 So we had started to talk about non-singular curves on non-singular points on curves. So we wanted to show that the local ring of a non-singular point of a curve has a special structure that's a discrete valuation ring. So curve, for us, is still a variety of dimension one. And we will look at a non-singular point. So P in C will be a non-singular point. And we want to study the local ring. So we had to recall the definition of a module. And we had stated the lemma of Nakayama, which is one of the most fundamental lemmas about local rings. So let A be a local ring, and with maximal ideal, M. And let large M be a finitely generated A module. So then there are two statements. The first one is, if it is true that this module does not change if we kind of multiply it by M, so it doesn't get smaller, this is equal to M. This can only happen if M was 0 to begin with. And the second statement was, and this statement we proved the last time, the second statement is in some sense a corollary. So write K for the quotient, which is the field. So residue class field. So let F1 to Fr be elements whose classes, so the elements in M, such that the classes F1 bar to Fr bar of these elements, generate the quotient M mod M M. Well, either you could say as an A module, but it's an A module on which multiplication by M acts by 0. So therefore, it is also just a K vector space. So basically it just means, if we take elements in M whose classes generate the quotient here, then they actually generate M as an A module. So we had proven the first statement by some indirect argument, and now we want to prove the second. So we put N, the module generated by F1 to Fr. Our task, obviously, is to show that N is equal to M. So for this, we consider the quotient module. So consider mod N. So this is also an E module. And we want to show that this module is actually 0. So we know that F1 to Fr generate M mod M M. So in other words, so obviously what is generated by F1 to Fr is N mod M N. Because N is generated by these. If I take the quotient, I get this. So thus it follows, if I take N plus M M modulo M M, no. So if I take this N plus M M, I claim, what am I saying? So we know that N modulo, so N plus M M modulo M M is equal to M modulo M M. Because this is precisely, these things generate precisely N plus M M modulo M M. No, one would think it generates this N modulo M M. But we do not precisely know whether N contains the whole of this. I don't know, it's just we have N is precisely the set of things generated by these. If we divide by M M, we get the vector space generated by these. And we know that this is M mod M M, so we get this. And then obviously we can, if we have this, if we add, if we make this N plus M M plus M M, then this will be equal to M plus, so it follows that if we take this, this is equal to M. No, because we have just the inverse image of this. And that dividing by it, and it's the same. I mean this contains M M, as we have put it like this. And we undo the dividing by it, so we get that this is equal to this. So now we can look at, if we look at M, we have the maximal ideal. And we multiply it by M mod N. So we want to know what is this thing. We want to show, we now want to use the first part, that if we have a module, which multiplied by the maximal ideal stays the same, it was zero to begin with. So we have our module, of which we want to show it's zero. And so we multiply it by the maximal ideal, we want to see it doesn't change. So by that, what I just showed, this is N plus M M, modulo N. M, what? Let's see. What am I doing? I seem to be a bit, I'm confused. I can see what I wrote originally, because what I wrote on my notes doesn't seem to make complete sense. Ah, okay. Yeah, indeed. I'm not, if we do this. So, I mean how does one do this multiplication? This thing is, will be M times M. This is a corresponding submodular N, except again, you know, if we want to divide by N, we have to make sure that what we have is at least contains N. So we have to add N and then divide the whole thing by N. So, I mean to take this quotient, if you multiply by N, it just means we multiply this element by the M. So this is just the equivalence class of something multiplied by this, is you first multiply the something with it, and then you take the equivalence class. So it just means you have M times M, modulo N. But if you want to make such quotient, it always means we have to add this thing. Well, and now the line before, I have just written that this is M mod N. So, with this we indeed find that this module does not change when we multiply by M. So it follows by the first part of Nakayama's lemma that this M mod N is equal well. It follows if the assumptions of Takayama's lemma are fulfilled. So it follows that this is equal to zero because M mod N is a finitely generated A module. So Nakayama's lemma, the first half says if you have any finitely generated A module, which multiplied by the maximal ideal stays the same, it is zero. Now M is a finitely generated A module. So M mod N also is because a set of generators of the quotient is the same. It's just the classes of the generators of what we had