 This lecture is part of an online commutative algebra course and will be about associated primes of a module M. So the set of associated primes of M is a set noted by S of M and it's a collection of primes that sort of control where the module is geometrically, as we will see next lecture. So as motivation, let's look at the case for modules M of finite length. So if a module is finite length, then we can write nought equals m nought contained in m1 containing m2 and so on, contained in mn equals m. With each quotient mi plus 1 over mi, a simple module, this means that there's no submodules other than a zero or itself and is therefore isomorphic to something of the form R modulo a maximal ideal. Well that's fine for modules of finite length but the trouble is most modules we want to deal with are not of finite lengths and most of them tend to be finitely generated. So can we do, so can we find an analog for finitely generated modules? Well over arbitrary rings we can't really do an awful lot. So what we're going to do is assume that R is notarian and you remember that finitely generated modules over a notarian ring are also notarians, in particular the module M will be notarian. So let's start by looking at some examples just to see what's going on. Let's take our ring to be the integers z. Then modules are just finitely generated abelian groups and we know what a finitely generated abelian group looks like. It can be written as a sum of copies of z plus a sum over or p of a sum over various values mi of z over p to the n, i and z. It's just a sum of cyclic groups of various orders and we can build this out of the following modules. First of all we have the modules, the simple module z over pz which is z over the prime p so these are the simple ones but we can't build z over simple modules so we should also add the module z which is just equal to z modulo the ideal zero. So what's going on here well if we look at this we see that these ideals here are well these ones here are maximal but this one here is prime so really we're just getting prime ideals in the denominator so this suggests the following question. Can we break up m finitely generated over a notation ring into modules of the form r over p for p a prime ideal? In other words we want to find a sequence nought which is m nought contained in m1 contained in mn equals m with mi plus 1 over mi isomorphic to r over pi for some prime ideal pi. Well the answer is yes we can so well if the module is zero we obviously don't anything to do so let's assume the module m is non-zero and we first show m has a sub module isomorphic to r modulo p for p prime and for this what we do is we pick an ideal p in r which is maximal among the ideals so that such that r over p is isomorphic to a sub module of m and as usual I'll just be a bit sloppy with notation and pretend that r over p is a sub module of m. Now if r over p is a sub module of m this is the same as saying p is the annihilator of some element a in m so we're looking for an ideal that's maximal among the set of annihilators of course this doesn't mean p is a maximal ideal of the ring just means it's maximal among this set of ideals and we show that p is prime so you remember I mentioned before that if you've got a randomly chosen collection of ideals the maximal elements of that set have a really strong tendency to be prime ideals well if p is not we can find x y not in p such that x y is in p and we will get a contradiction from this well let's look at the image x bar of x in in m so you remember we've got r over p which we're pretending is a subset of m and x is in here and x bar is going to be its image there and let's look at the annihilator of x bar so this is an ideal of r and let's think it contains or it contains the ideal p rather obviously and it also contains the element y and it contains the element y because x y is in p so it's strictly bigger the ideal p and it's strictly bigger than the ideal b p because y is not in p by assumption and it's not the ideal r and it's not the ideal r because x is not in p which means x bar is not equal to zero so we've used the three conditions x y is in p y is not in p and x is not in p so this contradicts the choice of p because we assume that p was a maximum element of the set of ideals that annihilated something in m but here we found a bigger ideal that annihilates something in m so we've got this contradiction by assuming that p is not prime so p must actually be prime so now that we show that m contains a sub module of the form r over p we can very easily show that that m can be split up into some such sub modules so we start with m and if it's zero we just stop and if it's not zero then we can find a module m one such that m one is a subset of m and m one is isomorphic to r over p one with p one prime so we can carry out the first step now we've got to find a module m two how do we do this well we look at the module m modulo m one and apply the previous result to this so it contains something of the form r modulo p two a sub module and now we just take the inverse image of r over p two and take m two to be it's to be this inverse image it's the inverse image of r over p two where we have m goes to m over m one and this contains r over p two and then we see that m two over m one is then isomorphic to r over p two and now we can continue doing m three and m four in the same way and this must eventually stop and it stops because m is notarian so if we've got an increasing sequence of ideals of sub modules like this it can't go on forever well so that's all very well but there's a little problem that comes up so we can ask how often does the module r over p occur in m so we might call this the