PSE Lectures Ep 27 - Zeros and poles of the coordinate functions of [n]

Transcript

uh last time um we proved I mean we we we are interested in studying the function um bracket n from E to E um that sends a point P to to n * p and um we we simply denote

um NP it has two coordinates right so um this this is two fun two functions two rational functions G&P and h&p um so so GN and hn belong to the field of rational functions and we we defined of course n um it to be

um all the points in E over K Bar such that um NP is zero the zero element and we would like to study GN and hn in order to understand ER the uh the group of of inion points um and what we showed we had a a a theorum for um algebraically closed

field um GN and H are irrational functions and the pulse for um um precisely at um n um and moreover n is finite for every

n and um what I want to do now is start um um doing the formulas for G G2 and H2 uh let's let's write um G2 and H2 um so I I I remind you we had this the general formulas for how to H add

two points um should I okay so let me just write the formulas this is um this is I mean you can you can extract I mean you can um get these formulas by by the general formulas of doubling of a point G2 of of a point P for p is some X comma

y on e um if you do G2 of P you get the following expression this is um x to the 4 - 2 a x Cub minus 8 B X+ a s divided by 4 s x where s X is as usual is this um X Cub plus a X+

B and the formula for H2 is um y times x to the 6 minus 5 a x to the 4 plus 20 b or I need more space H2 of p is um y * x to the 6us

5 a to the 4 + 20 b x cubus 5 a² x² - 4 a b x - 8 b² - a cub divided by 8 s

x squar so you will see soon that the the the details of the formula formulas don't matter that much and um but I would want to to uh to consider the leading terms Mo most of the time ER and um to study well we want um

information on a um G&P h&p um for some you know we need some kind of anchor um um and the only the only ER point in which we can hope to H to get information is is the point at Infinity because the rest we we have very little access to um

so um so this is the next proposition if um n is prime to um a p where p is the characteristic of the field then um if I take GN GN is is a function of X X and Y right and divide it by X and evaluated at

o I get 1/ n² and if I do the same for h n but now divided by Y and evaluated at O then I get one over n Cub okay so and this this kind of give you give you a hint of what we are going to um use it for because this n square is is essentially what we want we

want [Music] um we want to prove that N is a um isomorphic to ZN time ZN and it would follow um very easily from the the classification of finite bilan groups if we know that n has order n Square okay so um this General go um and okay so how do we

um of course the difficulty is that we I mean the formulas the explicit formulas for GN and hn are not accessible I mean they explode very quickly this this is if you want it's it's one of the reasons that elliptic curves are used in in cryptography I mean if you you really want I give you like a generic Point XY and ER and you want to ER to calculate I mean to calculate n * X comma

y ER the formulas become um quite involved so so if you try to do Brute Force procedures it becomes very expensive and and that's why that's one of the reasons elliptic curves are are useful in cryptography so we don't have a closed formula I mean okay I guess maybe um at least for us at this point we don't have a good Clause formula for GN and hn it's not accessible it's very complicated but we can hope that um evaluating

some expression from GN at at the point at Infinity is is kind of accessible so the proof is um by induction on n and sorry uh before I I proceed with the proof note this condition so we we really need that n is is a um is not divisible by

P H because otherwise the um these functions behave very differently um and we will see also this this go I mean this um this isomorphism is only true for um n that is prime to p otherwise it's not true and the the torsion points H behave badly so this is

kind of um the usual phenomena that um when you take a um something in the the characteristic of the the field for which the the elliptic curve is defined over ER you get unusual behavior um but but for um for integers that are um um I mean a prime to the characteristic you get the the typical Behavior this is how usually people see it okay so by induction on

N um first assume H that n is smaller than P um and we we need a base for the induction so for nals 2 I I take the formulas and I I divide um let's say I do GN ided by X and then evaluated it all so what do I get s x is um of degree

three right and if I divide by X I mean the the the leading monomial here becomes a x Cub so evaluated it all I have like a course of two pols of the same degree um and evaluation at o simply means you take the um the leading coefficient I mean the the coefficient of the the leading monomial divided by the coefficient of the leading monomial in

the in the denominator um and in this case it's like it's one over four right right so I get um for nals 2 G2 um D2 over X at O is a 1 over 4 which is indeed a 1 over n² and now I do H2 divide by y evaluated at

o um here what will happen is that y will drop out and s x square is um I mean the the the leading monomial is X to the 6 so again evaluated at o I simply take the the coefficient the leading monomials and and divide it by each other it's it's one over8 and now um

