 this lecture is part of an online commutative algebra course and will be about a class of rings with the rather cumbersome name of local complete intersection rings. Now as this is a bit of a mouthful to pronounce they're often called LCI rings for short. So we've been talking about several classes of local rings so I just briefly recall where these fit in. So we had regular local rings which correspond to non-singular points of varieties and these are a special sort of local complete intersection ring which are in turn a special case of Gorinstein rings which are in turn a special case of Kern-McAuley rings which are a special case of local rings. So in previous lectures we've covered Kern-McAuley rings and Gorinstein rings and regular rings so now we're going to do the last case which is local complete intersection rings. So first of all what is a local complete intersection ring? Well it's given by you take a regular local ring R and the local complete intersection is R modulo the ideal generated by x1 up to xi where x1 up to xi is a regular sequence. So for example x1 is a non-zero divisor and a non-unit and x2 is a non-zero divisor and a non-unit and R over x1. When I first came across this definition my reaction was it was looked like this completely artificial and rather silly definition but it does turn out to be really fundamental and natural concept although possibly not a completely intuitive one. The geometric meanings as follows. Suppose you've got some sort of variety in n-dimensional affine space and suppose it has co-dimension k then it needs at least k equations to define it and you can also talk about how many equations do you need to define it near a point so that there's a sort of difference between the number of equations you need to define it globally and the number you need to define it locally. If we can define the local ring at a point with k equations then we get a local complete intersection ring. So local complete intersection rings are something to do with being able to define things using the minimal possible number of equations. So let's have a few examples. So first of all all regular rings are obviously local complete intersection because you can just divide out by the empty set which is certainly a regular sequence so regular implies local complete intersection. More generally any hypersurface singularity is a local complete intersection local ring because here we're just taking a regular ring and quotient out by just one element. So this ring R will be the local ring of affine space actually calling it x1 to xn maybe a bit misleading as how you use that for a regular sequence. So we localize this at zero and quotient out by some hypersurface function of fy1 up to yn and then this is a non-zero divisor because this ring doesn't have any zero divisors so any hypersurface singularity is already local complete intersection in particular all singularities of say plain algebraic curves are local complete intersection singularities. If we want an example of something that's not a local complete intersection singularity we can have a rather clumsy example where we just take a variety consisting of the union of a plane and a point and at this point this is not a local complete intersection and intuitively you can see that's true because this variety has co-dimension 1 in three-dimensional affine space but there's no way you can define it with just a single equation because it's got this co-dimension 2 line there. That's not a terribly good example because it's not just not local complete intersection but it's not even Cohen-McCawley which is a much weaker condition so what we really want to do is find a rather better example. So first of all we recall that being a local complete intersection implies that the ring is Goronstein. I'm not going to prove this you can prove it in a way that's not too different from the way I sketch for showing that regular implies Goronstein you just take a regular sequence and sort of mess around with long exact sequences involving X's and so on to calculate X's so local complete intersection rings are Goronstein and the problem I want to discuss is find an example of a Goronstein ring that is not local complete intersection and this is actually I mean there are plenty of examples but they're not quite trivial to find it turns out that most obvious examples of Goronstein rings are automatically local complete intersection rings and in fact the by far the easiest way to write down Goronstein rings is just to construct them as local complete intersection rings and I think that the way I first came across the difference between a Goronstein ring and a local complete intersection ring was when Andrew Wiles proved Fermat's last theorem back in the 90s and one of the things in his proof I mean one of the steps in his proof was roughly speaking he had a he had a Goronstein local ring and he had to show that it was a local complete intersection ring and you know this sort of rather blew everyone away because I think most people rather like me had kind of come across Goronstein rings and local complete intersection rings as these rather technical obscure concepts and commutative algebra that we haven't paid much attention to and all of a sudden these were appearing in this really fundamental theorem that Andrew Wiles proved and he actually needed to understand the difference between them so what I'm going to do is just to start on this by by by just giving an example of a ring that that is Goronstein but not local complete intersection we're going to check the Goronstein and local complete intersection conditions and so the example is going to be the following ring we take the ring of polynomials in three variables and we quotient it up by the following ideal we we quotient up by the ideal generated by x squared x y y z z squared and y squared minus x z okay you see what I mean by the fact that these these examples aren't completely obvious you really have to think about it a bit to come up with examples like this so first of all let's explain why it is a Goronstein ring so first of all we work at dimension well the dimension is equal to zero in fact the ring obviously has length five and now we're going to try and draw a picture of it by drawing a sort of block for each each sub quotient isomorphic decay and picture of the ring I want to draw looks something like this so here I'm going to draw the whole ring as a sort of union of five blocks so each of these blocks corresponds to a sub quotient isomorphic decay and what we're going to do is we're going to take the three ideals x y and z to look like this now the ideal x doesn't consist just of this box it consists of this box together with the stuff underneath it because the ideal generated by x actually is length two and this thing down here is the ideal generated by y squared which is equal to x z and you