PSE Lectures Ep 24 - Rational maps between elliptic curves

Transcript

you look very summery yeah well it's a it's a summer is it hot in Berlin ER it's not too hot but I uh I came from the gym so uh I still I still nice I also have my uh summer clothes on all right cool so it's working then I wish you a good lecture guys and yeah I um I'll see you around then yeah thanks m

okay um I'll just connect with the tablet that's okay so um

last time we finished the proof of ab aobi theorem which um is probably the the biggest H result we proved so far and it will be maybe only secondary ER to ER to the way well pairing theorem um and the the next thing I want to talk about today is ISO isogenes um basically

what we want and this um is compatible with the things we did for groups and and Fields ER we want to talk about we want to have a notion of a a map between two elliptic curves I mean so far we we always fixed an elliptic curve and um we studied properties of this uh of this curve but ER in this in general well if you want to understand an object in mathematics you need to have a good notion of what

it means to to move between two objects of these types um so we have a notion of a a rational function rational function um call um a rational function um it it can be viewed is a map

M fun function um from the K rational points of e into the projective line and if you think of it the projective line is just a a another type of a of a curve right I mean it's h it's a it's a onedimensional er er algebraic object and um we we can we can use this this notion of

rational functions to well I mean of course um the key the key point of a a rational function it's it's a map um defined by um quotients of pols of pols on e um and in general in algebraic geometry um I mean we we always want to talk about either of pols or uments of pols

and I think this should should be enough of a motivation to H to the following definition um so let e a b and be an elliptic curve so this is a over k and

a rational map H say raw raw is um is is is a pair row and being R comma s where R and S are rational maps on E and

for all p um if I substitute P into um into SNR and I treat SNR is um is the the the y coordinate and the x coordinate respectively then I get that the the equation of the elliptic curve is satisfied

that is s p² is equal to r p Cub plus a r p plus b and and here I just um er to to to make it completely clear um if if p is the point at Infinity um then I remind you euristic we think of of the point at Infinity as as a as

being um Infinity comma Infinity so when I write this equation for the point at Infinity I I really mean that um s of O is equal to Infinity if and only if R of O is equal to Infinity okay so this equation is a you can just um

er think of it as as always always valid for all points when the point at Infinity you just substitute Infinity in both coordinates okay so this um this is rational map is as a in contrast to a rational function but ER it's not a standard terminology I I'm using it at the moment just to to disambiguate um the two Notions but in fact um

what we as I said what we called rational functions before is just a rational map between the curve e and the and uh the curve of U of the projective line uh is this clear any questions about the definition Lisa what do you think do do I understand it correct that uh can informally describe this map as a substituting

a aent that we had before for points for rational functions um it's like a um I mean we we we said we we write the notation maybe it would clarify we we write raw from E to e h right it's it's because um I I give you a point XY on E I mean on E of K right

and raw Maps it to a point of um of e of K again and the way I mean if I want this property so um so if m p is a point on E of K then um row of P by definition is um R of P s of P all right so we map a point on elliptic

curve on into a point on elliptic curve and rational function is exactly the map yes exactly oh yeah yeah sorry go ahead no no then I think it's clear okay now what do you think H yeah thanks and and this this was careful so thanks Lisa for for for asking it uh then this means that if we had sorry another kind of like uh cve shape the the rational map would follow that too yeah let me let me generalize the definition to to maybe help clarify the the notion I

mean more generally suppose um e over K is H this Y y² equals X Cub + a x + B and E Prime over the same field K is y² = X Cub + a prime X plus b Prime then a rational map

Row from E to e Prime and is a pair row equals r s where R ns are irrational functions on E and and such that for all p in e or in E of K let's say

um R RP square equals s h sorry it should be S and S P squar is equal to r p Cube plus a prime r r p plus b Prime and and this really means that with with the same convention for the point at

