 This lecture is part of an online commutative algebra course and we'll be talking about how to draw pictures of rings by drawing a base for the rings. So in the previous lecture, we drew pictures of rings by drawing a point for each element of the ring. This time we're going to draw a point for each basis element of the ring. Here we assume the ring is something like a ring of polynomials, which is a vector space over K. So we just sort of draw a basis for the ring. So for this example, we might just draw a point for one xx squared and so on. Now for polynomial rings, this isn't all that much use because polynomial rings are so easy to deal with anyway, but it works quite well for dealing with rings of polynomials and two variables where you can kind of imagine the ring as looking like some sort of two-dimensional array where the basis kind of looks like this. You can do it for three or more variables, but it gets a bit more difficult because it's harder to draw things in three dimensions, of course. So we've seen some applications of this already. For example, if you draw this diagrammatically, you can see that this is an example of an ideal such that the corresponding ring is not finitely generated as a ring or rather as an algebra over K. And furthermore, it's not notarian. So sub-algebras of notarian algebras need not be notarian and or finitely generated. So you can also see that the ring of polynomials and two generators really has rather a lot of ideals of finite co-dimension, finite co-dimension. So again, suppose we take the ring of polynomials and two variables and you might want to do something like, well, maybe it would be easy. You think it might be easy to classify finite dimensional algebras over K generated by two elements. Well, for that you take an ideal of finite co-dimension in the ring of polynomials and two variables. And the problem is there are rather a lot of such ideals. For example, you can find ideals by sort of drawing a diagram like this. And then you can see that this is an ideal of finite co-dimension. So you can see there are plenty of ideals because you can draw all sorts of different diagrams like this where all the monomials and green generation ideal. These are only very special sorts of ideals. They're called monomial ideals and which are ideals generated by monomials in X and Y. General ideals are of course considerably more complicated although there are techniques involving Groban of Bases where you can quite often reduce the case of monomial ideals. Anyway, the point is if there are already huge numbers of monomial ideals then they're gonna be even more general ideals of finite co-dimension. So this very strongly hints that finite dimensional algebras in two or more variables are going to be really rather too complicated to classify. Another example, let's have the following exercise find the dimension of the algebra K X Y and it's quotient out by the ideal generated by X cubed X squared Y squared Y to the five. So what's the dimension of this algebra? Well, this is incredibly easy to do if you just draw a picture of it like this. So all we do is we write out a basis element like this. One, two, three, four, five. I need to go a bit further. So this is gonna be one. Here we have Y to the five. Here we have X cubed and here we have X squared Y squared. So these are the basis elements of the ideal and therefore the ideal is kind of looks like this. So this is the ideal in blue and now we can see what the quotient by the ideal is. So we get these points here are going to be a basis for this quotient here and you can just count them. There are five, 10, 12. So this is 12 dimensions. So it's very easy to calculate just by drawing a little picture of it. Of course, it's not a particularly difficult example it wouldn't be that difficult to do without drawing a picture but whatever. So the next example, let's take the ring K X Y and I'm going to act on it by a group of order two taking X to minus X and Y to minus Y. And then I want to show that the ring of invariance is a two-dimensional free module over a polynomial ring. So this tells you something about the invariance. And again, if you try and do this algebraically it's not entirely clear what you do. If you just draw a picture of it then the result becomes almost obvious. So let's do that. So let's draw a picture of polynomials and two variables by writing down a basis as usual. So here we have one X X squared Y Y squared and so on. And now let's find the polynomials that are invariant under changing X to minus X and Y to minus Y. Well, that's pretty easy. All you do is you want the powers, you want the monomials whose total degree is even because these will be fixed by this automorphism and all the others will be minus Y. So the ring of invariance is this green ring here. And now we want to show it's a two-dimensional free module over a polynomial ring. So what's the polynomial ring going to be? Well, we're going to take the polynomial ring to be these blue ones. So we're just going to take polynomials and X squared and Y squared. So this polynomial ring is going to be K X squared Y squared. And now it's completely obvious it's a free module in two generators because we've got