 This lecture is part of an online course on commutative algebra and will be a sort of historical summary of what Hilbert was doing in invariant theory. So in the last lecture, I gave some examples of computing invariants, but these examples were kind of rather misleading and they were all rather easy. I mean, there's a selection bias there because obviously I was selecting the easy examples to present. What I want to do this lecture is to say something about the examples Hilbert was actually working on which were considerably more complicated. So the first examples are binary quantics. So what is a binary quantic? Well, binary means two variables. And if you want to say three variables, you would say ternary. And if you want four variables, which isn't very common, you would say quaternary and so on. And then quantics sort of indicates what the degree is. So quadratic means degree two, cubic means degree three, quartic means degree four, and quintic means degree five and so on. So what does quantic mean? Well, quantic means you don't know what the degree is. Notice there are three words, quadric, quartic and quantic, all of which sound very much the same, but are rather different. So binary quantic means you have two variables and you don't care what the degree is. Let's tacitly assume that these things are homogeneous. So binary quantic is going to look like a n x dn y to the naught plus a n minus one x dn minus one y to the one. And so all up to a naught y to the n. Some people like to put binomial coefficients in here. So instead of a n minus one, you would have n choose one a n minus one and n choose two a n minus two and so on. So there are different conventions for these things. And the idea is that you take a basis of c to the n plus one essentially consisting of these numbers a n a n minus one down to a naught. And we're going to have an action of the group SL2 over whatever field you're working on. I guess this would be the complex numbers acts on this space c to the n of all the coefficients. So how does it act on these? Well, it acts on x and y by mapping a b c d. And actually on x and y just is a x plus b y c x plus d y. So it's acting on the plane with coordinates x and y in the usual way. And if you think of these as being functions on the plane, then we get an action of this group on on this space of functions. If you want to look at it more explicitly, it kind of looks like this. What you do is you take a n and then you replace x by a x plus b y to the n times c x plus d y for naught. And then you have a n minus one times a x plus b y in minus one c x plus d y for one and so on. And if you expand this all out, you find it's equal to a n a to the n plus n a n minus one a to the n minus one and so on times x to the n plus something worse times x to the n minus one y and so on. You get some really complicated expression here. I guess that's not an a, that's probably a c or something. And what the action of SL2 is doing is it's taking a n to this thing here and it's taking a n minus one to this even worse expression here. So this gives the action of SL2 of c on c to the n minus c to the n plus one. And we want to know what are the invariants of this? Well, these are going to be polynomials in a naught up to a n invariant under this action. So what do these invariants actually look like? Well, here are some examples. First of all, we can have the discriminant. So the discriminant is more or less, it's really some constant times the product over alpha i minus alpha j. I guess we probably want to square it. Alpha i and alpha j are the roots of y a n x to the n plus and so on plus a naught equals naught. So we had the discriminant when we were talking about invariants of the symmetric group and we saw this discriminant could be written more or less as a polynomial in these numbers here. So that gives you some invariant. For example, if you're looking at binary quadrics of the form a2x squared plus a1xy plus a naught y squared, the discriminant is just a1 squared minus 4a2a0 or b squared minus 4ac in the more common notation. Other examples, if you're looking at degree 4 polynomials, then one famous example of an invariant is the so-called catalecticant. Most of these funny words were invented by Sylvester, which is something like the determinant of a naught and then probably want a1 over 4, a1 over 4, a2 over 6, a2 over 6, a2 over 6, a3 over 4, a3 over 4 and a4. So that's why people sometimes like to put binomial coefficients in front of these. So the basic problem of invariant theory for binary quantics is to show that you can find a finite number of invariants such that all other invariants are polynomials in these. And this was solved by Paul Gordon. So he showed the invariants of binary quantics are finitely generated. Gordon was sort of known as the king of invariant theory at the time because he was so much better at doing this than everybody else. And this is a really impressive result. And Gordon was rather unlucky because he did this fantastic work. It was just all overshadowed by Hilbert's work, which just sort of was so much better that it wiped it out. The invariants are difficult to compute explicitly. I think classically the invariants of binary quantics were computed