I can I can write NP as n minus one p plus P um and so if I do GN over X I can write it as um minus GN minus1 over X and minus one plus 1 / x h n

minus1 - y / GN -1 - x squar what I did here is is um I mean everything needs to be divided by X right I mean I do GN divided by X so this is um divided by X um I have um uh here I got one because I had um I also had an X um

this hn minus one minus y okay [Music] so let me just remind you we had it um when we studied um rational Maps as as points of an elliptic curve I mean just recall ER if um if f is R comma S and G is um what was it t comma V then f+

G which is let's say k um and K is W then you get the formulas Z or rather ah this is a w Comm Z sorry and um the formulas for um W is minus t + r plus Lambda

squar and the the formula for Z is a minus V minus Lambda W minus t and this is assuming that I mean Lambda could be um either s minus V divided by R minus t if R minus if R is different than than T and otherwise it is a 3 r² + a divided by

2s okay these are the formulas um for adding two um uh two points and what I did here is um I took these formulas and I I divided everything by X so let can just write it slightly better so um I have now n minus1 p plus P um and what it is so I need to take

um minus this is this this part this part is um minus t + r and of course divided by X so what is minus t + r T is um is the guy in the the um the first coordinate of the second um the second point I mean so so T is X here and R is um is the first coordinate of the ER the other thing so this is um so R

is um us sorry is is GN minus one okay so I did minus t + r / by X is this okay this is is it clear okay and and then in the second term I took this Lambda this this needs to be Lambda square right but what is Lambda Square Lambda square is I mean Lambda is s minus V ided by R minus t

so and S here is of course h n minus one and V is um is simply y okay so this this is a and I divided everything by X okay so um so this is a formula for GN over X and now I develop it a bit I can write it as um

- GN -1 / x - 1+ Y 2 / X cub and here I have h nus1 / y -1 and and this is divided by gn-1 / x -1 this all is square so what I did here I I just um in

in the second term I mean in this ER blue part I multiplied by y sar from outside and um divided by y Square from the inside but dividing by y s from the the ins I mean I need to divide by y because I have this a square here so and of course I did it because I want to use the inductive step and in the inductive step tells me um things about hn minus one divided by Y and GN minus one / X okay so now I evaluated o now I do GN /

X evaluated at o this is just evaluating this expression this orange expression in each term I mean if if I can evaluate in each term and it it gives me a finite thing then it means I can evaluate each term separately so now by induct by induction gn-1 / X evaluated at all is 1 / nus n-1 squar so I get the first term is minus nus1 to Theus

2 the second term remains one I evaluate it oh I I get one or minus one and and then y² ided X Cub evaluated at o is is what you remember I have what is y y 2 / X Cub evaluated oh yeah Lisa go ahead yeah but my question is earlier

yeah okay so go ahead I I didn't understand this formula that wrote for GN overx um I this one the the first one or the second one yes the first one I I don't understand what like I understand that I don't understand what stands for T and uh what stands for R okay okay let let me try to explain so um I have NP is equal to n minus one p+ p this means that I I can say that g n h

n is equal to GN minus1 h n minus one plus XY right I I I mean I didn't write the variables of GN and hn and and so on okay but so but but this is addition of a a rational Maps right so take this one to BF this one to be G okay M and now we have a formula for

how to add two rational maps I mean it's basically the formulas of of addition um of points on an curve right yes um so the the formulas are written here in in Orange um so I take um R to be um to be GN minus one right and S to be hn minus one and then T and V are X XY respectively and I have a formula for what is the the result is like a wz wz

is is this GN agent right and um ah I understand okay okay and then I need to divide by X because I have division by X here uh uh okay okay I understand okay so that's what you get now M there is a subtle Point actually how do I know that the Lambda is s minus V / R minus t and not the other thing the reason is that um we had a

proposition last time that um GN minus X is not identically zero so GN which is R is different as as a rational function than T now the reason GN minus X is not identically zero this we proved but I can repeat quickly the proof it's like if it was identically Zero then it would