before, by definition. So if M was finitely generated, then also M mod N is finitely generated. Okay, this proves the second part. So this lemma is a rather simple fact, but one comes across it very often when one deals with local rings. Somehow the most basic thing studies. Somehow the idea is that you can just, in order to find out what the generators of a module are, you just need to look at what happens after you divide by the module times the maximal ideal is somehow very useful. And also this simple criterion to show that the module is zero. Okay, now we want to come to the definition of a discrete valuation ring. So let A be a local ring, and assume that A is also an integral domain. So there are no zero divisors. So we call A a discrete valuation ring, and from now on this will always be noted just like this. DVR, because it's a bit long. If the following holds. So first, the maximal ideal in A is a principal ideal. So that means we can write M equal to T for some element T in M. So the maximal ideal can be generated by just one element. And in such a case, such a T will be called a uniformizing parameter. Well, I mean it was supposed to be small M, in particular it's also in A. There's no big M now, so now all M's are the same. Such a T is called a uniformizing parameter. Okay, the second statement is that, well, you can write any element in A as a unit times the power of T. So when T is uniformizing parameter. So if T is uniformizing parameter, so by part one, there exists a uniformizing parameter, namely the generator of M, and if T is such a thing, there might be many, but if I take any such, then every element A in A can be written. Uniformizing, yeah, or the orthography. I write it with a Z. Now you might object. Then every A in A as F equal to A times T to the N for A in A unit and N a non-negative integer. Okay, so we know that A is a local ring, so every element in A which does not lie in the maximal ideal is a unit. So then we have the case that we have F is equal to A with A unit and N is equal to zero, so in that case it holds. But now we want to claim that every element can be just written as a unit times the power of this uniformized parameter. That's supposed to be the definition of this key evaluation. So let's make some remark about this. So if T is uniformizing parameter, then if I take the Mth power of the maximal ideal, the Nth power of the maximal ideal is just the ideal generated by T to the N. So this somehow, and in fact, so let's see. So this is for all N bigger than zero. So by definition it's clear that T to the N is contained in the maximal ideal to the N because T is in the maximal ideal and T to the N is in the maximal ideal to the N. The maximal ideal to the N is the ideal generated by all products of N elements. Maybe you remember, maybe I could maybe write it. So by definition M to the N is equal to the ideal generated by all F1 times FN with Fi in N. That's how you define the power of an ideal. And then obviously if you have T to the N here, this lies in M to the N, therefore we have this. And by definition, it is clear also that T to the N is equal to M to the N for N equals zero and one. So the case N equals zero just precisely says that the units, so the first one says if I take T equal to one, it says that the maximal ideal is generated by T which is precisely our assumption. I mean on discrete variation ring and well the ideal generated by one obviously is the whole of the ring. So that's clear. So now we make induction on N. So we assume that T ideal generated by T to the N minus one is equal to M to the N minus one. Well, so by induction we have every element in M to the N minus one can be written as say A times T to the N minus one with A unit. And every element in M can be written as B times T with A unit. I mean this is not our, so just like this. Every element in M to the N minus one can be written as A times T to the N minus one because where A is in A, this is just a statement that is generated by, that the ideal is generated here by T to the N minus one despite T. So thus if I, an element in M to the N can be written as a sum of products of one element here and one element here. So every element in M to the N can be written as sum of elements of the form A times B times T to the N for A and B in A. So it is in T to the N. So we see, so this is not really such an exciting argument. One just sees that essentially trivially from the definition it follows that the maximal ideal to the N is generated by T to the N. And we see one other small thing more. So if I have an element of the form A times T to the N, it will be in the maximal ideal to the N. And it will be in the maximal ideal to the N plus one if and only if A lies in the maximal ideal. So therefore we find, if I take, so if F in A is of the form, so F in A is of the form F is equal to A times T to the N with A a unit if and only if F is an element in M to the N which does not lie in M to the N plus one. Because if it's a unit I can multiply with the inverse. So I see that it is, that this element actually will be a generator of M to the N and in particular it doesn't lie in M