 So this lecture is part of an online course on commutative algebra, and will be about Cohen-McCauley rings. So, more precisely about Cohen-McCauley local rings. So, remember last lecture we were looking at regular rings, or regular local rings, and we mentioned a couple of lectures ago that there was a sequence of other conditions. We could have local complete intersection rings, or Gorinstein rings, or Cohen-McCauley rings, or arbitrary local rings. So the last couple of lectures we were discussing regular rings, what we're going to discuss this lecture are the Cohen-McCauley rings. So, Cohen-McCauley rings, first of all, they're named after two people, Irvin Cohen and Francis Sowerby-McCauley. McCauley was actually a high school teacher. I think this was early 20th century when, if you were really interested in mathematics, there probably weren't any university jobs available for you, so you quite often became either a Church of England vicar or a high school teacher. So there are a lot of English mathematicians around that time who were either vicar or high school teachers. Anyway, so let's motivate the definition of a Cohen-McCauley ring. So first of all we recall, here we're going to take R to be local ring, and M it's maximal ideal as usual, and I guess I should say this is going to be notarian as well as local, and we recall that if X in R is not a zero divisor or a unit, then the dimension of R modulo X is the dimension of R minus one. And this suggests an obvious way to work out the dimension of a ring. We keep repeating this. So we can take X1 not a zero divisor or unit in R, and then we look at the ring R over X1, and we pick X2 not a zero divisor or unit in R over X1, and we continue like this. We then quotient out by X1 and X2, and we take X3 to be not a zero divisor or a unit in R over X1, X2, and so on. And any sequence like this, with this property, we say X1, X2, X3, and so on is a regular sequence. So this is yet another example of the word regular being heavily overused in mathematics. And we notice the dimension drops by one each time, so the length of the sequence, so the length of the sequence X1 up to Xn, so the length is of course just n, is at most the dimension of the local ring R. And we say the ring is cone McCauley if we can find a regular sequence of length equal to the dimension of the ring R. In general, the largest length of a regular sequence is called the depths of a ring, so a cone McCauley ring is one in which the depth is equal to the dimension. And I'll start by giving a few examples of rings that are or aren't cone McCauley. So first of all, any ring, any local ring of dimension zero is obviously cone McCauley because we can find a regular sequence of length zero, which is just the empty sequence. Any regular local ring is cone McCauley. To see this, we notice that regular implies that the ring is an integral domain, at least for local rings, so that's not true for non-local rings. And the reason for this is that if we've got a regular local ring R and we've got the corresponding graded ring R over M plus M over M squared plus M squared over M cubed and so on, this is isomorphic to a polynomial algebra in generator, you just take a minimal set of generators for M or rather a set of elements of M's that form a basis for this vector space and then we know that this is now a polynomial algebra over R and in particular it's an integral domain. And now if we take a filtered ring like a local ring and look at the corresponding graded ring, if the corresponding graded ring is an integral domain this implies that our original ring is also an integral domain and that's very easy because if AB equals naught in R and suppose that A is in M to the i and B is in M to the j but A is not in M to the i plus 1 and B is not in M to the j plus 1 but this implies that the image of AB in M to the i plus j over M to the i plus j plus 1 is non-zero because that's just like multiplying two homogeneous polynomials in a polynomial ring, which is an integral domain. So regular local rings are integral domains and now we can easily prove that the ring of Cohen-McCauley because we pick X1 to be some element of M not in M squared and then we know that the ring R over X1 is also regular because quotient out by X1 drops both the dimension and the dimension of the cotangent space by 1 so we can just repeat and at each step we can find some X1 that's not an integral domain because regular, sorry, that's not a zero divisor because regular rings are all integral domains so any non-zero element will do. So that shows regular implies Cohen-McCauley. Well, what about rings that are not Cohen-McCauley? So let's have an example of a local ring that's not Cohen-McCauley. So here's one, we can just take the ring, let's take power series in two variables which is a local ring and quotient out by the ideal Y squared XY. So let's try and think what this ring looks like. First of all, if we just took a polynomial ring and quotient it out by this, its spectrum would look a bit like this. So here we have why it's mostly the line Y equals naught, except it sort of sticks out a little bit at the origin. I mean it doesn't really have an extra point at the origin but you should sort of think of the point at the origin of sticking out slightly in one direction. So if we take the localization at the point zero what it looks like in form is we've got a sort of one-dimensional point of the spectrum that's got a sort of nought-dimensional point sticking out a bit. Anyway, this is reasonably typical of what's happening for non-Cohm-McCauley rings because you see we've got bits of different dimension kind of mixed up. We've got a sort of nought-dimensional bit intersecting with a one-dimensional bit in some sense. Of course that doesn't quite make sense but this is a sort of informal idea of trying to explain what's going on. Anyway, if we look at this, the maximal ideal M is just generated by X and Y and we notice that all elements of M are zero divisors. So the depth is equal to zero because we can't find any regular sequence in M of length greater than