mean that um a p plus minus or sorry n+ minus one p plus - one p is equal to zero for all p and this would mean that either e e n minus one or e n + 1 is is the whole e and this this is not possible because H um we proved that n minus one and I mean n in general e k is is

finite okay so this this kind of this is the the small pieces of the puzzle yeah yeah it's okay thank you all right um so we did this manipulation and now we substitute um we substitute o in in every point and I need to see what is Y2 over X Cub evaluated at o do you have an idea anyone yeah uh yeah what do you mean by substituting uh zero to every Point

um um I have a sum of rational functions here right in the yes and I need to evaluate this this sum I need to evaluate it at um at o h so if everything is we had like a proposition that if everything is finite you can evaluate each of the terms separately each of the sum and you can evaluate it o separately it's not true it's not always true I mean if you if one of the terms becomes Infinity then you need to be more

careful but if all of them are finite then you can you can evaluate each of the summons separately and then sum it sum it up and you mean yes sir you mean that we evaluate just every term at zero not a different point yes exactly I I messed up something sorry yeah no no it's like uh the terminology is we haven't fixed it well maybe um but also here in the blue part I have a um a product of two rational function functions and I need

to evaluate this product at o if one of the functions evaluated at o is not finite then I need to I need to be careful but if both of them evaluated at o are finite then what I can do this we had like a proposition I can evaluate each of the the um the factors of the product evaluated it all separately and then I I multiply it and so I mean I'm implicitly claiming that y Y2 over X Cube evaluated at o is

finite but what is it how do we evaluate a rational function at at o you remember like the the exponents of of the numerator and denominator er no what it's it's about so you have basically two o you can think of O as this euristic it's it's

very useful to think of it as Infinity comma Infinity yes okay so it means in the x coordinate I substitute infinity and in the y coordinate I also substitute Infinity but remember that the degree of X is not the same as the degree of Y so I I need to know I mean if I do a um I need to know what what is the the degree of the numerator and what is the degree of the denominator if the degree of the numerator is bigger than the degree of

the the denom denominator then evaluated at o I get infinity okay if the I mean if the degree of the top is is bigger than the degree of the bottom then I I plug I plug Infinity everywhere then I get infinity it's like if you evaluate x squar divided by um I don't know x square + one divided by X you plug Infinity everywhere you get infinity right this just tically you don't need to I mean there's no

a you follow now yes so and but on the other hand if I evaluate if I want to evaluate um X ided x² + 1 evaluated at Infinity this I get I get zero right because this x square is is tending to Infinity ER faster than than x are you with me Lisa now Nicholas yeah yeah yeah yeah

okay so so if so I what I need to do is to um examine the degree the degree of the numerator and the degree of the denominator now if the degree the last case is if there degrees is is equal if you have like a x² + 1 and you have 2 x² + 3 x and + 2 and now you evaluate at Infinity what what do you get if you evaluate this expression at Infinity let me write it somewhere else if I do

x² + 1 ided by 2 x² + 3 x + 2 evaluated at Infinity what do I get remember you you do this yeah Nicholas one half one half exactly I I simply take the the leading coefficients I mean the leading Co the the coefficients of the leading monomials and I I divide it by by each other because you can do a a lupal and then you have like a one plus

you divide everything by X squ you have 1 + 1/ x² and then in the bottom you have a 2 + um 3 over x + 2 over x² and all these guys they become zero when when X is infinity right so you you are left with um with just the the the coefficients of the leading monomial so this is in one variable but in fact in two variables you do something very similar you but we have a convention that the degree of

Y degree of Y is three and the degree of X is two is two um so so in fact y^2 over X Cub is a a a a quotient of two pols of the same degree right because this this the top has degree six and the bottom has degree six as well so so to evaluate it all I simply take the coefficients of the leading monomials and divide it by

each other this is one okay so now I um I evaluate GN over X at o I need to evaluate the last expression but the last expression I can write as um the first Factor namely Y2 over X Cub evaluated at o this is one and then the second the second thing evaluated at o but this I know by the inductive assumption so I get um a one

over um n minus one or let me write it as just n minus one to Theus 3 -1 divided by nus 2 to the minus 2 minus one and this all is is squared okay and now I got some uh some expression you do the algebra and and you get that this is n s

sorry sorry um 1/ n² okay ER I guess yeah um it's yeah it's it's a I think from here it's really just like a simple manipulation you I hope you can believe me I mean I believe the the book took it from um and similarly we can