up to a degree about six and a bit later last century, they were computed up to a degree about eight. And last I heard was they've been computed up to a degree about nine by computer and the invariants of degree nine are really rather hairy. So for example, just how hairy they get, let's look at the case of ternary cubics. So you remember ternary means three variables and cubics means degree three. So the possibilities you have are x cubed, x squared, y, x, y squared, y cubed, x squared, z, x, y, z, y squared, z, x, z squared, y, z squared, z cubed. So if you count up, there are 10 of these. So there really should be 10 variables, giving the 10 coefficients of these numbers. And this is acted on by the group SL3 of C, which acts on x, y, and z in the obvious way and therefore acts on ternary cubics. So you can ask, what are the invariants? Well, the invariants are generated by two of degrees four and six. And what do these invariants of degree four and six look like? Well, I've got them written out here. This is Bernsternfeld's book on algorithms in invariant theory. If you want to see how these things are actually computed, this is a really great book to look at. And here is the degree four invariant. Let me just magnify it opposite. You can see it in all its glues and detail. So there you see it goes on for about six or seven lines. And these a, three nought, nought, for example, would be a coefficient of x cubed, y to the zero, z to the zero. So that's the degree four invariant. The degree six invariant is even more impressive. It looks like this. So here's the degree six invariant of the ternary cubit and it goes on for the whole page. And then it goes on for a fair amount of the next page as well. So how do you compute these? Well, he says so here. Here we solve 540 linear equations in 103 variables. So I'm guessing he didn't do that by hand. Well, as impressive as that is, this is nothing compared to what people in the 19th century used to do. So here I have a book on higher algebra by Salomon. This is Lesson's Introduction to Modern Higher Algebra. So this is a 19th century book, which as far as I can figure out was an undergraduate textbook. And he has something on invariant theory. Here is this calculation of a certain invariant. This is invariant e. Let me show you it in its full glory. So it goes on for the entire page. Well, actually it doesn't go on for the entire page because it also goes on for the next two pages and the next two after that and the next two after that and the next two after that and the next two after that and the next two after that. And then he continues with another invariant and so on. What is really impressive about this is that Salomon did this entire calculation by hand. This was the 19th century when they didn't have computers around to do these invariants. So how he did that, I have no idea. He must have had a lot of spare time on his hands. In fact, looking at 19th century papers on invariant theory you sort of get the impression that you were having a sort of informal contest with each other to see who could write down explicitly the most complicated possible invariant. You sort of have more street credibility. If your invariant was even worse than the other guy's invariant. In fact, 19th century mathematicians were really, really good at computing invariants and quite often beating modern computers. There have been some occasions when some mathematician used a computer and a lot of computer calculation to compute invariants and it was slightly embarrassed to discover that some guy in the 19th century had already done it by hand. So anyway, so as you can see these invents are so complicated that trying to show the rings of invents are finitely generated by explicitly calculation is quite out of the question. Even for reasonably small things like ternary cubics, the invariants are really too much to write down by hand and degree three and three variables is really not very much and you could have much higher degree and much more variables and you'd like to prove finite generation for all these. So there's an even apparently even worse problem in that we not only want invariants to be finitely generated but we want synergies which are relations between invariants to be finitely generated and if invariants are bad then first order synergies are even going to be even more of a nightmare and the second order synergies are going to be even worse than that. So showing synergies of finitely generation seems even worse than showing invariants of finitely generated. And this is where Hilbert had his first spectacular idea. He showed that if you can show invariants are finitely generated then synergies are also finitely generated. So the apparently harder problem of finite generation for synergies turns out to follow from the apparently easier problem of invariants. Well how did he do this? Well suppose invariants of finitely generated well that means you've got a map from some polynomial ring and a finite number of special invariants onto the ring of invariants and the first order synergies are going to be some sort of ideal. So these are the first order synergies. So this is a homomorphism of rings. The synergies are generators for this ideal because a