zero. However, the dimension is equal to one so the depth is less than the dimension and the ring is not Cohen-McCauley. Well, that one is a little bit fishy because it's got nilpotent elements. So can we find an example of a Cohen-McCauley ring with no nilpotent elements? So this is not... Here we want an example that's not Cohen-McCauley and there's no nilpotent elements. I mean, it's not really surprising we can make things go wrong with nilpotent elements. So for this one, let's think about the following algebraic set. We take a plane and we take a line going through it and we look at the local ring of this point here. So if this plane is Z equals naught and this line is X equals Y equals zero then the coordinate ring of this is KXY modulo XCYZ. And now I'm just going to complete it to get a local ring. I mean, we don't have to complete it we could just take the localization at the origin but it's easier to write down a completion than to write down the local ring at the origin. So we take the... Sorry, there should be a Z in there. So we take the power series in three variables and quotient that by XZ, YZ. And now what we have to do is to pick a non-zero divisor. Well, that's quite easy. We can pick the element, the non-zero divisor to be the first element of our regular sequence with X plus Z. And now we have to work out what is our ring KXYZ modulo XCYZ and then we also have to quotient out by this element X plus C. So that's just setting X equal to what? X equal to minus Z. However, if we do that we get the ring KXY modulo and now we're just setting Z equal to minus X so we get X squared YX. And this is the ring of the previous example and all elements of M, all elements of the maximal ideal are now zero divisors. So we find the depth is equal to what? Actually we haven't quite found the depth is equal to one because I said here we pick some non-zero divisor and we found we couldn't continue this and you may well wonder, well maybe we chose a stupid non-zero divisor and maybe if we'd chosen a better one we'd be able to continue it. Well, in fact it turns out it doesn't really matter which non-zero divisor you pick because all maximal regular sequences have the same length at least for notary and local rings and we will prove a little bit later but for the moment let's just assume it and we find the depth is equal to one. On the other hand the dimension is obviously equal to two because we've got a two dimensional bit there. So it's not Cohen-McAuley and it's got no nil potent elements I mean it's the coordinate ring of a perfect good algebraic set although when we quotient that by zero divisor then it does pick up nil potent elements. So again this is an example of this ring is not equidimensional so equidimensional means that all the irreducible pieces passing through a same point have the same dimension and here we've got pieces of dimension one and dimension two meeting at a point so that's not very good. So this is an example of a ring this is not equidimensional in general Cohen-McAuley rings are always equidimensional and I'm feeling too late to prove that so you can just look it up in the textbook. And you can ask is the converse true is an equidimensional ring Cohen-McAuley? Well not quite it's not far off being Cohen-McAuley but you need to say not only must the ring be equidimensional but it must also be equidimensional if you sort of slice it a few times. So let's see an example of so the next example is going to be an example of an equidimensional ring with no nil potents that is not Cohen-McAuley so being Cohen-McAuley really is slightly stronger than being equidimensional and what we're going to do is we're going to take a plane and then we're going to just take a second plane and these two planes are going to meet at a point and they're going to meet at a point in four-dimensional affine space so obviously in three-dimensional space it's a bit tricky getting too cold and at planes meeting at a point because two planes meet in a line in four dimensions we can do this. So in four dimensions we're going to take one of the planes to be w equals x equals zero and the other plane to be y equals z equals zero so the coordinate ring is kxy sorry missed out the w w x y z and then we should mod out by w y w z x y x z so we want to find what its depth is so we have to pick a non-zero divisor and if we pick a non-zero divisor we can pick the first element of our regular c3x1 which is w, let's just pick w minus y we have to be a little bit careful we can't pick a polynomial in that's just a polynomial in w and x because then it will be killed by y and we can't pick a polynomial that's just in y and z because it will be killed by x so we have to pick a non-zero divisor that has something in these two variables and something in these two variables and here we're going to take our ring r to be the completion of this so it's k w x y z modulo w y w x x y x z and if we take x1 equals this then we take the ring r modulo x1 and now we're just identifying w with y so we're going to get the ring k x y z modulo y squared y z x y x z and let's try and think what to the spectrum this ring sort of looks like well we've got the plane y equals 0 and we can have these planes either x equals 0 or y or z equals 0 but so the spectrum contains two lines but it also contains a sort of little bit sticking up at the origin so we can think of this plane as being where y is equal to 0 but there's a sort of infinitesimally little bit sticking up because if x and z are both equal to 0 then we don't quite get y squared y equals 0 we only get y squared equals 0 which means y can sort of be a sort of infinitesimally small non-zero number in some rather meaningless sense so if we take our modulo x1 we've suddenly picked up a sort of little nil-potent bit in our ring and just as in our first example this now means that all elements of the maximal ideal of r over x1 are zero divisors so we find the depth of r is equal to 1 but its dimension is obviously equal to 2 so again this is an equidimensional local ring that's not cone Macaulay well we've mentioned that we