PSE Lectures Ep 25 - Pointwise sum of rational maps is rational

Transcript

now what I want to to say here is is there is an abstract way to view rational Maps just maybe before we do this um it's just a word on terminology I say rational map as opposed to rational function right so rational functions are these guys they correspond to a um to Maps between e and and the projective line

and Russ so rational map is supposed to distinguish the case in which we talk about a map from E to itself or even more generally it could be from E to another elliptic curve e Prime um and generally in mathematics the reason we want to I mean even if we are only interested in an object e we would like to understand all the self maps of of e and rational map map is uh is a reasonable notion for a a

self map of an elliptic curve um if you want one justification is that it induces this this map for any I mean on all h on all points and there is an abstract way to to view rational Maps is as follows so recall um K Bar e is a field I I remind you what it is it's like

a you take all expressions of the form F over G where f and g are pols over e and G is not the zero polinomial and here I write it slightly informally because on these quotients we have a an equivalence relation just like with rational numbers I mean just like one half is a is is two two quarters and so on um the same same equivalence

relation um um holds on IR rational functions and this this is a field I mean you have an additive inverse and you have a multiplicative inverse for any nonzero element I mean the this zero and one are the constant constant rational functions I mean so this is like I claim that this is field

um so the the the zero element is simply um zero over some over ngg basically H one is is simply one over one and the inverse I mean the additive inverse of a um f over G is is minus F over G and the the multiplicative inverse of f over G is G over f sorry okay so this is a field now this

field contains a copy of K because you can view elements in K as constant rational functions right is like k k by K Bar e um contains a copy of k um thought of is constant rational functions that

is K Bar e and also K I mean I do it for K Bar because we need it for K bar but it's true also for K of e um K Bar of e extends um extends scale this is this is a field extension and this is maybe an interesting example that we we didn't ER we didn't discuss so much ER until now I mean we we saw that if K is FP if K is

FP then we have this a a tower infected it's like a partial Tower um FP is contained in FP squ and it's also contained in FP Cube and and both of these are contained in F FP 26 and so on and at the top of this H this chain there is FP bar which is like the the union the union of of everything um

but a k of E I mean given an elliptic curve um K of e does not does not belong to any any part of this chain I mean k of e is another way to extend FP which is not part of this chain and um we didn't discuss the general construction of algebraic closure but you could also do algebraic closure for for the field K of

e and it will result in in another chain and a k of e bar is at the top of this chain so you see I mean there there are in fact a um many other fields that can extend FP and the algebraic closure is not the is not the the last word in in in field extensions but for us what is interesting you know what okay so what is interesting is the following we can consider

e over K of e right because the the equation of e has coefficients in K and and a k of e is an extension of K is a field extension so I can consider the the the set of solutions for the equation of e but in K of e then I can consider the k e irrational

points let's let's put K bar here um because we we usually consider the IR rational functions over the algebraic closure and and what are the points e of K Bar of e these are pairs R Comm s such that R and S are rational functions on E and they satisfy the equation of e but when you consider the equation of e um over over the the the new field of of

rational functions over it that is um s s² = RB plus a r plus b and here A and B are considered as a um a constant rational functions and we have have the the point at Infinity which I want to distinguish it from the the the usual point at Infinity so I will write o of M so the finite

points are simply the rational Maps um is this clear this is um this a bit abstract but soon you will see why it is useful or let me let me say immediately why it is useful this means that for f and g um rational Maps we

have f plus G which is a the addition in this in this elliptic curve okay and I mean let let me write it like this f and g are simply elements in the curve e defined over the field of rational functions and and hence I have a um I have plus G that is Plus in the elliptic

curve so f plus G is another point on the elliptic curve which is a rational function I mean since f and g are finite points um f plus G is well I mean it could be the um f is equal to minus G or something like that and then they become the the um the element o of M by convention h o of M um the

constant um m on o um is considered the rational function a Russi yes hi one question so um here you say that the the the plus of f plus G this is like you say over the Cur but they are not points right they are rational

functions over the curve or or or is equivalent to being point so they are points um in in the elliptic curve considered over the the field of um I mean the field k k or K Bar e so it just like to to try to rra my my mind around it so in in the real world this would be just like points it would be no each of f and g is like a pair of functions a pair of rational functions s RNs right