can think of this bit here as being the whole ring and this bit here is the maximal ideal m and this bit at the bottom is the maximal ideal m squared and to explain where this construction comes from we more generally you can take v to be a vector space over k which may as well be finite dimensional and then you can take a ring consisting of k plus v plus another copy of k and this k is going to contain the identity of the ring and this vector space is going to have a symmetric bilinear form from v tensor v to k and the product on this bilinear form is going to be the product on v to this bit here so so this is this is going to be the maximal ideal m squared and this is going to be the maximal ideal m and the map from v times v to this k is the bilinear form so you can do this with any vector space in any bilinear form and you'll get a you'll get a little zero-dimensional local ring and now if the bilinear form on v is non-degenerate this implies that the ring is Goronstein and that's easy because Goronstein just means home over r from k to m r has dimension one and if this bilinear form on v is non-degenerate then any homomorphism from k to r must have image inside this bit here if the form on v is degenerate and has something in the kernel then then the image of k could also be something in v and it would no long be Goronstein so you can construct Goronstein rings from any vector space with a non-degenerate symmetric bilinear form that gives you lots of Goronstein rings here we're taking v to be three-dimensional reason we're taking it be three-dimensional is that if v is two-dimensional then this turns out to be a complete intersection ring so so for a ring that's not complete intersection we just have to take v of dimension at least three and and the simplest cases v being three-dimensional anyway now we have to see that it's not a complete intersection not a local complete intersection ring and let's just give an informal argument and we can write r as a quotient of k x that the power series ring x y z modulo sum ideal and what's this ideal going to be well that the ideal is going to be a sub ring of let's say m squared where m is going to be the ideal generated by x y and z in r in sorry in k x y z and you notice that m squared over m cubed has dimension six because it's spanned by x squared x y y squared x c z squared and whatever the last one is y z so in order to in order to get our ring we need to kill off a five-dimensional sub space of m squared over m cubed so this this is now just a vector space over k and we need to cut it down from being six dimensions to one dimension and obviously to do that we need to pick a five-dimensional subspace of that well this needs at least five generators sorry at least five relations because we need at least five we need to kill off at least five elements to cut this down however k x y z obviously has dimension as a ring equal to three and r has dimension zero so if r is a complete a local complete intersection we should use at most three relations because more or less the definition of a local complete intersection ring is that you have to use the minimum number of relations to cut it down so this sort of strongly suggests that the ring is not a local complete intersection ring and it's not quite approved because you know we have to fuss a little bit about this I mean we've written r as a quotient of this particular ring but maybe there's some clever way of writing r as a quotient of a ring with power series in 15 generators and using some very clever ideal or whatever I mean it seems unlikely but but proving you can't do that is really you know I mean you can do it it's just a little bit fussy so so this is a sort of sketch of a reason why it's not a local complete intersection but it's not quite a complete proof and so we're going to introduce another criterion for showing that something is not a local complete intersection and this is going to use the fitting ideals that we talked about last lecture and for this we're going to use the following theorem a nought-dimensional notarian ring is a local complete intersection if and only if the zero fitting ideal of the maximal ideal is not equal to zero so this is a nice criterion for local complete intersection rings well before we use that let's say what happens if your ring isn't zero-dimensional well there's an extension of this theorem to higher dimensions and one version of this was actually introduced by Wiles in his work on Fermat's last theorem and Wiles's version of this has since been simplified and generalized quite a bit so so there are there are theorems that will tell you about whether or not a high-dimensional notarian ring is a local complete intersection ring we're not going to worry about that instead we're going to use a rather rather simpler criterion which which says that a local complete intersection sorry a ring R so local ring R of dimension greater than nought is a local complete intersection if R over x1 is a local complete intersection where x1 of course is not a unit or zero divisor so you can reduce the case of testing where the positive dimensional rings are local complete intersection to zero dimensional rings by killing off by a regular sequence and then you can test the zero dimensional ones by looking at fitting ideals as usual in this lecture we're not actually going to prove any of this I'm just going to work out an example so let's work out the example we had earlier where we take R to be the ring of polynomial power series in three variables modulo x squared x y y z z squared y squared minus xz and we're going to take m to be the maximal ideal and we're going to work out the fitting ideals of m and we're going to work out not just the zero fitting ideal but all the other fitting ideals just for practice and working out fitting ideals so m is a module with three generators x y and z and actually it gets a bit confusing because the generators for module m are also elements of the ring R that we're working over and it's sort of rather confusing trying to remember whether you're thinking of x as being a generator of this module or as an element of the ring anyway let's write down a matrix of relations for the module m so it's got three generators which we think of as x y and z and these are generators of it as a module and we have the following relation x zero zero this comes from the relation x squared equals zero you see x times this generator is zero so we get this as a as one of the relations and then we have the relation y x equals zero and this just gives actually gives us two rows because y times this generator is zero but also x times this generator is zero so x yx gives us two generators and similarly z squared gives us zero zero z and y z equals zero gives us zero z zero zero zero y and y squared equals xz gives us a couple more generators zero yx z y zero so this this