multiplicity of r over p in m and for m is at finite length we saw last lecture the multiplicity is the number of times r over p occurs in any maximal chain and more importantly it's additive so if we've got this sequence of finite length modules what goes to a goes to b goes to c goes to zero we know that we saw the multiplicity of r over p in a plus the multiplicity of r over p in b is equal to the multiplicity of r over p that should be a c i guess r over c in b so what happens if we try doing this for finitely generated modules well let's take a look at a module m being say z plus z over 2z over the integers say and then the multiplicity of z over naught in m well that's how many times does z occur in m well that's obviously one and how many times does z over 2z occur in m that's pretty obviously just c z over 2z just occur the once so that multiplicity is one okay well maybe not there's a little problem with this which we will see in the next next piece of paper the problem is the following we look at the universal counter example to everything which is naught goes to z goes to z goes to z over 2z goes to naught now let's look at what is the multiplicity of z over 2z in this well it doesn't occur there doesn't occur then it occurs once there so it's not additive so something has gone wrong and the problem is that you know we were saying z over 2z doesn't occur in z well in some sense it does because it occurs in this as a quotient so so this claim that the multiplicity of z over 2z in m is one is really dubious because z over 2z occurs in z and there's really no limit on how often it occurs in z because you know if it occurs in here once well the kernel of the map is z and then we could pick out another factor of z over 2z in here and so so z over 2z really seems to occur an infinite number of times in z and this really rather messes everything up well there are cases when the multiplicity is defined for instance if we want to know the multiplicity of z over zero in m this is sorry in m this is well defined because we can just define it to be the dimension over the rationals of the vector space m tensed with q and if m is equal to z to the n plus some finite group m tensed with q is just equal to q to the n plus well we've killed off the finite bit we just get zero and the dimension is just n and this is called this is just of course the rankover of the finitely generated group and it's additive and behaves really well and so on and you may think to yourself well if we define the multiplicity of z in m like this why don't we define the multiplicity of z modulo 2z in m like this so we could just define the multiplicity to be the dimension over the field with two elements of m tensed with z over 2z so what's wrong with this is a definition of the multiplicity well the problem is it's not additive that seems a bit odd because the um additivity of the dimension that the dimension of a vector space is certainly additive the problem is when we turn this into vector spaces over the field with two elements we run into the problem that tensuring with z over 2z does not preserve exactness as we've seen about 15 gazillion times in previous lectures so um we really run into a problem of trying to define the multiplicity of z modulo 2z in and in in a finitely generated abelian group so let's just summarize when the multiplicity as well behaved so the multiplicity of z over p in m so this is additive for p equals zero here m is finitely generated abelian group of course it's not additive or even clearly well defined for p equals um um sorry for p not zero it is additive for p even if p is zero for finite groups rather obviously so sometimes the multiplicity is additive and sometimes it isn't and you know you can ask when is it additive and when isn't it and we will see later but the answer is the multiplicity is additive for modules such that the associated primes of m has no elements strictly smaller than r over p that should say the multiplicity of r over p in a module so we're about to define the set of associated primes of m which among other things controls when the multiplicity is additive um actually um just before defining the associated primes i want to give an example which is even more disturbing than the previous one this time i'm going to take m to be the ideal in the ring of polynomials in two variables generated by x and y so as usual we can just draw a picture of this ideal to keep our um get some idea of what's going on so here i'm just drawing a dot for each basis element so this is one xx squared y and so on and we can think of the the ideal m as looking something like this you see the it's got a basis of all the monomials inside this blue region so there's a picture of m and now we can find several decompositions of m so we can first of all have the sequence north goes to r goes to m goes to r over y goes to zero and we can draw a picture of this we can understand this by most easily by drawing a picture and what's going on is just this so here we've got the module m which looks like that and then we can take a sub module isomorphic to the ring r consisting of all these bits here so it's just a sub module generated free sub module generated by that element there and then we can look at the quotient which looks like this green area here by the way i this this green area isn't a sub module it's a quotient it's the sort of sub quotient of r we're taking the um we're taking m and if we quotient out by this orange bit then then this green area is a quotient of m and you can see that it's actually isomorphic to r modulo y um this is really confusing at half the time i accidentally write it down as r over x um and this corresponds to the