uhu but these pairs the this pair of rational functions can be thought of as a pair a pair of coordinates of a point in e but over a different field over the field of rational functions okay okay it's a bit it's a bit abstract but this kind of um this kind of view is very useful because then I can I can immediately see that I can add him I can add the two the two rational functions I mean then what you should consider I mean there is a naive a naive way to add a um a two

rational functions I mean if I give you R comma comma s and t comma v um so naively you could say I mean I I I could definitely add a two rational functions so I could Define this to be r + t s+ v um but this is not a this is not a good definition because there is no reason to expect that now s + V squar would be equal to r

+ T Cub plus a r + t plus plus b okay right yeah okay so you constrain it to the so I I need to do a different I need to do a different addition I mean the naive addition doesn't work so how do I find the the the appropriate addition that gives me for a pair of rational Maps gives me a canonical a a a third map I I view the rational Maps as elements in an elliptic curve of you

know in this elliptic curve I mean it's it's basically it's the original equation of the elliptic curve but if interpreted over a field extension of K this you can always do right I mean y thanks yes okay so and once I view FNG G as elements in this in this new curve then I know what is the addition and this is and I can write the formulas but I will avoid it for now yes Lisa I'm not sure I understand why we jump between uh F set of finite points

and set of finite points plus this zero M well I mean finite points is here it's like a because rational function rational map as I defined it last last week is is always a pair of functions RNs right so it it is it is really the finite points as I defined it last week it's really the finite points of um um of of this curve right finite Point simply mean not the point at

Infinity yeah yeah okay I mean why we should consider point at Infinity as but we want to consider well I mean this this is this is a curve this is a curve so it I mean if you want to consider it properly you need if you want to have you know the a billian group structure you need the zero element ah I see okay but I I I denote the zero element here differently than the zero element in E over in E of K right e of K also has a zero element

which we typically denote as o yeah and now I want to disambiguate den notation so so I call the the zero element here in the the The Orange Box a o of M and now I say okay I mean since rational Maps as we defined it last week are the finite points of this um orange curve um let me extend the definition and say that o of M is I also call it rational map because it's convenient for a for talking about this

right okay and the the a zero m i can you can think of it as as as the map from E to e that sends everything to zero p is going to zero to okay okay so this is this is also I mean you know whether you we call it a rational map or not this it's by convention but you will see that it's convenient to to consider it as a as a rational map okay makes sense Okay so

now we have a um we have an issue to um to resolve and that is the the following theorem and sorry before I I move on and Nicholas are you are you following yes okay so so um it may the following Theory may look trivial but it's in fact not so trivial

um let e over K be in elliptic curve with um okay algebraically closed then for rational Maps f and g um um

and a k k being the the result of the the elliptic curve addition of f and g we have K of p is equal F of P plus G of G of for for any p in okay okay

so this may may seem trivial but it's not because um a prior the opp operation that I do here is is happening in I mean in a different in a different elliptic curve right it's in the the operation here plus is in the elliptic curve e of K of e and the operation here happens in E of K so um so when I do K of P I first do the

operation in E of K of e and then I evaluate what I got I mean I get a some rational function and I evaluate the rational function I get at p and more concretely um maybe before the proof okay let's say um note that even if um f is not equal to minus

G um uh to to minus G or to G um or to or or to the the zero rational man it may be that for a given for a specific point p FM F of

p is equal to H say minus G of P okay so so you know in this case if if F of p is equal to minus G of P for example then when you add them to together you get the you get o but K is not is not the zero rational map K is not o of M because if if f is not

identically equal to minus G as a rational map then K would not be a a would not be equal to the the zero rational map you see so so a prior we would need to kind of make sure that even in this case um K of p i mean k would not be the zero rational map but K of P would still be zero so you see this is the the kind of tension that we need to resolve um is this point clear I mean that's I think that's maybe I mean to to really

see why why we need to prove this is is um is a big portion of the the proof and now are you are you following yes sorry okay so it's important I mean this kind of stuff that may look um may look very easy at the beginning they turn out to be quite involved once you really TR try to pin down the the details