Infinity I mean the point at Infinity says that um R of R of O um basically R of O is infinity if and only if s of O is infinity and what what this this notion gives you is um for any e in E of K I can apply the rational function row on P this really means that I apply the rational function r on P and the rational function s on P

and the equation here star this this means that row of p is a point on E Prime right because it satisfy I mean the X and Y coordinates of r p satisfy the equation of e Prime ER is this helpful have now I think um I'm digesting uh so

let me try to to explain it from a a a another angle I mean or maybe just in other words I mean you can really start from saying that you want you want some um you want some notion of a map between e of K and E of K Prime sorry e e Prime of K right you have two curves e and e Prime they are of the over the same field but they have different coefficients for their

equations right yeah quick question do they have the same order not necessarily no you know nothing about um they're completely General the only thing you know is that they are defined over the same field okay so the coefficients a b and a prime B Prime are are in the same field k so if you want such a notion I mean of course I mean this is a set this is another set there could be many

functions between the set between these two sets but we want functions that are defined by pols or or really quotients of pols so I mean obviously I need to give you at least if if I'm looking at the the finite points I mean finite points have two coordinates right um so if if I if I I want to give you a a a map

RW that um for any finite Point P gives a a finite Point row of P then row of P must be a um must have two coordinates so so I write it as R of P and S of P right whatever Ro row of p is it has to have two coordinates you you follow or now yes yes thanks okay so and moreover if if I want that R of P would be um a

point in in in the in the Target curve e Prime then the two coordinates of r p need to satisfy the um the equation of e Prime what does it mean it means R of P sorry I mean so it means that I regard R of p as the ex s of p is the Y and I want that y² equals all the rest of it right so so s of P squar needs to be R of P Cube well what is what is written in Star

yeah okay so this um and for and for the point at Infinity I I can say that I mean it could be that um S and R um map the point at Infinity H to a finite to a finite element in the field this this could be um so I require um this equation to be satisfied also for for the point infinity and I mean if I I mean if I require this I

mean of course a s of P where where p is the point at Infinity so s of O and R of O are defined this we when we talked about rational functions we said that you can evaluate a rational function at the point at infinity and you get something on the projective line but the projective line is just a a a k together with a an additional point so um I I interpret this equation star I mean if R and S are sorry r r of O and S of O

are are simply elements in the field k then this equation star needs to be satisfied as it is it's just an equation in K and if I mean if one of them is the is the I mean if if one of R of o or S of O is infinity then it um satisfying this equation means that the other one must be Infinity okay ER tell me tell me what is missing now

no I think thanks because it was very helpful this I think what what it's like uh my head not wrapping around is the fact that you can have different orders on those curves and then this mapping uh is like it canot it not will always be one to one or um it will not always be one to one but as we will see it will always be subjective okay okay so this I think your intuition is right that you have a certain rigidity um but in general also for maps of groups right I can map

the group a the group of integers to the group of integers modul n by taking a a to a modu n right this this is not an injective map but it's still a group map so in general Maps between objects um I mean for example between groups do not need to be a injective um and remember that you always have the zero map right I can take any Group G take any

Group H and let's say they are both a billion and written additively you can map all X and G g into the zero element in ag this is also a group homomorphism so that's not a problem that something is not injective we we actually want to be able to um to be able to move between two elliptic curves or two objects also non- injectivity okay thanks okay now

a main example um let and be a positive integer in the map bracket n from E to e defined by P

goes to NP here a rational map right um just taking a point and adding it to itself this NP I remind you is is like P plus p n times and it making a point and adding it to itself end times I mean just you can think of it you can I mean

ER prove this this claim of the the example inductively so what really matters is that I take a point um and let's say I add it to itself so adding to itself this um requires to solve a a um some some equations I mean I I draw the line and I intersect the line with the Curve and then I uh um I negate the the y-coordinate but all of this I mean I I will not write the

formula although you can find it in the notes and in other places um but obviously all these formulas are are just a rational functions I mean mostly they are pols but when we um when we do p plus P we need to take the tangent line to p and the tangent line has a the numerator right so so I actually get a rational function and not just a polom in each coordinate um

is this clear Lisa do you follow or now Nicholas yes and now what do you think yes yes it's clear thanks it's clear okay thank you okay so now this is a very important map right um because we we would like to to understand um we would like to understand for example all the points um

that satisfy n p is equal to zero this this is a generalization I mean we have seen the how useful was to understand exactly what are the points of order two um and um and and we also saw in the in proof of ab aobi in the classification of divisors of line that we needed to understand a a some points of order

three and in general in order to really um er have a a systematic treatment we want to understand uh points of order n or what is called n torsion points so um let's say the goal understand what is called osion points these are the points on E

and here I um I take the algebraic closure of k um so sorry so this is the ENT points are denoted e bracket n and they consist of all the points in E over the algebraic closure such that NP is equal to the point Infinity so let's say this is one of the goals one of the goals of of

talking about rational function rational Maps is to understand and torsion points and and torsion points they are basically a a the it's like the kernel of of the the multiplication by n map that goes from E of K Bar to e of K Bar okay so note that um I can view a rational map

um is Mark if a raw from let's say from E to e is a rational map and and let's say e and e is defined over some field k then for any for any extension L of k um