to the N plus one. And if it's not a unit it lies in the maximal ideal. This element A and so this product actually lies in the maximal ideal MN plus one. Okay, so this is, so in some sense, so this was just this. So now I want to sketch the proof that the local ring of a non-singular point of a curve is a discrete valuation ring. I will call this exercise because I don't do all the details and you could fill them in if you want although I basically give the complete argument but without giving all the details. So we want to prove that the local ring that for P in a local ring for C non-singular curve for C curve and P in C non-singular point the local ring OCP is a discrete valuation. Okay, so I just say it's all rather standard maybe some of it you already had in algebra somewhere. So first if A is a ring and I an ideal in A and you look at the projection, the canonical projection from A to A mod I then I claim that if I take the inverse image of an ideal here. This map which associates to an ideal here, its inverse image is injective. So what I mean is, so the map from the ideals of A mod I, the ideals of A which associates to some ideal J, its pulver, its inverse image, this map is injective. It's actually quite straightforward to see. I think it is also in, you also had it in the algebra course. I'm not quite sure but at any rate it's quite simple. It's almost by definition but you can do it. Then one can use this. So if A is a nuterian ring and I is an ideal, then we deduce from this statement here that also A mod I is a nuterian. Well, that's very simple. So if we have a chain of, if you have an ascending chain of ideals, so we use the ascending chain condition. So if J1 contains in J2 an ascending chain of ideals in A mod I, then it follows that if I take the inverse image under this thing, it's also an ascending chain. So P to the minus 1 of J1 is an ascending chain of ideals in A and it therefore becomes stationary because A is nuterian. So after a certain point all the ideals downstairs here become equal. But as the pullback map was injective, it means that also the original chain becomes stationary. So with this we have, so therefore we find that thus if X in the end is a close-up variety, then it follows that A of X is nuterian. Because we know that A of X is KX1 to XN divided by the ideal of X and we know by Hebert's basis theorem that KX1 to XN is nuterian, so also A of X is nuterian. And then the third statement is, so let X be a variety and P be a point in X, then you want to show OXP is nuterian. Well, you know, this is just the local ring, it depends on any neighborhood of P. So we can take an affine open neighborhood, so we can assume that X is affine. On a neighborhood of P we can assume first that X is affine and so then we can also assume it's a close-up variety of affine space. And we know therefore that A of X is nuterian and then, so I just claim that by a somewhat similar proof to what is happening here, the map which sends an ideal, so I say map from the ideals in the local ring to the ideals in AX, which just takes an ideal and maps it to its intersection with AX. AX, so OXP is some subring of the quotient field of AX, so elements in OXP are quotients of two elements in AX and I can view AX as a subring of those elements where the denominator is one. And so if I have an ideal, I can look at all the elements in the ideal, whose denominator is one, and I claim, and this can then easily be seen to again be an ideal in AX and I claim that this map from the ideals in the local ring to the ideals in AX is injective. And if you have this, it follows that the same proof as the one of two will show that OXP is nuterian. In fact, it's a more general statement. If you have a ring and you do localization, then if the ring was nuterian, it's nuterian afterwards. But here you can see, I mean, I can maybe repeat what the same proof is. Assume you have a chain of ideals in OXP, in ascending chain of ideals in OXP, then I can take the intersection here for each of the ideals in the chain. I can take the intersection with AX, so I have a chain of ideals in AX because the X is nuterian, this chain becomes stationary, so from some point they all become equal. But S's map was injective, it means that also here the corresponding elements become equal, and so the chains constitute here. And this is the nuterianity. So the exercise was actually to show that this thing is a nuterian ring, and now we want to use this to show that it's a discrete variation ring, but that I actually will prove properly. So let P be a non-singular point on a curve C, then OCP is a discrete variation. So we know that OCP is a subring of the function field of C, which is a field. Now, I mean it's defined as a subring of the function field. The local ring is always defined as a subring of the function field, so a field. So if you have a subring of a field, it certainly is a discrete variation ring. It certainly is an integral domain because a field certainly does not contain zero devices. So we have seen that OCP is nuterian. So let M be the maximal ideal. Then we know to begin with that M is finitely generated