okay so um let us let us write um recall the addition on M K yes um um I don't get the remark can you can

you say it Lisa it's not important it's not important just you thought that like it might look like trivial or something and I just thought that it looks as trivial as like thinking that like if um oh I I wrote it I wrote it in the wrong way but like never mind it's it doesn't matter it doesn't matter yeah I mean really the point is that we have different different additions here it's like one is the addition in E of K of e and one is the addition in E of K I mean first and in what happens in in this this equation double

star is that we on the left side we first do the addition in E of K of e and then evaluate at p and in the right hand side we first evaluate both both both functions at p and then we do the addition in E of K and it's not clear a prior that these two things coincide of course we would like them to to coincide because it will make our life much easier okay so um how does addition look like like in um in E of K of E I mean I'm going to write the formulas that we discussed

before I'm not going to develop the formulas again um so so say f is is R comma s g is um T comma V and K which is f+ G um is um W comma Z okay all all the the small letters here are irrational functions on

E and A the formulas are a dou is umus t + r + Lambda squ and Z is um minus V minus Lambda wus T where the Lambda it has two cases um if R is different than

T again different than T in the the rational fun in in the the field of rational functions um so if R is different than T then Lambda is given by um s minus V divide by Rus T and if R is equal to T then um Lambda is a 3 r² + a

ided by 2s okay what is formulas do they do exactly the same thing as if you would have I mean this RS T and V if there would be elements in in K instead of functions you would you would have the same for formulas right it's like a that's that's the way to define addition on an elliptic curve okay so

um suppose first that m r is different than T this means that um let me call it [Music] say one and two so we are Case Case one so in case one we use Lambda as in equation one and still there there are things to um to verify so okay um

if um r&p R of p and t of P are are a finite and not equal then we are okay because um then it means that the when you do the addition of um of f f of p and g of P you will use the same the same type of

Lambda right I mean the if you want the the the whole complexity follows from the fact that AR of you know and T may be different rational functions but they may be equal at a certain point P if they are equal at a certain point P then we need to check something because the the way I mean we always have w and z given by these formulas but um if R is not equal to T then the Lambda is is as in equation one and then when I when I do the the F of p

and g of P if R of P happens to be equal to to um to T of P then I will use a different kind of Lambda I will use Lambda equal to this this guy where you substitute P everywhere so I I would need to check that these two things are equal but if r and r r of p and t of P are are both finite and they are not equal then if they are not equal and

finite then I use a for for the formula of f of p and plus G of P I use this Lambda this Lambda when you substitute p so so everything works um if R of p and t of P are M finite and equal then

um either I mean if you have the same x coordinate R R and P and T are the x coordinates right um so if you have the same x coordinate the um the y coordinate can only defer by a sign so this means that um s of p is equal to V of P or S of p is equal to minus V of P right that's the only the only two options if you want the point to be on

on the curve um so let's do a case one um okay so if a s of p is equal to minus V of P then um clearly F of

p is simply the the additive inverse of G of P um so when I do F of p plus G of p i get the the pointed Infinity so this is the the the right hand side of the the the equation we always need to check and on the other hand um remember we are in case one case

one means that um um R is not equal to T if R is not equal to T then I use I use Lambda as in equation one and um we see that this Lambda will have a pole at at P um Lambda equals um s minus V Rus T and we have pole at at P so everything is fine right

I mean so so K of p i mean I only need to to calculate the x coordinate of K of P to know that it's I mean if the x coordinate is infinity then K of P must be Infinity remember if we have this um a pair of rational functions RNs that um become together a rational map then they they have exactly the same pulse I mean one is infinity if and only if the other one is infinity this is

kind of the by by the the condition of of being a rational map um okay so this is if s of p is equal to V to minus V of P um I guess the the second case is a bit um bit invol so um if um r

ofp t of P finite and um I mean we are we're in the case that they're they're equal and um s of p is actually equal to V of P um

so the formulas for for f of p plus G of p and for K of P they only they only defer possibly by the definition of Lambda um so since a a a s is not equal to a minus v um we have I I can write Lambda as follows I write Lambda as vus s divided T minus

r this is I think I wrote above s minus V divided by R minus t but this H this is the same and now I I multiply and divide this by a V plus s and I can do this because s is not is not equal to minus V so I'm not dividing by by the zero rational function um okay but then I I open it up and this is is a tbus

