 This lecture is part of an online commutative algebra course, and will be about some applications of the theorem of dimension theory we proved in the previous few lectures. In particular, we'll be talking about Krull's principal ideal theorem. So let's just quickly review what we did in the past few lectures. You remember we had three different definitions of the dimension of a notarian local ring. So first of all, we could define the Krull dimension, and we also had a definition using Hilbert polynomials, and we had a definition using a system of parameters. And what we did was we showed these three apparently completely different definitions all gave the same answer. And this is very useful because we can often switch between these and use whichever of these definitions is easiest. For example, the Krull dimension gives an easy lower bound for the dimension of a ring. All you need to do is to find a chain of ideals, prime ideals, of length n, and that will give you a lower bound. On the other hand, if you can find a system of parameters, this will give you an upper bound. So if you've got a system of parameters x1 to xd, then d is an upper bound for the dimension, and it's quite hard to use Krull's definition to give an upper bound, and it's quite hard to use the system of parameters definition to give a lower bound, but since they're equal, you can sort of combine them to get the dimension. Hilbert's definition gives upper bound and a lower bound, but it takes a certain amount of work because you have to figure out what the Hilbert polynomial is in Sun, so it sort of requires a bit of effort. Anyway, we're going to give a few applications of this. So the first application is very easy. So the first application says that if you've got a notary in local ring r, then this has the same dimension as the completion r hat. This is not so trivial to prove using Krull's definition of dimension or the system of parameters definition, but it's completely trivial if you use the definition using Hilbert polynomials because you notice that r over m to the n is just isomorphic to r hat over m hat to the n, where this is the completion of r. And since the Hilbert's definition of dimension depends only on the size of these rings, we see that r and r hat have the same dimension. For the next application, well, all the applications of the rest of this lecture will be along the theme of the zeros of a function have co-dimension one. And I'll explain in what sense this is true. So the first version of this principle says that suppose x is in a notary in local ring and x is not a unit or a zero divisor, then the dimension of r over x is equal to the dimension of r minus one. And let's explain what's going on here. So the idea is that if you've got a function, then normally zero should have co-dimension one. So let's look at some cases when the zeros of a function do not have co-dimension one. Well, the first case is the function x has no zeros at all. In this case, the set of zeros is just the empty set, which is dimension minus one or something. So this corresponds to the case when if x is a unit, for example, then it will have no zeros. So we don't expect this to hold if x is a unit. There's a second slightly more subtle case. Suppose you've got a space which is two components, then you might have a function that is zero on one component but none zero on the other component. And then the zeros would have co-dimension zero, not one. Well, in this case, x is a zero divisor, at least in reasonable cases, because you could also take a function y that was zero here but none zero here, and then x, y would be none zero. So in the case when a function vanishes on one component, we don't expect zeros to have co-dimension one. So this case corresponds to this other thing we excluded where x is a zero divisor. There's a third case where a function doesn't have zeros of co-dimension one, which should be somehow x squared plus y squared on R2. So the zeros of this is just a single point. So we seem to have a function with zeros of co-dimension two. Well, this doesn't really, this example doesn't really occur in ring theory because you notice that R squared is not the whole spectrum of the ring R and x, y. So you remember the spectrum of the polynomials over R in two variables. Well, it sort of contains a copy of the affine plane over R, but it also contains lots of other points. For instance, it contains a one-dimensional point corresponding to the ideal x squared plus y squared, and in some sense the function x squared plus y squared vanishes on this one-dimensional ideal. It also vanish on a lot of maximal points which aren't defined over the reals. For example, it should vanish at the point plus or minus i1 in, well, this is in C squared not R squared, but this corresponds to a perfect good ideal. So this corresponds to the ideal where x squared plus 1 and y. So you see that although at first sight the zeros of this function seem to have co-dimension two, that's because we didn't