 This lecture is part of an online commutative algebra course and will be about some examples of art in rings. So last lecture, we had this rather hairy theorem saying that artinian rings are notarium. And we also showed as a consequence of the proof that any art in ring is a product of local art in rings, which makes them particularly simple to study. A typical example of this might be something like the art in ring Z modulo 60Z. And the Chinese remain the theorem from number theory says that this is just Z modulo 4Z times Z modulo 3Z times Z modulo 5Z. So here we've got an art in ring written as a product of three art in local rings. And this is sort of fairly typical of what happens in general, except it gets more complicated. So now what we're going to do is have a look at some examples of art in rings and we can classify art in rings in terms of their length. So you remember every art in ring has finite length. And let's just do art in local ring. So if you've got an art in local ring, it's got a single maximal ideal M and R over M is going to be some field K. And if R has length N over K, it is in some sense about the same size as an N-dimensional vector space over K. Although as we'll see in a moment, it need not actually be a vector space over K. In fact, you can see it right here. So this is about the same size as a two-dimensional vector space over a field with two elements, but it's not a vector space. Anyway, so if we look at length zero, this is trivial, we just get the zero ring. Length one is equally trivial. We just get one example, which is the field K itself. Length two starts to get a bit more interesting. So one way of doing this is to take a ring, polynomial ring K of X and quotient it out by some polynomial F of X of degree two. So we might take it as to be K of X modulo, say X squared plus AX plus B. And now we get several slightly different cases. We might get KX over X squared. And this is a non-reduced ring. You see that the element X is nil-potent. On the other hand, you might take K of X modulo X minus alpha, X minus beta for some alpha and beta in K. And by the Chinese remainder theorem, this is just isomorphic to the product of K with itself. So here we have an art in ring. That's the product of two art in local rings. Finally, we might have KX over F of X where F of X is an irreducible polynomial. So this will just be a bigger field containing K. So a fairly typical example of this, you might just take the field Q of I, the sort of Gaussian numbers, which is just QX modulo the ideal X squared plus one. So it can be a field or product of two fields or it can be this funny non-reduced ring. So these are examples that are vector spaces over K. We can sometimes find examples that aren't vector spaces over K. So a fairly one we had up there was just the ring Z over P squared Z, which is length two. And here the quotient field K is just Z modulo PZ. And for higher length, things like this that aren't vector spaces over K become a little bit tricky to classify. In fact, even ones that are vector spaces become tricky to classify. So let's just look at some length three ones. Until nothing terribly new happens, we can get things like K times K times K and analogous to this. We can get things like K of X modulo X squared and then we can take a product of that with K. We can take a field of degree three or the product of a field of degree three and the field of degree two and we can get something that's slightly new. We can take polynomial ring in two variables and quotient it out by the ideal generated by X squared, X, Y and Y squared. So now this is a three-dimensional vector space with basis one, X and Y and any product of two of X and Y is zero. And in dimension four things become even more complicated. So let's just have a quick look at some examples of length four. Well, here for example, we could take our ring R and it might have a maximal ideal and this might have a, has the square and the cube with the maximal ideal and we can take this to be zero and we might take this to be dimension one and we might take this to be dimension three. Now you notice we've got a map from M over M squared to M squared over M cubed and this is dimension one. So we can identify it with the field K and this is dimension two. So and our map could take X to X squared and now what we've got is more or less a quadratic form on a two-dimensional vector space. So there's a sort of quadratic form encoded in this and over various fields K there can be quite a lot of quadratic forms. For instance, if K is the field of rational numbers there are a fairly large number of quadratic forms in two variables. And again, instead of taking this to be this question be two dimension high length this could have dimension higher than two. So the classification of these is going to be more complicated in classifying quadratic forms over the field. For length greater than or equal to five and six the classification becomes starts becoming really rather complicated. In fact, what happens is it becomes so complicated you can't really describe the classification. And this is something whenever you try and classify