kernel of any homomorphism of rings is an ideal and what Hilbert showed is that any ideal of kx1 up to xn is finitely generated as an ideal. So if you show the number of the ring of invariants is finitely generated it's going to be some sort of quotient of a polynomial ring and finitely many variables and all its ideals are going to be finitely generated whether or not they come from a map to a ring of invariants. So this is Hilbert's fundamental theorem about polynomial rings that we're going to prove in a couple of lectures. Rings with this property. So if all ideals are finitely generated the ring is called notarian. This is after Emmy Nerter. So what Hilbert did was he proved that polynomial rings and finite numbers of variables over a field on notarian. Of course the name notarian came later than that. Hilbert sort of was the first person to introduce and use notarian rings. They were named after Emmy Nerter because she studied them very deeply and showed that a lot of theorems that people have proved about polynomial rings could in fact be proved much more easily in the more general case of all notarian rings. So she greatly simplified the theory. So when you talk about things being finitely generated you have to be careful to distinguish between three different meanings. So there are three meanings of finite generation which are easy to get confused because you quite often use two or three of these meanings in the same proof. First of all we can talk about being finitely generated as a module or an ideal, which an ideal is of course a special case of a module. Secondly you can talk about being finitely generated as an algebra over some field or more generally a ring. So in case I forgot to say it an algebra over a ring just means a ring together with a homomorphism of your base ring to it. I don't know why it's called an algebra. Or terminology is a bit of a mess. Thirdly we can talk about being finitely generated as a field. So let me give some examples. First of all if you take a polynomial ring this is finitely generated as an algebra over k because it's generated by this one element x but not as a module. So if you want to generate it as a module you might take a basis one x squared x cubed and so on. So for modules you can multiply by elements of the base ring which would be k in this case but you can't multiply generators together. So if you take x as one basis element of a module you still have to add x squared as another basis element. If you take kx this is rational functions this is finitely generated as a field over k because it's finitely generated by x but not as an algebra. For example we've got one over x one over x minus one one over x minus two and so on. So if k is infinite you can't get all of these with only finitely many elements of this because there will only be a finite number of things in the denominator. If k is finite it's still not finitely generated because you get one over various irreducible polynomials. So for example you have to distinguish carefully between being finitely generated as an ideal or as a ring. So let's look at the ring kx y for example and we can write down a basis of it of things like one x x squared x cubed y y squared y cubed x y x squared y. So this is a basis for this as a module over k and now I can look at the following ideal I'm just going to take all these things here so this is an ideal and we notice that it is finitely generated by the element x because everything in this is x times some polynomial. On the other hand it's if we think of this as a ring we can think of this as a ring without an identity element because its identity is there and this is not finitely generated because if you take some finite number of polynomials then there's going to be a maximal degree of y in these polynomials say all polynomials in your set of generators have at most y squared then your generators might be you might take your generators to be these blue things here and if the maximum degree is y squared then you see that x y cubed is not going to be in the ring generated by these I mean you can get it by multiplying x y squared by y but y isn't in your ring you're not allowed to do that if you want this to be a ring with identity that's quite easy you can just add the identity to it and you can take this set here so these things here form a ring with an identity and again it's not finitely generated as a k algebra so when I say this is not finitely generated I mean it's not finitely generated not just as a ring but as a k algebra I sometimes say finitely generated as a ring what I mean as an algebra so Hilbert's theorem that ideals of finitely generated doesn't immediately help you with showing that sub-rings of finitely generated for instance you might imagine the ring of invariance of some group turned out to be this ring here in which case it wouldn't be finitely generated and Hilbert's theorem says that at least for the groups he was considering that can't happen so we're first going to prove Hilbert's theorem that says all ideals of polynomial rings are finitely generated and then we're going to prove Hilbert's theorem that rings of invariance are finitely generated as this example shows not all sub-rings of this are I'll mention a few