 This lecture is part of the online commutative algebra course and will be about the topology of the spectrum of a ring R. So we'll just very quickly start by recalling what this is. If you've got a ring R, reconstruct topological space from it called the spectrum of R, whose points are just the prime ideals. And the topology can be defined in one of two ways. You can either define it to the basis of open sets consisting of the sets U of F, which are the set of prime ideals, such that F isn't in the prime ideal. You can think of these informally as being the places where F is not zero, although that shouldn't be taken too seriously. Alternatively, you can define it by saying the closed sets are the sets Z i, where i is an ideal. And this is just the primes containing i. So this is called the Zariski topology on the spectrum of R. And it's highly non-Hausdorff. So you remember that we worked out the spectrum of Z and it had a sort of one, none open point zero, which we think of as being one dimensional and closure contains the various maximal ideals of spectrum of Z. So what we want to do is to go through some basic topological properties like connectedness and compactness and see how they apply to the spectrum of a ring. So the first result is that the spectrum of a ring is quasi-compact. So what does quasi-compact mean? Well, it means the same as compact, which means every open cover has a finite sub-cover. The reason for the name quasi-compact is for some strange historical reasons that back in the 50s, some people used to use the word compact to mean compact and Hausdorff. And then when they needed to work out with not compact non-Hausdorff spaces they had to invent this word quasi-compact for it. But most people use compact to mean compact but not necessarily Hausdorff. So these two terms are equivalent. So let's show that the spectrum of R is compact. So suppose the spectrum of R is covered by open sets and we want to show there's a finite sub-cover and to do this we may as well assume that they're open sets Ui of our basis of the topology. And what does it mean for spectrum of R to be covered by open sets? This means there's no prime ideal or no maximal ideal that contains all the Fi. And this is just the same as saying the ideal generated by the Fi is by the set of all the Fi is the whole ring R because if it wasn't there would be some maximal ideal containing these. And this is the same as saying that the element one is a linear combination of the elements Fi. However, if one is a linear combination of the elements Fi then it's a linear combination of a finite number of them. So this implies one is finite sum and we can then work backwards and discover that the spectrum of R is covered by finite number of the elements Fi. Notice this depends on the fact that we can't add an infinite number of elements together because we were doing algebra. If we were analysts and allowed infinite sums and this argument wouldn't apply and you find the analogous result doesn't really hold anymore. So next we can look at the property of connectedness and we can ask is the spectrum of R connected? And the answer is sometimes yes and sometimes no. So let's just quickly recall what connected means. This means it's non-empty and not the union of two disjoint non-empty open sets. Well, if R is an interval domain this implies that the spectrum of R is connected. So this follows because the spectrum of R is just the closure of the ideal zero. And you notice that this is prime because R is an integral domain. And if the spectrum is the closure of a single point then any open subset containing zero must contain the whole of the spectrum of R. So R can't be the union of two disjoint proper open sets. Now let's see some examples where the spectrum of R is not connected. Well, one easy example is when R is a product of two rings A and B. So if we denote the units of the rings A and B by one A and one B, we notice that one A times one B is equal to one. So any prime P of R contains one A or one B so it contains A or B. And from this you can see very easily that the primes of R form a disjoint union of the primes of A and the primes of B, well, more or less. So we can write the spectrum of R is equal to the spectrum of A union the spectrum B and you can check that these are both open subsets. For example, you can look at the open sets U one A and U one B and one of these is spectra of A and one is spectra of B. And I'm not gonna say which is which because I always get muddled up about it. So an example of this, we could take the ring Z modulo 120 Z the integers modulo 120. And we recall there's the Chinese remainder theorem from number theory, it says that Z over MNZ splits as a product of Z over MZ times Z over NZ whenever M and N are co-prime. And by using this, we see that this thing factorizes as say Z over 8Z times Z over 3Z times Z over 5Z. And since there's a product of these rings that spectrum is just the union, this joint union of the spectrums of these three rings and it's obvious what the