calculate um h n divided by Y at o to be 1/ n NQ so this if you want you can take it as an exercise um okay we are not done yet um but maybe it's good to ER to take a break here and and and come back to the rest of the proof do you have any questions so

far I think I have one more question may uh maybe it's simple but I don't understand um how do you know that by induction uh g n minus one is n n minus one to the to the power of minus 2 well this is like induction you assume that the claim is true the claim is that GN minus I mean that GN divided by X evaluated at o is one over n s ah okay right so I assume by induction I mean I need to have a base

case this we had like the base case was n equals 2 ah okay okay and then yeah and then I of course I deliberately I I wrote GN over e um is is a a stuff that are a g GN minus one over X and H N minus one over y because then I can use the inductive assumption but how did we come up with the proposition like this uh this assumption where did we take it from that GN /x is

1/ n squ i I want to prove it so I want to prove it for all n so I'm saying I mean for all n bigger than one okay so we just want to prove it and we took nals 2 as a base case we showed that it is correct for nals 2 and then we assume that just correct then we assume it's correct for n minus one yes okay yes and that's what we prove it for n but we didn't do it for n yet no we I mean look we did a

calculation GN over X GN over x i first wrote the the general expression and then I eval uated it at o right using the inductive assumption the inductive assumption is that when you evaluate GN minus one over X at o you get one over I mean sorry yeah 1 over n minus one squ right ah we used this assumption to calculate GN minus one and we got the result for GN that is exact L what to assume I mean we no we how how does

induction work you have you want to prove a claim um for any end then you you first check the base case that the claim is true for for n equals 1 or n equals 2 it's some finite n and then you you the inductive process is you assume that the claim is true for n minus one and then you prove the claim for n right you know this yes so I assumed my inductive assumption is that GN minus one over X evaluated at o is 1 / nus1

s yes yes I understood okay thank you okay and now I want to prove I want to I want to ER prove by induction I mean using this assumption that GN GN over ex o is 1/ n s so I write okay you you follow now yes yes I understood I understood thank you okay all right let's take a break and return to the proof you can think in the break you can think what is missing because let's say apart from the the exercise I mean this hn over y is is um it's a similar manipulation but let's

say we proved that GN over X at o is 1/ n s so what else do we have to prove H this I Le I leave you to to one during the break e

e e e

e e e

e e e

e e e

e e e

e e okay um so what is think what is

missing I mean look at the the statement of the theorem has um that n is prime to P I mean it's not the proposition is not supposed to to be true if n is not prime to p and we said in the beginning that we first assume that n is is smaller than P ER and we we proved by induction I mean that for all n smaller than P the claim is true I mean we had like a a base n equals

two and then we prove that if it's true for n minus one then it's true for n but we get stuck we apparently get stuck when n is equal to P because when n is equal to P the claim is not supposed to be true so H how can we proceed with the induction you see the point this this is a this is a gap to to fill but that's not it's not a big problem um so it appears we get

stuck when n is equal to P but um um we can circumvent it as follows um instead of writing NP equals to n minus one + P I can write NP = nus 2

p+ 2 p so then then I I need to assume then then I can kind of skip I can kind of skip P or skip n equals P because I do the same thing that I did here I mean I I open up I have this H these formulas here instead of X comma y it's going to be a um whatever the formulas for doubling the point x XY and then since I know by induction

the the claim is true for um for p minus one then I would I so I would take a um so we take um n equals um I want I want now to prove the claim for nals p + 1 so for nals p + 1 I know the claim for I mean n minus 2 is going to be P minus

one in which the in which case the claim is true you see and then I would um I I I would get the claim for n by by the same procedure is this clear Lisa what do you think and sorry I don't understand today things from first from first I understand the first row that when you

rewrite n p as n minus 2 p plus 2 p this is clear yeah transformation and now so we want to prove obviously we we we don't want to prove the claim I mean for n equals P it's not it's not supposed to be true uh so to prove the claim for n equals P I want to prove the claim for n equals p + 1 and then I can procedure by induction up until the point up until n equals p + 2 sorry up

up until n equals 2 p right M um so how do I prove the claim for nals p + one H we know the claim the claim for n equals sorry for um for n minus 2 because n minus two is is p minus one right MH so we know the claim IE we know that

gnus [Music] 2 ided by X evaluated at o is 1 / N - 2 2 and we also know know that h nus 2 / y evaluated at o is 1/ nus 2 cub and we also know the claim for for n equals 2 right yes okay so now I have all the ingredients I need right

