 This lecture is part of an online commuter algebra course and we will be discussing notarian topological spaces. So, in particular, if R is a notarian ring, then the spectrum of R will be a notarian topological space. And this allows us to describe its closed subsets more precisely. So, let's first define what we mean by notarian topological space. For topological space X, the following are equivalent. So, first of all, we can say that every non-empty set of closed sets has a minimal element. And secondly, we can say that every non-empty set of open sets has a maximal element. And these two are obviously equivalent properties. You just take the complement of each closed or open set. The property says that every increasing sequence of open sets U1 contained in U2 contained in U3 stabilizes. So, this is equivalent to property two by the discussion we had about the different conditions for ring to be notarian. These are actually equivalences for any partially ordered set. And we can say the same for closed sets. I'm going to say the same condition for closed sets. It means every decreasing chain of closed sets stabilizes. And fifthly, we can say that every open set is quasi-compact, which means the same as being compact. So, we notice that condition three is equivalent to condition five because if you've got an increasing sequence of open sets and their union is compact, then it must be covered by a finite number of these so this stabilizes. So, condition five implies condition three and condition three implies condition five can be proved in a fairly similar way. Next, we notice that if R is notarian as a ring, this implies the spectrum of R is notarian. And that's kind of obvious because closed sets correspond to certain ideals of the ring R, not all ideals. And if the ring R is notarian, then ideals have a maximal element which corresponds to closed sets having a minimal element. Because you remember when you go from ideals to closed sets, it actually changes the order of inclusions. Well, you notice that the condition that every open set is quasi-compact seems a bit bizarre because in a Hausdorff topological space, it's quite common for no open sets to be quasi-compact apart from the empty open set. In fact, the condition of for being notarian is almost incompatible with being Hausdorff. You can check that if X is Hausdorff and notarian, this implies X is finite and discrete. So apart from rather trivial examples, Hausdorff spaces are never notarian spaces. And we can ask about is the converse of this true? So if the spectrum of R is notarian, we can ask is R notarian? And the answer is no. It's very easy to give a counter example. We can just take R to be the following ring. We take the ring of polynomials over a field in infinitely many variables. And then let's quotient out by the ideal generated by X1 squared, X2 squared, and so on. Then the spectrum of R is a point. You can see that any prime ideal must contain X1 squared and X2 squared and so on. So the only prime is the ideal generated by X1, X2, and so on. On the other hand, R is not notarian because the ideal generated by X1, X2, X3, and so on. You can easily see is not finitely generated. So notarian rings have notarian spectra, but the converse isn't necessarily true. And what can you do if your space is notarian? Well, you can apply property called notarian induction, which is a common way to prove things about notarian topological spaces. So suppose P is a property of closed sets. C contained in your space X. And suppose X is a notarian. Also, or proper closed subsets of C have property P implies that C has property P. Then all closed sets have property P. So this gives us a way of proving things about closed sets of notarian spaces and the proof of this is easy proof. If not, so if there's some closed set not having property P pick minimal closed set without property P. And we can pick a minimal one because our space is notarian. So any collection of closed sets has a minimal element. So let's pick this minimal closed set C. So all proper sets or proper subsets of C have P by minimality. As we said C was minimal. So anything contained in it has property P. So C has P by assumption. As we assumed that if all proper closed sets of C have P then C has P. Well, this is a contradiction. So all closed sets must have this property. Okay. Well, let's have an application of this. So let's prove the following theorem in a notarian topological space such as the spectrum of a notarian ring. Any closed set is the union of a finite number of closed irreducibles. So the key point is we've got a finite number of them. If we allowed an infinite number then this would be a kind of boring theorem because pretty much any point is irreducible. And here we just apply notarian induction. We take the property P to mean C is the union of a finite number of irreducibles. And what we have to do is we have to check that if every, so suppose every proper subset, proper closed subset of C has property P. We want to show that C has property P. Well, there are two cases. C might be irreducible and this is okay because C is a union of one irreducible, just C itself. Or it might be not irreducible. Well, this means C is equal to D union E where D and E are proper closed subsets. So D and E are both the union of a finite number of irreducibles because they're proper closed subsets of C and so C is because it's the union of D and E. So we've shown that the property of being a union of a finite number of irreducibles is true for C if it's true for every proper closed subset. And now we can just apply notarian induction and we see that every closed set is a union of a finite number of irreducibles. So this means for a notarian ring, we know all closed subsets of the spectrum of R. They're just the union of a finite number of irreducibles. Well, we know what the irreducibles are. They're just closures of points. So provided we know all the points and we know the closures of the points. We know all the closed subsets of a notarian topological space. Sorry, that of a spectrum of a notarian ring. So let's see some examples of this. So the first example, let's just take R to be the ring R to be say the ring of polynomials in two variables over C. Then in algebraic geometry, we define an algebraic set is the set of zeros, set of common zeros of some polynomials. Well, we can define it in terms of spectrum to be the set of prime ideals containing some polynomials. So if we draw the complex plane as a real plane pertaining to complex numbers of one dimensional, a typical algebraic set will look something like a union of points and a union of irreducible curves. And since the spectrum of C, X, Y is notarian, so this is a notarian ring, we see that every closed subset of this. In other words, every algebraic set can be written as the union of a finite number of irreducible algebraic sets. And of course, a similar thing holds in higher dimensions. So in other words, to understand algebraic sets, all you need to do is to understand the irreducible ones, which are called varieties. Another example, let's look at the spectrum of C of X, where X is compact and house dwarf. And remember, this just means continuous real functions. Well, now it's not true that every closed set is the union of a finite number of irreducible. So closed sets of the spectrum are not usually the union of a finite number of irreducibles. Well, first of all, we have to figure out what the irreducibles are. And for that, we need to know