 So this lecture is part of an online algebraic geometry course about schemes and will be on differential operators on a ring. So a typical example of a differential operator is the operator x squared d squared by dx squared plus x d by dx plus x squared minus alpha squared. So this is Bessel's differential operator. The functions killed by this differential operator are just some sorts of Bessel functions. And we can think of this as being a map from if we want, we can think of this as being a map from polynomials over the reals to polynomials over the reals. You notice that this operator is are linear, but it's not linear over r of x. So if we let this operator d be that, then d of fg is not equal to f of dg in general, though it is if f is real. So what we want to do is, suppose we're given a ring A and an A algebra B. So we think of A as being a sort of base ring. And we want to define the concept of a differential operator from B to B. More generally, we could take M and N to be B modules and ask what is a differential operator from M to N. So that's what we want to define. So let's start by just having a look at a very simple differential operator. D is differentiation with respect to x and here we'll take A to be a field. Let's take A to be the reals and B to be just polynomials over A. And then what properties does this differential operator have? Well, we have Leibniz's rule which says that d by dx of x times f equals x times d by dx of f plus f. So we see that it doesn't commute with x. The difference between putting d by dx and x in a different order is this operator here. So what we can do is we can say d by dx of x equals 1. Here we mean by x we mean the operator multiplication by x. And if we've got two linear maps x and y then by the Leib bracket of x and y we usually mean x, y minus y, x. So this formula here is just the same as this formula here. It says that if we apply both sides of this to f then they're the same which is that formula there. And similarly we see that d by dx of f where f is the operator of multiplication by x is just d by dx. So this is an order, so this is just a linear map multiplication by this function. So we can say the order 0 linear operators, sorry, the order 0 differential operators are just linear operators. And in order 1 differential operators such as d by dx has the property that d by dx of f is order 0. It's just the operator df over dx. And this gives a strong hint about how we should define order n differential operators in general. So we have a definition, d is order n differential operator if, well first of all, we want it to be linear over the basering a, but it's not necessarily linear over b. And secondly, the commutator of d with f for f in b, that should be order n is of order, less than or equal to n minus 1. And we also say that order less than or equal to minus 1 differential operators are just 0 operator. So this sort of defines differential operators inductively. So order 0 are just those that commute with multiplication by elements of b. So they're just linear maps. And order 1 differential operators are 1 whose commutator with an element of b is a linear map and so on. So for example, let's find all linear operators, all differential operators of the ring b equals r of x. Here we're working over the ring a, which is just the reals. So we don't want to worry about differential operators, which might not commute with elements of r. They're all going to commute with elements of r by definition. Well, there are some obvious examples. If we just take x to the m times d by dx to the n, then you can see that this is an order n differential operator. And of course, we can take linear combinations of them. So we can take sum over aij x to the i d by dx to the j for aij in the reals. And the question is, are there any others? Well, this is quite easy to prove that there aren't any others because if we take d by dx to the n and commute it with x, we get n d by dx to the n minus 1. So from this, it easily follows that any differential operator of order n is of the form some order n plus 1 operator commuted with x because you can get any order n minus 1 operator by commuting some order n operator with x. So suppose e is a differential operator of order n, then ex is order n minus 1. So this is of the form dx where d is one of these operators here because by induction on n, we know that any operator of order differential operator of order n minus 1 is of this form. And therefore, by this observation, we can write it like that. So e is equal to d plus an order 0 operator. And the order 0 operators are just linear. They're just multiplication by elements of b. So we've shown that any differential operator e must be written as one of these operators here. In other words, the differential operators using this funny definition by commutators is just the same as the obvious definition of differential operator you get from analysis. Well, now let's try looking for differential operators of z of x. And there are some obvious ones. We can just take sum of aij dx to the i d by dx to the j with aij in the integers. And are these all of them? Well, no, there are some extra ones. So this is not quite like over the reals because there's extra differential operators. We can take a half of d squared over dx squared, for example. So this takes x to the n to n n minus 1 over 2 x to the n minus 2, which is a perfectly good differential operator of order 2, even though it's got this factor of a half in it. In general, you see that all that's more generally, we can take things in the form sum of aij x to the i d by dx to the j over j factorial. And this will be a differential operator with aij and now still integers. And does this give all of them? Well, yes, it does. It's fairly easy to check because just as before, we can now show that d by dx, the n over n factorial bracketed with x is just d by dx, the n minus 1 over n minus 1 factorial. So we see that every differential operator of order n of this form is the commutator of some operator of order n plus 1 with x. And now we can just copy the proof that we gave for differential operators