I I like I you see this I I want to prove the claim for n equals p + one but I I cannot rely on the claim for n equals p i mean for n minus one because the claim is not is not supposed to be true for n for n equals P yes so instead I do a different decomposition I and and for n minus 2 I do know the claim and for and for two I also know the claim so here I I would get this is right this is GNH

hn but here I will have a a I will be able to express GN hn as some kind of a I mean elliptic curb sum or sum of rational functions where I get this GN minus 2 hn minus 2 and and here I have a G2 and H2 right and I will do the same very similar manipulation I will write a the the sum on on the right side

is a expressions of gnus 2 / X and H nus 2 / Y and then I would want to evaluate it at o right I mean I so yes yes okay you see you see the point uhhuh okay I'm not I'm not writing the details but ER I'm trying to convince you that it's a it's in this the same I mean it's the same way that we did the the previous thing we can do this

okay and we do it to ensure that n and p are prime copre we we do it to to be able to keep the induction going I mean what is this induction you have like a [Music] um you have a sequence of a propositions a raw let's say indexed by the natural numbers and you want to prove that all the propositions are true so you prove that Row one is true and then you prove that for any

n if row N minus one is true then row N is true this gen this like generalities on induction yeah and if you if you prove these two claims then you know that all row N is true that row N is RO if you want Ro K is true for any K right so you you kind of I mean you

the the induction procedure is like climbing climbing a letter um of proving that each step of the letter is um is true yeah so our problem was that when we went one by one for every n some of them might be not prime to P yeah and we know exactly what are what are the ends that are not prime to P you have a um it's all multiplications of P it's like yeah P times k for for for various KS so we go from one up to P minus

one here we we CL we do the the induction ladder um we we have no problem but then we cannot proceed proceed by by like trivial induction to to n equals P right so but we prove that if the claim is true for p minus one then it's true for p + one and then we climb the letter the lad up to 2

p 2 p sorry 2 p minus one okay and by by the Same by by the same procedure I mean we had in the the inductive step is is is okay as long as we know that n is prime to P but we cannot we cannot move up the induction kind of naely if n is equal to if if n is equal to some um multiple of P so we changed the size of the step in a way that we are sure that they are always Prime yeah so what

we did here is like we proved yeah we proved that all the the the propositions Row one up to uh row P minus one are true and then we prove that if row P minus one is true then row P + one is true and and then the same the same argument that we had before proves that if Row p+ one is true then Row 2 p minus one is true and here we will need we will need

to do another jump we will need to prove I mean but this it's basically the same it's the same that if Row 2 p minus one is true then row 2 p + one is also true this is the same inductive I mean it's basically the same trick as here yeah okay so this this is a it's kind of a sketch it's not a it's not a foolproof but uh I think it's it's a rather solid uh sketch it's like um and it will be ended to p+ one or should go further

yeah we want to prove it for all n Prime to P so we need to be able to do the jump infinitely many times we every every time that that we get to a multiple of P we need to do a jump I see okay all right so this is yeah this is kind of uh there are various H there are various um versions of induction to collections of statements that are not er um not so easy to prove in in in the standard induction like for example just

to maybe trigger your curiosity I could prove if I want to prove all the the the the claims R in I could prove that um for any okay I I need a base so I I could prove that Row one is true and that row of 2 to the K is true for any K and that

if row of um I don't know T is true then row of T minus one is true this is also this this a three I mean if I prove these three issues then I can conclude that row of n is true for all for all n this this is called a

um er ascending descending induction you see why why this procedure would work I mean because now you you give me an n and you want um I want to show you that that raw n is true so what do I do I pick um so Fix N so to show that um row N is

true choose K such that um 2K is bigger than n there there must be some some K like this um we know that row 2K is true and then Row 2 K minus one is also true because if I I have that if row is row of T is true then row of T minus one is

true and I mean I go down I have to I have to reach at some point into to a row N and I include that raw and is true you see the this this like a just maybe a revealing a something about induction that you you didn't notice ER yeah it's clear Nicholas you're are now you you're fine with it okay