I can talk about um a a function I guess you can H you can add a subscript L from E of L e of L right because raw is um I mean row is some some pair of rational functions RNs R ands are defined over K if they are defined over k I mean their

their coefficients are in k then you can view them as rational functions um o over L because L is bigger than k h and so you have the same thing I mean r of sorry this equation s of P so for any P now in in E of L I can do s of I mean I know that um

s of p² hang on I I don't want to confuse you okay so um I would have um okay let let me let me erase this remark this there is a problem here um I I mean if I have a Russian I I I'll explain if I have a rational function defined over k

then I cannot necessarily extend it to a a map between e of L and E of L Prime because if I have a point in E of L its coordinates may not lie in K and and then I don't know that um I don't know that this equation r p q + a x sorry plus a RP that's B um I don't know the this equation is satisfied because the co the

coordinates of um of P may may lie outside of K and then I am not guaranteed any of this so often when we talk about rational Maps um we want that the russal map will be um even if e is defined over a non- algebraically closed field we want that the rational map will um will be valid for every I mean will be defined over the the algebraic closure um so I'll amend this remark Mar um we usually

want um er that for any point in E over K Bar the equation s um SP P Cub is equal to r e sorry SS p² equals to R RP Cub Plus AR RP plus b will be satisfied

um and if I have if I have this property I mean if I have um let's call a double star a double star is usually the property you want from rational Maps um and once I have double star then for any L that is an extension field of of k um this equation will be will be true so I get so I get um a map Row from E of l e of L

okay um I'm sorry for the confusion maybe uh maybe you have questions on on that I can I can further clarify so if you want I will simply amend the definition the remark is maybe amending the definition of a rational map so that um for this equation double star needs to be true for all points over the algebraic closure not just um overk er and

once you have this I mean then you can regard the multiplication by n map this bracket n um as a rational map I mean obviously you take you add two points over the algebraic closure over I mean two points of e over any field um you you you can Define the the addition of two points by a quotients of pols I mean by rational functions so so multiplication by n map is is a rational map and the kernel of a rational map I mean

is simply all the points that go to to zero to the point at Infinity H So So eventually the end torsion points will be the kernel of um of a rational map so if we understand rational Maps we we have a good chance of understanding toring points this is this is kind of the high the the um a glimpse to the future of why we are interested in in such things uh maybe it's it's a lot to take so uh

Lisa what do you think um do I understand it correct that what you were trying to to tell is is that uh the map will hold for K only if it holds for the algebraic closure for K er no I I want to say that so let me just correct the definition um sorry I had a feeling that I'm following but then when you started to H remove part of the explanation I'm I'm not no not sure

yeah okay so let me just I I'll correct the definition this is something I should have done um Oh you mean we now are dealing only with algebraic closure of K and not with K on its own yeah although the the the the elliptic curve may be defined over a non- algebraically closed field but the rational functions that we consider will be um always um over the algebraic closure and want sure that they are well defined because we want we want the

rational functions to induce maps from E of L to e of L for any L between I mean for any extension of a of K that is we need this equation to be satisfied for any extent for any um for any extension L of K but if this equation needs to be satisfied for any extension it means that RNs need to be defined over um over the biggest extension of K of K which is

the algebraic closure because a point in P of L sorry in E of L I mean a point p in E of L may may have a a coordinates that are outside of K so if the equation star is only satisfied for points with coordinates in K we do not know that it is satisfied for points in h in e of L yeah that makes sense okay so we what we really want is um is that the the um the equation star will be satisfied

for points with coordinates in potentially in K Bar and then we we get a we get an actual function function from the the L rational points to the L rational points for any extension L of K okay okay so this this I think uh this this one maybe a good clarification this is the content of the remark I and so okay so we have this um um we have this example of a um rational

map we have another more General ER example let um e over K be an elliptic curve and um let's say Q is a point in E over K Bar then the map um

