 This lecture is part of an online algebraic geometry course about schemes and we'll be covering finite and quasi-finite morphisms. So if you have a morphism of rings, say R to S, there are two finiteness properties you can put on S. The first one says that S is finitely generated as an R algebra. The second one says that S is finitely generated as an R module. So we've got these two different words here and these are quite different. First of all, if it's finitely generated as an R module, it's obviously finitely generated as an R algebra, but the converse isn't true at all. For instance, if S is equal to the ring of polynomials in X, this is finitely generated as an R algebra by the element X, but it's obviously not finitely generated as an R module because you need an infinite number of generators, one XX squared X cubed and so on. Now, for schemes, we've given an analog of being finitely generated as an R algebra. This corresponds to morphisms of finite type or locally a finite type. This one is going to correspond to finite morphisms. And you see the terminology is a bit confusing because finite type and finite morphisms are quite different. So if X to Y is a morphism of schemes, it's called finite if Y is covered by open affines VI, so this is going to be the spectrum of some ring VI so that F to minus one VI is open affine in X and say this is equal to spectrum of AI and AI is a finite VI module. So we're saying that AI is finite as a VI module. If it was finite as a VI algebra, then this morphism would be a finite type. So this is a much stronger condition. So we've said this is true if Y is covered by open affines. It's actually a local condition, so you can check this is, if it's true for some cover but by open affines VI, then this condition is actually true for all open affine sets VI. So it doesn't matter whether you take it covered by affines VI or take all open affines VI. So the first property says that if X to Y is a finite morphism, then all fibres are finite and discreet. So we'll first prove this. First of all, it's enough to prove for the case when you have the spectrum of S mapping to the spectrum of R. In other words, R maps to S. That's because Y is covered by open affine, such as the image is open affine, so you only need to check it for these open affines. So what we've got to say is suppose we have a morphism of rings F from R to S and P is prime in R and S is a finitely generated R module, then only a finite number of primes Q in S with F to the minus one Q equals P. So just to draw a picture, here we've got the spectrum of S and here we've got the spectrum of R and what we're doing is we're picking a prime P here and we're looking at the inverse image and saying that only a finite number of primes Q in the inverse image. And this is quite easy to prove. First of all, we can assume P equals zero. You just replace R by R over P and S by S over PS and you can check that this is still a finitely generated module over this ring and you're not affecting the primes whose inverse image is P very much. Secondly, so this means R is an integral domain and P is equal to naught. Second, we can take the quotient field, the field of quotients of R and again, this doesn't affect the primes who's in S whose inverse image is zero. You're just inverting all non-zero elements of R which you can assume as a subring of S. So we can assume R is a field but then S is a finite dimensional vector space over R so is an artinian ring. So remember, artinian rings are one that satisfy the descending chain condition for ideals and if R is a finite dimensional vector space then obviously it satisfies the descending chain condition. And the basic result about artinian rings is that it has only a finite number of prime ideals. Moreover, all these prime ideals are maximal and they're all open in the spectrum of R. In the spectrum of R, so the fiber of this map from spectrum of S to spectrum of R is finite and discrete. There's a bit of a problem here because this property that all fibers of finite is related to the property of quasi-finiteness and we're going to give a definition of quasi-finiteness. Well, we're going to give three definitions because unfortunately there are several in equivalent definitions of this in the literature. So growth and Dick gave a first definition in I think it was SGA1, which says that quasi-finiteness mean all the fibers are discrete of finite and discrete. He later added, he later gave a second definition of this in I think EGA2, where he added the condition that the morphism is of finite type. Hart-Shorn in his book gives a third definition, he just says the fibers are finite. And these three definitions are in equivalent so whenever anybody is talking about quasi-finiteness, you have to sort of worry about which definition you're using. In other words, you've heard, you've seen there's a problem with algebraic geometry and scheme theory, there are far too many definitions. You not only have to keep track of all the definitions, you have to keep track of everybody's version of these definitions. So the three different versions of quasi-finiteness Fortunately, in practice, most of these three definitions turn out to be equivalent to most examples. So we've shown that a finite morphism is quasi-finite. Actually it's quasi-finite according to any of these three definitions. The converse isn't true, and it's quite easy to give a counter example to this. Suppose we look at the spectrum, of