that, well, they're just the closures of the points. Now, it turns out, when I said we can know the irreducibles of a topologic of the spectrum of a ring, I was lying slightly because it requires knowing what the prime ideals are. And as we saw last lecture, figuring out the prime ideals of this space is actually kind of rather tricky. However, we don't really need to know because if you draw X as a space here, it's got maximal ideals corresponding to points. And each prime ideal is contained in a unique maximal ideal. So although it's not quite clear what the irreducible subsets actually are, we can see that each irreducible subset has at most one maximal ideal in it. So each irreducible subset contains at most one maximal ideal. So since the maximal ideals form a copy of the compact house or space X, it has huge numbers of those subsets that are not finite union of irreducibles. This ring here is, of course, non-notarian, so its spectrum is a highly non-notarian space. And this theorem of decomposing closets into irreducibles doesn't apply to it. So for the final example, let's do some representation theory of non-Abelian groups. So let's look at the group S3. And it has a group ring. Let's take the group ring over the integers to get a slightly more interesting answer. And there's a problem with this group ring. It's not commutative. And this is a course on commutative algebra, so we're not really allowed to think about group rings of non-commutative groups. So what do we do? Well, we take the center. It's the set of elements of the ring that commute with everything else. And it's not difficult to figure out what the center is, the center of any finite group ring. So the center of the group ring of a finite group just corresponds to conjugacy classes. So the elements of the center are spanned by one. Let's call take the element A, which is equal to one, two plus two, three plus three, one. And this is the three elements of order two in the symmetric group S3. Then we can take the element B to be one, two, three, plus one, three, two. So you remember S3 has three conjugacy classes. There's the element one. There's the triple of involutions. And there's a conjugacy class of three elements, which are the two elements of order three. We've now got a, let's call the center R. And R is isomorphic to Z plus Z plus C as a group, as an Abelian group, where the basis for this is one A, one A and B. Now we need to know what is the product of A and B and so on. And you can easily check that A squared is equal to three plus three B. And B squared is equal to two plus B, and AB is equal to two A. So now we've got a completely explicit commutative ring, and we want to draw a picture of its spectrum. And we want to look at the irreducible components of the spectrum. Well, let's suppose P is prime. And then in R over P, we say that B squared minus B plus two equals zero. So B minus two. So B plus one B minus two equals naught. So B equals minus one or two. That's because R over P is an integral domain is just in R over P. Of course, that's not true in the ring R. And so we've got two cases, B equals minus one or B equals two. And now we have a look at the equation A squared equals three plus three B. And here we find A squared equals zero. And here we find A squared equals nine. So A equals plus or minus three. So we've got three cases to consider. And this gives three homomorphisms from R to Z because we can take B to one, B to minus one, A to zero, or B to two, A to three, or B to two, A to minus three. And this means there are three maps from the spectrum of Z to the spectrum of R. And we've also got a map from Z to R because we've got a map from any ring. So we've got a map from the spectrum of R to the spectrum of Z. So with this information, we can sort of now draw what the spectrum of R looks like. So let's draw the spectrum of Z down here for reference. So we've got this non-closed point zero. And we've got these closed points corresponding to prime ideals, two, three, five, seven, and so on. And now let's draw the spectrum of R above this. So here we've got a copy of the spectrum of Z inside it. So this corresponds to A equals minus three, B equals two. And we've got, it actually gives a copy of spectrum of Z and we've got a point naught. And we've got a map from this to there. And then we've also got the another map where A is equal to three and B equals two. And this sort of looks the same except it doesn't quite because these two points coincide. Because you notice if we quotient out by the ideal two, then these two homomorphisms become the same. And similarly, if we quotient out by the ideal three of Z, these two become the same. And then we've got a third homomorphism where we take A to zero and B to minus one. And this is the same as these two prime three, but not at the prime two. So it ends up looking like this. So we see the spectrum of R has three irreducible components, corresponds to the closures of these three non-closed points of Z. And we can see what all the irreducible closed subsets are. They're just all these points here except we only get two points over the prime two and one point over the prime three. So we've got a complete picture of this space. And this is also in some sense a sort of picture of the representation theory of the symmetric group S3. So if you remember, the symmetric group S3 has three irreducible representations of dimensions one, one and two. And these actually correspond to these three homomorphisms from the center to Z. Because if we've got an irreducible representation S3 to some vector space, K to the N, then we get a homomorphism from the center of S3 to the field K we're working over. So prime ideals of this are very closely related to representations of S3. In fact, if you work it out, these three points sort of correspond to the three complex or rational representations of S3. This corresponds to the identity to the trivial representation where everything maps to one. This sort of corresponds to the sine representation. You remember S3 is a sine representation taking odd elements to minus one. And this thing here corresponds to the two dimensional representation of S3. And then if you look at this picture, you can see that over primes other than two or three, there's not really any difference. And that's sort of because the representation theory of S3 over fields of characteristic not two or three is much the same as in characteristic zero. But if you look here, there's something very weird going on at the primes two and three. And this is a sort of picture of the modular representation theory of S3. So modular representations of a group of representations of the group over finite fields. And they can be quite complicated and difficult. Anyway, this way of drawing a picture of the center of the group ring of a finite group. The irreducible components and the way they intersect each other kind of gives you a way of drawing what the modular representations are doing. It doesn't give you a complete description of modular representations. Roughly speaking, this picture is more or less showing what the so-called blocks and modular representation theory are, but it's any rate. So representations of finite groups over varying fields can be visualized by drawing the spectrum of the center of the group ring and figuring out what the irreducible components are. Okay, that's enough about drawing pictures of the spectrum of a ring. Next lecture, we'll be talking about localization of a ring.