 This algebraic geometry video will be about the Axe-Grotendick theorem. So the Axe-Grotendick theorem says that if you have a regular map from a variety v to another variety v over an algebraically closed field, field k, if f is injective, it is surjective. Of course, if a map from a set to itself is injective, there's absolutely no reason why it should be surjective in general if the set is infinite. And this theorem has an amazingly simple proof found by Axe using some rather odd techniques from model theory. So let's first of all just have a couple of examples. So if the field is not algebraically closed, then the theorem is false because we can take, for example, k equals q, and we take the map f to be the map taking x to x cubed from a1 to a1. So this map is injective, but it's obviously not surjective because not every rational number is the cube of something. You can also ask if a map is surjective, is it injective? And this is obviously false. So there's example one. So example two, let's just take k be the complex numbers and f of x equals x squared. So this is surjective, not injective. Okay, we can just give the proof of the Axe growth in the theorem in a few, almost a few minutes. So first of all, it's trivial if k is finite. Because if k is finite, then v is only a finite number of points. And if you've got a function from a finite set to itself, it's injective if and only if it's surjective. So over finite fields, if it's surjective, then it is in fact injective. Next, it's true over any algebraic extension of a finite field. And the key point is that the coefficients of all equations defining a variety and the coefficients of a point all lie in some finite field because they're only a finite number of coefficients of equations. They're only a finite number of coefficients of a point and they only have a finite number of coordinates. So you can find some finite field containing it all. And that means that if you've got an injective map from an over an algebraic extension of a finite field and you want to show it's surjective, you take some finite extension containing all the coordinates of some point and just apply the theorem of that finite extension. Notice by the way that over an algebraic extension of a finite field, surjective no longer implies that the map is injective. If you look at the argument it's going from finite to finite extensions, the argument for injective implies surjective still holds but the other way around doesn't essentially because showing something as surjective involves finding a point which you can find over a finite extension. So if a map from the problem is that a map may be surjective for an algebraic extension of a finite field without being surjective for the individual finite fields. Well now it's true for algebraic extensions of finite fields. So what you can immediately deduce is true for all algebraically closed fields. So why can't you do this? Well we know it's true for algebraic closures of finite fields and now we just use a result from model theory that if a statement is true for an algebraic closure of an algebraic closed field in characteristic zero if and only if it's true for algebraic closed fields of all sufficiently large characteristic. And the second result from logic says that if a statement is true for one algebraically closed field of given characteristic it's true for all algebraically closed fields of that characteristic. Well this is obviously false as I've stated it. For instance the complex numbers are uncountable and you can find algebraically closed fields of characteristic zero that are not uncountable. So it's not true for all statements. Well it's true for all statements. These have to be first order statements. So I'll explain what first order means in a moment. So this statement here is sometimes called the left shet's principle. So it says the left shet's principle sort of says that if something is true over the complex numbers and it's true over all algebraically closed fields of characteristic zero. And this is nice because to prove things over the complex numbers you can use all sorts of things like analysis and differential geometry and then you find it's true for all algebraically closed fields of characteristic zero. And again not all statements have this property. It applies to first order statements or statements that can be reduced to first order statements. So now I should explain what a first order statement is. So first order is a statement that can be stated in first order logic. And in first order logic the things you're used to make statements are the usual logical connectives like and or not and implies. You're allowed to make statements that use the operators of your field. So we're allowed to make statements like a plus B a times B a minus B and so on. And you're allowed to quantify over all elements of whatever field or model you're working in. So you can say for all X in your model something rather is true. So this is usually abbreviated as for all X. And you can say there exists an X in your model such that something or other which is usually abbreviated as there exists an X. So for example, how do you say a field is algebraically closed? So you might want to say for all polynomials a n X dn plus a n minus one X dn minus one and so on. There is a root. Well, this isn't allowed. This is not a statement of first order logic. And the reason is you're saying for all polynomials and you can't say for all polynomials. You're only allowed to say for all elements of your field. So what can we do? We can say for all polynomials of degree n as this is just saying a n X dn plus a n minus one X dn minus one and so on. And this is a perfectly good statement for first order logic because it's the same as saying for all a n a n minus one up to a naught something or other is true. This polynomial has a root. So how do we say a field is algebraically closed in first order logic? Well, you can't quite. In order to say a field is algebraically closed. It takes an infinite number of statements of first order logic. But taking an infinite number of statements of first order logic is generally a rather harmless operation. All that means is that the axioms for an algebraically closed field are an infinite set of first order axioms. You have all the axioms for field and for every