as an ideal because every ideal in the nuterian ring is finitely generated. The nuterian condition was two equivalent ones. One was with the chains, the other one with the ideal. So then M is finitely generated ideal, or also if you want, which is the same as being finitely generated as an OCP module because OCP is nuteric. So remember that in Nakayama's lemma there was always the assumption at the beginning that whatever module we want to look at should be finitely generated. Now the module we are looking at is actually the maximal ideal. So we know that this point is a non-singular point. So the dimension of the tangent space P at x is equal to 1 of this vector space. And we had seen that this is equal to the dual vector space M mod M squared dual. That was how we had the intrinsic definition of the tangent space was the dual vector space to the vector space M mod M squared. And we now know that the dimension of this thing is 1. That's what I mean. So this thing is equal to M mod M squared. And so then this is the same as this. Yeah, yeah, yeah. So this is for any x, but now our x is c. Yeah, I mean. OK, so finally I got rid of some of this. So that's this. So we know that the dimension here is 1. So this is a k vector space of dimension 1. So we can take a basis, which will not be very large. So let t in M be an element. It's class, say t, is a basis of M mod M squared. So I should say the dimension of the dual vector space is 1. So the dimension of the vector space is also 1. So we also have therefore the dimension k of M mod M squared is equal to 1. OK, we take this element. I want to claim that this t will be a uniformizing parameter. So now here we are precisely in this situation that we have an element, t in our module, which is M, such that if we take the class of the element, module or the module times the maximal ideal, then it is a basis. So if you have a set of generators, which in this case contains only one element of... So in the other situation it was we had a big M, then it was M mod M M. But in our case the M was a module, but in our case M is just the maximal ideal. So M mod M M becomes M mod M squared. So we have an element in the module which is a basis such that its class, the module times M is a basis of it. So it follows by the second part of Nakayama's lemma that we have that t, the ideal is a generator of M. So in other words t is equal to M as OCP module. Okay, which is precisely saying that t is a uniformizing parameter. So we have found our uniformizing parameter and with this therefore we have checked the first property of the skeet valuation ring, that the maximal ideal is a principal ideal. Now the second statement was that whenever you have a uniformizing parameter, you can write any element as a power of the uniformizing parameter times a unit. So in order to prove this we will show something which doesn't look so very related. We'll show if I take a new module or a new ideal M by the intersection over all n bigger equal to 0 of the maximal ideal to the n. So define this module, well then this is 0. So we want to show that now. Then you will see why it shows the second property of the skeet valuation ring. So now in principle obviously it could be that if you multiply by higher and higher powers of n obviously this thing becomes smaller and smaller but it could be that there is something which all of them have been common. But the claim is they don't. And this is now actually very simple because we are, so M is an ideal in OCP. This is kind of clear. I mean if you just look at the definition of ideal, if you make an intersection of ideals like that, just you know you need to see the sum of any two element slides there and the product of any element with an element in the ring and it's straightforward to see that it's an ideal. And now we still remember that OCP is a Neuterian ring so this is finitely generated. And it is finitely generated because OCP is Neuterian. And here it seems really clear that one wants to use this Neuterian property because you know it's difficult, otherwise be difficult to see with such an infinite construction to be finitely generated. But it's finitely generated. And you know if we take M times M, well this is you know just, it's clear to see that this is intersection over L or N bigger than 0, M to the N, or you have M to the N times M. But you know if you, whether you, this is the same thing, this is equal to M. An element, if an element lies in all powers of the maximal ideal, it also lies if you want in all powers to the one power higher. So this ring does not, this ideal or this module does not change if you multiply it with the maximal ideal. And other, so by Nakayama's lemma the first part it is 0. And now, so, but what does it mean? So this means every element say F in OCP lies in M to the N minus M to the N plus 1 for some N. And you know it cannot lie in the, you know, if all N be equal to 0. It cannot lie in the intersection of all of the M to the N. So there's a last one in which it lies. And then if it's the last one, if the last one is M to the N, it doesn't lie in M to the N plus 1. So, and we have just