 This lecture is part of an online course on commutative algebra and will be about notarian rings. So we remember from last time that the condition R is notarian is equivalent to the condition that all ideals of R are finitely generated. So this is the definition of a notarian ring. So what we're going to do next lecture is prove Hilbert's theorem that all polynomial rings over a field are notarian. But before doing that, we're first going to give some examples of notarian and non-notarian rings. And we're also going to prove a couple of conditions equivalent to being notarian that we will use when proving Hilbert's theorem. So the third condition says that every strictly increasing chain of ideals so if we take I1 contained in I2, take an I3 and so on is finite. And the fourth condition says that every non-empty set of ideals has a maximal element. So maximal doesn't mean it contains all other ideals of the set. It just means there's no other ideal in the set that strictly contains it. So we'll start just by proving these four conditions are equivalent. So first of all, conditions one and two are equivalent by definition. So let's prove that conditions two and three are equivalent. So let's prove two implies three. So suppose all ideals are finitely generated and then given a chain, I1 contain I2 and so on form the ideal I which is just the union of all these ideals I and you can easily check it's an ideal because the change is increasing. In general, the union of two ideals need not be an ideal but it is if one of the ideals is contained in the other. Now I is finitely generated by say I1, I1, opt to AK and because this is a finite set and all these elements are in one of these ideals, I means all of them must be in some, one of these ideals, IN. So all AI in IN for some N. But then I equals IN so the chain stops at IN. And then we want to check that three implies two. So if every strictly increasing chain is finite, suppose I is an ideal, we pick an ideal. So if I is not equal to zero, pick A1 not equal to zero in I. If I is not equal to eight, the ideal generated by A1, we pick A2 not in A1 and of course A2 should be in I. If I is not equal to A1, A2 then we pick A3 not in A1, A2 with A3 in I and so on. So what we do is we get a chain not contained in A1, A2 and so on. And this is a strictly increasing chain. This chain must be finite so it stops. And if it stops, then the only way it can stop is if we can't find any other element not in I. So I is finitely generated. Next we want to show that three and four are equivalent. So three being equivalent to four is really nothing to do with rings. It's a general property for all partially ordered sets which are sometimes called the post sets. So if we've got any set with some sort of transitive and symmetric order on it, then strictly increasing infinite chains exist if and only if not all non-empty sets have maximal elements. So what we're really going to do is to prove three implies four for any post set. So I suppose every strictly increasing chain is finite. Well, we want to show that a non-empty set, suppose S is non-empty and we want to show that it has a maximal element. Well, what we do is we pick I1 in S. If I1 is not maximal, pick I2 with I1 contained in I2. If I2 not maximal, pick I3 with I1 strictly contained and I2 strictly contained in I3. And so if no elements are maximal, we would get an infinite chain. And we assumed that all strictly increasing chains are finite, so there is a maximal element. By the way, you notice this argument and also one of the previous arguments we had involves making an infinite number of choices of ideals. And this means we're implicitly using the axiom of choice in mathematics, which says that you can always make an infinite number of choices. And this action was rather controversial at the beginning of the 20th century and people argued about whether it's true or false and it turns out that whether it's true or false depends on which model of set theory you're working in. Anyway, we're not going to worry about that because this isn't a course on set theory and we will just assume the axiom of choice because we're lazy. Now you want to check that four implies three. Well, condition four said that every nonempty set of ideals has a maximal element and three says that every chain is finite. Well, if you've got a chain of ideals, all you do is pick a maximal element, which you can do by condition four. So if this maximal element is in and if in is maximal, this implies the chain stops at I n, so the chain is finite. So we've now got four or three or four conditions for a ring to be notary and this is really useful because we quite often switch between these two, these three different conditions when proving things about notarian rings. So I'll first give an example where all three conditions fail. So you remember last time, we had an example of a non-notarian ring which had a polynomial, which was polynomials infinitely many variables. And let's see how all these three conditions fail for this ring. So first of all, the ideal X1, X2 and so on is not a finitely generated ideal. Secondly, we've got an infinite strictly increasing chain because we can just take X1 contain an X1, X2, contain an X1, X2, X3 and so on. So this gives us an infinite chain. And next we notice that this set here has no maximal element. So we've got a set of ideals with no maximal element. Now we've been talking about increasing chains of ideals and sets of ideals having maximal element. You can flip that and ask about whether a ring has infinite decreasing chains of ideals and whether every set of ideals has a minimal element. The condition that every set of ideals has a minimal element turns out to be a much stronger condition. This is the condition that the ring should be artinian that we will study later. So I just want to give you an example to show that even the easiest examples of rings kind of infinite descending chains of ideals. So let's just take the ring of