I I call it to q a to Q from E to e um defined by to Q of a given p is um p+ Q so to is like a translation it's a translation by Q um so for any for any p in um

in E of K Bar in particular P could be I mean p could be in E of L for L some some intermediate ER extension of K I mean when I say p in E of K Bar this is the more the most General situation um and this is is a rational map again okay so this is a again the formulas of of addition of points on on E

tells you that a translation by a fixed point is a is a rational map ER AR now what do you think do you follow yes okay so you you understand this modification we did to the the definition I mean the the K Bar yes I think so okay so um okay so now we have a notion of how to move between elliptic curves and

if I'm about to move between elliptic curves then I want a I want to be able to compose two rational maps and still get a rational map this is the next proposition and let a e over K lipic curve and a row and Tow two rational Maps um

Russian and Maps can be composed um and I claim that let's say a tow composed raw from E to E is a rational Mar I didn't exactly Define how to compose rational Maps but I will do it in the proof it's it's going to be very easy you just H substitute so um

let's say that raow is um R Comm s and to is U comma V and again I I remind you with our amended the definition of rational Maps um I know I mean I assume that RS u and v are rational functions over KB um then how do you compose you take um to

compos row applied on some um it can be either a finite point or a I mean let's say a finite point so it has two coordinates each and Y and you do um U of r x y um or you know what let me let me do it straight ER on the level of points so for any p in E over K Bar

and to composed row of p is you take um R of p s of p and you plug it in into U and in the second coordinate you do the same thing for V so you take R of p s of P and you plug it into um into V this is by definition this is composition and you need to remember that these guys may

take the the value of infinity at some point but if one of them takes the value of infinity then the other one and then the two coordin Ates if the two coordinates are Infinity comma Infinity we interpret it as as o right and we

assume that R of p s of P um like this is simply the point at Infinity okay and then everything is is defined I mean I can always evaluate a a rational function say U or V at the point at Infinity instead of assume what we we mean um

okay so this is composition I I basically I didn't prove anything I just um um I just wrote what is the the definition of composition but um this is a pair of rational functions I mean I have a rational function in each coordinate one r function is you take p and you you take the evaluation U

of R of P comma s of P this this guy I mean this coordinate is a is simply a rational function from E of K Bar say into H into P1 and also the second coordinate is a rational function from um E I mean this is a this is because I substitute I mean what is a rational function is is just a quotient of pols you substitute quotient of pols into a quotient of polinomial you get a quotient of pols

okay so so this is a pair of rational functions um and now I need to to check that um they satisfi the the equation of E I mean um we want or we need to check let me Den note these guys is a uh what we have RS UV and

uh let's say AB so we want to check that um b² equals a uh maybe I write it with with p this is a of P this is B of p and we want to check that b of p square is equal to a of P Cub plus m

a a of P that's me but um we know that um our of sorry that you know that and row

is a rational function from E to e so row of P which is R of p s of p is always it's guaranteed to be a point on E so and this is this is row of P right and also this guy is I mean the in the other argument I also have Roy

so um and we also know that M to from E to e is a rational map i e for any Q um to sorry to is is is U comma V so for any q u q v Q

is guaranteed to be a point on E so take Q = row of P then we get that um a of P comma B of p is the point we want to check H that is on E is is is guaranteed this is this is simply um sorry um this is simply to

of roow of p and a which is to of Q well whatever this is guaranteed to be a point in e okay so this I I try to be very explicit here but the content of the statement should be fairly intuitive I would say um any questions about the okay oh now you you're fine um then let's uh let's just uh take a 10 minutes

break un let say Ana you wanted to say something it was me ah Elisa question yeah we have like somewhere uh upwards um the statement that uh if s over point of infinity is infinity it implies that R over point of infinity is also infinity and vice versa I'm not sure I understand why um this you can because this equation cannot be S I mean let's say if you take this equation yeah uh yeah if one of them is infinity then

you cannot have this equation satisfied if the other one is not Infinity yeah of course of course it makes sense okay only for this equation it's not for all rational functions no no only for this so okay okay this is It's a condition we further require from a the rational function the rational functions RNs I mean hypothetically you can find and not not just hypothetically I mean also in practice you can find IR rational functions RNs in which for which you know there is a point P such that r r of p is infinity and S of

p is not Infinity h and for all other I mean but but this equation may be satisfied for all other cases this pair of functions will not be a a rational function according to our our definition okay okay I got it I got okay all right so uh let's meet again in 10 minutes e e