over the reels and see that every differential operator over the integers is of this form. So we see that differential operators over polynomial rings are more or less what you would expect, except if you're not working over a field of characteristic zero, you have to watch out for the fact that you sometimes get these slightly unexpected denominators allowed in differential operators. Moreover, there's one other thing we can do with differential operators, we can normalize differential operators. So if d is a differential operator, it's called normalized if d of 1 is equal to zero. And now we can write d, any differential operator d is d naught where this is normalized plus some constant k with some constant c with c and a, where c is just equal to d of 1. So this is normalized order n, and this is just order zero where this is order n. So any differential operator can be written as the sum of normalized operator and a constant. Differential operators, by the way, form a filtered ring. So if we get this sort of ring of operators and so on here, fi is the order less than equal to i differential operators from b to b. And it's fairly easy to check that fi times fj is a differential operator of order at most i plus j. So we get a sort of filtered ring. By the way, normalizing shows that we can write f1 as normalized operator plus something in f naught. So f1 sort of splits as f naught plus something. You might wonder if f2 splits as f1 plus something. And in general, it doesn't in any natural way. So there's no way to extend normalization to splitting off it. We can't write order two operators canonically as the sum of an order one operator and something. Incidentally, although we won't need it this lecture, I'll just remark that we can form the corresponding graded ring where we take the sum of fi over fj. So fi over fi minus one. So this is a graded ring. And this, in fact, commutative, which is fairly easy to check. So although differential operators don't commute with each other, they're not too far from commuting. You can turn them into a graded ring by taking the associated graded ring. So we can also define the universal differential operator. So suppose you've got b is an a algebra. We can define a map d to a module b. That is a universal order n differential operator. If any other order n operator factorizes through it. In other words, suppose we've got a map n and an order n operator here, then we can find a linear operator from m, a unique linear operator from m to n, factorizing it. The existence of a universal linear operator of order m is almost obvious. So m is generated by elements d of b for b and b, modular relations forcing d to b order n. If you look at the relations for d to b order n, which we will actually write down fairly soon, you can see that just a lot of linear relations between these elements here. So it's obvious that a universal linear operator exists. And similarly, there's a universal normalized linear operator of any given order. Now we find the module of differentials omega b over a is the universal, well, actually, we don't really just have a module. We also have a module with a map d to it is the universal order one normalized a linear differential operator. In other words, if we've got any module m and some order one normalized a linear operator from b to m, then there's going to be a unique map from this module of differentials to m. And let's see a few examples. Let's take b to be polynomial rings in two variables over a ring a. You can take a to b. Let's just take it to be the real numbers. Why not? Then what's omega b over a? Well, it's generated by the elements d of f for f in in b. And you can see it's generated as a b module by dx and dy because d of f is just partial derivative of f with respect to x of dx plus delta f over delta y of dy, where this formula follows just using the fact that d is a first order normalized differential operator, which means that d of fg is equal to fdg plus d plus g df. And you can also see that there are no relations between dx and dy because if we take the free module with basis dx and dy, then the map taking df to this thing here is, in fact, a first order normalized differential operator. So omega b over a is just isomorphic to b plus b with a basis dx and dy. As another example, let's see what happens if we've got a field. Suppose we take a to be the rationals and b to b. The rational is extended by the square root of minus one. And let's ask what is omega b over a? Well, it's generated by d of i. So we need to work out what d of i is. Well, i squared equals minus one. So two i di is equal to naught, because d is normalized. So di is equal to naught. So omega of b over a is just zero. And the same thing works for any separable extension. So suppose we've got some a as a field k and b, let's write it as big k is a finite separable extension. So we can write b is equal to big k is equal to k of alpha, where alpha to the n plus a n minus one alpha to the n minus one and so on plus a zero equals naught for some alpha. And we need to figure out what is d of alpha. Well, this says that f of alpha equals zero. So we find that f prime of alpha df is also equal to zero. And now f prime of alpha is not equal to naught because the extension is separable. So df must be sorry, that's f prime of that should be a d alpha. So d alpha is equal to zero and therefore the ring of differentials omega b over a is again just zero. Well, I said that's what happens for a separable field extension. What about an inseparable field extension? So let's take an inseparable extension where k is k of big k is little k where we adjoin a to one over p is some a in big k. Well, this time if we try and work out d of a to one over p, let's put b equals a to one over p, we see that b to the p equals a. So we find p times b to the p minus one dp is equal to zero. And now this doesn't imply that dp is equal to zero because p is equal to zero. So we don't get any conditions on d of p. In fact, we find that omega of big k over k is one dimensional generated by d of p, not d of p, d of b, that should be a b. So this is one difference