KXX minus one, and this maps to the spectrum of KX. So this is just an affine line, and this is an affine line with the origin removed. Okay, this certainly has finite fibers. Obvious, the fiber is either a single point or it's empty in the case of this point here. So this is quasi-finite. But not finite because KXX minus one is not a finitely generated KX module. Finitely generated KX algebra, but not finitely generated as a module. The other slightly disturbing thing is about this example is that the morphism from KXX minus one is not a finitely generated KX module. What's interesting about this example is that the morphism from KXX to itself is finite, rather obviously it's just an isomorphism. So this is actually in some sense smaller than the finite morphism because you've taken a finite morphism and removed points from it, and it's no longer finite, which is a slightly disturbing property, but anyway. So let's just sum the relations between these various properties. So finite implies quasi-finite and this implies finite type. Here we're using growth index second definition. In growth index first definition or Hart-Schmidt's definition, this implication doesn't hold. And finite type then implies quasi-compact. And locally finite type. So finite and quasi-finite are rather strengthenings of finite type. So let's have some examples. So suppose you've got a plane curve and you project it to the x-axis. Is this finite or quasi-finite or whatever? Well, let's have a look at some examples. So if you take the curve x equals y squared and project it to the x-axis, this is a finite morphism. This is because the coordinate ring here is kx and the coordinate ring here is kxy over y squared minus x. And this is a module over kx generated by two elements. So this is generated as a module by one and y. So it's finite. So this morphism from this parabola to the x-axis is finite. On the other hand, we can take a different curve. Let's take the curve xy equals one and project it to the x-axis. And this curve is not finite because here we have kxy modulo xy minus one is essentially just the ring of polynomials and x and its inverse, which as we saw is not a finite morphism. So this is quasi-finite but not finite. So if you project a curve in the plane to the x-axis and ask whether it's finite or not, that's a little bit tricky. You have to think about what the curve is. Another example of a finite morphism is you just take the ring of integers of an algebraic number field, say zi, and map it to the spectrum of z. So here the spectrum of z has a generic point and it is primes two, three, five, seven and so on. So here's the generic point zero. And the spectrum of zi we saw, the map to spectrum of z looks a bit like a curve. So it's got a sort of ramification point and then has a point three two points above five, two plus i, two minus i and so on, point seven and so on. So you can see that this is, that the inverse image of each point of spec z is a finite discrete set. So in terms of a number theorist would say that each prime in the spectrum of z splits as the product of a finite number of primes in spec zi. But this is a geometry course. So what we do is we say this map from the spectrum zi to spec z has finite fibers. In other words, it's quasi-finite. And it's also finite because zi is a finitely generated z module, rather obviously it's generated by one and i. So if we take the spectrum of k of x, so this is rational functions and we map it to the spectrum of k of x. So this is polynomials. So it looks like the affine line. So this gets mapped to the generic point. Then this is, it is finite fibers. So it's quasi-finite according to growth and its first definition and according to Hart-Shorn. So it's quasi-finite by Hart-Shorn and growth and its one, but it's not quasi-finite by growth and its second definition. Finally, we will just show that Hart-Shorn's definition and growth and its first definition aren't the same. You have a rather silly example. Let's take the spectrum of kx. So this is just the affine line and let's have a rather stupid scheme which is an infinite union of a countable number of points. So this is another example that's quasi-finite according to growth and its first definition but not according to growth and its second definition. And the example I meant to give quasi-finite according to Hart-Shorn Hart-Shorn's definition, but not growth and its definition is where you take the homomorphism from the spectrum from k to the localization of the polynomials at zero. So in other words, you're taking a map from the spectrum kx at zero to the spectrum of k. Now the spectrum of this has two points, a closed point and an open point and the spectrum of k just as one point. So the inverse of every point here is finite because it consists of just two points. However, it's not discrete. This is not a discrete topological space. So this is finite but not discrete. So it's quasi-finite by Hart-Shorn but not by growth and its first or growth and its second definition of quasi-finiteness. So the point of this is that if you know the fibres of homomorphism are all finite, this doesn't really tell you very much useful unless you've got some other conditions like being a finite type. In particular, it's rather weaker than asking for the morphism to be finite. In next lecture, we'll see more precisely what the relation between having finite fibres and being a finite morphism is when we discuss open and closed notions.