 This lecture is part of an online commutative algebra course and will be about irreducible subsets of the spectrum of a ring R. So you recall that if X is an element of any topological space, big X, then the closure of X is irreducible. Conversely, we show that in the spectrum of a ring R, any closed irreducible subset is of the form X bar for some X in the spectrum of R. So what this means is that we can classify all the closed irreducible subsets. In next lecture, we will see that for many rings, every closed set is a finite union of closed irreducible subsets. So this will classify all closed subsets of the spectrum of a ring. So let's give the proof of this. Suppose we have an irreducible closed subset. Now, any closed subset is of the form Z of i for some ideal i. So let Z of i be some irreducible closed subset. So here i is an ideal. And what we want to do is to show that Z of i is equal to the closure of p, where p is some ideal, prime ideal, considers a point of spectrum of R. So what's p going to be? Well, we could try taking p to be equal to i. The problem is i need not be prime, even if Z of i is irreducible. For example, i could be the ideal for in the integers and then Z of i, which is a single point and is irreducible. So we first of all have to fid with i a bit in order to make it into a prime ideal. And what we're going to do is we're going to replace i by its radical. So this is called the radical of i. So I'll just quickly recall what the radical of ideal is. So the radical of i is just the set of elements such that some power of the element is an i for n greater than or equal to one. And it's perhaps not immediately obvious that this is in fact an ideal. It's obviously closed under multiplication, but to see it's closed under addition, you have to think for a few seconds. So let's just take it's closed under addition. Well, if R and S are in the radical of i, we want to show that R plus S is in the radical of i. Well, we know R to the m is an i and S to the n is an i for some m and n. And now what about R plus S? Well, if we take R plus S to the m plus n and expand it out, it's R to the m plus n plus some binomial coefficient times R to the m plus n minus one S plus some binomial coefficient times R to the m S to the n plus something, one, I guess, times S to the m plus n. And now we see that all these bits here are divisible by S to the n, which is an i. Whereas all these bits here are divisible by R to the m, which is again an i. So all the terms in this are in i. So this is an element of i and this is indeed an ideal. So the radical of i is an ideal. Now we notice that if i is contained in an ideal p, then the radical of i is also contained in p because R to the n in p is equivalent to R being in p. So the prime ideals containing i are the same as the prime ideals containing the radical of i. So we may assume that i equals its own radical. So now we show that if i is the radical of i and z i is irreducible, then i is prime. And so pick a b with a times b in i. So we want to show that a is an i and b is an i because that's the definition of a prime ideal. So suppose not. Well, we have i a times i b is contained in i because a b is an i. So the prime ideals containing i a union, the prime ideals containing i b is equal to the set of prime ideals containing i. Now we notice this is irreducible by assumption and this is closed and this is closed. Now if an irreducible set is a union of two closed subsets, one of those subsets must be equal to for that set. So one of z i a, z i b, say z i a, is the same as z i. However, we're now going to get a contradiction from this because we notice that no power of a is in i because we assume that a was not in i and i is prime. So by the lemma below, which we're gonna prove shortly, there is a prime p disjoint from set one a, a squared and so on and containing i, but this contradicts z i a equals z i because we found a prime ideal p which is contained in that but p is not contained in that one there. So we've reached a contradiction from assuming that a and b are not in i, so we see that i is prime and now it's easier to check that z of i is just the closure of i where i is considered as a point of the spectrum. So that shows that our irreducible closed set z of i is in fact the closure of some point. Well, in the proof of that, we used a lemma that we postponed the proof of. So let's just state the lemma. Suppose s is a multiplicative subset of r. So just remember this means one in s and a in s, b in s implies a, b in s. And suppose it's a multiplicative subset disjoint from the ideal i, be any ideal of r. Then we can find a prime ideal p such that p is disjoint from s and p contains the ideal i. And proof of this is very easy. We pick p maximal among the ideals disjoint from s and containing i. So how do we do that? Well, whenever you're picking a maximal element of something, it's a pretty good guess you're using zon's lemma. So I'm just going to put the word zon here to show that p exists. So the set of ideals that are disjoint from s and contain i is not empty because it contains the ideal i. So by waving zon's lemma, this we see there is such a maximal ideal. And now we show p is prime. So suppose a, b is in p, then we want to show one of a or b is in p. Well, the ideals p plus a and p plus b cannot both contain elements of s because otherwise, because s is multiplicative, p would contain an element of s. So say p plus a is disjoint from s. By maximality of p, p is equal to p plus a because otherwise p plus a would be a bigger ideal that contained i and was disjoint from s. So a is in p. So we have shown that a, b is in p, implies a or b is in p, so p is prime. So that's the end of the proof of the lemma. By the way, this is fairly typical example of the fact that if you've got any set of ideals and you pick a maximal element