a basis here consisting of this element here and this element here. So these two elements form a basis for the invariant ring as a module for the two-dimensional free module. For the invariant ring as a module over the polynomial ring K X squared Y squared. And just looking at this diagram is obvious because the blue circles form a basis for this ring. And if you shift them, you get a free module in one variable generated by this element here. So we can see the ring of invariance is just polynomials in UV W where U equals X squared, V equals X, Y and W equals Y squared. Modulo, well, it's a free module in two variables over polynomials in U and W. So V squared is equal to some polynomial in U and W. And you can easily see that polynomial is just U times W. So we see that the ring of invariance is isomorphic to this ring here. And you can see that's just the coordinate ring of a cone. V squared minus U W is some sort of cone. So this makes it obvious what the structure of the ring of invariance is. So the next application is a little bit more complicated. So all the previous ones, the result wouldn't have been too difficult to do without drawing pictures of the rings. This time we're going to show that K, the ring of formal power series in X and Y is a unique factorization domain. So let's give some background. First of all, showing that the formal power series in X is a unique factorization domain is trivial because this is just a discrete valuation ring. The only ideals are either nought or powers of X. So it's in particular, it's a principle ideal domain and therefore a unique factorization domain. So for one variable, same power series or unique factorization domain is not terribly interesting but two variables is kind of trickier. Now you may recall that if R is a unique factorization domain, this implies the ring of polynomials in R is a unique factorization domain. This was essentially proved by Gauss. And by iterating this, we see that the ring of polynomials in any number of variables is a unique factorization domain. And you may think you can try doing this for power series rings. The problem is that if R is a unique factorization domain this does not imply in general that the ring of formal power series in R is a unique factorization domain. Although counter examples are slightly messy. There's one in Samuel's book on unique factorization domains if you want to look one up. So we can't use sort of induction on the number of variables to show this is a unique factorization domain. What we're going to do instead is use Weierstrass polynomials. So this is the Weierstrass preparation theorem which says that any power series in k x y. So I'm doing power series in two variables which is let's take it not to be zero can be written uniquely as power of x times a unit times a Weierstrass polynomial. So what do I mean by Weierstrass polynomial? Well, a Weierstrass polynomial is going to be something that form y to the n plus y to the n minus one times something plus y to the n minus two times something. So where all these are going to be power series in x which are divisible by x. So the point is units aren't too difficult to deal with. So what this says is for most purposes we can pretend that this is only a polynomial in y not a power series, which makes things a lot easier. So this is an element of it's a polynomial in y with coefficients that are power series in x. And you've got to be really rather careful here fiddling around with power series and polynomials is a bit confusing because we noticed that this ring is not equal to the ring of power series in y with coefficients in x. So you can't swap these two bits here. And to see this, we notice that this ring is actually rather bigger. It contains a ring such as one plus x y plus x squared y squared plus x cubed y cubed and so on. So you see this is a power series in x whose coefficients are polynomials in y but it's not a polynomial in y whose coefficients are power series in x. So you mustn't get these two confused. So we're saying this is in a smaller ring here. So how do we prove the Weierstrass theorem? Well, it's quite easy to prove by drawing a picture of it. What we drew is we draw a picture of a power series. Now you can indicate a basis for polynomials as a sort of rectangular array. So we have one x, x squared, y, y squared. Now this is not quite a basis for power series because power series are infinite sums of these but it's sort of close enough. So I suppose we've got a polynomial here. Well, let's take out the factors of x so we assume it's not divisible by x. And this means it's got a non-zero coefficient somewhere in this column. Suppose, for example, that this coefficient here is non-zero. So here we've got some coefficient, something times y cubed and these coefficients here are all going to be zero. Now the Weierstrass power series says that we're allowed to have terms like this but we want to kill off everything in this region here. So how do we do that? So we might have a non-zero coefficient here. Well, we can kill this off by multiplying by something of the form one plus something times y. So by multiplying by one times something times y we can kill off the coefficient of y to the four