spectrum of these are. They all only have one prime ideal. So the spectrum of Z over 120 Z just looks like this. It's just three distinct points where this corresponds to the prime ideal two, this three and this to five. And the spectrum of Z over NZ for any N is similar. It's just a sort of disjoint union of one prime for each prime divisor of N. So you can sort of see the Chinese remainder theorem visually as just saying the spectrum of this is the union of distinct points. Another example is the group ring of an Abelian group. So let's look at the group ring QG where G is an Abelian group and Q is the rational. So this is a basis of elements G and G and the multiplication is defined like this. You say G of H is equal to G of H. Well, this means multiplication in the ring and this means multiplication in the group. So we just define the multiplication of elements in the ring to be the same as it is for the group. And this is obviously a ring. So let's just do an example and let's take G to be the line for group with elements one, A, B, C. So A squared equals B squared equals C squared equals one and A, B, C equals one. And let's try and work out the spectrum of the group ring Q of G. Well, that's quite easy because this splits as a product. In fact, this group has four id impotence. Let's just follow this. We take one plus A plus B plus C over four and the other three look like this. One minus A minus B plus C four and there should be one more, which is one plus A minus B minus C over four. So let's call these E naught, E one, E two, E three. Then we see that the group ring just splits as a product of these because E i squared is equal to one, E i, E j equals naught for i not equal to j and E zero plus E one plus E two plus E three is equal to one. So if we put R equals Q G, then R is equal to E naught R plus E one R plus E two R plus E three R. So R splits as a product of these rings here and each of these is isomorphic to the rational numbers Q. So the spectrum of Q of G just looks like this. It's four points corresponding to these four elements here. In fact, you can think of these as being the characters of G. So you remember a character of G is just a homomorphism from G to, well, it could be the nonzero complex numbers or the nonzero rationals. I don't really care in this case. So there are four characters of G and these correspond to the four idempotents. For example, the idempotent one minus A minus B might plus C over four correspond to the character taking A to minus one, B to minus one and C to one, which are coefficients that appearing here. If you want to know what happens for an arbitrary group, let's just work over the complex numbers to make things a little bit simpler with the complex group ring. Then this splits as a product over the characters of G where now a character is a homomorphism of G to the complex numbers. And we call the characters by the Greek letter chi which looks a bit like an X. And if we've got a character chi, then we can define an idempotent just by summing over G of chi of G times G over the order of G. So in the example we had before, if we had the character where chi of A equals chi of B equals minus one and chi of C equals one and G has order four, then we get the character one minus A minus B plus C over four which is one of the idempotents we had earlier. Well, as all disconnected rings are not really all that common. So the property of being connectedness is not all that important. What's more useful is a rather stronger property called being irreducible. So a set is called X is irreducible. That this just means X is non-empty and X is not the union of two proper closed subsets. So this condition here just says that any two non-empty open sets intersect. So this is a very strong property. You notice that if X is house-dorff and irreducible, this just implies X is a point because if it's got two points, we can find non-empty open sets, disjoint open sets containing these points because it's house-dorff and therefore it's not irreducible. So if you're an analyst and all your topological spaces are house-dorff, then the concept of being irreducible is completely useless because nothing is irreducible. But for the spectrum of rings that they quite often are irreducible. For example, the spectrum of Z is irreducible. And for that, we just notice that the spectrum of Z is the closure of the ideal, is the closure of the ideal zero. And if X is a point, the closure X bar is irreducible. And that's because and if the closure is the union of two closed sets, one of them must contain X. So it must be the whole of the closure of X. So let's see an example where a ring has a spectrum that's not irreducible. For this, we're going to take the group ring over the integers of G and let's take G to be the climb four group again. And let's ask what is the spectrum of this integer group ring? So last time we did the rational group ring and found it was just four points. The integer group ring is a little more interesting. First of all, we notice we've got