seen before that this means, that this means that F can be written as A times T to the N for A, a unit. So we had this matrix remark where I had said in the end that precisely if the element lies in M to the N minus M to the N plus 1 it means I can write it as a unit times T to the N. Okay. And so this is precisely the second statement of discrete valuation ring. So we find, we have a discrete valuation. So I mean, if one thinks of it like in complex analysis, so a local parameter, a uniformizing parameter at the point P, you know, you somehow should view this as a function which vanishes to order one at this point. If you are in complex plane, you know, for instance the function Z would be, you know, would have this property in complex analysis. And in, you know, then for a holomorphic function on the plane you can say it vanishes to a certain order at a given point. And so, and this would correspond to maybe in our language that it can be written as A times T to the N where A is a unit. Anyway, so what I want to say is that in such a discrete valuation ring we can somehow say what the order of vanishing of an element in the, so we can associate to every element in the ring a number which we can kind of think of as the order of vanishing. And so let's me do this a bit more formally. So definition, let again P in non-singular point on a curve C we define new P, a map new P from the local ring at P without zero to the non-negative integers. Namely, we put new P of F equal to N if and only if N, if and only if F can be written as A times T to the N where A is a unit. And as I said, and new P of F is the order of F at P you can think it of the order of vanishing of F at P so where T obviously is uniformized parameter T, uniformizing. So this function has certain properties which are the defining properties of something that is called valuation in mathematics so that's why it's called the discrete valuation ring. So we have the following properties as properties, the following properties. So the first one is, I mean they are all very simple so if I take new P of a product this is the sum. The second is that if I take the valuation of a sum this is at least as big as the minimum of the valuations of the two summons and we have that this is actually in equality if the two valuations are different. So with equality if new P of F is different from new P of G. So again, if you think of this as, you would think of this as order of vanishing of functions so you see that the sum, the order of vanishing of a sum of two functions is at least the minimum of the order of vanishing of the functions and if this order of vanishing is different then the corresponding lowest order terms cannot cancel and so therefore it is equal to the minimum. That's somehow the same as you have in complex analysis or whatever if you look at polynomials. Okay, the third statement is F is a unit if and only if new P of F is equal to 0. Now this I have already said several times F is a unit if and only I can, I mean here by definition I write this thing as a unit times T to the N and so this is trivial. The other two are also quite simple. So I write F equal to E times T to the N and I obviously assume that A is a unit G equal to B times T to the M then obviously F times G is equal to A times B times T to the N plus M. It's not very surprising. And if A and B are units then AB is a unit. So this proves one. In other words it shows that one is completely trivial. Now the second one is ever so slightly less trivial. So we have assume one of the two numbers must be smaller equal to the other. So we again use the same F and G and assume that say N is smaller equal to M. One of the two must be smaller than the other. So then it follows if you take F plus G. What is it? We just put it together. So it will be divisible by T to the N. So this is T to the N times A plus B T to the M minus N. And so it certainly is a multiple of T to the N. Similarly it follows that a new P of F plus G is bigger equal to N. Now if these two, so which was the first statement. And now we have to see if these two are different. I have to show that we have equality. If N is strictly smaller than M then I claim that A plus B T to the M minus N is a unit. Because otherwise if I take A I can write this as A plus B T to the M minus N minus B T to the M minus N. That's not very surprising. The point here is so if it's not a unit then it would be in M. So this is a different M but I think you can manage that. So we know that everything which is not a unit lies in the maximal ideal. So this thing would be in the maximal ideal. This thing is in the maximal ideal. So the sum is in the maximal ideal. Which is obviously a contradiction because A was supposed to be a unit. And so thus this element is a unit. And so then this formula says that the order is precisely N, which is the minimum. Now we want to see, so we have to find this valuation, this order on the local ring at P. Now we want to see that this valuation can be extended to the whole of the function field of C. So I can say for every, which is again kind of obvious in some sense, for every rational function