RB plus a tus r divided by uh T minus r v + S and I can further rewrite it is t² plus RT + r² + a divided by um s

+ right this t t Cub minus r Cub I I can um I can divide I I I mean I can cancel T minus r from it um and and and this is what I'm left with um if um s of P which is equal to V of p is non zero not I mean ER not the element zero in in k

then um then I can I can evaluate everything ATP I can um this becomes um you check it out I mean p uh so so r r of p is T of P so I have um three 3 r² plus a divided by um 2 SP

okay and so so the the Lambda so this is the Lambda of K of P right so K of p is actually equal to f f of p plus G of P right I mean the only thing I need to check in this this equation is that the two lambdas are equal so if um if R is equal to T sorry if R of p is equal to T of P then the Lambda I use the Lambda I use is the um is the

second Lambda right this this Lambda so I need to check this is the Lambda I use for f of P plus G of p and I need to check that the Lambda I use for K of P which is this one number one because R of T R and T are assumed to be different rational functions I need to check that when I evaluate this Lambda at P this this is the Lambda of K when I evaluate it in p i I actually get this guy evaluated at p is this clear I um I know it it may be a bit hard to do the

bookkeeping maybe can you repeat the second case of B yeah so um the second case is is the case in which s s of p is equal to V of P right because I mean if R of p is equal to T of P it means that a f of p and g of P have the same x coordinate right and if if two points on the curve have the same x coordinate the y-coordinate can be either equal or um or the opposite

sign right so this is the reason that we we I mean that we have only these two cases one and two okay so I assume that the y-coordinate of of f of p and g of p is equal but of course the the rational functions s s and minus V are not equal um so you know otherwise if they they would be equal the claim would be trivial um right right because if if the rational functions s is equal to minus

V and um are you know then then F itself is minus G you see yes okay so so I assume s is not equal to minus V and um and then I take the Lambda of K and I multiply and and divide by v+ S I get these equations that can be um reduced into the last one this last fraction MH now now I I substit I I want to to

calculate K of of p in fact I don't really need to calculate all of K of P all I need to know is that the lamb I mean the Lambda of K that when you substitute it in when you substitute p in it is equal to the Lambda that you use when you do F of p plus G of P okay okay what is the Lambda that I mean when I do F of p plus G of P what is the Lambda I use I mean in our case we assume that s s of p is equal to G of to V of

p and sorry I mean we assume that R of p is equal to T of P so the Lambda we use is this three the Lambda we we use is this 3 r² + a ided by 2 s when you substitute P it okay so I need to check that the the the this Lambda the first l so I take the Lambda of of K and I substitute a a p in it but I I got this expression right I got this H blue this blue

expression and now I substitute um a p in it let's suppose that s s of P which is assumed to be V of p is not zero then I can substitute a a p in in the blue box and what I get is 3 R p² because everything on the in the numerator everything beside the constant is is equal to R right you have like this T is equal to R so I get I get a oh I see yeah I get three r r of p² plus a divided by h 2sp

so so I get I get what I want I mean yes yes I see okay I miss this step before yeah so as long so this is like what we need to do here is to compare the lambdas I mean the rest the rest has to be the same but the lambdas may may cause some problem um okay now finally if um s of P which we assume to be equal

to to V of p is equal to zero then um then F of p is a point of order two both F of p and g of P right the only points on the curve

that have a um a zero y coordinate are the points of order two and if F of p is a point of order two um then um Lambda I mean this lamb I should write I guess Lambda of K evaluated at P um has a has a pole or P let's

say p is a pole of Lambda K Lambda K is like this h s minus v ided r minus t which which is equal to the blue box right so um but but in fact that's that's what we want I mean it has a poll this me if it has a poll it means that um k at p is equal to

O and and and then um since G of p is also of order two um f of p plus G of P has to be equal to O if you add two points of order two you get um you get

oh I mean sorry you g sorry F of p is equal to G of P um so if you add a point of order to itself you get you get and so so we see that also in this case k of p is equal to F of p and plus G of Okay so we we covered um the ER the two cases of of