so what's next yeah okay so I um the the next thing we want to do um is to to calculate the divisor of um GM minus GN this we will not this this will take

a few sessions um but um one of the tools we we are going to need um a good notion of um derivation of rational functions on E and remember we are in an algebraic I mean

um there is no way I mean we work perhaps over finite Fields or over typically over fields of positive characteristic so there is no apparent geometry but nevertheless because we we work with pols pols we can differentiate formally I mean you know that H like the derivative um of x to the N uh the x is like

a I mean n x to the N minus one DX right um so we we can derive formally um this sorry the the notation here is bit misleading this usually you in analysis you write it like this um D of um of x to DN and um in elliptic curves we have a relation right we have um

I mean in e um Y 2 - x Cub - a x minus B is equal to zero so the derivative whatever I do the derivative of this thing needs to be zero right um so but if I take the the derivative as I know um I do let me do instead of small d i do Big

D um so if the I mean the the derivative of this whole thing y^ 2 - x Cub - a x - B needs to be zero and what I get is 2y time Dy H must be equal to 3 x² plus a DX um and how do we make it true I mean one

canonical way to make it true um definition um Define uh a derivation on M ke um by setting H the derivative of x is 2 Y and the derivative of y

Y is 3 x² + 8 so you see if I do it here the derivative of y is 3 x^ 2 + a and the derivative of x is a 2 y then I get an equality right I get here 3x^ 2 + a is indeed equal to to this two y okay so this is a natural a natural choice of a

derivative um and we now want to um to define the derivative for all rational functions rational functions on E they have two variables right X and Y and they are some quotients of polinomial but we want the usual properties of derivative um um and extending so I extend d h so um so that it

satisfies the the usual properties of derivative um are one that the derivative of a constant is zero um for any any constant in the field when I write constant here I mean the the constant rational function right um the second is that the derivative of the the sum of two functions is simply the derivative of the um the first plus the derivative of the second for any two rational

functions and third um I want the deriv of constant time a rational function is simply constant times the derivative of this function so for any c c in in K and for any R in in K of and the last rule this is the the most most important one how do I do derivative of a product R * s this needs to be R * D of S

Plus s * D of R okay for for any RNs in IR rational function on now when I say extend extend um I need to I need to kind of show you that there is a unique once I set up a a this this kind of anchor condition and I want these four properties then there is only one function we we want to show

that um there is a unique function D from K of e to K of E satisfying let's call this conditions I don't know I and 2 I so it needs to satisfy

I and 2 I plus one up to four but um this is this is not hard so um observe if if I have a um so I want to do D of um of x to the N what is D of x to the n um let's do D of x

s d of X squ is X time DX plus e x * DX which is 2 DX um sorry to XD X yes uh when when you tell that we want uh to show that there is a unique function that satisfi with this property H do we want to say that there is only like only one function that will satisfy all this property yeah that's what

unique means okay yeah unique this is why I mean we I want to I mean I could have kind of er just g g give you a definition and and forget about all the the stuff and then prove these properties but I I want to motivate you um so this is kind of a one approach to to motivate definitions is that I give axioms that I want I want the the the the new notion to satisfy these axioms they need to be kind of natural like

analogous from analogous to to Notions that we already know these properties for and and then I show that there is only one um one Function One notion that satisfy these these axioms this this is usually considered a good motivation for a definition because then you know then I know that whatever I I I wanted there is only one such thing I mean okay I just I just asked because it didn't sound intuitively for me that there would be only one such function yeah but it's but it is true and this this shows

you that this notion of derivation that we are using is is a is canonical I mean once you you kind of spell out natural properties there is only one thing that that does the trick um okay so I'm saying like um obviously I can I mean I start with something d i i i know how to derive X ER I mean by one and two so I simply start with something that um I mean I I know how to do D of X and D of Y and now I

want to show you that if the four properties one up to four need to be satisfied are satisfied then a there is only one way to extend D to to an actual function from um rational functions to rational functions and the reason is okay D of x² is 2x to 2x DX H then you do D of X Cub what is it you can write it by the again this

fourth property is is the the most important you can write it as x times D of x² plus x² * DX but D of x² we calculated this is 2x DX so this is x * 2x DX plus x² DX and I get 2 x² DX + x² DX which is a a a 3 X Square

