the the ring is non regular at every maximal ideal. So something has gone very badly wrong with our notion of regularity being non singular. We've got some variety over a none algebraically closed field that's regular at every point and when we moved to the algebraically closed field something goes completely horribly wrong. We start getting nil potent elements all over the place and so on. Actually to see what's going on it's easier to simplify it a bit and look at the nought dimensional case. So instead of looking at the variety x to the p plus y to the p equals a let's just look at the variety x to the p minus a equals 0. So this is just nought dimensional so making everything nought dimensional makes it a lot easier of course we're saying a is not power of p for p and k. And there's the the field that sorry the ring kx over x to the p minus a is now just a field. So it's only got one maximal ideal which is 0 and the local ring at that point is this field and it's certainly regular. There's no problem at all. But over the algebraic closure if we take k bar x over x to the p minus a this splits as k bar x over x minus b to the p. And now this is not a field and in fact there's nothing not a field this is nil potent elements as before. So it's definitely not a regular local ring or a regular ring or anything like that. So this phenomenon happens even in the nought dimensional case. And we can sort of see what's going on here because if we call this field L what we're doing is this field here is L tensed over k with k bar. So what we're really doing is taking a tensor product of two fields over another field and in characteristic p this can behave a bit oddly. So let's look more generally at a tensor product L tensed over k of m where L, k and m are now all fields with k contained in L and k contained in m. And let's suppose that L is a finite extension of k then in characteristic nought L tensed over k m is a product of fields. Since it's a finite product of fields every point of it is still going to be regular because its local ring is just a field. And you can see this as follows. So right L is equal to k of x over f of x where this is now irreducible over k. Now if it's irreducible over k and the characteristic is equal to zero this means all roots are distinct over k bar. That's because the roots, the multiple roots are given by the common zeros of f and its derivative and if f is irreducible the derivative has smaller degrees so it can't have any common roots in characteristic zero. So f splits as f1x, f2x and so on over the other field m with the fi distinct. So L tensed over k of m becomes a product m of x over f1 of x times m of x over f2 of x and so on by the Chinese remainder thing where these are all fields. So in characteristic nought there are no problems if we've got a funny sort of nought dimensional variety and we look at it over the algebraic closure it just becomes its coordinate ring is still a product of fields and it's regular at every point. In characteristic p things can become much stranger. So here we take L to be k of x over x to the p minus a and we're going to take this to be m as well. And now if you work at L tensed over k of m this becomes kxy modulo x to the p minus a y to the p minus a and this is not a product of fields. The proof that it's a product of fields in characteristic nought breaks down because x to the p minus a can now have multiple roots over the algebraic closure in fact it's derivative is identically zero. In fact it has nilpotent elements as usual in fact we can see there's a nilpotent element would be for instance x minus y is nilpotent because x minus y to the p is just equal to x to the p minus y to the p which is a minus a which is equal to zero. So what is going on here is in characteristic p greater than nought the product of two regular so I mean the tensor probably two regular rings can be non regular. So in terms of varieties this means a product of two non singular varieties can be singular if by singular you mean the local rings are not regular and this is extremely disconcerting because a non singular variety should be a really nice one that sort of looks I don't know locally like a manifold or something like that. And if you if you take a product of two manifolds without singularities this really ought to be a manifold without singularities so in characteristic p over non algebraic enclosed fields this breaks down and there are various ways around it. What it's saying roughly is is that the notion of a regular local ring is really doesn't really quite capture our intuitive concept of non singular variety in characteristic p. So there are various ways to fix that. Zariski introduced the concept of geometrically regular local ring and this is a local ring r over a field k so that r tensor over k with k bar is regular. For example our previous example k of x over x p minus a is regular it's just a field but it's not geometrically regular and geometrically regular rings behave rather better than regular rings if you're working with in characteristic p over non algebraic enclosed fields. There's actually a closely related concept called smoothness which is closely related to rings being geometrically regular. This was introduced by growth and Dick and he has a much more complicated concept of a map between rings r to s being smooth and in the special case when r is a field this is related to the concept of being geometrically regular. Anyway I won't be discussing smoothness in this particular course. The other example I wanted to cover in this lecture of regular local rings was the example where you take z of root minus three. So this is an if you've done algebraic number theory you know this is an order in the field q root minus three. And we also notice that it's not a unique factorization domain because one plus root minus three times one minus root minus three is equal to two times two is a is a non unique factorization. Now if you've done algebraic number theory you know that's easy to fix because you take its normalization. In other words it's integral closure in the quotient field and the normalization is z one plus root minus three over two and this is a unique factorization domain. And what we're going to do is we're going to look at a certain local ring in two of these and in one of them this local ring will be non regular and the other be regular. So here's a maximal ideal of z root minus three. Well one maximal ideal is two one plus root minus three. You can easily check the quotient of this by if this is m and this is r then r over m is just a field with two elements. On the other hand if we look at r over m squared we can work out this as let's just localize z to make things a little bit easier. z root minus three and then we quotient out by the square of this ideal which is generated by two squared and two plus two root minus three. And you can easily check that this has length equal to three. In other words it's got three composition factors of this. So m over m squared has dimension two. Sorry I should have said we're going to localize this ring at the ideal at the at the maximal ideal m and forgot to localize it. Now this ring r has dimension one so it's cotangent space is dimension two so r z root minus three. So z root minus three is not regular and it's not regular at the ideal two one plus root minus three. On the other hand if we take its normalization then z one plus root minus three over two is regular. In fact it's regular at all maximal ideals. You can check this in several ways for instance this is a unique factorization domain and you can easily check that at all maximal ideals the localization is a discrete valuation ring. And discrete valuation rings are certainly regular because they're one dimensional and not only the maximal ideal but in fact all ideals are generated. So what we have is we have two rings here we've got the ring z root minus three it's contained in the ring z one plus root minus three over two. And this is not regular whereas this ring is regular and in algebraic number theory you need to do a very similar thing for any order of an algebraic number field. You really want the ring to be regular and to do that you do what is known as taking its normalization in other words just taking the integral closure of the ring in its quotient field. And geometrically what this corresponds to is sort of making all the points of its spectrum regular or you can think of this informally as making them non-singular. By the way there's one slightly disconcerting pitfall here so as we take the ideal two in this ring here. Now this is a prime ideal in fact you can easily check that z one plus root minus three over two modulo the ideal two is just the field with four elements so it's a maximal ideal. On the other hand if you take the ideal two in this ring here it's not prime. In fact you can check the quotient of this is just a sum of two fields of order two and we see in a slight paradox because the inverse image of a prime ideal should be prime. But if you take the ideal two here the current spawning ideal two here isn't prime. Well what's going on? Well this is a sort of subtle trap because this the ideal two is not the inverse image of the prime ideal two in this ring here. It's a sort of easy mistake to make if you take a set of generators for a prime ideal the inverse image prime ideal need not be generated by the inverse image of all the generators which is kind of obvious when you state it but it's a sort of easy trap to make. The inverse image of the prime ideal two is in fact the ideal generated by two and one plus root minus three that we had earlier. Okay that's enough about regular local rings. The next lecture will be about Cohen-McCawley local rings which is about the mildest condition you can push on a local ring which makes some behavior reasonably.