B so we covered a if r r of p and t of P are finite and and not equal then um then this this is by definition B if they are finite and equal and now we need to deal with some other cases but I um I suggest that we take a break here and we continue in 10 minutes of course um unless you have any question David I I see that you just joined us okay so uh let's let's meet in 10

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so we continue with the cases H the next case um so we do the case in which exactly one of um f of p and g of P

um say F of p is a is a the zero element um recall that f is is R comma s h so by assumption s and and R have po at P now if rational functions have P it P

whatever p is I I can take uniformer um at p and um and write R is equal to R1 ided by U to the D and S is equal to S1 divided by U to the e where U is a uniformize

ATP right and and D and E are the the multiplicities of the of the pole P um now remember that a r comma s is a rational function that means it a rational map it means that it it satisfies the the equation of the elliptic curve and I can substitute instead of R and S these these things R1 divided by U to the D and so on and what I get is that

um S1 squar is equal to U the 2 e ided u a to the 3D times r 1 Cub plus a u the 2 e divided by U the D times R1 plus u to the 2 etimes

B I mean what I did here is um S1 square is simply sorry s square is simply S1 Square divided by U the 2 e and i multiplied all the equation by U to the 2 e okay um now this means that um 2 e has to be equal to 3D because s S1 is is finite right I mean S1 is finite and non

zero h u is U is zero at P so um if a 2 e is not equal to 3D then either this expression when you substitute P either it becomes a zero if if 2 e is bigger than 3D then it becomes zero and and everything else on on this uh right hand side BEC zero or if a um 2 e is is smaller than 3D then this expression becomes Infinity but S1 is not zero and not Infinity at P so um so

the the only option is that 2 e is equal to 3D um and furthermore um I must have so now that that I know that 2 e is equal to 3D I also know that S1 Square evaluated at P has to be equal to R1 Cub evaluated at P right because when I evaluate at p this

equation triangle everything else vanishes I mean when because 2 e is bigger than D then this guy when I substitute P it's like u u to the positive power at P this is zero and this is also zero and so um this this thing uh cancels out U to the 2 e / U to the 3D and I get um this the second condition s S1 s p equals to R1 Cub at p is this

clear Lisa Nicholas [Music] some so tell me tell me what is not I mean maybe one thing that I forgot to say I mean when I wrote here in the The Orange Box ER this is this a uniformized business um it means that if I have a poll if if let's say R has a poll at p then I can write r as R1 divided by a uniformize to the

D where R1 is um is not zero and not Infinity at P this you remember uh yeah okay this is like this uniformer stuff I mean we we Pro that uniformis exist for All Points yep um and so now I I have the equation s s squ equals to R Cub plus a r plus plus b right and I substitute in this equation I substitute you know R1 divided by U to

the D instead of R and and the same for S H I substitute and I I mean so s um s Square for example would be S1 Square divided by U the 2D right this guy would be S1 Square divided by UD squar it's U to the 2D yep okay and so I multiply all the equation by U by U to the 2D and I get a this this equation triangle okay yeah okay this this is an

equation of a of rational functions so it has to be true for any point that you substitute um now S1 at p is is not zero and not Infinity right this is by again this uniform business I mean I can always write a function that has a poll at P by a um you know a function that does not vanish and does not have a poll at P S1 divided by this some power of the uniformer so S1 at p is not zero and not

Infinity but if two of e is different than 3 of D then U to the 2 of e ID U to the 3D is um is either zero or Infinity right because U to the U at P when you evaluate U at P this this is another property of the uniformer U at p is always zero yep okay so so I get that a a a 2 e has to be equal to 3D but if 2 e is equal to

3D that means that 2 e is bigger than D yeah which which means that when I substitute p in this in this expression this um blue guy I I get zero because u u of P or u u to the positive power at p is zero so it means that this guy becomes one this guy becomes zero and this guy also becomes

zero you see H Nicholas yeah yeah I El now okay and now so so then I I get that S1 Square evaluated at P has to be equal to R1 squ evaluated at P um okay and now we we run a computation I mean it's just a manipulation of symbols um let's do a w I want to calculate W remember K is W comma

