 This lecture is part of an online commutative algebra course and will be a continuation of the previous two lectures where we were trying to show that the three different definitions of the dimension of a local notarian ring are all actually equivalent. So let's just quickly recall what we did in the previous lectures. So we had three ways of defining the dimension of a notarian local ring. We had the crawl dimension and we had a definition using degrees of Hilbert polynomials and we have had a definition and the minimal number of elements of a system of parameters. And in previous lectures we proved this inequality and we proved this inequality. So in this lecture we're going to prove that the definition using a system of parameters is at most equal to the crawl dimension. So this is the inequality we're going to do now. And of course once we've proved these three inequalities this will prove that these three different definitions are all in fact equivalent to each other. So before proving this we need a lemma called the prime avoidance lemma for a reason that will appear in a moment. And the problem is the following. Suppose we're given ideals, finite number of ideals p1 up to pn contained in a ring R. And suppose we're given an ideal i of the ring R. And the problem is find some value x in i where x is not in any of these other ideals p i. Well there's an obvious necessary condition which is that i is not contained in p i for any i. Because if it was then obviously we couldn't satisfy this condition. So the picture is something like this. We've got this ideal i and we're trying to cover it by a finite number of ideals say i1, i2, i3 such that none of the ideals i1, i2 or i3 contain i. And this can sometimes happen. So here's an example. Let's take R to be the ring f2xy. That's the field with two elements and we're having an algebra generated by x and y. Let's just kill off all products of two of these. So this is an artinian ring. It's a three-dimensional algebra over the field with two elements with basis 1, x and y. And if we take the ideal i to be generated by x and y then i has just four elements 0, x, y and x plus y. And if we take the ideals p1 to be the ideal generated by x and p2 to be the ideal generated by y and p3 to be the ideal generated by x plus y, then each of these contains just two elements which is 0 and its generator. And i is contained in p1 union p2 union p3. So we can't always find an element x like this. However, we can find x with x in i, x is not in the pi, if all the pi are prime. So this is why it's called prime avoidance. We're sort of finding an element x that avoids all these primes. In the previous example, we note that p1 is not prime because say for instance y squared is in p1, because this is 0, but y is not in p1. So none of these three ideals are prime ideals. So let's show how we can prove this. So here's a proof. We can assume that pi is not contained in pj for i not equal to j, because if it was, one of these ideals would just be redundant and we could throw it out. Now we're going to use induction on n, where n is the number of ideals p1 up to pn we have. So n equals 1 is trivial because we can just pick x is in i, x is not in p1 because we assumed that i is not contained in p1. And now by induction, let's pick x is in i, x is not in p1 up to pn minus 1, which we can do by induction. Now if x is not in pn, we're done. We found an x that's not in any of these pi's. So we can assume that x is in pn. And now what we do is we pick y not in pn, so yi that's not in pn, but yi is in pi, which we can do because pi and pn neither is contained in the other. We can also pick an element y in i with y is not in pn, again because i is not contained in pn. And now let's look at the element x plus y times y1 up to yn minus 1. And we're going to show that this element satisfies the condition we want. So first of all, we notice that x is not, so x is in pn. On the other hand, this element is not in pn. The reason it's not in pn is that y and the yi are not in pn and pn is prime. So if none of these factors are in pn, then the product isn't in pn because pn is prime. So we conclude that since this is in pn and we're adding something that's not in pn, this is not in pn. Next, we observe that xi is not in pi for i less than n, but this is in pi for i less than n because yi is in pi. So again, we've got the sum of something that's not in pi and something that's in pi, so this is not in pi for i less than n. Finally, we need to show it's in i. Well, that's easy because x is in i and y is in i, so this is in i. So we're adding two things that are in i, so this is in the ideal i. So we found an element that's in the ideal i but not in any of these prime ideals that we started with. If you're a little bit careful, you don't need to assume that all these numbers are prime. For instance, we never actually assumed that p1 was prime, so we can allow one of these not to be prime and there are various slight strengthening of this you can do. Okay, well now that we've done this lemma, let's get on to proving the main theorem. We want to prove the minimal size of a system of parameters is less than or equal