and then we can kill off the coefficient of y to the five by multiplying everything by one times something times y squared and so on. So we start off by killing all these coefficients in order. And once we've killed them off, then we can start on these ones so we can take our power series and all these coefficients are zero and we can now multiply by one plus x times something and that will kill off this coefficient and then we can multiply by one plus something times x and then we can multiply one plus something times x y and we end up killing off all these coefficients in order and then we repeat this. So we end up killing all coefficients other than the coefficient of y to the n. Here I'm taking n equals three and what this ends up with is a power series whose only nonzero coefficients are this power of y we had here and these terms here and this is just a via stress polynomial and we've multiplied by an infinite product or things of the form one plus something times some sort of monomial in x and y and if we multiply all these together we get a unit in the ring of formal power series. So our power series x commutant as a power of x times a unit times a via stress polynomial and you can also show quite easily this decomposition is unique. I won't bother proving uniqueness I'll just leave that as something for viewers to do. So now we're going to just show how to use this to show that k x y ring of formal power series in x and y is a unique factorization domain. And first of all, we know it is notarian which means that everything can be written as a product of in decomposables. I'm going to skip the minor routine details and just talk about the key point. We need to show if f is irreducible it is prime. And saying it as prime means if f divides gh then f divides g or f divides h. So we're just going to prove this part and leave all the rest as a routine exercise. Now the point is we use via stress we can assume f, g, and h are via stress polynomials because we can multiply f, g, and h by units which makes absolutely no difference to divisibility. So we may as well assume they're via stress polynomials. So we have f divides gh so f r equals gh for some r in k x y. And we want to show r is a polynomial in y. And notice that if f and g, f, g, and h are polynomials then the fact that f times some power series is gh doesn't imply that r is a power series. For instance, f might be one plus x, r might be one minus x plus x squared minus x cubed and so on and g might be h, might be equal to one. So here we have f, g, and h are all polynomials but r doesn't have to be a polynomial. However, using the fact that f, g, and h are via stress polynomials we can show this. So we know r is equal to r naught times r one where r naught is a via stress polynomial and r one is a unit. So if f times r naught times r one is equal to gh and now this and this are both via stress polynomials. Now this gives us two decomposition of g and h as a via stress polynomial times a power series. Here we can just write it as g of h times one or we can write it as this via stress polynomial times r one. So r one equals one by uniqueness because we said the decomposition into a via stress polynomials was unique. So in fact r, if f, g, and h are via stress polynomials it implies that r must be a via stress polynomial. So r is a polynomial in y but not in x. So what have we found? We have found that f divides gh in the ring of polynomials with, sorry, that should be a x, ring of polynomials with, try and get this the right way round. The ring of polynomials in y with coefficients in x is extremely confusing to remember which of these is a polynomial and which is a power series. So f divides g or f divides h because this is a unique factorization domain. Here we're using the fact that k of x is a unique factorization domain and the ring of polynomials over unique factorization domain is a unique factorization domain. So here this happens in kx, y and if something divides something in this ring we see that f divides g or f divides h in kx, y which is what we wanted to prove. So this implies with a small amount of extra reasoning that the ring of polynomials in two variables is a unique factorization domain. Should have a warning here, the ring of convergent polynomials is not a unique factorization domain. For example, we can take f to be a convergent polynomial with an infinite number of zeros, z1, z2 and so on. And then we might write f of x is equal to x minus z1, x minus z2, x minus z3 and so on. So it doesn't have a decomposition into a product of a finite number of primes because there are infinitely many zeros. So this doesn't work. In fact, this product won't even converge. There is actually a way of writing convergent power series as infinite products due to via stress but that's really doing analysis rather than algebra. So you've got to be really rather careful because talking about whether power series rings or unique factorization domains is rather tricky. There's a difference between whether they're formal power series or whether they're convergent or not. Okay, so next lecture we'll be discussing the third and most powerful method of drawing pictures of rings, which is to draw a point for each.