 This lecture is part of an online commutative algebra course and will be about proof of some theorems about dimensions of local rings that we stated in the previous couple of lectures. More precisely, we're going to show that the three definitions of dimension that we gave, so we gave a definition due to Krull, a definition using Hilbert polynomials, and a definition using a system of parameters. So we've got three notions of dimension, and what we're doing is proving the following inequalities. So here we have the Krull dimension again. And if we prove these three inequalities, we'll prove that all these three notions of dimension are the same. So in the previous lecture, we proved this inequality. What we're going to do this lecture is prove this inequality here, and in the next lecture we will prove that inequality. So before proving this inequality between the Krull dimension and the Hilbert dimension, we're going to have a lemma about that quotienting out by an element will often reduce the dimension by one. So suppose X is in M and is not a zero divisor. Here M is the maximal ideal of a notarian local ring R. So then the Hilbert dimension of R over X is less than or equal to the Hilbert dimension of R minus one. Here I write Hilbert dimension for the dimension defined by Hilbert polynomials, which will later turn out to be the same as all the other dimensions. In fact, equality holds as we will see a little bit later, but for the moment we just want this inequality. What this says is that by quotienting out by an element, we'll usually drop the dimension by one, provided it's not a unit, in other words in the maximal ideal, and not a zero divisor. If it is a zero divisor, then this need not hold. So let's just have a quick look at an example. Suppose you take R to be the local ring of kxy modulo xy at the point at the prime ideal where x and y are both equal to zero. So diagrammatically you're taking the local ring at this point. You can see that there are zero divisors because this variety is reducible. Then if we take x to be the element x here, x is a zero divisor. But if we quotient out by x, it's just the local ring of ky at the ideal where y is equal to zero. And both of these rings have dimension one. You can see if we divide quotient out by x, we're just setting x equal to zero. In other words, we're basically looking at the y-axis. So what this is saying is that if you've got two lines, the two coordinate axes, and you take the sub space where x is equal to naught, the dimension doesn't actually drop by one. So here these two dimensions would be the same. So that shows that why we need to add this assumption that x is not a zero divisor. Well, the proof of this, what we do is we write down the following exact sequence. We take naught goes to R, goes to x. This is multiplication by x goes to R, and this goes to R over x times R goes to zero. And this is exact as x is not a zero divisor. So this is where we use the condition. And then we quotient out by some power of m. So we take R over m to the n, and this maps to R over xR intersection m to the n. Just zero, and sorry, not intersection, it's generated by those two. And what we get here is R over xR intersection m to the n. And we just put a warning as usual, this is not in general R over m to the n. In fact, you can see that that wouldn't be exact by taking very easy examples. For instance, you take naught goes to the integers localized at two. You could multiply by two, the integers localized at two. The quotient is zero, two z goes to zero. And if you look at this example, you can see that if we took R modulo m here, so let's just take m to be the ideal two and n to be one, then R over m under multiplication by two is not a submodule of R over m. So we really do need to be a little bit careful about what we quotient out R by there. However, although this isn't equal to m to the n in general, the filtration xR intersection m to the n is stable by the strong art in re-slammer. So what he's saying is that if you're taking the filtration by powers of m and intersecting, then with a submodule, you don't quite get the intersection, you don't quite get powers of m times R here, but you get something pretty close. So what this means is that roughly speaking, saying this stable means it differs from m to the n by almost a shift. So we have m to the n contains xR intersection m to the n, but this in turn contains m to the n plus k for some fixed k. So up to a finite shift of k, these two are essentially the same. Well, what this means is that the Hilbert polynomials of these two terms, so we've got nought goes to R over xR intersection m to the n, goes to R over m to the n. So we're looking at these two terms. So the Hilbert polynomials are each bounded by a shift of the other. So if the two Hilbert polynomials are f and g, we have sort of f of n is less than or equal to g of n, is less than or equal to f of n plus k for some fixed k or something like that. So they have the same degree and the same leading coefficient because this condition for polynomials implies that f and g must have the same degree and same leading coefficient. So f minus g has smaller degree. Well, if we extend this sequence a bit longer, we get to R over xR goes to zero and because the Hilbert polynomial is additive in short exact sequences, this is the Hilbert polynomial of R over xR. So this proves the lemma that the Hilbert polynomial of X over XR has degree less than that of the Hilbert polynomial of R, which is what the lemma stated. Now that we've got this, we can prove that the Krull dimension is less than or equal to the Hilbert dimension of a local ring. So suppose the Krull dimension is greater than or equal to n. We then want to show that the Hilbert dimension is also greater than or equal to n and this will be enough to prove this inequality. So what we do is we pick p0 contained in p1 up to pn where these are primes in the ring and we can find the chain of strictly increasing primes of length n because the Krull dimension is at least n by assumption. And we may as well quotient by p0 so we can assume p0 equals zero, so the ring is an integral domain because p0 is prime. And now what we do is we pick x in p1 but not in p0. We notice that x is not zero divisor because we've quotient it up by p0 so we're working with an integral domain. So this is actually x is just a nonzero element of p1. And now we have the Hilbert dimension of R is strictly greater than the Hilbert dimension of R over xr by the lemma that we proved which is greater than or equal to the Krull dimension of R over xr and this follows by induction on n and this is greater than or equal to n minus 1 so we've got a chain p1 up to pn, sorry p1 over x up to pn over x so we've got a chain of n minus of length n minus 1 and the key point here is that this inequality here is strict so this shows that the Hilbert dimension of R is strictly bigger than n minus 1 so the Hilbert dimension is at least equal to n which is what we wanted to show so this shows that the Krull dimension is at most the Hilbert dimension so what we're going to do in the next lecture is complete this by showing that the dimension defined by system of parameters is at most the Krull dimension