Z and I want to I mean I know how to calculate F of p and g of p and now I want to calculate K of P so in order to calculate K of P I need to calculate what is w and what is z so w is minus t + r plus Lambda square but Lambda in our case we are in the case that R is different than T So Lambda is s minus V / Rus T squ okay so I I write it down um I mean it's in

the notes you you can manipulate the symbols you get um t r² plus m t² + a r um+ v² - t Cub plus b minus 2vs and all of this is divided by

m r² minus 2 t r um Plus t^2 okay and now remember I have this h i I mean r is R1 ided U to the D and S is S1 ided U to the E so I can plug it in the last fraction that I got and what I get is um m

t r 1 s * U to Theus 2D plus m t² + a R1 U to Theus d um plus uh v² minus t Cub A + B minus 2 V

S1 U to the minus E and and uh in the bottom I get R1 squar time U to the minus 2D um - 3 T R1 U to the minus D Plus t^2 okay I I mean I hope you you you can

believe me that this this calculation is a is is okay and now I I I simply write it a bit more clean as tr1 squar plus UD R1 M - 2 v u to the 2D minus E S1 plus u to the

2D a t + 2 b and then divided by R1 squar plus u to the D R2 plus u the 2D t² where um R1 and R2 um our finite at

p and now I evaluate the whole thing at P when I evaluate the whole thing at P many things vanish because you with a positive power evaluated the p is zero so w evaluated at P this this whole fraction is supposed to be w w evaluated at p is um let's see what vanishes this vanishes this also vanishes and this also vanishes so I

get um T evaluated at P times R1 squ evaluated at p and in the bottom this vanishes and this vanishes so I get r s sorry R1 Square evaluated at p and then these things cancel out so I get that W of p is T of P but that that's exactly what we want right because remember we are in the case we're in the case in which F of p

is equal to to O and that means that F of p plus G of p is equal to G of p and the x coordinate of G of p is T of P so we get that the x coordinate of K of p is is the same as the x coordinate of G of p and that's H that's part of what we want um similarly we or one can do one can show

that um the y coordinate of a k of P equals um that of G ofp and and as a result K of p is equal to F of p f of p plus G of okay so this is our case and and we are

we are almost done I mean now it's going to be the easy part um so D suppose both F of p and g of P are a are the point at Infinity o um of course I could I could do this this calculate like a similar calculation to what what I did here in in the previous case but that's

a that's a that's a very long way and I have a shortcut and the short cut is to use a a we use the associativity in E of K of e to deduce [Music] um I mean I I I I know that K is equal to f f+ g and this means

that um f + G minus K is equal to the the zero element in the in the elliptic curve e of K of e and in fact by by associativity I can do um f plus G minus K is equal to O of M

now um uh denote um K of p is equal to to Q if m q is equal to O we are done because then clearly I mean our assumption is that both F of p and g of P are o and if K of p is O we we um we did what I mean we proved what we

have um what we what we wanted and and otherwise um otherwise um minus minus K of P is equal to minus Q which is not o hence by the previous Case Case Case C I mean if if I have two rational

functions and I want to evaluate them at a at a point um and exactly one of them is is zero then I can I can evaluate the the sum of the rational functions at at the point this is this is K C so um I get that g minus k at p is equal to G of P minus K of P which is equal to minus Q right I mean equ Al G of p is equal to o k k of p is equal to Q so I get 0 minus Q This is minus

Q um um but I also know that um um also by by KC um now I know that g minus K EV G minus K evaluated at the point is non zero it's minus q and F evaluated at the point is e zero f of p is zero so I use KC again and I get that f plus G minus

k at p is equal to um f of p plus G minus K of P which is a a 0 minus Q um and and and and this this contradicts this has to be equal to0 M evaluated at p 0 m evaluated at p

is zero that's a contradiction okay so um so you see this is a this is a shortcut using this associativity on this higher level um I I get that um so so we must have that K of p is equal to F of P we must have a Q is equal to O

hence h k of P which is Q is equal to f p + G okay so we are almost done um now if okay and so case two is that a r is actually equal to T

if s is equal to minus V then then f is equal to minus G as a as a rational as a rational map hence K which is f plus G is equal to the zero rational map and um and this is this is indeed a compatible I mean so