homomorphisms of rings going like this. So this is just the obvious one taking the identity to the identity. And then we've got four homomorphisms to the integers which are the four characters of Z of G. So these take A to plus or minus one and B to plus or minus one. Now, if we've got a homomorphism between rings, we get a homomorphism between the spectra going in the opposite direction. So we have spectrum of Z is the image of the spectrum of Z of G and we get four copies of the image of the spectrum of Z mapping to it. And from this, it's fairly easy to work out what the spectrum of Z of G looks like. First of all, we draw the spectrum of Z down here. So this is going to be down here and we've got these points two, three, five and so on. And this is the generic point zero. And then spectrum of Z of G is actually the union of these four copies of spectrum of Z and they all map down to this. So here we've sort of got a copy of spectrum of Z. Again, we have the generic point zero and these points two, three, five and so on. And this, we draw this map in red. So we have a sort of red map going down here. And then we've got a second copy of the spectrum of Z which comes, well, then it sort of intersects this point here and the others do that as well. So what is happening is that we've got four copies of the spectrum of Z, but they're not disjoint. They overlap at this point too. And you can see that because if we take the spectrum of Z modulo two Z with coefficients in G, you can easily check it's only actually got one prime ideal. So these things all coincide. And now what's going on is the spectrum of Z of G is not irreducible because it's the union of these four proper closed subsets. These are called its irreducible components. And as we'll see fairly shortly, it's quite common for the spectrum of a ring to be the union of a finite number of irreducible components. And as before, first of all, these irreducible components as before correspond to the characters of this finite group. And secondly, if you sort of look carefully here, you can see there's the spectrum of the rational group ring of G sort of sitting with these four points here. So the spectrum of Z of G is a kind of refinement of the spectrum of Q of G. And the spectrum of Q of G is disconnected. It's just four components. But if you look at the spectrum of Z of G, these all become connected, but the spectrum of Z of G still has four irreducible components. Another example of irreducible components is something like, let's take a variety in two-dimensional affine space given by product of two polynomials. I'll take my polynomials to be X and Z. So the spectrum of this ring can be pictured as a union of the X-axis and the Y-axis because any prime ideal of this ring must contain either X or Y. So we can think of it as being either a prime ideal of C of X, which is basically a line, well, pretending the complex numbers are a line, or it's a prime ideal of C of Y. And the spectrum of these two spaces are just complex lines, which I'm going to draw as real lines because I don't know how to draw four-dimensional space. So the spectrum of this thing consists of two lines and you can sort of see that these are two irreducible components. So the spectrum of this ring is reducible, but it's the union of two irreducible subsets. You've got to be a little bit careful with drawing pictures of curves. For example, if you look at the an elliptic curve, so its coordinate ring will look something like this. So we can just think of this as being polynomial functions on an elliptic curve. So this might be the curve Y squared equals X cubed plus X. Let's put a plus there. We've got a sign over here somewhere. So you might look at this. So the spectrum of this, well, it doesn't really look like this, but this is the sort of first approximation to it. And you might sort of carelessly think, well, this elliptic curve has two components. Therefore, the spectrum is disconnected. And this is completely wrong because the elliptic curve is disconnected in the Euclidean topology, where we embed it into Euclidean space. However, this Euclidean topology doesn't have much to do with the topology of this space as the spectrum of R. And in fact, you remember that the close sets of this in the Zariski topology are just going to be finite unions of points. So this, although it looks disconnected, it's really not only connected, but it's even irreducible. By the way, I've missed out all the complex points of this, which should also, I guess those would make it connected too, but anyway, you've got to be a bit, I'm just adding a warning that just because the graph of something looks disconnected in Euclidean topology does not mean it's disconnected in the Zariski topology. Um, next, so next lecture, we will say more about irreducible sets. So we're first going to show that the irreducible sets of the spectrum of a ring correspond exactly to the points. And then we're going to study the decomposition of close sets into a finite union of irreducible subsets.