nil potent objects. So if we've got an art in local ring with maximal ideal m, you know the ideal m is nil potent. So m to the n is equal to nought for some finite n. So we've got a sort of nil potent thing lying around. And nil potent things include art in rings and they include, well, nil potent ly algebras and they include things like finite p groups. And what happens with nil potent objects is if there are any generation by one or two things you can usually classify them. But as soon as they start being generated by too many things and being a sufficiently high dimension, the classification goes completely wild. For example, for finite p groups, I looked up the number of order two to the 10 which doesn't look all that imposing turns out to be 49487365422. So the number of nil potent finite groups grows very, very, very rapidly with the dimension and it's sort of hopeless to classify them beyond a rather low point. And what I'm going to do next is to show that for art in rings, the classification again goes a little bit wild. As soon as the length and number of generators becomes a little bit too big. So we can ask what is the dimension of the space of art in rings that are over a field k that have dimension n and have m generators So what we're doing is we've got m generators. So we take the field k of x1, x2, xm and we quotient out by an ideal of co-dimension n and we can think of these ideals as forming some sort of space. And we can ask what is the dimension of this space? Well, one obvious way to get ideas like this is to take n distinct points in affine space a to the m or should just k to the m and then take the ideal i to be all polynomials vanishing on these n points. And we've got n points in m dimensional space. So that sort of gives us an mn dimensional space of art in rings. And in general, you can try and imagine a general art in ring like this to be some sort of limit of n points moving around an m dimensional space. And this would suggest that the dimension of the space of art in rings is mn. And this is a very plausible argument which everybody thinks of when they first start studying this question. And it works for m equals one and two. So m equals one, it's sort of trivial because there you're just looking at i is just going to be an ideal generated by a polynomial of dimension n. And this obviously has dimension n times one. And for m equals two, it's still true. But for m equals three, it just fails totally. In other words, there are more ideals of co-dimension n that you might guess than you might guess by using your geometric intuition about this. And it's not very difficult to see that there are ridiculously large numbers of ideals and dimension m. I'll just write down a few of them. So let's just look at the ring kx one, x two, x three. So we're just taking m equals three. And let's take the ideal m to be the ideal generation by x one, x two, x three. And then we can look at m to the i. And I'm going to take the ideal to lie between m to the i and m to the i plus one, where i is some integer greater than zero. And now you notice that any subspace i like this is in fact an ideal. So we can generate lots and lots and lots of ideals of finite co-dimension by taking any subspace of m to the i over m to the i plus one. And let's just try and estimate very roughly how many of these we're getting. So what's the co-dimension of this? Well, the co-dimension is, I don't really care exactly what it is. Something of order i cubed. I mean, it's roughly bounded above and below by some constant times i cubed. And similarly, this also is co-dimension i cubed up to some constant. This is also a co-dimension that's roughly i cubed times some constant. And then we notice that the dimension, oops, the dimension, let me try again, that worked. The dimension of m i over m i plus one is roughly some constant times i squared. So the dimension of the subspace of subspaces of m i over m i plus one is about order i to the four. So the space of all subspaces of a vector space is called a grass manian. And if a space is dimension n, then the dimension of the space of the grass manian is some constant times n squared very roughly. So we've got enormous numbers of ideals, but the co-dimension, so the number of space of ideals has dimension something times i to the four, whereas the co-dimension of the ideals is something times i cubed. And this number here is going to be approximately the number n, which is the co-dimension, and the number m is going to be three. So the co-dimension of this ideal i is going to be order i cubed. So m n will also be order i cubed, and this is very definitely less than the dimension of the space of ideals, which is order i to the four, order i to the four. So what this is saying is that art in rings with at least three generators tend to be a bit weird and they're rather hard to imagine. There is actually a space parameterizing them called a Hilbert scheme that was constructed by growth and dick. And people study these quite a bit, especially for spaces of dimension one and two, but as soon as you stop looking at spaces of dimension three or more, the Hilbert