some background for extensions of Hilbert's theorem first of all Hilbert showed that invariant rings of g acting on k x1 to xn are often finitely generated here he did the case k has characteristic zero what this means is he did it for g reductive and he asked if this was true for all groups g and not just the ones he had covered which were the reductive ones included most cases of interest in invariant theory like the special linear group and all finite groups and so on so he asked is the ring of invariance finitely generated as an algebra for all groups and it took several decades to answer this question the answer was finally found by Nagata who found a counter example of a rather complicated group well actually it was a billion it's not that complicated such that the ring of invariance was not finitely generated incidentally Nagata is responsible for a very large number of bizarre counter examples in mathematics as well as a large number of rather good theorems the story I heard about him was that the way he worked was he would sit down with a conjecture and every morning he would spend trying to find the counter example of the conjecture every afternoon he would spend trying to prove it when he published a counter example and half the time he published a proof anyway next Hilbert showed that polynomial rings have the property that every ideal is finitely generated this isn't true for all rings so we can ask are all ideals finitely generated in any ring and the answer is no and this is quite easy to say all you do is take a ring of polynomials in infinitely many variables and then if you look at the ideal of all polynomials with constant term zero this is not finitely generated so this only works for polynomials in a finite number of variables not for infinitely number of variables next what Hilbert showed is that if you've got a ring of invariants which you write as a quotient kx1 to xn modulo some ring of invariants Hilbert showed that the first order of finites so if you call this ring r we get some map from r to the n1 mapping onto this and it follows easily from Hilbert's work and as we'll see fairly soon that the second order of residues are also finitely generated and the third order ones are finitely generated and what Hilbert showed is this chain also comes to an end so not only are all these finitely generated but the entire chain is finitely generated by the way this is an example of something called a right free resolution so what we have here is you think of the ring of invariants as being a module m and if you've got a module m over ring r you can try and understand this module by sort of writing it as the tail end of a sort of sequence of free modules like this where the sequence is exact at every step the things that map to 0 which would be the nth order of residues are exactly the image of the next module in the sequence so what Hilbert showed is that if you've got a module over a ring of polynomials over a field then it has a finite free resolution and finite free resolutions are going to come up quite a lot in the course so I'm going to ask does every module over a notarian ring have a finite free resolution every finitely generated module so we can ask does every finitely generated module over a notarian ring have a finite free resolution well this is true for the integers if we take a module over the integers well we can have some free abelian group mapping onto it and now every subgroup of a free abelian group is also a free abelian group so we can find a subgroup that's also a free abelian group that maps to that so every module over the integers has a finitely generated module over the integers of two it's also true for rings of polynomials in n variables as Hilbert proved however it is false for some notarian rings so here's an example where it fails so I suppose we take our ring R to be a ring of polynomials and then we quotient out by the ideal x squared so it is a basis over k of two elements one and x squared is equal to zero and now I'm going to take our module to be R over x which is just isomorphic to k with x acting as zero and now let's try and find a resolution of it well so we first of all map R sorry my R seems to look rather like k as we try and distinguish them we map R onto k where we can just map them and the kernel of this is just x so we need a second copy of R that kills off this x so it maps one to x well what's the kernel of this map well the kernel of this map is going to be all multiples of x so we need to kill those off by having another copy of R mapping to this where we map one to x and it's kind of obvious that this process is sort of going on forever so the obvious way of finding a finite free resolution of this fails you just go on forever and with a little bit more thought you can show there's no other more cunning way of doing this I'm not going to bother because this becomes trivial when we do homological algebra so I'll leave it till later so that's the end of the summary of classical invariant theory what we're going to do next lecture is discuss notarian rings those such that every ideal is finitely generated in more detail and then we will go on to prove Hilbert's theorem 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