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okay um let's H let's start again um now that we we know that composition of rational functions is is a rational function ER we we can move a bit uh a bit better ER between elliptic curves I mean imagine you can uh you can you can take the um the doubling of a point

um say um we we can we can write it as as a bracket two this is from E to e and it Maps um p in e in E over K Bar into a 2p and I can um if um if n is even

then um bracket N is a a is simply a composition of two actually um for I don't need n to be yeah if if if n if n is even then this is a composition of a doubling of a point um a let's say k times so the next proposition

uh may be a bit surprising and maybe not so much if you think of it um because well let me just state it so let let e over K be an elliptic curve um and row which is R comma s is a rational function a rational map between e to itself H and here

um okay so if row is not constant it induces a rejective map between e over K Bar to itself okay so this this proposition tells you

that um rational maps are ER rigid I mean you cannot H you cannot have a um nons subjective maps that are um defined by rational functions RNs so yeah uh do subjective in this context means that is subjective or bjective or it is strictly subjective no just subjective it h it can certainly be not not

injective subjective means that for any ER Point Q here there exist some point P here such that um p is mapped into p is always mapped into row of p and and I mean for any Q There exist a p such that row of p is equal to Q yeah but why why I'm not sure understand why it can't be injective no it could be injective but it could also not be injective what the the proposition guarantees that

it's always subjective oh okay and it doesn't tell us anything about injectivity or no this this you cannot guarantee ER why you will soon see why okay ER I mean the the answer to why is is the proof really okay so let's first um let's show the the analog statement and for um irrational functions

and say again raw from E to P1 so if um okay so maybe instead of row I I I do it for r um that is R is simply a rational function over K bar so

and if R is not constant then it then it must have a zero a zero um over um e of K Bar and what is the reason why why does a non constant rational function must have a zero

because the polinomial and polinomial always has a root H it will be either yeah I mean it will be a quotient of pooms and pols always they have to have a root because they have a a pole at at the point at Infinity right so so essentially it must have a zero because of ABI I mean I mean the the the numerator of I mean the numerator of R is a polom

this polinomial must have a zero by AB aobi or I mean some similar statement um for a point in K Bar um the same argument well um the the rational function um

Rus x0 again going from e to the projective line must have a zero um but if Rus x0 um has a zero so say um p is in the set of zeros of Rus Z

then R O I mean r of p is equal to x0 in the I can write it a bit more detailed I mean r minus x0 evaluated at P this is by definition r of P minus x0 this is equal to Z this is the same as saying that R of p is equal to x0 so R

this R is rejective I mean it's it's subjective on on the projective line of course R um if it has a zero it um it must have a poll because so so R is rejective well so our um

let's say is subjective on K bar but but since since R has zero it must have a poll again this is ABIA and um

IE there exist some Q in E over e over K Bar such that R of Q is equal to Infinity and this means really that a r from E of K Bar into the projective line over K bar is rejective um is this clear this is the the main

part of the proof actually um now what do you think I'm processing I mean digesting the but I mean so when when I when I watch your explanation it like seems uh mat and make sense in my mind but when giving a second read is when I find out that not let let me just go over the argument again I mean at the moment I'm just I I just want to show

that a rational function um r on E on E over K bar is I mean if it's not constant it must be subjective okay so I'm saying if R is not constant then it must have a zero why does it why does it have to have a zero because if it is not constant then um it is a quotient of pols right yes yes and the the the numerator of the

quotient of pols I mean the the um the polinomial at the top um it must I mean the the polinomial at the Top If it's a um okay so so the the polinomial at the top let's say it's not constant then it must have a zero right I think I'm not seeing this but I think okay so um so I'm saying if it's not constant it must have a zero