integers, then we have the ideal two contains the ideal four, contains the ideal eight, it's contains the ideal 16 and so on. So here we have an infinite decreasing chain and the set has no minimal element. So the integers has the property that is no infinite increasing chains in all sets of ideals of maximal elements but the analog fails for decreasing chains or minimal elements. Now I'm going to give a series of, I think it's seven different rings and we're going to try and figure out which of them are notarian. So the first one is just going to be the ring of polynomials over the reals. And this is contained in the ring of analytic functions on the reals. So analytic functions are just those that can be expanded as power series at any every point. If you've done a complex variable course and prefer you can think of these as being holomorphic functions instead. And this is contained in the ring of functions that are analytic on closed interval nought to one. So I want them to be analytic at the points nought and one as well as everywhere in between. Actually, I think it'd be easier if I take the interval from minus one to one because then that will make the rest of this, these examples go a bit better. And this is contained in the functions that are analytic on the open interval minus one to one. So this is closed and this is open. And this is contained in the space of functions that are analytic at zero. This means that there's some small interval around zero that they can be expanded as a power series in. And this is contained in functions that are smooth near zero. We're really looking at the ring of germs of smooth functions. In other words, we consider two functions to be the same if they're the same in some neighborhood of zero. And this is not contained in the ring of formal power series, but maps to it. So we've got seven rings and these are all quite similar. So the polynomials and formal power series are really fairly similar. Formal power series is just a sort of infinite polynomial and most of these other rings sort of lie between these two examples. And what I want to do is ask which of these are notarian. So first of all, polynomials are notarian. In fact, polynomials form a principal ideal domain as you remember from an earlier algebra course. This means that all ideals are generated by one element. So they're certainly generated by a finite number. Now, analytic functions on the reels are not notarian. And to see this, what we do is we take a set Z, let's take a set Z with no limit points and let's take I to be the functions vanishing at all but a finite number of points of Z. And then you can easily check that this ideal is not finitely generated. So you need, I mean, it has sort of functions vanishing at all but one point of Z and there's a bigger ideal of functions vanishing at all but two points of Z and bigger still you could have functions vanishing at all but three points of Z and so on. On the other hand, if we restrict analytic functions on the closed interval, this is notarian. And the reason is if you've got any analytic function on the closed interval, it's only got a finite number of zeros on the closed interval. So this means we can write F as a polynomial times an analytic function with no zeros at all and that's a unit. And from the fact that every function can be written as a polynomial times a unit, it's quite easy to show this is notarian essentially by reducing to the case of the ring of polynomials is notarian. I mean that these functions are so close to polynomials you can get from one to the other. This one, although it looks very similar to analytic functions on the closed interval is not notarian. And to see that, what we do is we were working with the integers here, but instead of working with the integers, we can take the numbers one minus a half, one minus a third, one minus a quarter and so on. So these are, this is an infinite set with no limit points in this set. And again, we can do the same trick. We just look at all analytic functions vanishing at all but a finite number of points here. So this set is not notarian. What about functions vanishing at zero? Well, this is notarian. And again, we can write any function that's analytic at zero as a power of X times a function which is non vanishing at zero. And therefore, it's inverse is also analytic at zero. So again, we can write any function as a power of X times a unit. And this means the only ideals are the ideals X to the n for n, ratio equal to zero and also the ideal zero. So every ideal is finitely generated. This is an example in fact of something called a discrete valuation ring. Well, what about smooth functions near zero? You may think smooth functions are kind of rather like analytic functions but they're not in particular, this isn't notarian. And what we do is we take F with an infinite order zero with F not being identically zero. For instance, F could be e to the minus one over X squared for X not equal zero and F of zero equals zero. And then we can look at the following ideals. We can look at the ideal of all multiples of F and we can also look at the ideal of all multiples of F the half and the ideals of all multiples of F the quarter and so on. So we have infinite increasing chain of ideals and F is not notarian. Well, what about formal power series? Well, this is easy to figure out because if you look, you can see it's alternating between being notarian and non-notarian. So obviously this pattern is going to continue and the ring of formal power series is indeed notarian. And formal power series is very much like ones that are analytic at zero. We can again write F is any formal power series is actually n times a unit. So again, it's a discrete valuation ring. So you can see from these examples that being notarian is actually really rather subtle property. We keep on increasing the ring slightly and this keeps on switching it from being notarian to non-notarian and back. If you try and figure out what the non-notarian