between separable extensions and inseparable extensions that inseparable extensions have non-trivial modules of differentials. So another example, it's usually fairly easy to calculate omega of c over a. So for example, suppose a is some ring, and let's put b to be a ring of polynomials over a in x1 up to xn. Then omega b over a is pretty easy to work out. It's just a free module with basis dx1 dxn. Now suppose c is equal to b over some ideal i, and we want to work out omega c over a. Then you can fairly easily see that c is still going to be generated by dx1. So omega c over a is generated by dx1 up to dxn. But this time it has some relations where whenever f, if f is an element of i, then we get relations saying df equals nought, where df is just equal to sum of df by dxi times dxi. So for example, if we want to work out omega c over a, where c is say, let's take an elliptic curve. Let's take a to be a field k, and let's c to be k of xy over y squared minus x cubed minus x. So here we've got the coordinate ring on an elliptic curve, and the ring of differentials omega c over a is generated by dx dy, and we're quotient up by the relation 2y minus 3x squared 2y dy minus 3x squared plus 1 dx. So we just get a resolution of omega c over a by b squared with kernel b. So this maps 1, 0 to dx and 0, 1 to dy, and this maps the element 1 to whatever it was, 3x squared plus 1 times 1, 0 plus 2y times 0, 1. So here we've got a resolution of the module of differentials by two free modules. So more generally, if you do this for arbitrary rings, you find that if c is equal to b over a, b over i for some ideal i, then we find we get this sequence i goes to omega b over a, tends with c, goes to omega c over a goes to 0. So in the example we had earlier, omega b over a was generated by dx and dy, and i was generated by 2y, sorry, i was generated by y squared minus x cubed minus x, and this methods d. So the image of this element i would be 2y dy minus 3x squared plus 1 dx. And doing this for more general, the proof for more general rings, c and b and a is pretty similar to the case, the example we did earlier. Well, Hart-Shorn gives a formula for omega b over a as i over i squared, where i is the kernel of the map from b tensor over a b to b. And the map from b to omega b over a takes b to b times 1 minus 1 times b. And the question is, where on earth does this come from? So I want to explain why this is true. And what I'm going to do is I'm actually going to do something a little bit more general. So what I want to do is to look at a map from b to m, where d is going to be any order n operator. So here this map d, we're taking little d to be order 1 and normalize, but I'm going to forget about normalization and look at all orders for the moment. And we can have a map from b tensor over a b to m, which takes b tensor c to b dc. And we're going to make this into a b module by letting b act on the left. So this defines the b module structure. You notice there are two ways of making this into a b module. We could have b acting on the left or b acting on the right, and we're going to take b acting on the left to define the b module structure. So this is now a homomorphism of b modules. And we're going to define a map from b to here by taking b to 1 tenths of b. And we notice that this is not b linear. So this map is b linear and this map isn't. And what we're going to do is try and make this into a universal order n differential operator. So, well, we can't do that quite yet. So first of all, we have to look at what relations does an order n operator have. So suppose d has order zero. Well, this means dA minus AD1 is equal to zero. What about d having order one? Well, this means the commutator of d with multiplication by b is order zero. And if you figure out what this says, it says that d of AB minus ADB minus BDA plus ABD1 equals zero. What about order two? Well, if you work it out, you find this says dABC minus ADBC minus BDA minus CDAB plus ABDC plus BCDA plus CDAB minus ABCD1 equals norton. You can now see what the pattern is. For order n, we're getting two to the n plus one terms where we sort of sum over all subsets of n plus one variables. Anyway, you can write this as follows. If d has order n, then d vanishes on i to the n plus one, where i is the ideal of B times over AB, sorry, is the ideal. i is the B module generated by B times one minus one times B. And you can also check that this is equal to the kernel of the homomorphism from B tensor A to B taking B times C to BC. And so you see from this that if we take B times over BA and we quotient out by i to the n plus one, then the map from B to this taking B to one times B is now is an order n differential operator. And furthermore, we can also see it's a universal differential operator, because if we've got any differential operator d mapping to m, then the fact that d is of order n means that the image of i to the n plus one in m is just zero. So we do indeed get a map from B tensor B to m taking B times B or B times C to B times BC. So B goes to B times over AB over i to the n plus one is the universal order n differential operator. Well, we haven't said anything about the normalization yet. So now let's put in a normalization. Well, B goes to one tensor B and B tensor B over i to the n plus one is not normalized, but we can normalize it by mapping B goes to one tensor B minus B tensor one, because this now takes one to one tensor one minus one tensor one equals zero. So this is normalized. And the image of what one tensor B minus B tensor one is now in the ideal I. So we get a normalized differential operator. If we take i over i to the n plus one and map B to B tensor one minus one tensor B, this is universal normalized order and differential operator. Now the case, Hartshorne is using in his book is just the case n equals one. So the universal order one differential operator whose image is the space of differentials is just i over i squared with this map here. Okay, so that's all we say for the moment about differential operators on rings. Next lecture, we will generalize this and discuss differential operators on schemes.