really, that's because it's got a lot of zeros that you don't quite see in R squared. So this example, although at first sight it looks like a counter example, isn't really a counter example because you have to be a little bit careful about what you mean by the dimension of the zeros. And this result is actually quite easy to prove. So let's have a proof that he was showing the dimension of R over x is equal to the dimension of R minus 1 where x is not a zero divisor or a unit. Well, we've already shown that the dimension of R over x is less than or equal to the dimension of R minus 1 a couple of lectures ago as part of our proof that the three definitions of dimension are the same. So we have to show that this is greater than or equal to this. What we do is we pick x1 up to xd where d is the dimension of R over x. That's a system of parameters and in R over x. So these generate an m over x primary ideal in R over x. And then x1 up to xd and then you add x as well is also m primary. So it's a system of parameters as we're using the same argument in the previous lecture. So the dimension of R is less than or equal to 1 plus the dimension of R over x because we found a system of parameters of this cardinality. So that's one version of the principle that the zeros of a function should have co-dimension 1 if you look at the zeros carefully. There's a more precise version of this which is Kroll's principle ideal theorem. So this says suppose that x in a notarian ring R is not a zero divisor. So x is in R where R is notarian. Then all primes minimal among the ones containing x have co-dimension 1. So you recall the co-dimension of a prime is the maximal length of a chain of primes ending at p, sometimes called the height, but it's very difficult to remember what the height is. Codimension 1 is explains what is going on geometrically. So let's just explain the geometric meaning of this. So what happens is you think of x as a function. So we think of R as being some sort of space of functions and x is some function and we look at the zeros of x and the zeros of x will form some sort of subset of say the spectrum of R. As you remember elements of a ring aren't really functions of the spectrum of R but we're going to sort of pretend they are just to explain what's going on in this theorem. And then the prime ideals containing x, so primes containing x correspond to irreducible subsets of the zeros of x equals zero. So one prime containing x might correspond to this. There might be a non-minimal prime containing x which kind of corresponds to this. So this would be a non-minimal prime. You remember that when primes get smaller the sets that they vanish on get bigger. So it looks a bit funny saying this point corresponds to a bigger prime than this pink thing because it looks smaller but that's because everything gets reversed. So what this theorem says is you look at the minimal primes containing x correspond to the maximal irreducible that they correspond to the irreducible components of the zeros of x. So Kroll's theorem is saying that the irreducible components of zeros of x all have co-dimension 1 which is obviously just another way of saying, of making strict sense of the informal principle that zeros of a function have co-dimension 1. So we'll just finish this lecture by giving a quick sketch of the proof of Kroll's theorem. So proof. All we do is we let P be minimal among the primes with x in the prime. So as we said before P sort of corresponds to a maximal irreducible component of the set of zeros of x. Now the key point is we localize at P. In other words we invert everything that's not in P. And this means we can assume P is maximal. We need to do a bit of work to check that localizing at P it's enough to prove this result for localizing at P which I'll just skip. And then we notice that the ideal x is then P primary. So P is its only... so that's because P is the only prime containing x. So x, this is a one element set, is a system of parameters for this ring R. Since we've localized at P we can assume that our ring R is local and P is its maximal ideal. So we found a system of parameters for our local ring. So this implies that R has dimension less than or equal to 1 because the dimension is bounded by the size of a system of parameters which is 1. So P has co-dimension 0 or 1. And if it is co-dimension 1 then we finished which is what we are trying to prove. So we just need to check that the co-dimension of P is not 0. Well if the co-dimension of P equals 0 then P is minimal. So all elements are 0 divisors and this is false because x is in P and x is not a 0 divisor. So you see if x is a 0 divisor then the prime P doesn't necessarily have co-dimension 1 it might have co-dimension 0 as we saw in some earlier examples. Okay so that's more or less the end of the part on dimensions of local rings. What we're going to do next is discuss various properties of local rings such as being regular or a local complete intersection or Gorinstein or Cohen Macaulay.