sequence nought equals m nought contains an m1 contained in m2 equals m where this quotient is r and this quotient here is r over y well that's fine what's the problem well the problem is we can also do things the other way round like this so we can take nought goes to r goes to m goes to r over x goes to nought and this time m looks like this and r um we're going to take r to be a slightly different sub module of m we're going to take it to be this one here and r modulo x of course looks like this stuff that's left over which looks like this so now we see that r modulo y doesn't really occur in m by looking at this sequence and r over x doesn't really occur in m by looking at this sequence so the only module that really occurs in m is r itself but the problem is we can't build m just out of copies of r you can easily check that if we try to write nought goes to r goes to m goes to r goes to zero we just can't do this um um so we seem to be kind of stuck that that doesn't seem to be a well defined notion of which modules occur in m well what we do is we define we now finally come to defining the associated primes of m so the associated primes of m is the set of prime ideals let's put that in a big box so you notice it um that um that are annihilators of elements of m so um there will be plenty of ideals that may be annihilate elements of n but aren't prime but we don't count these as being um elements of the associated primes of m we can also think of this as ideals p for p prime such that r over p is isomorphic to a sub module of m um so if m is none zero we've seen that the associated primes is none none empty so if m is not zero the associated primes of m is also not empty this is as you're assuming m is finitely generated r is notarian informally the associated primes of m is the set of p such that r over p is definitely occurs in um the module m um so for let let's just see some examples of this um so for example the associated primes of z is just the zero ideal of z the associated primes of z plus z modulo two z is now z and z and prime two and the associated primes of the funny module x y we had last time is just the zero prime um so um we can now ask what properties does the set of associated primes have so suppose we've got an exact sequence what goes to a goes to b goes to c goes to naught we can ask how do the associated primes of b compare with the associated primes of a and the associated primes of c let's ask the obvious question is set of associated primes of b equal to this union the most obvious guess and we answer this by using the standard counterexample to everything naught goes to z goes to z goes to z modulo two z goes to naught and here the associated primes of z is naught the associated primes of z is naught the associated primes of z modulo two z is naught so this is definitely wrong so what else can we do well we can at least say the associated primes of a is contained in the associated primes of b and this is completely obvious because if r over p is a sub module of a it's obviously also a sub module of b slightly more subtle fact is the associated primes of b is at least contained in the associated primes of a union the associated primes of c so although that these two aren't equal at least one side is contained in the other and this is easier to show so suppose r over p is isomorphic to a sub module and it's called a sub module x of b and there are two cases either x intersection a is empty or x intersection a is not empty and if x intersection a is empty this means x is isomorphic to a sub module of c which means that p is an associated prime of c on the other hand if x intersection a is non-empty let's pick so when I say non-empty I mean non-zero so let's pick a not equal zero in x with a in a and now we notice the annihilator of this element a is equal to p and the reason is the annihilator of any element x for x in r over p is p and we notice that because r over p is an integral domain so the annihilator of any non-zero element is is the is is is just zero which would be p so in this case p is an associated prime of a so that there are two useful consequences of this so here are some consequences first of all the set of associated primes of m is finite as usual m is finitely generated over a notarian ring if you drop those conditions there's no reason why this shouldn't be finite and what we notice is that if we have the sequence nought equals m nought contained in m1 so n contained in m n equals m then if we have mi plus one over mi equals r over pi we notice um the only associated prime is pi and now since um we see that if we've got any sequence like this then the associated primes of m must be among the associated primes of the quotients by induction and since the quotients all have only one associated prime we see that the associated primes of m must be contained in the set p1 p2 pn which is finite you remember we had some examples where the set of associated primes doesn't have to be equal to this set it might be strictly smaller but it's still finite the second property is that if we have a chain nought equals m nought contained in m1 and so on with all the mi plus one over mi of the form r over pi pi prime then any element p that is an associated prime of m um occurs in this chain and this again follows because um any associate is more or less what we said earlier um it must be one of these primes pi so so the set of associated primes is finite and they are primes that have to occur in any um decomposition of m into ideals of the form r over p so next lecture we'll be looking at um the the geometric meaning of the associated primes of m and in particular how it relates to the support of m in the spectrum of r