sort of says that z times x plus y times y is equal to zero so I guess I should put in some minus signs there so here we've got a largest matrix giving us the relations defining m as a module and now we can work out the fitting ideal so let's first we'll work out the zero fitting ideal of m well for this we have to take all three by three minors of matrix and work out the determinant well the determinant is going to be a product of three things each which is x y z or zero and the product of any three elements of these generators is automatically zero so the zero fitting ideal is just zero so by the criteria and we see R is not a local complete intersection ring because if it was a local complete intersection the zero fitting ideal would have to be zero while we're about it let's work out the first fitting ideal well this is given by two by two determinants and here you can see there are various two by two determinants you can get for instance if we take this row and this row we get xz well xz is actually a generator for the ideal m squared so you remember this this ideal has length 5 so we have R contained in m contained in m squared this is length 1 this is length 4 and this is length 5 now all two by two determinants will lie inside m squared and m squared is only one dimension so soon as you've got one non-zero element you've got the whole of m squared so the first fitting ideal of m is just m squared which is equal to say the ideal generated by xz what about the second fitting ideal of m well this is going to be generated by all the one by one determinants and the one by one determinants so just the entries so we get x y and z so this is the ideal x y and z which is just the whole ideal m and what about the third fitting ideal of m well this is going to be generated by zero by zero determinants and if you sort of think about it a bit the zeroes by zero determinant always has value one so this ideal is going to be the ideal generated by one which is the whole of R so we've now worked out all the fitting ideals of this module and in particular by using the criterion we mentioned earlier we see that the ring is not a local complete intersection ring and well that's a zero dimensional example with lots of nil-potent elements and let's have a sort of more geometric example without nil-potent elements so here's a one-dimensional reduced example you remember reduced means no nil-potent elements so it actually looks like a something coming from a classical geometric object and for this we're going to take one of the examples we did last time so you remember we had the one of the rings we had was the Goronstein ring where we take all polynomials generated by t to the 5 t to the 6 t to the 7 and t to the 8 so you remember this is essentially the the local ring of a certain embedding of the affine line into four-dimensional space with some sort of weird cusp at the origin and so we want to we saw its Goronstein some for some earlier lecture now let's show that this is not a local complete intersection well what we have to do is a quotient by a non-zero divisor and unless you're trying to make life difficult for yourself you're going to choose the non-zero divisor to be t to the 5 and we worked out the quotient by t to the 5 last time and it it had length 5 and we saw there was a basis of it consisting of one t to the 6 t to the 7 t to the 8 and t to the 14 and let's show that this is not a complete intersection ring well that's easy because if we call these elements 1 x y z then we see this element here is x z and it's also equal to y squared so this ring is just k x y z quotient out by all this stuff we had last time x squared x y y z z squared x z minus y squared which we just seen was not a local complete intersection ring so so here we've got a nice geometric example of a ring that is Goronstein but not a local complete intersection so to finish off with we can just sort of summarize how to test a local ring for being Goronstein or local complete intersection or whatever so so we're going to have a sort of flowchart for what you do if someone presents you with a ring and demands to know which nice properties it has so this is going to be let's assume it's notarian and local as I can't be bothered to work out whether we need it to be notarian or not so first of all we work out what is the dimension and suppose the dimension is greater than zero then we ask is there a none zero divisor in the maximal ideal m and if there's no then it's not Cohen Macaulay because we've got a positive dimensional ring and we can't find any regular sequence of length greater than zero suppose we can then what we do is we quotient by it and when we've quotient out by it we go back to case zero so we we keep reducing the dimension by quotient out by a by non-zero device until it becomes dimension zero so suppose the dimension is zero what do we do well then we can test to see whether it's Goronstein so we work out the dimension of home over K so home over R from K to R and there are two possibilities first of all it's greater than one then it's not Goronstein we should have said here when it's zero dimensional it's certainly Cohen Macaulay so and we can then the next step is to work with it was Goronstein then what we do is we work out the fitting ideal of the maximal ideal m as we mentioned earlier and there are two cases if it's equal to zero then our ring is not local complete intersection I should have said in this case if it's equal to one the ring is Goronstein and we then test to being local complete intersection so now we found that it is a local complete intersection and now let's look at the dimension as a vector space over the field K and if the dimension is one then the ring is regular what if the dimension is not one or greater than one well you might say well then we can say the ring is not regular it can't really because you can take a regular ring and quotient it out by something and get quotient out by regular sequence and get a ring that isn't regular in fact that's more or less the definition of a local complete intersection ring so what do you do if this dimension is not one well then you have a look and you see you go back to these non-zero divisors were any of the non-zero divisors in M squared and if the answer is no then the ring is not regular because what you were doing was quotient out by non-zero divisors that were actually non-zero elements of M modulo M squared well what if some of them weren't M squared well then you did something stupid and you should go back and pick a collection of non-zero divisors that are in M but not in M squared so you should go back here and start all over again so that's a sort of rough flow chart of how you how you test local rings for these properties okay the next two or three lectures are going to be about an application of commutative algebra to the Bernstein SATO polynomial