 This lecture is part of an online algebraic geometry course about schemes and will be mostly giving a few more examples of schemes continuing from last lecture. So the first example we're going to look at is the spectrum of r of x, the ring of polynomials with real coefficients. Now we saw earlier that the spectrum of c of x just looks like c together with a generic point. This generic point comes from the ideal 0 and the points of c correspond to the ideals x minus alpha. So you might guess that the spectrum of r is going to look like the real numbers are together with a generic point. So we have a line of real numbers and then we have some sort of weird generic point whose closure is the whole thing. So this would be the ideal 0 and a point like 3 would correspond to the ideal x minus 3. However, that's not all the prime ideals of r of x because the prime ideals correspond to irreducible polynomials and there are two sorts of irreducible polynomials. We can either have things that form x minus alpha or we can have x squared plus bx plus c with two complex roots. So this will have a complex roots beta plus or minus gamma times i. So we should really add extra points that correspond to pairs of complex numbers. For instance, there's a sort of single point here corresponding to the ideal, I don't know, x squared plus 3x plus 5 or something like that. If I've got that with imaginary roots. So this corresponds to this ideal and it corresponds to a single point in the spectrum which corresponds to a pair of complex conjugate numbers. So the points of the spectrum really consist of orbits of the complex numbers under complex conjugation. And you notice complex conjugation is really the absolute Galois group of the real numbers that you take the algebraic closure of reels and take the Galois group of that over the reels. And similarly for other fields that points of the spectrum of polynomials over a field are actually going to correspond to orbits of elements of finite extension of the field. And of course there should be a generic point. I guess I should really draw this generic point a bit bigger because it should also include these points here. Next, we're going to look at an example of a curve mapping to the real line so here I'm going to take the curve y squared equals x cubed minus x. And it's coordinate ring is just going to be say r of x, y over y squared minus x cubed plus x and its spectrum is going to look a bit like the following. So the real points of the spectrum are going to look like the curve. And then it's also going to have various complex points which are sort of pairs of complex conjugate numbers. For instance, we could take say that the points x equals minus two and then y is going to be plus or minus the square root of minus six. So we'll get a pair of points. Well, this ring contains the ring r of x. So there's a map from r of x to this string. And this corresponds to a map of this, the spectrum of this ring to the spectrum of this ring. So let's draw the spectrum of the real line. So here we have the spectrum of r of x, which is we did in the previous sheet. So it's, it's, it's got some real points and it's got a generic points, and it's also got some pairs of complex conjugate points. And we can map the, we map this spectrum, in other words, this curve to the real line. And there are three things that can happen. First of all, there are these funny points that kind of ramify. You see there are sort of, there's a, there's only one point mapping to this real point, but it's a sort of double point. So we call these ramification points. So ramification means branching out. And you think of this blue curve as kind of branching out from here. It's got two branches there and there. Next, we can have points that where a point on the real line just splits into two points in this curve. So that's fairly common. And finally, we have examples where a point on the real line only splits into one point consisting of a pair of complex conjugate points. So if x is minus two here, then it's that there's only a single point of the spectrum here, but it's not really a ramification point. It's really sort of a double point rather than a single point occurring with multiplicity two. So that's three things that can happen. Now I'm going to look at a different example. This time I'm going to look at the ring z of i. This is just the Gaussian integers. And there's a map from z to this. And now I need to figure out what the spectrum of z of i looks like. Well, z of i is a principal ideal domain. So you may remember from number theory. And it is three sorts of primes. What you can do is you can take every prime of z and figure out how it splits in z of i. So that's three things that can happen. First of all, prime two splits as one plus i squared times a unit, possibly minus i. Three just remains a prime, but five splits as two plus i times two minus i because five is equal to two squared plus one squared. So whenever a prime is the sum of two squares, it factorizes. So this is typical for the prime two. This is typical for primes that are three mod four. And this is typical for primes that are one mod four. So there are three different things that a prime can do. Now the fact that we've got a map from z to z of i means we've got a map from the spectrum of z of i to the spectrum of z. So here I have the spectrum of z and this is going to contain points like two, three, five, seven, eleven, thirteen and so on. And it's also got this generic point zero and we've got a map from z of i to z. Now what the fibres are and there are three sorts of fibres. First of all, the prime two sort of is the square of a prime here. So we're going to draw that as a one plus i. Next three kind of remains as a single point. So we're going to draw that in yellow, I believe. And finally, we've got various points like five, which split as two points. So here we have two plus i, two minus i, nine above five. And similarly for thirteen, we'll get two points, three plus two i, three minus two i. And for seven and eleven, we're only getting a single point. And what you should do is you should think of this as being a single curve. So we have a sort of blue curve which corresponds to the generic point and it sort of kind of goes like that. It sort of splits up a bit. And now if you look at these two diagrams, you can see they're very, very similar. We've got this generic point corresponding to the prime zero. And above each point of spec z or spec of r of x, it can do one of three things that can ramify. We should really have drawn this as ramifying, so