of that set of ideals, then there's a pretty good chance that that maximal element is a prime ideal. So maximal elements of some set of ideals have a very strong tendency to be prime ideals. Anyway, so that completes the proof that any irreducible subset of spectrum of r is the closure of a point. For the rest of this lecture, I'm going to give another application of the useful lemma. This is a sort of slightly nasty example. The application is the spectrum of c of x is weird. So what is c of x? Well, here I'm going to be taking x to be compact house or space and c of x just means continuous functions, continuous real functions on x. So this is a ring under point-wise multiplication and we can ask what its spectrum is. Well, some bits of the spectrum are easy to figure out. So the maximal ideals of c of x are just the ideals mx for x in x where this is the functions vanishing. And as we saw earlier, you can in fact recover x from c of x by taking the maximal ideals to be the points of the space and then putting a suitable topology on it. And this is quite easy to see because if i is any ideal, suppose it's not contained in any of these maximal ideals. So such that for each x, we can find f of x in i with f of x none zero. Then we want to show that i is equal to the whole of the ring r. So let's write r for this ring. Well, that follows because we can cover x by the sets where f of x is not equal to zero. And these are open sets. So we can cover x by finite number of them because x is compact. So we've got a finite number. This produces a finite number of functions f of x such that x in x implies some f of x so for any y of x, some f of x is none zero at y. And now we just take the sum over this finite number of x of fx squared and we see that this is greater than zero at all points on x. So is a unit in i. So i is just equal to the whole ring. So the maximal ideals of c of x are very well behaved. They're just the points of x as you would probably guess. What about prime ideals? Well, this is where it gets a little bit stranger. First of all, every prime of c of x is contained in some, is contained in a unique maximal ideal mx for x in x. And that's very easy to see. If some ideal p is contained in mx and also contained in my, then we've got these points x, y in our space x and all we need to do is to pick a function that vanishes, that is none zero on x and another function, let's call this function g that is none zero on y. So we're taking f of x is none zero and g of y is none zero. And we also take fg to be the zero function. It's not too difficult to do that on a compact house or space. So we find fg is an i, so it is in p. So if p is prime, then it couldn't be contained in both mx and xy, so p is not prime. So every prime ideal is associated at some point. Next, it's quite easy to see that if p is closed and p is contained in mx, then p is not prime. It's not equal to mx. And this is quite easy to see because if y is not equal to x, p contains an element f with fy, none zero by primeness. So, and from this, you can easily check that p, that if p is closed, it's equal to m of x. I won't bother with the details since this is an algebra course, not an analysis course. However, and this is the point of this example, the none closed prime ideals can be weird. In fact, they're not only just can be weird, they are indeed weird. The problem is to show they exist. And if you try and construct one explicitly, you won't be able to. You can only show they exist sort of indirectly by using the lemma we had earlier. So what we're going to do is to pick a point x and we suppose that this is none isolated. And we're going to let i be the ideal of functions vanishing in the neighborhood of x. And we're going to pick the multiplicative subset s to be the powers of one f, f squared and so on, where fx is equal to zero, but f does not vanish in any neighborhood of x. And you can do this because x is not an isolated point. And then we see that by the lemma, we've got a multiplicative subset s disjoint from a. So we can find a prime p such that p is disjoint from s and p contains i. And the fact that p contains i, this implies that p is contained in the maximum ideal. Mx, because that's the only maximum ideal that can be contained in, but p is disjoint from s. And this implies that p is not equal to Mx because Mx contains this element f, but f is not an element of p. So we found none maximal and for that matter, none closed. Prime ideal. Well, except we haven't really found one because the proof of the lemma involves the axiom of choice. And it turns out that this is one of the cases when you can't really give an explicit construction of the ideal p without using some sort of, maybe some weak version of the axiom of choice. And the moral of this example is that if you're willing to use that topology when you mess around with ideals and you shouldn't even be thinking about none closed ideals because if you start working with none closed ideals, you get all these weird examples. And anyway, the whole point of an ideal is that you can quotient out by it and get a subring. And if you've got a none closed ideal and tried quotienting out by it, then you would get a none-house-dolphed ideal. And if you're using that topology when you mess around with ideals and you start out by it, then you would get a none-house-dolphed topology on the quotient, which is really extremely ugly. I mean, none-house-dolphed topologies are okay on the spectrum of a ring, but you really don't want your ring itself to have a none-house-dolphed topology. Okay, so to summarize, this lecture we showed how to find the irreducible subsets of a ring. And next lecture, we're going to use that to classify all closed subsets of the spectrum of a notarian ring by showing that they're just unions, finite unions of irreducible closed subsets.