um this is basically the the content of Abel aobi um you you cannot I mean you have to have a a um the the the number of zeros with multiplicity has to be equal to the number of poles right with multiplicity so it has to have a now a polinomial a polinomial always has a pole at the point at Infinity right because if you substitute Infinity comma Infinity in in a in a a two variable polinomial you get you get infinity

okay I see now thanks okay so so I mean you know that a polom every polinomial that is not constant has to have a zero this by abob and similarly every every non- constant rational function has to have a zero now if every a non- constant rational function has to have a zero it means means also that I take any x0 in K Bar and I take R minus x0 this is a non constant rational function

right ER it has to have a zero but what does it mean that it has a zero it means that this this equation star is satisfied this equation star is the same as saying that I mean it's satisfied for for some P right okay so so it means that I can reach I can reach every X Zero by by some point p and so so now I'm almost I almost proved subjectivity but I need to show that I I

I also get the I mean P1 I remind you is like K Union Infinity so I need to show that I I also get Infinity but that's not a problem because if I have a if R has a zero it has to have a poll the this is the the symmetry of of ab yobi gives us all this H all this information so and what is a poll of r a pole of r is some point Q such that R of Q is equal to Infinity so I get I get that R um covers the projective line completely thank you very much okay so

okay so this is for um for rational function and now I I want I want to prove the statement for a rational Maps so now consider m p sorry row row is this R comma s this goes from e e to e um so

if R is constant then since for any p in in e e over K Bar we have to have s of p square equals R of p q that's a of p plus b we get that s of P

can take at most two values namely plus minus square root of r p Cub plus a r p plus b i mean remember r r is I mean if R is constant then R of p is equal to R of Q for any P for any p

and Q so this expression is is constant since R is constant and S has to um has to satisfy I mean we have to satisfy this this equation star so it means that s can can take at most two values namely the the plus or minus of the square root of this constant expression okay so um but of course if s um if s can take at

most two values so s from E of K Bar into P one of K Bar cannot be subjective right because the projective line over K bar is always infinite and if s can take at most two values it cannot it cannot cover an infinite set H so by the above

s must be constant okay so if so the so what we proved here is that if R is constant then automatically um s must be constant and if R and S I mean if RNs are constants then then uh then raw itself is constant um

otherwise R must be subjective um if it is subjective it has a pole hence it has a zero or you know you could go the other way

as well if it's rejective it has to have a zero and hence it has to have a pole um in particular there exist a p in E over K Bar such that R of p is equal to Infinity ER

in this case it follows that s of P must be Infinity as well right this this is the defining property of rational functions rational Maps is that if one of the coordinates um I mean one of the coordinated functions either r or S is a is infinity at some point then the other one must be Infinity um

and this means that row evaluated at P which is R of p s of p is simply the point at Infinity okay so let q a be an arbitrary point in E over K Bar

then um translation by minus Q is a rational map from E to itself um and and thus you take translation by minus Q and you compose it with Rock and composition I mean row is is a

rational map translation by minus Q is also a rational map so the composition is also a rational map um it it is it is a non constant um to minus Q compos

is non constant hence um it has um it has a poll there is some P Prime in E over K Bar such that to minus Q compos with row evaluated at p sorry at P Prime is the point at Infinity but then

I mean what is to minus Q composed with row it means that um minus Q plus row of P Prime is equal to O which is the same as saying that row of P Prime is equal to 2 Q so raw from E of K into e of K Bar is reject

okay I took an arbitrary look I took an arbitrary Point Q in E over K Bar and I have found a a p Prime that a satisfy this box equation that is um a point P Prime such that row of P Prime is equal to Q so I I proved the surjectivity uh Lisa what do you think