on C I can say to which order it vanishes or has a pole at my point P. I can extend this thing to the whole of KC but now it will no longer be a map to Z, to the non-negative integers but to all the integers. If I can find it there. So again definition. So you should note that if I take K of C this is also the quotient field of O, C, P. By definition you know it's O, C, P is by itself a sub ring of K of C. But if we assume for instance that we have some fine neighborhood here then we look at A of X then KC will be the quotient field of that. And here we have a sub ring which lies in between the two and we take the quotient field of it then they will be equal. You know you have just, if you take quotients of quotients it's the same as just taking quotients. I mean anyway you can work this out that this is what it is. So it somehow we can view element of KC also as a quotient of two elements of O, C, P. And so we want to extend this valuation from KC again without zero to Z. Well just in such a way that it is compatible with taking quotients namely by new P of F divided by G. So if F and G are elements in O, C, P we take the quotient is defined to be new P of F minus new P of G. And so you have to see that this is well defined. So it doesn't depend on how I represent F divided by G as a quotient of elements here. So if F divided by G is equal to F prime divided by G prime. So then it means that F G prime is this equivalent to F G prime is equal to F prime G. Then I can say if I take new P of F plus new P of G prime is this. This is then new P of F G prime is equal to new P of as these things are equal F prime G. This is new P of F prime times new P plus new P of G. And we can now bring this to the other side. So it follows that new P of F means new P of G is equal to new P. So this subtracted this and I subtract this new P of F prime minus new P of G prime, which precisely says that this thing was well defined. Okay, so this is again straightforward. So we have for any element in the function field except for zero, we can associate this number new P, which is an integer. So and then I really want to really want to have this integration. So let H be an element in KC. So N equal new P of H. Then we say H has a zero of order N if N is bigger than zero and H has a pole at P of order minus N if N is smaller. Okay, we just say this is just the words we want to say and it somehow is like you have them. And maybe I should say, you know, so the simplest case of a curve is obviously if C is a one. And so then it is easy to see that to take our point P to be zero. Then the maximal ideal will be equal to the ideal generated by X. So the thing which vanishes there, so all the class in the, okay. And so P is the point zero. If you want, it's one pupil, but anyway. And then we see that. And so if then, if F is a rational function in X. So we can write X, F is equal to G of X divided by H of X. Then X, then obviously we can say what the order of zero of G at zero is. It's by what power it is divisible. So G is equal to X to the L times G prime with G prime of zero is different from zero. I mean G prime doesn't mean derivative, it's just another G. And H is equal to X to the M times H prime with H prime of zero is different from zero. Then obviously we would get according to this definition that F, that new P of F is equal to L minus M. In this particular case, we can obviously just also, we can pull out the factor X to the M minus X to the L, and the rest would be a unit at the point zero. And that's actually the next thing we will see. Okay, anyway, we can see that in this trivial case, it does precisely everything the way you expect. It's just the order of zero, the order of pole in the obvious sense. But one thing that we have just seen here, which we will show, is that we obviously also have that F can be written as X to the L minus M times G prime divided by H prime. We just have pulled this out. And now these are both non-zero. So this actually is a unit at zero, is a unit in O A1 comma zero. So that is, we have found that in this particular case, if the order of an element in the quotient field is something, we can write it as a unit times the uniformizing parameter to the something, where this power can be both positive or negative. And the claim is that this holds in general, and this finds obvious. Namely, the following corollary, the corollary to the definition in some sense. So let P be again a non-singular point curve C. So first, so if we take an element in the function field, then the order of this element is if and only if we can, there is a unit, there is a unit or we can just write it directly. So we can write F is equal to A times T to the new P of F for T uniformizing parameter and A a unit. So before, in the definition, we had to require this for any element in the local ring. Now the same statement is true for any element in the function field, but this power can be negative. And the second statement is that if I write such a thing, an element here, so that if I know the order of the element, it will be in the local ring if and only if the order is non-negative. So OCT now is equal