were cheating a little bit because we haven't yet shown that if we picked a different x1 here we might be able to find a non-zero divisor in the quotient so now I want to fix this so we want to show that any two regular sequences any two maximal regular sequences in a notary in local ring have the same length and the easy way to do this is to use a little bit of homological algebra so more generally we can define depth for a module m so in other words a regular sequence for m is a sequence x1, x2 and so on xn with xi in the maximal ideal of our local ring and x1 not zero divisor of m and x2 not zero divisor of m over x1 and so on so you can define depth for modules over a local ring as well as the depth of the ring and the following are equivalent first of all there is a regular sequence for m, x1 up to xn of length n secondly x to the i of r over m m is equal to naught for i less than n and thirdly any regular sequence can be extended to one of length so if we can prove that these three are equivalent then all maximal regular sequences have the same length because part three says that any regular sequence can be extended to one of length n and you see this second condition here is a condition about module m that says nothing at all about elements of a regular sequence so the maximal value of n doesn't depend on which regular sequence we're choosing and if you choose m to be the module r then this shows that the regular sequences for the ring r have the same length so we're going to prove this as follows we're going to show that condition one implies condition two and condition two implies condition three and we don't need to show that condition three implies condition one because condition three implies condition one is completely trivial so we've just got to show that one implies two and two implies three so to show that one implies two so that means we have a regular sequence x1 to xn which is regular for m and we want to show that x to the i of r over m and m equals naught for i less than n so by induction this, meaning this bit, is zero for i less than n minus one and now what we do is we look at the following exact sequence and we know that this is multiplication by x1 and this is exact because x1 is not a zero divisor so this is where we use the fact that x1 is not a zero divisor and you know if you've got a short exact sequence of modules there are some rather bewildering long exact sequences for x that you can never remember properly well I'm just going to write down the one we need says that x of i minus one r over m m over x1 m maps to x to the i of r over m m maps to x to the i of r over m m so now by induction this is zero so as this is exact it means this map here is injective however this map here is multiplication by x1 and we notice that x1 is zero on r over m because x1 is actually in m so this map is injective and it's also zero but if the zero map from a module to itself is zero this implies the module itself must be zero so x to the i of r over m m is equal to zero which is what we wanted to prove so sorry I should have said we're taking i to be n minus one here so now we want to prove that condition two implies condition three so you remember condition two says that x to the i of r over m m is equal to naught for i less than n and condition three says that any regular sequence can be extended to one of length one of length m so we're trying to prove that this condition implies that condition so we first suppose n equals one then this condition here becomes hom r over m to m equals zero so this means that m is not an associated prime of m because if it wasn't associated prime m would have a sub module of this form so this map here would be none zero now the zero divisors is just the union of the associated primes and the union of the associated primes can't be the whole of the ideal m because m is a maximal ideal and there are only a finite number of associated primes and the maximal ideal can't be the union of a finite number of primes by prime avoidance so m has a none zero divisor so there is a regular sequence of length one so for n equals one to do this now for n greater than one, I'm just going to be very brief here we pick a none zero divisor one in m and apply induction to m over x one m and what you can do is you can now show that this module here has a similar property except this only vanishes for i less than n minus one and for that you use exact sequences of x and I'm feeling a bit bored of writing out exact sequences of x and I'm going to get them wrong anyway so I'll leave this as an exercise so we've got two corollaries of this first of all, any two maximal regular sequences of m or r have the same length so if you find a regular sequence it doesn't really matter very much which element you pick as your first element because you'll still be able to extend it to a regular sequence of maximal length so all that worrying about it we did earlier was unnecessary another corollary is that any quotient of a regular local ring by a regular sequence possibly non maximal is Cohen-McCawley that's because if you take your regular local ring r and you quotient it out by x one up to x i then this is dimension is equal to the dimension of r minus i and you can extend this to a regular sequence of length dimension of r so this has a regular sequence of length dimension of r minus i which is equal to its dimension so for example any hypersurface singularity is Cohen-McCawley so you probably haven't come across many singularities that aren't hypersurface singularities so most of the singularities you've seen will be Cohen-McCawley singularities in particular for example any singularity of a plain curve is automatically a Cohen-McCawley singularity as a Cohen-McCawley local ring okay the next lecture we're going to be talking about something you can construct out of a regular sequence called the causal complex and as one application of this we will show that if you've got a regular sequence for a notarian local ring then any permutation of that is also a regular local sequence which is not at all obvious from the definition