to the curl dimension. Let's call the curl dimension d. So given a curl dimension d, we need to find the system of parameters x1 up to xd of size d. Well, we only need less than or equal to d but we'll actually find one of size d. And we're going to construct this as follows. So we're going to construct x1 up to xd with the following property that every prime p containing x1 up to xi has co-dimension greater than or equal to i. You remember the co-dimension of a prime is the largest length of a chain of proper inclusions p0 containing p1 up to pi which is equal to p. Well, it's easier to think about what this condition means if you're trying to think about it geometrically. So we think of the xis as being functions on some space or variety and then we might have x1 equals 0 or what is going to be some sort of hypersurface and then we might have another hypersurface say x2 equals 0. So if I'm taking i equals 2, I've got two hypersurfaces. And what this condition says is that every prime containing x1 and x2 is co-dimension at least 2. And the primes correspond to irreducible sets contained in the intersection of these two hyperplanes. So what it means is that if you take two hypersurfaces given by functions xi equals 0 then all components of that have co-dimension at least 2 and so on. If you take three hypersurfaces all components of intersection at least 3 and so on. So we're eliminating the degenerate case where you might take an intersection of three hypersurfaces and it only is co-dimension 2 or something. So we're going to try and find xi with this property. So suppose given x1 up to xi minus 1 and the problem is we want to find xi. So in this particular case suppose x1 and x2 kind of look like this. What we're trying to find is some sort of x3 such that the hyperplane x3 equals 0 intersects all components of x1 and x2 equals 0 in some set of higher co-dimension. So for instance we don't want x3 to vanish on this component here. So that's the geometric intuition. Now we actually need to construct it algebraically. First of all we notice that only a finite number of minimal elements of primes p1, p2 and so on containing x1 up to xi minus 1 of co-dimension i minus 1. Well what does this mean? Well the primes p1, p2 and so on correspond to the irreducible components of the intersection of these hyperplanes of largest possible dimensions. They're going to be of co-dimension i minus 1. Here in this picture we would have i equals 3. So we've got x1 and x2 and we're trying to find x3. And the reason they're only a finite number of these is that these just correspond to the associated primes of r over x1 up to xi minus 1. And they're only a finite number of associated primes of a finitely generated module. Now none contain the m which is the maximal ideal. And the reason for this is that this has co-dimension equal to the curl dimension which is greater than i minus 1 otherwise if i was equal to d we would have finished. So d is greater than i minus 1 but all the ideals pi have co-dimension equal to i minus 1 so they can't be equal to m. So by prime avoidance we can find xi with xi is not in p1 not in any of the pis but xi is in m. So here we're taking m to be i in the prime avoidance theorem and m is not contained in any of the ideals pi and all the ideals pi are primes so we can find some xi with this condition. And now we notice that all primes containing x1 up to xi have co-dimension greater than or equal to i. That's because x1 up to xi minus 1 kill off all those of co-dimension less than i and then xi has killed off all the ones of co-dimension i. So as I said geometrically what's going on is that we're just choosing xi to reduce the dimension of each of these components which correspond to these primes pi by at least one. So to summarize we've constructed a set of numbers x1 up to xd such that it satisfies this property here. So to finish off with we now need to show that x1 up to xd is a system of parameters. So how do we show that? Well that's quite easy because we know that any prime containing x1 up to xd has co-dimension at least d because that's how we constructed x1 up to xd. Well this is equal to the crawl dimension of the ring r so the only prime ideal, the only prime of co-dimension d is the maximal ideal m because d is the crawl dimension. So x1 up to xd is m primary if you remember what m primary means from earlier this course. So it contains the power of the maximal ideal m so x1 up to xd is a system of parameters. So what we've shown is that if d is the crawl dimension then we can find a system of parameters of size d. So the minimal size of the system of parameters is less than or equal to the crawl dimension of a local ring r. So putting this together with the two results we had in the two earlier lectures here the crawl dimension is equal to the dimension defined using Hilbert polynomials which is equal to the dimension defined using a system of parameters. Okay that's the last three lectures proving the equivalents of this are probably the most technical result of this course. Next lecture we will be giving some applications of the equivalents of these three different ways of defining dimension.