so a k of p is O and and clearly F of p plus G of p is also wrong so we are okay um and the last possibility is that um s is not equal to minus v um so

assume s is not equal to minus um but of course s of P may still be equal to to minus V of P um now since s is not equal to minus V

it means that f is not equal to minus G as a as a rational man but um but remember we are in the case where R is equal to T So f and g have the same x coordinate if they have the same x coordinate and there they don't have the same um y coordinate um sorry if they have the same x coordinate and they are not a a additive inverse of each other then they must be equal and

since R is equal to T F is equal to g h so so now we have two two possibilities if R of p is finite then a a k of p is equal to F of P that's G of p and because they are a identically

defined um if if R of p is equal to O then we have something to check um but remember F of p and g of p i mean f is equal to G so f F of p plus G of p is simply the the formula for doubling a point in this case f of P um the duplic duplication formula is um w w is the x coordinate of K is minus

4 r s + 3 r² + a divided by 2 s um but if R of p is O this implies automatically that s of p is O because we're talking about um a rational map R Comm s and um this this formula shows that W has a

poll at at all at sorry at P which means that K of p is um I mean if the x coordinate of K of p is is infinity then so so is the y-coordinate and so K of p is equal to O um and this uh let me just check yeah so this should complete the proof

but I uh I'm missing something so oh yeah so if R if R has a pole at at P yeah if R of p is is is infinity then s of p is infinity and this means that F of p is infinity I mean the point at Infinity which is equal to G of P

so and indeed in this case k of p is also the point at Infinity so it is um it it checks out um okay so this is a um this is the the the this completes the proof basically um this covers all the cases um that ER that show that you can do a pointwise addition um this theorem in

Orange you can do pointwise addition of H rational functions any questions I I know it's been uh a long one I'm not quite sure I understand when we were going through Case C um and we were making some U transformations in the equation we uh like there is a large r that popped up

from nowhere and then dis appeared uh disappeared again I didn't understand yeah I I just wanted I mean it's not crucial you can you can even do the same ER I mean I just wanted to write it so you you see you have TR R1 SAR um I I multiplied all the equation by U to the I mean all the fraction by U to the 2D M right so then I get a um um I mean in I in the second

expression the blue one here I had something times U the minus D multiplied by U the 2D is something I mean U to the D times times something this R1 is this something R1 is t t ^ 2 + a Time R time small R1 okay R1 is t² plus a oh I see small R1 and the just confusing because R was large and before that we used large letters for other

notations ER but not for the RNs yeah yeah yeah sure okay but I'm sorry for that but that's just to simplify it was just to simplify the the the whole point is like it's U to the D times something that does not vanish at p i mean that is a finite and and does not vanish at P okay so this is a and the same for R2 so when I multiply it by actually it could vanish but it's it's finite so when I multiply by U to the D and I evaluate it at p i get

zero okay now um well I guess yeah I guess that's a that's a good place to stop unless and now maybe you have a other question Nicholas questions for my side could you could you follow or now uh so yes but my feeling is that I need to do like a second round by myself because like I mean it was very um I mean it's a lot of steps and a lot of things to be kind of unfolding uh along many pages so yeah I guess this

weekend I will go over it again yeah okay no problem but I mean the general idea yes it just that I would like to to go like in theep for each step to just to make sure that I fully follow yeah let me just say what what I will do next time but it's a coroller of this um I mean we want to show that the map n from E to e um I mean I said that it's a rational map but I didn't prove prove it right I mean this this it takes a point

p and send it to NP right and I I want to say that this is a rational map so how how will I do it I I I will decompose n as n minus one plus one and I will prove by induction I mean but but I need to know that when I evaluate this at P I can do I mean when I evaluate the right hand side at P I can I can uh I can separate the

terms right so this would be one application of the theorem this this is because of the theor okay and once I know so once I know the theorem I can use this equality to prove by induction that um multiplication by N is a rational map and it has consequences I mean if multiplication by N is a rational map remember we showed that a rational map is either constant or subjective okay so this

this this all all things are um aiming towards understanding points on on elliptic curves but they sometimes they go a bit wild ER all right that's it so uh see you next week thank you very much have a nice weekend by e

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