it should sort of come round like that. Or the point here is the image of two different points of the spectrum. Or the point here can be the image of the double point of the spectrum. So the point of this is that the Gaussian integers behaves very, very much like an algebraic curve. And the homomorphism from z to the Gaussian integers corresponds to the projection of a typical algebraic curve onto one of the axes. So we can think of spec z as sort of similar to an affine line. So what is going on is that r of x and z are both principal ideal domains. And if you've got a principal ideal domain, its spectrum rather resembles that of an affine line. And you can do the same thing for all algebraic number fields. You can think of them in this geometric way. So for example, if you've got a geometric object, there's something called the Picard group that we may discuss later, which is just a group of line bundles. So you won't worry too much about what that is just now. So roughly speaking, a line bundle is a way of assigning a one-dimensional vector space to each point. For instance, you could take the tangent bundle of this that would assign a one-dimensional vector space consisting of the tangent space. And they form a group, which is the Picard group of a geometric object, which is really important. And if you've been to an algebraic number theory course, you know that for a ring of integers of an algebraic number field, there's something called the ideal class group, which consists of fractional ideals, modular principal fractional ideals, and is really important in number theory because it vanishes if and only if you have unique factorization domain. And it turns out that you can define Picard groups of schemes. In particular, you can define the Picard group of a ring of integers of an algebraic number field. And it turns out to be exactly the same as the ideal class group. So these two apparently completely different things, ideal class groups and groups of line bundles, turn out to be really the same thing if you're working with schemes. Now I'll give another example. I'm going to look at the spectrum of a local ring. So I'm going to take CXY, localised at the ideal XY. This means it consists of all rational functions g over h with h naught naught and not equal to naught. We've got that the right way around. And we want to figure out what this looks like. Well, first we recall what the spectrum of CXY looks like, which we did last time. And you remember it's got some maximal ideals corresponding to closed points. And it's got some not so maximal ideals corresponding to various lines. And it's got even less maximal ideal consisting of the ideal zero consisting of a generic point. So now here XY is going to is the ideal correspond to this point in it turns out the spectrum of this localisation will discuss localisations a bit next lecture turns out to be a sort of blown up view of this bit of the spectrum. So let's look at the ideals of this. First of all, we've got the ideal zero, which corresponds to a, it's going to be a generic point. And then we've got ideals of the form that then we've got the maximal ideal. Let's do the maximum ideal next. I think I was using purple for this. So the maximum ideal is XY. And it's maximal because any anything not at this is invertible. And so we just get a single closed point. And then we've got various ideals of the form f, f XY equals zero. And these are just going to be little lines passing through the, the point of the maximum ideal. And if you look at this, it sort of as if you took a magnifying glass and looked very closely at this point here. So all you're seeing are the tiny, I mean, you're looking at this so closely, you can only see one point, one closed point. But you can still see the sort of vestiges of these lines. I mean, I'm not quite sure how you can see a line while only seeing one point on it. But somehow that's what the, that's what the spectrum is doing. So this operation of taking the local ring at a point geometrically corresponds to kind of really focusing on in on this point. That's why it's called localization. By the way, you're sort of localizing at this point looking very, very closely at it. Now we will have a trick question. What is the spectrum of the zero ring? So the zero ring has only one element zero and one is equal to zero. Well, the points are the prime ideals. So what are the prime ideals of this? Well, you might think zero is a prime ideal, but it isn't because if we question out by this prime ideal, we just get the zero ring and the zero ring is actually not an integral domain by definition because an integral domain by definition has one not equal to zero. So the spectrum of zero is empty. And it is just one open set, which is the empty set. And in order to find that sheaf, we need to say what is the sheaf of this empty set. And the sheaf of the empty set is just the zero ring are with just one element. So that's the spectrum of zero. It basically corresponds to the empty topological space. Finally, we'll finish by asking what is the spectrum of a product of two rings? Well, we need to know what are the prime ideals? Well, the prime ideals are going to be homomorphisms to integral domains. And if you've got a homomorphism of r times s to an integral domain, then either s maps to zero or r maps to zero. So we can see the primes of s of r times s are the primes of r union the primes of s. And it's quite easy to see from this spectrum of r times s is just a disjoint union of topological spaces spec of r times spec of s. And we can check that this is an open subset and this is an open subset. So it really is a disjoint union of topological spaces. And similarly, if you've got an open set of spec of r times s, it's the union of an open set u union and open set v. And the ring of functions on this is obviously just going to be the ring of regular functions on u times the ring of regular functions on v. So the geometric interpretation of this is that if you've got to, if spec of r is here and spec of s is here, then these are disjoint and nothing to do with each other. And all you're saying is that a function on the union of these two spaces is given by function on this space and a function on this space. In other words, you're taking the product of the functions on this space and the functions on this space. Okay, the next task is to show that the spectrum of a ring as we've defined it really is a ring space, and we haven't yet checked the sheaf condition. And to do that, we are going to first review the operation of localization of a ring.