schemes become kind of wild. The rule of thumb is that anything that possibly can go wrong will go wrong for a Hilbert scheme. So let's have a few more examples of art in rings. One way in which art in rings often come up is as a tensor product of two fields. For example, what happens if I take the tensor product of the complex numbers with the complex numbers over the reals? So this is a algebra of dimension four over the reals. So it's an art in ring and it splits as a product of art in local rings. So in fact, it splits. It turns out to be just a product of C and C. By the way, don't confuse the tensor product here with the ordinary product here. It just happens that the tensor product of C and C happens to be the ordinary product, same dimension as the ordinary product. That's because two times two happens to be the same as two plus two, but you really mustn't confuse these. Well, if it splits as a product of two things, these should be generated by idempotence because the unit of this will be an idempotent in this ring here. So what are the idempotence? So in this ring here, we should be able to find two idempotence. You can write them down like this. It's one times one plus i times i over two and one times one minus i times i over two. You can check that both of these square to themselves. So they give this decomposition of this art in ring into a product of art in local rings. We can do another example. Suppose we just take q of root two, tensor over q with q of root three. Then this is actually a field. q of root two, root three. So sometimes a tensor product of rings will be a field and sometimes it splits as a product of fields. So you can see that's the tensor product of r. We join the square root of minus one on both sides and here we join the square root of two and the square root of three. So we can ask, when does a tensor product like this become a product of fields and when is it just a field? Well, that's fairly easy to answer. So let's look at k tensor over k with l with k and l separable finite extensions. I'm gonna say k and l separable. I mean k or l is going to be separable for the moment. So let's suppose that l is separable. So we can write l is equal to k of x, modulo f of x where f has no multiple... So f is a separable polynomial. So it has no multiple roots in the algebraic closure of l. And then k tensor over k with l is easy to figure out. This is just the same as k of x, modulo f of x. And now we notice that f may split in k. So we're assuming f to be irreducible here. So it's an irreducible polynomial in little k of x, but it might not be irreducible in this. So it splits as a product of factors, f equals f one of x, f two of x, and so on. And the key point is these are all co-prime. And the reason they're co-prime is that f has no multiple roots. So f has no multiple roots and f has no multiple roots because of the separability condition. And this implies these polynomials are co-prime. And since these polynomials are co-prime, the Chinese remainder theorem implies this is k of x one over f one of x times k of x two over f two of x and so on, which is equal to product of fields. So we see that if l is separable, then this art in ring decomposes a product of art in local rings and all the art in local rings are just fields. Well, I've been emphasizing this separable condition several times. So obviously you can ask what happens if it's not separable. Well, if it's not separable, something a little bit weirder can happen. So let's take k equals k and let's join the pth root of a where k has characteristic p greater than zero and a is not a p's power. So we're assuming that x to the p minus a is irreducible. And let's take l to be k times the pth root of a as well. So l is equal to little k of x and then we're quotient out by x to the p minus a. So k tens of over k with l is just equal to k of x to the p, so k of over x to the p minus a. And let's say big k is equal to little k of b. So b to the p equals a, and I'm calling it b rather than x so we don't get confused. And now you notice this over k, this factors as x minus b to the p. So this is equal to k of x over x minus b to the p. And now you notice this has large numbers of multiple factors and this ring here has nilpotent elements. So it's got this nilpotent element x minus b for example. So in particular, this isn't a field at all. It's some sort of art in local ring with nilpotent elements. So this is still an art in local ring but it's not a field or a product of fields or anything like that. So in characteristic p, if you're working with inseparable extensions some slightly weird things can happen when you look at tensor products. Just to finish off by remarking that in general the tensor product of two art in rings need not be an art in ring. For instance, if you take the tensor product of k of x that's the field of rational functions and x over k with k of y you can check this is not an artinian ring. So these are infinite dimensional over k so perhaps that's not too surprising. Okay, that's enough about artinian rings for the moment and next lecture we will be discussing associated primes of a module.