 This lecture is part of an online commutative algebra course and will be more about the dimension of local rings. So we have just given four different definitions of the dimension of a local ring. First of all, there's the definition due to Brouwer, Menger and Uri's son, which is, you remember you define the dimension of a space x to be less than or equal to n, if every point has a neighbourhood with, so every point has arbitrarily small neighbourhoods with boundary of dimension less than n. And we say that a local ring has dimension n if its spectrum has dimension n. Then we have the dimension due to Krull, where what you do is you look at chains of irreducible closed subsets, where Zi is irreducible closed and Zi is not equal to Zi plus 1. And you look at the length of this chain and that's the dimension of the topological space. Well, rather the supremum of the length of the chains is the dimension. And as before, you define the dimension of a local ring to be the Krull dimension of its spectrum, consisting of prime ideals. Thirdly, we had the definition using Hilbert polynomials, which I'll describe a bit more precisely. So if you've got a local ring R with maximal ideal M, we can look at sum of M to the i over M to the i plus 1. So this is R over M plus M over M squared and so on. So this is now a graded ring and we're assuming R is notarian, so we can apply the previous theorem about Hilbert polynomials to this and we see that the dimension of R over M to the N is a polynomial in N for N sufficiently large. This means the dimension is a vector space, not its dimension as a ring. And the degree of this polynomial is the dimension of the local ring R. And the fourth definition we're going to look at is the minimum number of parameters. So a system of parameters is a set of generators for an ideal i with M contained in M to the R for some R. So a set of generators for M will certainly be a system of parameters, but we may be able to find a smaller system of parameters that doesn't generate the whole of M but still generate some power of M. So the dimension is the minimum of the cardinality of a system of parameters. So let's just give some quick examples of these definitions. So let's just take the ring R to be, say, take a ring of power series in two variables and quotient out by the ideal y squared equals x cubed. So the spectrum of this ring sort of looks like a ghost of a cusp. So if this was polynomials then R would be just a cusp but we sort of taken a completion so all we get is a tiny little bit of R. And then for the curl dimension we have the following ideals. We've got the ideal zero and we've got the ideal generated by x and y. These are both prime ideals and this corresponds to a point and this corresponds to the whole spectrum of R which is one dimensional. So we've got z naught, we've got a chain of length one and it's not difficult to check. This is the maximal chain, in fact this ring only has two prime ideals so that's rather easy to check. So here we find the dimension is equal to one. If we look at the definition using Hilbert polynomials what we do is we look at the dimension of R over M to the N where M is the maximal ideal xy. Well we can also write R as being a set of formal power series in T squared and T cubed where T squared is x and T cubed is y. So the maximal ideal M is just essentially all non-zero powers of T. And you can see that R over M to the N has dimension two N minus one for N greater than zero and this polynomial has degree one so the dimension is equal to one using the Hilbert definition. Finally we can ask the definition using a system of parameters. Notice that M needs two generators. However we can find an ideal contained in M and containing some power of M. If we just look at the ideal generated by T squared equals x then M contains this and this contains M squared. So x is a system of parameters for this ring and this contains only one element so the dimension is equal to one using the definition via a system of parameters. And what we're going to do is to prove that all these three definitions of a ring, so all these four definitions of a dimension of a ring are equivalent. And what we will do is we will show the Krull dimension is less than or equal to the definition defined by Hilbert polynomials. And we'll show this is less than or equal to the definition using a system of parameters. And we'll show that's less than or equal to the Krull dimension. So if we prove these three inequalities that will prove that these three are all the same. And we will prove that the Krull dimension is equal to the Brouwer-Menger-Urissen definition. So this is all for notarian local rings. Before starting on the proof of this I'll just make a few comments about this. First of all you can ask what happens for non-notarian rings? Well for non-notarian rings everything gets weird. The dimension using one of these is quite often just infinite which isn't terribly useful. But we can have other weird things. For instance the Krull dimension of r of x might be greater than one plus the Krull dimension of r. So r of x is basically sort of affine line over the ring r. And that really ought to have dimension one plus the dimension of r. And it does for notarian rings but for non-notarian rings it doesn't. So everything is going a bit weird. I should have mentioned we can define the dimension of any ring r to be the supremum of the dimension of rp where these are the local rings of points. And it's an easy exercise to show that this is just another way of defining the Krull dimension of the spectrum of r. If you want the definition using Hilbert polynomials or systems parameters then it's not so easy to define the dimension directly for a non-local ring. Well what I'm going to do is I'm going to break up the proof of these three inequalities into several different lectures because the proofs are all maybe a little bit technical and if I put them all in one big lecture it would just kind of be a bit too much. So what we're going to do in this lecture is just show this equality here showing that the Krull dimension is the same as the standard definition of dimension in general topology and in the next few lectures we will show these three inequalities. So let's first of all show that the Krull dimension is less than or equal to the Brouwer-Menger-Urisson dimension. So suppose we have a chain z0 containing z1 containing z2 and so on of irreducible closed subsets. So the Krull dimension is going to be given by a supremum of chains like this and what we're going to do is that the BMU dimension of zi is at least i and that would be enough to show the Krull dimension is less than or equal to the BMU dimension and this is quite easy because we just pick some point in zi that's not in zi-1 and this has BMU dimension greater than or equal to n-1 by induction sorry greater than or equal to i-1 and we also note that zi-1 is in the boundary of any neighborhood of some point p in zi-ci-1 you can see that using the fact that zi is irreducible so zi has BMU dimension at least equal to so it's strictly greater than the BMU dimension of zi-1 and by induction that's enough to show the BMU dimension of zi is at least i so this equality holds so this holds for all topological spaces finally we want to show the other implications so we're going to show that the BMU dimension of a space is less than or equal to the Krull dimension this is only for notarian spaces and it fails in general for instance if we take our space x to be just the real line the b-brow and manga orison dimension is 1 whereas the Krull dimension is 0 in fact the Krull dimension is 0 for any house or space as we mentioned earlier so and the idea is that we show for any closed subset the Krull dimension of x is at least the BMU dimension of x here I'm using this just to mean the Krull dimension and this to mean the Brow and Manga orison dimension otherwise pick a minimal closed subset violating this and here we use the fact that space is notarian because in a notarian topological space any collection of sets any non-empty collection of sets has a minimal element and we can assume this is x by otherwise we just replace x by this minimal closed subset we can assume x is irreducible otherwise x is the union of two spaces two closed subsets y and z and you can check that either y or z has to violate this inequality and suppose the Brow and Manga orison dimension of x is greater than or equal to n then what we do is we pick a point p and an open neighbourhood u so that the boundary of u has BMU dimension at least n minus 1 so the Krull dimension of the boundary is at least n minus 1 by notarian induction because this is true for all proper subsets and now we can get a chain z0 contained in z1 contained in zn minus 1 in the boundary and we can extend this to a chain z0 contained in z1 contained in zn minus 1 contained in x which now has length n so if the BMU dimension has at least n then the Krull dimension is at least n so this contradicts our assumption that x violated this inequality so this shows that the Krull dimension of any notarian topological space is at least the Brow and Manga orison dimension okay so next lecture will not be terribly long and we'll just be showing that the Hilbert's definition of dimension is at most equal to the dimension defined using a system of parameters