integer n you have the axioms stating that polynomials degree n have a root. So what you can't do is do things like state for all integers n something rather is true that the point is you are not allowed to quantify overall integers. You can't say all for all integers something has such and such a property. You're not allowed to quantify over all elements of the field. So that's roughly what a first order statement is. And next we have rather basic results that the theory of algebraically closed fields of given characteristic P. So P can be 0, 2, 3, 5, etc. It's either a prime or zero is complete. So complete means that every statement you can write down in first order logic is either provable or disprovable. So everything provable or disprovable. Well, if you've heard of Gerdel's incompleteness theorem, this might be possibly a little bit puzzling because Gerdel's theorem says that there are always statements that you can't prove or disprove. Well, if you look at the statement of Gerdel's theorem, it only applies to logical systems that can encode the positive integers in some sense. And the axioms for an algebraically closed field can't actually encode the integers. There is no way to define the integers using the axioms for an algebraically closed field. I mean, if you take the complex numbers, you can certainly define the integers as a subset of the complex numbers, but you can't do it using just first order logic. So you can't actually encode the constructions needed in Gerdel's theorem and you can find complete theories provided they can't encode the integers and the algebraically closed fields of given characteristic are a very well known example of this. The proof that this is complete is actually rather easy that there is a unique algebraically closed field of given characteristic and cardinality equal to the continuum. And it follows fairly easily from a bit of first order logic that if there's a unique model of some theory and some characteristic, then that theory is complete. You can actually deduce this from Gerdel's completeness theorem. He actually has a completeness theorem as well as an incompleteness theorem. The completeness theorem says that if any first order theory is consistent, then you can find a model of it. And if the theory of algebraically closed fields were not complete, then you could find two bigger theories, one of which some statement was true and one of which was false, and then you could find two different models of any given infinite cardinality, which should be a contradiction. So the theory of algebraically closed fields is complete because algebraically closed fields are unique up to isomorphism in uncountable cardinalities. Encountable cardinalities, they aren't unique up to isomorphism because they can have a transcendence degree, which varies. Well, now that we've seen that algebraically closed fields are complete, we can now see that anything true for all algebraically closed fields of large characteristic is true for algebraically closed fields of characteristic zero. And the reason for this is you stop and think about how can you say something is of characteristic zero. Well, you can't say a field is of characteristic zero with just one statement of first order logic. So you might say nx is not equal zero for n greater than zero and being an integer. Well, you can't state this in first order logic because you can't talk about integers in first order logic if you're working in the theory of fields. So that's not allowed. What you can do to define characteristic zero is you can say 2x is not zero, 3x is not zero, 5x is not zero and so on. So we need a countable number of statements in order to say a field is of characteristic zero. And now suppose you've proved something for algebraically closed fields in characteristic zero. So any proof uses only a finite number of axioms because a proof is finite. So it can only use a finite number of axioms. So it can only use a finite number of these axioms. So if you've proved something about algebraically closed fields of characteristic zero in first order logic, it's automatically true for all algebraically closed fields of sufficiently large characteristic because your proof can only eliminate a finite number of characteristics. So true for algebraically closed characteristic zero implies true for algebraically closed and for large characteristics. And conversely, if something is true for all algebraically closed fields and large characteristics, it's also true for an algebraically closed field in characteristic zero just by applying this to the converse statement. And this now implies the ax growth in the theorem because the ax growth in the theorem again, you can't state it using a single statement of first order logic because it's talking about arbitrary varieties and a variety could be of any dimension and defined by any number of equations. But for any family of varieties defined in given affine space with a fixed number of equations, you can state the ax growth in the theorem for that in first order logic. And that's now true in characteristic zero because it's true in all sufficiently large finite characteristics. So the ax growth in the theorem is can be produced as a conjunction of a countable number of statements of first order logic. So when you prove it in characteristic p greater than zero that automatically proves it in characteristic zero. So this is a rather unusual technique because you're proving something in characteristic zero by first proving it in characteristic p and then kind of waving your hands and applying this weird idea from model theory and first order logic. So in characteristic p that there are some powerful techniques you can use like you can use counting arguments as we did for the ax growth in the theorem. You can use the vague conjectures in characteristic p. There are things like Frobenius endomorphisms and so on. So you can use these indirectly to prove things in characteristic zero by first proving them over all finite characteristics and then applying model theory, a famous example of this is Maurice end and break argument where he first proved it in characteristic p by counting and then extended that to characteristic zero. Okay, that concludes the part of the course on regular morphisms. The next lecture will be on rational maps.