 So this lecture is part of an online commutative algebra course and will be about Zariski's lemma. In fact, it will be almost a repeat of last lecture. So last lecture, we were talking about the Null-Stelenzatz for fields and there were two forms of this. A weak form said that for K an algebraically closed field the maximal ideals of K x1 up to xn are just the obvious ones of the form x1 minus a1 up to xn minus an, in other words, they just correspond to the point of affine space. Then we have the strong Null-Stelenzatz which said that if i is an ideal of K x1 up to xn and V is at zero set, then f equals naught on the zero set implies f to the K is an i for some i, which I guess these are equivalent. And what we did was we proved the weak Null-Stelenzatz when K was the complex numbers by kind of using the fact that the complex numbers had an uncountable number of elements. What we want to do is to give a slightly more precise proof of the Null-Stelenzatz that works for all algebraically closed fields, not just the uncountable ones. So the key point of the proof is Zariski's lemma. This says, suppose the field K is finitely generated the field K is finitely generated as an algebra over the field K. Then it is finitely generated as a module. And you remember there are three notions of being finitely generated or at least three notions. You can talk about K being finitely generated as a field or finitely generated as an algebra or finitely generated as a module. And for rings in general, these three notions are different. All those Zariski's lemmas says that two of them are the same in this particular case. So last lecture, we proved this when little K was the field of complex numbers and there we used the fact that it was uncountable. In fact, you can sort of prove the Null-Stelenzatz for all algebraically closed fields using just the notion from last lecture. If you're willing to use a certain amount of model theory. The point is the argument for previous lecture proves it for uncountable algebraically closed fields. And there's a principle in model theory called the Lefschitz principle, where it says that any first order statement true for one algebraically closed field of given characteristic is true for all of them. This essentially follows from the fact that the first order theory of algebraically closed fields of given characteristic is complete. To understand what that means, you have to spend a semester going to a course on logical model theory and really it's rather easier just to spend about five minutes doing a little bit of extra work on commutative algebra, which is what we're going to do now. So let me first give you the idea of the proof of Zariski's lemma. Let's pretend that K is just a field of rational functions in little K. So let's say, sorry, this should be rational functions, X1 up to Xn. Then we can ask, can it be finitely generated? Well, the generators of the algebra K, big K, are all of the form Fi X1 up to Xm over Gi X1 up to Xm. So these XI generate big K as a field, but of course that doesn't mean they're generators of algebra and the algebra generators might be like this. So all poles are factors of the Gi, that's all poles of all the generators. Well, that only gives you a sort of, that there are only a finite number of possible factors of all the Gi. So all poles of these generators must be among a finite number of irreducible polynomials, but this ring has an infinite number of irreducible polynomials if M is greater than zero, so it can't be generated by finite number of elements. Well, that's okay for an idea for a proof, but you notice it's not quite complete because maybe the field big K is not the field of rational functions over little K, it might be something bigger. So now we've got to give a more complicated version of this proof where we don't assume K is just generated by algebraic independent elements, but the idea is going to be the same. We're going to essentially look at poles of generators and show that there just aren't enough poles of a finite number of generators. So let's give the proper proof. Now, let's assume K is generated, by X1 up to XM and XM plus one up to XN. And we're numbering them in this funny way because we're going to number them so that these are algebraically independent, whereas all the remaining ones are algebraic over KX1 up to XM. So you remember from field theory, we can do this. We can pick out the maximal algebraically independent set of generators, and then the remaining ones must all be algebraic. We sort of renumber them to get them in that order just for notational convenience. Okay, well, if M equals naught, we're done because these are all algebraic, so they all generate a finite extension. So we assume that M is greater than zero. And now we notice that the ring KX1 up to XM, so this is polynomial ring has an infinite number of irreducibles up to units. Well, I mean, it means we count two irreducibles as being the same if they're a unit times even then there's an infinite number of them. And this is kind of fairly easy. If K is infinite, we can just take X1 minus alpha for alpha in K. If K is finite, so if K is finite, you can, and we've only got one variable, you can copy Euclid's proof that there are infinitely many primes over the integers and find there are infinitely many irreducible polynomials over a finite field. So the point is that infinitely many irreducibles and that's going to mean there are, informally that there are infinitely many places where functions here can have poles. And now we notice that each XM plus one to XN, so these are the leftover elements is a root of a non-zero polynomial with coefficients that are polynomials in X1 up to XM because they're algebraic over it. And it will have leading coefficient, let's say the leading coefficients are BN plus one up to BN and the leading coefficients in some sense are the possible poles of XM plus one to XN. So informally, these are the, we can think of these as being the polynomials defining zero sets where these can have poles or all this discussion of poles is just informal and not part of the proof. So what we do is we can then see let's say that XM plus one up to XN are integral over the following ring KX1 up to XM and then we have invert one over BN plus one to one over BN. So integral just means the leading coefficient of the irreducible polynomial is one but obviously if we invert the leading coefficient then we can make the leading coefficient one just by dividing by it. So this is a localization of the polynomial ring. So K is finite as a module over this ring. That's because if something is integral then it generates an extension that's finite as a module. I mean, in fact, we'll be discussing integral extensions a little bit more later. Now what we do is we pick an irreducible polynomial F in KX1 up to XN not dividing any of BN plus one to BN. So what this is really is it's a sort of a new pole which is not a