 This lecture is part of an online course on commutative algebra and will be about symbolic powers of a prime ideal. So let's just recall, last lecture we talked about Alaska-Nurta theorem, which said that if you've got an ideal in a notarian ring R, so all rings will be notarian, then I is an intersection of primary ideals. And what we're going to do is to give some examples of primary ideals, and in particular we will instruct some of them using these things called symbolic powers. So the first question is what do primary ideals look like? Well, for the integers the primary ideals are just nought and the ideals generated by P to the n, which is the nth power of a prime ideal. And this kind of suggests that primary ideals should just be powers of prime ideals. So we can ask our primary ideals the same as powers of primes. And this is true for principal ideal domains such as the integers, but in general the answer is no. In fact, this fails in both directions and we're going to give some examples of primary ideals that are not powers. And we're also going to give some examples of powers that are not primary ideals. So first let's do an example of primary ideals need not be powers of prime ideals. And this is easy. There are huge numbers of them. We can just take R to be a ring of polynomials and two variables. One variable is a principal ideal domain, so this won't work. And we're going to take a prime ideal to be the ideal generation by X and Y. We're going to draw a picture of this as usual. We draw R by just drawing its monomials. And then we can see this ideal P is generated by all the monomials in this set here. And now let's just take I to be some random ideal. For instance, we could take I to be, say, generated by Y cubed X squared Y squared X cubed Y next to the six or something. So it's going to be generated by this element here. Then we have X squared Y squared. Then we have X cubed Y and then we have X to 1, 2, 3, 4, 2, 3, 4, 5, 6 or something like that. And then we're going to show that I is primary. And this is easy to see because R modulo I has a sequence of submodules. If we put M equals this, more equals M naught contained in M1. So I'm contained in, if I've counted correctly, I think it's about M10, but I may have miscounted. And that's quite easy because you can take, for instance, M1 is going to be generated by this. So here we have M1 and M1, M2, M2 might be generated by these two. And then you add a third point and get M3 and so on. So the only associated prime of M is R over P. So R over I is co-primary, so I is primary. There are lots of different ways to check I as primary. On the other hand, I is not a power of P. In fact, the powers of P are fairly limited. For instance, the cube of P just looks like this. So this thing here is P cubed. So there are huge numbers of primary ideals for this prime and only a very small proportion of them are powers of P. So in general, primary ideals, I mean, it's quite unusual for a primary ideal to be a power of a prime. And by the way, I should have a sort of warning here. Do not confuse the following two things. So do not confuse. First of all, we can think of P equals X, Y as a co-primary module. And here it's associated prime is just the zero prime. So we're using it as an example of a co-primary module. Here we're not using it as an example of a co-primary module. We're thinking of P equals X, Y as prime ideal, or if you like, a primary ideal. And its corresponding co-primary module is now R over P. And here the associated primes consist of the ideal P rather than the ideal zero. So there are two quite different ways in which this object is appearing as an example. Now we want to have a power P to the n that is not primary. And this is actually a bit more surprising because you would have thought that powers of a prime ought to be well behaved, but they're sometimes not very well behaved. So for this example, let's take R to be k X, Y, Z, modulo X, Y minus Z squared. And I'm going to take the ideal P to be generated by X and Z. And it's prime. And you can see it's prime because R over P, well, it means we're quotient out by X and Z. So this is just isomorphic to k of Y, which is an integral domain. And we should sort of draw a picture of this. So we can draw a picture of the spectrum of R informally just by drawing the the cone X, Y equals Z squared. So it sort of looks something like this. Here we've got the X axis. We might have a Y axis here. And the set of points where X, Y, Z squared ends up looking like a double cone. So this is X, Y equals Z squared. And this double cone is something to do with the closed points of the spectrum of R. And the prime ideal P corresponds to the points where X equals Z equals naught. Well, that's just the Y axis. So this is X equals Z equals zero. And it looks reasonably harmless. I mean, we're just taking one of the obvious lines lying on a double cone. And the only thing to be a bit suspicious of is the cone does actually have a singularity at this point here. Well, we haven't actually defined what a singularity is, but it's pretty obvious whatever a singularity is, this double point of a cone is going to be one. Now, let's show that P squared is not primary. Well, let's look at R over P squared. Well, this is K, X, Y, Z. And now we've got to question that by P squared. Well, P squared is generated by X squared, X, Z, and Z squared. And then we've also got to remember to quotient out by this ideal here. Well, we've already killed off Z squared, so we just have to kill off X, Y as well. So it's not very difficult to work out the base for this ring, because this ideal is nicely generated by monomials. And by doing that, you can notice that Y is a zero divisor because Y, X is equal to zero, but Y is not nilpotent. And this is because one Y, Y squared and so on are all linearly independent. I mean, none of them have anything to do with this ideal we are quotienting out by. Well, a primary ideal is one such that our modulo the ideal has the property that every zero divisor is nilpotent. And this obviously fails, so P squared is not primary. Well, if P squared isn't primary, then it must still have a primary decomposition. So what's the primary decomposition of P squared? So this will actually be quite a good example of a non-trivial primary decomposition. We find P squared can be written as X, Z squared intersect X, Y, Z, or squared. And now this is primary for the ideal P equals X, Z. In fact, you can see this because our modulo X, Z squared is isomorphic to K of Y, Z modulo Z squared. And this