DX okay so this is and um and more generally I mean you prove by very simple induction that d d of xn is n um x to the nus1 DX okay this is this clear this I mean this just uses the the property4 now Lisa

Nicholas yep okay and now I want to do D of um x to the n y to the k um how do I do it this is like um X by by rule number four this needs to be x to the N times D of Y to the K plus y to the K time D of x to the the N but I mean I did say it but of course

D of Y to the K by the same procedure the same reasoning is going to be K * x to the k minus1 d y right um so now I know how to do a a I mean you see there is only one way to do D of x to the n y to the k right um and if I know that then I know how to do D of any polinomial by rule number um what is is it a two Rule Number

Two by rule number two I mean now that I know I mean this equation star tells me how to do a d of a monomial and from rule number two I can do a a d of D of sum sum of rational functions is um the sum of the D so I I I get that there is only one way to do D of a polom and then um er I I also know how to do D of um D of R

over s is going to be D of m r * 1 / s h so so there would would only be one one way to do D of of a General rational function and of course okay I I didn't say um here to do D of a polom I used to I need to use rule number two and rule number um three as

well actually I need to use rule number one rule number two and rule number three because to do a deal a polinomial a polinomial is like a sum um x to the i y to the J times CI J and of course if I and J are equal to zero then I I have the the um the free coefficient I need to do I need to know how to derive the free coefficient but this by axum one this is a derivation if is is zero and I need to know how to derive a a this monomial c i j x x to

the i y to the J but I know how to derive x to the n y to the K and rule number three tells me how to derive constant times times a function okay so so this the all these aoms tell me that there is a canonical way to derive a a polom and now I want to derive a quotient of pols um that's that's also not it's not a big deal because a quotient of R / s is just the product of

um h of r with one / s and now if you want to do so I yeah I didn't uh yeah I guess

um to do the quotient you can do the following so you can show you just take um you take a you want to to to do the derivation of X over y but this this you can do as you you know in high school of the derivation of a quotient it's going to be y DX minus X Dy divided by y s um this this would follow from the

axum I I guess you could you could say because here n and K can be any integers they don't need to be positive necessarily you see this is um I mean when you when you derive for example you do D of one / X we know how to do it this is um one over this is sorry A minus one Time 1 / x² so it also fits it Al also fits um the x of course so it also fits this H

um this formula star okay so we actually once we know how to um how to derive er um um x to the the the power of any integer then we uh we get a a the derivative of a of a general rational function this is is going to be derivative of R time derivative of 1 / S one / s has various it's like a sum of monomals

so all in all what I did here I proved that there there is only one one function D that satisfies this uh these properties [Music] um I will come back to it next time so uh of course we will repeat but any questions for now erh is the it is it is it clear or you have problem to even ask um I have a question but more like

high level is where do we sit in the whole scheme of things like where do we see compared to pairings like like I'm a bit lost in where I we going and deration of this stuff yeah I I said it a bit in the beginning maybe you were not here what we want to prove is this this um this result we want to know that the end torsion points um when n is prime to P have have a structure of a I mean we want to

understand this group yep um and pairings because the whale pairing and in fact all pairings involve ENT torsion points yep okay so whatever we do about perings will will heavily rely on this on this Theory ER so this is in the the the next three or four sessions we will prove this H this result once we have it we are essentially ready for um defining and proving

the the properties of the well pairing and so the derivation real the derivation rules that we are starting is still the same so now the derivation thing we want in the way to so H how do we prove this what will happen is eventually we will um we will simply be able to say that en en has n Square points once you know that n has n Square points then you have this structure theorem of a billion groups that you can prove with it that n

has to be a ZN time ZN but how do you prove that n has exactly n Square points you need to examine the the functions GN and hen okay and we proved some properties of a and GN um this this property should should kind of raise a flag that we will eventually use it to H to count the points on of n but the the the current thing we want

to do is to calculate the divisor of GM minus GN and I mean of course you cannot see it at the moment but H in order to do it we we will need to use derivation all right so I yeah I mean of course it's it's hard to answer a to give the the full picture this this is happens a lot in mathematics ER to give like a a a full answer I give I need to give you like the full details and that that's like um opens up

time thanks okay uh so uh have a have a good week weekend and uh see you next week thank you very much by bye bye e

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