I think I need time okay it's fair enough but I can also repeat the the argument yes please okay so let me ask this do you understand the first part the first part yes H AR now do you understand the first part of the proof I think yeah yeah so the first part is just saying I mean every rational a a rational function on E

is either constant or subjective okay now we move to the um the the general claim so the general claim here starts the general claim I consider a rational map it has two coordinates RNs now I I look at I I look at the first coordinate R and I claim that if the first coordinate is is constant

then the second must be constant as well yeah that's clear this is clear right because I need to satisfy this equation star all the time I mean if this equation in the right hand side is always a constant expression our R of R of p is is always constant then also the right hand side the the left hand side must be constant so if a um row is is Conant it really means that I mean it's I mean it's enough to it's enough for R to be constant

to deduce that that raow is immediately constant as a as a function um so I mean or if you want we can I mean in the general case we consider a row which has two coordinates r and s and there could only be two cases either the for first coordinate of row is constant and this this automatically implies the second is constant or this the first coordinate of raow namely R is is not constant but if it is not constant then

the first part of the proof show that it is it is subjective Okay so um in fact um if I know that row is not constant then I know that R is rejective okay okay so if it is subjective it has to have a pole a poll and a zero because I mean subjective means that um there is some point Q such that r r of Q is is infinity

it remember surjective means subjective on the projective line right so it means that there is a point Q such that R of Q is infinity and there is a point Q Prime such that R of Q Prime is is zero is this clear yes okay so this this this is what is said here it has to have a poll and a zero now are you with me yes

okay so so I take a pole a poll of R let's call it p p so p is a point in E of K Bar such that R of p is equal to Infinity okay but since a row is a rational a rational map if R of p is equal to Infinity then also s of p is equal to Infinity right but if r r of p is infinity and S of p is infinity it means that row of P is the is O

right right okay so I I have found I found a a um a point P which is a I mean such that a raw maps maps speed to the point at Infinity now I take an arbitrary a a point Q in E of K Bar and I I want to show I want to find some P Prime that is mapped by row into into Q if I if I can find such a p Prime I finish my claim because I want to show subjectivity of raw

okay you with me or now yes yes yes okay so I took an arbitrary Point q and I consider now the um er the translation by Q translation by Sorry by minus Q translation by minus Q it takes an arbitrary point I don't know T and it Ms it to a T minus Q right and this is a rational map this this was one of the examples

um because adding or substracting points on eliptic curve you can do it with a um um quotient of pols is this clear yes oh now okay yes so so this is a rational map if it is a rational map I can compose it with another rational map so I have this is translation by minus q and I also have raw

and I can compose I can uh I can compose these two rational maps to get here translation by minus Q composed withraw this composition this this is what we um talked about in we had a proposition about it you remember yes so composition of rational function of rational Maps is again a rational map now this rational map um cannot be constant if it was constant then then row was constant right because this rational map is just

in Formula you take p and you map it into row of P minus Q if this is constant then it means that raow is constant right okay so since I know I I assume the roow is not constant then this map is also not constant oh is non constant it has it should have zero it must have zero so it has a the Reon Point yes this is what we proved I mean we proved

that um a a every non constant row we proved it on row but this is true for any any non- constant rational function has to have a point that maps to O right this the the the way we proved it we said if it is not constant then the first coordinate of it is not constant I mean the first coordinate R is not constant hence it is subjective as as a rational function hence it gets it gets a point it has a point that maps to Infinity but then s has to map this

point also to Infinity so so altogether raw would map this point into Infinity comma Infinity which is O so so the the the takeaway from this part is that every non constant rational rational map has to has to have a point that maps to O yes are now you with me okay so in particular also this rational map a a translation by minus Q composed with row has to have a has to have a point P

Prime that maps to O right right but but this is simply saying that row of P Prime minus Q is equal to O which is the same as saying that a row of P Prime is equal to Q okay okay so I get that row is subjective I I chose q i I didn't have any restriction on Q Q was arbitrary and I I have found a point P Prime such that raow of P Prime is q okay uh less questions before we end up

thank you for this recap like this uh repetition sure sure sure just uh feel free in in every session to to ask for a repetition and I'm happy to make it this is a very optimistic session because it feels more H much more clear than the last three or four that we had before ER okay I'm I'm glad and also a bit said that the the previous ones were not H not as good but uh no it's just this one is composed uh of um kind of like parts that are familiar and clear and that's

why it is easier yeah it should be easier it is definitely meant to be easier as I said the the ab aobi is the the biggest result that we we have proved so far so it's natural that it's more complicated to understand and again I'm I'm happy to revise it at at any point if you want even next session but ER you can also read it in the notes and I think yeah yeah I mean you know as you wish so just you can you can ask me at any time to to

repeat okay so um that's it for today and have a nice weekend thank you thank you very much bye bye e