does have nilpotent elements, but you can check that every zero divisor such as Z is in fact nilpotent. And this primary ideal sort of corresponds to this Y axis. It's a sort of doubled version of the Y axis. You can think of it as sort of being two copies of this Y axis in some sense. On the other hand, this ideal here is primary for the ideal X, Y, Z. So the ideal X, Y, Z is the set of polynomials which vanish at this singular blue point here. So now you can see that we've actually got another example of an embedded component. So this is a minimal associated. This is a minimal associated prime and this is an embedded prime. You can see that this blue point is sort of embedded in this orange line here. So the set of associated primes of P squared has two elements. It is this ideal here and this ideal here. So these two together form the set of associated primes of P squared. So as well as having P as an associated prime, it's sort of mysteriously picked up this extra embedded component. As I mentioned before, this is obviously something to do with this being a singular point. Well, we can generalize the first example we had as follows. So suppose M is maximal in the ring R. Then any ideal I with M contained M to the N for some N greater than equal to 1 is primary. And the proof of this is fairly easy because R over I has a maximal ideal and this maximal ideal is obviously M over I. But it is nilpotent because M to the N is an I which is 0. So we know M to the N and over I to the N is equal to 0 in R modulo I. So all 0 divisors are nilpotent because you can easily check that M is the unique maximal ideal of R over I. In fact, this is obviously just a local ring. So the only 0 divisors are elements of M and they're all nilpotent. So I is primary. There's all 0 divisors of R over I nilpotent. There's an important special case of this that the M to the N is primary if M is maximal. So in the previous example, you see here we have a maximal ideal and its square is automatically primary. And we didn't bother checking it was primary, but now we have checked its primary. Whereas this is a non-maximal, P is a non-maximal ideal and its square isn't necessarily primary, although sometimes it is. Well, now we can define symbolic powers of an ideal. So we do it like this. So we've seen that powers of ideal need not be primary, but as a variation called a symbolic power, which is primary. So let's pick a P to be a prime of R. And now we're going to look at the map from R to the localization of P. And notice that P, R, P is maximal. In fact, localizing at P is really a way of making P maximal by inverting everything not at P. So P, R, P to the N, which is equal to P to the N, R, P is primary. As we just showed that for maximal ideals, the powers are primary. And now we can take the inverse of this, which is the inverse of P to the N, R, P is also primary. And why is this? Well, the inverse of a primary ideal is always primary. Well, the reason for this is that a primary ideal is one such that if you take a quotient, every zero divisor is nilpotent. And if you've got a ring with this property, then every subring also obviously has that property. And since this property holds for subrings, this means that ideals whose quotients have that property are closed undertaking inverses. So you remember earlier on this lecture, we used this to show the inverse of a prime ideal was a prime ideal. So prime ideals have the property that quotient's integral domains and any subring of an integral domain is an integral domain. So here we've got a way of producing primary ideals. And this primary ideal isn't quite the same as P to the N. What does it consist of? Well, this consists of elements X such that X, Y is in P to the N for some Y, not in P. And sometimes this will be the same as P to the N. And sometimes, as we've seen before, it isn't. And this is called the N-symbolic power. And sometimes denoted by P to the N. So although P to the N isn't necessarily primary, we can make it primary by twiddling it slightly like this. And let's see how this acts in the example we did earlier. So you remember we were looking at R is K of X, Y, Z modulo X, Y minus Z squared. So we had this double cone. And we were looking at the ideal P equals X, Z. And now we want to know what is its symbolic square? Well, its square is X squared, X, Z squared. The symbolic square, on the other hand, is just X, Z squared. So this is strictly bigger than P squared because you see that X is in P squared. So P is the symbolic square, but X is not in the square of P. And the reason for this is that we see that X, Y is in P squared and Y is not in P. So it satisfies the condition for being in the symbolic part. The reason this is in P squared is, of course, that X, Y is equal to Z squared. This is some rather strange properties. For example, you notice that X, Z is contained in the ideal X, Y, Z. Well, let's look at the symbolic square of both of them. So if we take X, Z, take the symbolic square, this is just equal to X, Z squared. And if we take the symbolic square of this, this is just the same as its ordinary square, as you can easily check. And now you see that the symbolic square of this is not contained in the symbolic square of this ideal. So just because one prime is contained in another, that the symbolic square of the first need not be contained in the symbolic square of the second. So finally, we can ask, what is the geometric meaning of the symbolic square? Well, you notice that in this cone, X actually vanishes to order two along the Y axis, except possibly at the origin, where everything goes a bit haywire. But is not in the ideal, but is not generated by products of two elements vanishing to order one. So normally you might expect that a function vanishing to order two along some subset, you could write as a linear combination of products of pairs of functions vanishing to order one, but this actually fails in this example. So this gives a sort of geometric interpretation of a symbolic nth power. So the symbolic nth power of an ideal is informally functions vanishing to order n along the set of zeros of p. And if there aren't any singularities anywhere around, this is what you would expect to be. You can just take products of n functions vanishing to order one and take linear combinations of these. But as we've just seen, if there are singularities, things get a little bit more complicated and you have to use a symbolic nth power. OK, that's about all I want to say about primary ideals. Next lecture will be about Hilbert's Norstellensatz.