and notarian ones have in common, if you notice all the notarian rings, the zeros are really well behaved. They all have finite order and you can control them. Whereas the non-notarian rings, the zeros are functions are a bit bizarre. You can have infinite numbers of zeros or zeros of infinite order. So informally, you can think of notarian rings of functions as being ones where the zeros of the functions are maybe well behaved in some way. For the next example, let's ask, is a subring of a notarian ring notarian? And the answer is usually no. And there are plenty of examples of this. For instance, we had the space of analytic functions which was non-notarian was contained in, say, the space of formal power series, which was notarian. Another example is we can take any ring whatsoever and that's an integral domain, that means no zero divisors and we can embed it in Q which is the field of quotients and this is certainly notarian because all fields are notarian. And so we could take any non-notarian ring here which was an integral domain and that would be contained in a notarian ring. So subrings, if a ring has some property, there's generally no particular reason why any subring of that ring should have the same property. What about quotient rings? Well, this is quite different. If R is notarian, so is any quotient R over I for any ideal. And the reason for this is that ideals of R over I are the same as ideals of R containing I. Well, they're not quite the same as but they sort of correspond naturally. You know what I mean. And if the ideals of R containing I have this property, so if the ideals of R have the property that every set, every nonempty set contains a maximum and the same is obviously true for ideals that contain I. This is really useful because it means if R is a finitely generated algebra over S where S is notarian, then R is notarian because R is a quotient of a polynomial ring over S by some ideal. And we're shortly going to prove that the polynomial ring over S is notarian. So an awful lot of the rings who get a number theory in algebraic geometry are in fact finitely generated over a field or over the integers. For instance, the coordinate ring of any algebraic variety such as K X Y over Y squared minus X cubed plus X say is a quotient of a notarian ring and therefore notarian. And similarly, rings of integers of algebraic number fields are obviously quotients of polynomial rings by ideals. So they're also notarian. Next, we notice that a ring of polynomials in one variable over a field has the property that every ideal is generated by one element. And it's natural to ask, is the analog true for polynomial rings and more variables? So we can ask, is every ideal of K X Y generated by two elements? So Hilbert's theorem shows that they're generated by a finite number, but maybe we can do better than that. And the answer is no. And this is easy to see. Let's just write out some monomials. So we can Y, Y, X, Y, X squared, Y squared, Y squared, X, Y cubed. And now I can write down lots of ideals of this just by choosing some of these monomials. So here I could choose all these monomials and I can consider the ideal spanned by all these monomials. And it's obviously an ideal and it's also obviously not finitely generated because you can see it's generated by Y cubed, Y squared, X, Y, X squared, and X cubed. So we can certainly generate it with four elements and it's quite easy to see you can't generate it with any less than four elements. For instance, if you look at the ideal I modulo degree four, polynomials, this is just a vector space of dimension four and you can see that any set of generators of the ideal must include a set of basis for this four dimensional vector space. So this ideal takes at least four generators and obviously you can modify this and find ideals that take a minimum of a million generators and so on. So this fact about polynomial rings in one generator doesn't generalize in the obvious way to polynomial rings in two or more variables. So I'll just finish with one final example of a non-notarian ring. This is the ring of Prisa series. So this is sort of like a power series except you allow fractional powers of X. So we take the power series in a variable X. So it's a double square brackets, normally mean you're looking at the ring of all formal power series. So this would be the ring of all functions A, not functions all powers is A naught plus A1X plus A2X squared and so on where you allow an infinite number of entries. Back in the 19th century, people used to get very worried about whether formal power series converged or not. And this question is actually completely irrelevant because we're not thinking of these formal power series as functions, we're just thinking of them as being a sequence of elements A naught, A1, A2 and so on of R. And the only reason we write them as formal power series is because this makes it easy to remember the rule for multiplying and adding these series. Anyway, the ring of formal power series is contained in the ring of formal power series in a variable X to the half, which is contained in the ring of all formal power series and say X to one over six. And now we sort of go on with KX to one over N factorial and so on. So you can take the union of all these and this gives us the ring of all formal power series such that the exponents have a common denominator. And we can ask, is this ring notarian? And the answer is no. For instance, we can easily find an infinite increasing sequence of ideals. We can take the multiples of X. This is strictly contained in the multiples of X to the half, which is strictly contained in the multiples of X to the quarter and so on. So although pre-sourced series look a bit like formal power series, there's this fundamental difference that formal power series are notarian, but pre-sourced series aren't. Okay, that's enough examples. Next lecture, we're finally going to get round to proving Hilbert's Finiteness Theorem about polynomial rings.