to the set of all F in KC, such that the order of F is non-negative, because you have to back at back zero. So an element in the function field will be actually in the local ring if and only if its order is non-negative. So which is again, in some sense, if it matches with the intuition, it is in the local ring if and only if it doesn't have a pole at the point. So if these words have some meaning, then this should be clear. But as we have just defined these words, it's not like from the fact that we call something in some way, that we cannot deduce that anything is true. So we have this statement. So let's see. It's actually quite simple. So you see we have in some sense a simple description of what this order is. We can always write an element as a uniformizing parameter times unit, uniform parameter to some power, but also negative powers are allowed. We know by definition that mu P of F will be equal to N if and only if we can write F equal to G divided by H, where G and H are elements in the local ring. And we have the difference of the valuations are N. Well, but for elements in the local ring, we know if there's a certain valuation that we have this kind of statement. So it follows. So equivalently, yeah, I exchanged it would have been more natural to call this thing H and to call it F divided by G, but as I didn't, I have to match it. So equivalently, there exists some numbers such that G can be written as A times T to the say M plus N and H can be written as B times T to the M for A and B units and M some positive integer. We know that this would mean the valuation of G is M plus N. It must for some M and the variation of H is M and it must be true for some M because the difference is N. So, well, but then, you know, I can just, you know, I look at G divided by H, so why don't I just divide them? So thus is equal to G divided by H equal to what is it A divided by B times T to the N and A and B are units, so A divided by B is also a unit. So it's really very simple. And now the second statement is even a little bit simpler. So we know if F lies in the local ring, then its valuation is non-negative because we know that then it can be written as a unit times T to the N for some N for some positive non-negative N. Conversely, I mean it's actually to reality, but anyway, that F, so we assume that new P of F is bigger equal to 0. Then by the first part, we can write F is equal to A times T to the new P of F where A is a unit. But this means, you know, T is an element in the local ring. If you take a positive power of it, we are still in the local ring. And we multiply by a unit, we are still in the local ring. So this is also trivial. How much, ah, my time is actually up. Didn't notice. Okay, we are almost, there's not so much left, but I will just say what I want to do next time. So we want to use this to show, so use this. So if C is a, say, a non-singular curve and we have a morphism, P is a point in C, we have a morphism which is defined outside the point P to Y, where Y is a projective variety, then Phi can be extended to Phi from C to Y. So if you have a morphism on a curve which is defined outside one point, we can always extend it to the whole curve. And if the target is a projective variety, what? Yeah, well, except that, well, I mean, the homework was a bit, yeah, it was a bit different, I mean, in that, you know, it was a different statement that you have to prove. But so this is actually something that's a rather important fact for many applications. So it actually turns out that under some reasonable assumptions, this statement is equivalent to the fact, you know, so if you have the property that whenever you have a morphism from a curve without a point to a variety, so assume you have a variety with a property, whenever you have a morphism from a curve without, a smooth curve without a point to the variety, then it can be extended to the whole curve. This is essentially equivalent under some reasonable assumptions to this variety being proper. So because in some sense it's, you know, and it gives a more kind of nice, it somehow gives you a better feeling for what proper means, namely really that no point is missing. You know, if you have a map from this curve without a point to Y, then the idea is, you know, everything is mapping somewhere, except for this point. And, you know, this, the image point of this thing does not, you know, the point where you would want to map this point after having mapped all the other ones is not missing. You can put this one there. So, you know, if this would be a fine, it could for instance be that this point would have to, you know, that you have some, that this goes off to infinity and you cannot do it, no? So it really is, and, yeah. So it's actually the way, so in a slightly more abstract setting, in Hartzorn this is called the evaluative criterion of properness and it's a place also, you know, that's the way that I would usually check whether something is proper or projective, because it's, I mean, more useful for kind of in more complicated situations. Anyway, so this we will check next time, but that is, and that's more or less it.