pole of any of the generators. So informally we think of the zeros as F as being the location of some sort of pole of a rational function. And pole is just as a complex analysis means where the function is infinite. And now you notice that KX1 up to XM F to the minus one is contained in K. And this is a finite K module. So it's a finite KX1 up to XM module. So since this is a notarian ring this is also a finite module over KX1 up to XM. And now there's something very fishy going on because if this is a finite module we can very quickly reach a contradiction. For example, we notice we've got an increasing chain of ideals. So we have an increasing chain of sub modules of this. We can have the sub module over this ring generated by one or by one and F to the minus one F to the minus two and so on. So this is an increasing sequence of modules over this algebra that are contained in this ring. And so it must eventually stop. So we find F to the minus N must be in the sub module generated by all the smaller powers of F. So it must be a naught plus a one F minus one and so on plus a N minus one F one minus N for some N. And then we can just multiply by F to the minus F to the N minus one and we find F to the minus one is equal to a naught F to the N minus one plus plus a N minus one. And this is a contradiction because this is an element of the polynomial ring and this very definitely isn't. So our assumption that the sub module field K, so our assumption that M greater than naught has led to a contradiction. So that proves the weak Null-Stellen-Satz for, sorry, that doesn't prove the weak Null-Stellen-Satz. That proves that the risk is lemma for arbitrary fields as before this implies the weak Null-Stellen-Satz and the proof of this only takes a minute. So I'll just quickly repeat it. What we do is we have the field K X one up to X N and we quotient out by a maximal ideal and we want to show that this maximal ideal must be one of the obvious ones. So here K is algebraically closed. You just recall that this is algebraically closed. So Zariski's lemma works for all fields K but for Null-Stellen-Satz we're taking K to be algebraically closed. This is a maximal ideal. So this is a field and it's finitely generated as an algebra over K. So this is finitely generated as a module over K by Zariski. So it is K as K is algebraic. So this is an algebraically closed meaning it doesn't have any interesting finite extensions. So each X i is equal to some A i modulo the maximal ideal M with A i and K. So the maximal ideal K must contain X one minus A one X two minus A two and so on. So it must actually be equal to it. Okay, there's another useful corollary of Zariski's lemma which says that if A maps to B is a homomorphism of finitely generated algebras over K. Here we mean they're finitely generated as algebras, not as fields or modules. And then F minus one of any maximal ideal is maximal. Notice that if B is not a finitely generated algebra this fails. For example, if we take the ring of polynomials over K and map it to the ring of rational functions over K this is not finitely generated but F minus one zero equals zero and zero is a maximal ideal here but it's not a maximal ideal here. So this definitely fails for general algebras over K being finitely generated is essential. And this follows easily from Zariski's lemma. So we've got F A goes to B and we've got A over F minus one of the maximal ideal and this maps to B modulo the maximal ideal and we've got K mapping to there. So this is injective and this is injective. And now we notice that this is a finitely generated K module by Zariski because it's a field and it's finitely generated as an algebra. So this thing here is contained. So this is contained in a finite extension of the field K and you recall that if you've got a finite extension of a field K then any subalgebra of that is in fact a field. So it is a field. Well, if A over F minus one M is a field this implies that F minus one M is maximal. So this result is why people could get away an algebraic geometry for quite a long time without actually using prime ideals. People would define the maximal spectrum of a ring and use just its maximal ideals. And the point is people were only using finitely generated algebras over fields for a lot of the time. And for these ones, you can get away with just using maximal ideals because the inverse of a maximal ideal is a maximal ideal. And there are other nice properties like for these sorts of rings any prime ideals and intersection of maximal ideals. However, for more general rings such as local rings, these theorems break down. For instance, for local rings, prime ideals are not in general intersections of maximal ideals. And as we've seen, the inverse image of maximal ideals need not be maximal. So finally, I'm going to give a different proof sort of different proof that the weak Nelstellensatz implies the strong Nelstellensatz. And this time I'm going to do it using localization. So we recall we've got i as an ideal in kx1 up to xn and f is equal to zero on V, which is the zero set of the ideal i in k to the n. And what we do is we look at the localization kx1 up to xn f to the minus one. And we notice that the maximal ideals are just contained in the set of maximal ideals of kx1 up to xn by general properties of localization which are just of the form x1 minus a1 up to xn minus an. And more precisely, they're the maximal ideals of the polynomial ring such that f is not in the maximal ideal. Informally, we think of this as saying f is not equal to zero at the maximal ideal. So i of f to the minus one, which is an ideal of this ring here, is not in any maximal ideal of kx1 up to xn f to the minus one. Because the only maximal ideals this can be contained in correspond to points where f vanishes and f to the minus one can't be in the corresponding maximal ideal. So is the whole ring because it's not in any maximal ideal. So one is contained in i f to the minus one. Well, i f to the minus one is just the set of elements of the form i over f to the k for various integers k greater than or equal to one. Well, this just says that f to the k is equal to i for some k and for some i in the ideal i, which is the nullstellensatz. Well, I was actually slightly lying to you. I said I was going to give a different proof that the weak nullstellensatz implies the strong nullstellensatz. But if you look at this really carefully, you see it's actually the same as the Rabinovich trick and sort of explains what's going on in the Rabinovich trick. So the Rabinovich trick, you were adding a new variable x naught with x naught f and then quotient out by the ideal x naught f minus one and so on. But if you think about it, that's really just an explicit way of taking the localization. OK, I think that's quite enough about the nullstellensatz for the moment. So next lecture, we're going to be looking at integral extensions of rings in a bit more detail.