PSE Lectures - Ep 22 - Function fields & rational functions over EC

Transcript

all right so today I want to talk about ER function fields and um in the the previous uh version of this course in the seminar that I ran ER I I think H some people had difficulty to to understand this uh this subject so I hope to give a a better explanation today um and let me first start with the

um motivation I mean recall that um we want to think of of KX K is a field um is analogous to the integers right we um we can factor a a polom to a prime comp components and we have gcd and and so on this analogy is

actually um even stronger even deeper when when K is finite which is many times what we're interested in ER but for now ER let's try to push um so to push uh the analogy um we can imitate uh the construction of

Q I mean the rational numbers from a and from the integers how do we how do we construct Q from uh from the integers we say um we Define an equivalence relation on set of integers um sorry Define an equivalence relation on the

set of integers whose second coordinate is non zero um and we say that AB is equivalent to C D and this this AB represents would or the equivalence class of ab would represent what we want to think of as a modul b or a divided by B and this of course would be C ided by D so we want to say that they are equivalent if and only

if a d is equal to h c or CB okay and it's very easy to to check that this is an equivalence relation so so Q in the the [Music] fundamental mathematical definition it's it's simply the the set of integers whose second coordinate is non zero modul this equivalence relation okay so this is

a and and the the equivalence class of ab as I said is um is denoted as a a ided by B okay so so how do we uh how do we do it for pols uh the function field in in One V in in one variable is a um

simply the question pols or the cartisian product of a pols in x times itself but such that the second coordinate is not the zero polinomial modul the equivalence relation that you can guess f g is equivalent to F Prime G prime or maybe

h a i if and only if f * I is equal to H * G okay and we we Den note so in other words the function field in one variable is simply the H the set of quotients F ided by G such that f and g are pols in one variable over K and G is non zero and of course the the

implicit relation is is a is the one coming from the equivalence relation here okay and this is a um is is a field right because um I mean addition and multiplication is is a just like you know from fractions I mean from q and it's going to be asso ative and distributive and so on and the multiplicative inverse of a a nonzero element f f modu g is going to be

G divided by F okay is this a are there any questions on on this now um maybe let's remark on on the one variable case um we saw that M K bracket X is the minimal

containing okay and a new symbol X right because um any field that contains K and and the symbol X it must contain all pols in x and since it's it's supposed to be a field it must contain all the inverses of polom so one divided by a polinomial in X and and we get the the function now when we do elliptic curves

elliptic curves we want um the the two variable version so um that's that's not hard and we Define K XY to be the the set of all quotients

fxy divided by gxy such that f and g are pols in two variables over k and G is non zero and this is this is again um the field the function field or also it's called a um the field of rational

functions now um why should we care about rational functions in in two variables well if we we have an elliptic curve over k and a a rational point I mean a point on E of K then we we defined

the doubling of the point doubling of the point it had some formulas which I will not repeat because first I don't remember and and second it's it's not relevant but it is some pair of rational functions where R and R Prime are rational functions in two variables right and I I cannot it's not just R and RP Prime are not simply pols because when we Define for example the

tangent line we divide by a polinomial I mean we we took the derivative we have we have some some rational function not not just a polinomial so so if we want to for example understand this doubling of the point when P varies then we need to understand um rational functions in two variable now we call it rational functions but um it's not actually functions um so every element are

let's say it has a representation f modu g um or let me even just pick a specific representation f f modul g it defines a partial function ER this partial function is from the the aine plane

A2 into K and this goes as a um you take a point x x node y node and you send it to f of x node Y no g x node why no but of course this this is a partial because when X node y node um nullifies G or when when it is a root of G then this function is not defined

um so so this is not defined it's not defined on on the curve CG which is the set of solutions to the the equation gxy equal Z but there is a subtlety so this representation is not unique and it could be that I I take a

different representation let's say if if F module G is a equivalent to F Prime modu G Prime and this of course means that um f * G Prime minus um f Prime G is equal to zero to the zero polom ER then it could be that

ER that I mean the the set of of roots of G of G is different than the set of roots of of G so you know on one representation this would Define a partial function defined on a a certain subset of the Aline plane and on another representation it will be a partial function defined on another subset of the the Aline plan for example we can

consider H the two functions two rational functions x² time sorry modu XY and XY modul y s these are two rational functions in kxy and they are equivalent I mean they are equal in in K XY um but H but of course you know one of them is not defined on uh on the line

x equal Z and the other is not defined on the line Y equals z so we have a different domain of definitions we will say that um a rational function R XY is regular it's

p if there exist uh a rxy equals fxy modu gxy such that g of p is non zero okay so this is a this is rational functions on in two variables so on the

the Aline plane you can H you can view rational functions in one variable as as partial functions on the the aine line but what we are what we really want to to look at is um rational functions on elliptic curves so now yeah I'm sorry could you remind what is a partial function partial function is um is simply a uh it's something that is defined or I

mean it is a function that is defined on a subset of the of the domain it is not defined on all the points okay okay okay so in our case I mean it is not defined on any any time that is aan anytime that you have a a point that is um I mean if R is not regular at P then you cannot Define H the resulting um at at p i mean maybe I should add to the previous

definition in this case in this case we Define the value of of r at p is simply um FP modu GP for the specific representation the appropriate representation okay and you can see it's very easy to see that if in one

representation um g g of P does not vanish then in any other representation such that that the the denominator is does not vanish in P you get the same value you get I mean if you you substitute p in the the the other representation you will get the same value as as FP modu GP this is simply because of how we defined um the equivalence relation on on the the set of I mean rational yeah um maybe I'm messing up different

things but is it like uh when we say that P is regular uh is it this R is regular as B um is it the same that uh before we called like well defined at P ER you could say well defined is a is a general term in mathematics that refers to cases where um where you're trying to define something on repres I mean Representatives but you want to you want to claim that H your definition actually does not depend on the choice of

Representatives oh I see this is a general so it's it's related I mean but a in in algebraic geometry this this case is called regular I mean regular is again is some generic terms that is used all over mathematics when you want to say that something is nice they are like a regular normal and so on so this it's just a standard terminology and so in this case it will mean regular will mean that it is defined in this particular point B yes and again but what is defined in

particular part it means that there is a representation ER for which it is it is well I mean for which the denominator does not vanish at P of course you can always multiply I can always multiply this representation by e say x minus y or whatever I I can multiply this representation by a polinomial that does vanish at at X at XY and then it would not be defined right so just like in this example here right I can always

uh I can always multiply by a whatever I want I mean say y y and so on so what I'm looking for is is a a a rational functions that are that admit a representation that is defined at P okay so now let e over K be an elliptic

curve with um let's say the defining equation I call it f fxy um so e is is um fix Y is is just H say Y 2 - x Cub - a x minus B and is is the set of set of solutions of this this polinomial um for

any in finite Point p in the the K rational points finite I mean simply not the point at Infinity consider

um any rational um well any find sorry any finite Point um we can consider it as as a point in the Aline plane so we can consider any rational function say f modu ER

say with the ER is restricted to H to the finite points right uh so to to EK minus the point at Infinity and this this is of course a subset of the the Aline plane now

note that if Fe divides the polinomial um gxy then this this rational function R is a a is not defined on any point in ER in the elliptic curve

minus the the point at Infinity right this H so this is this is kind of a a very bad case we want to avoid um so otherwise yes sorry I'm not sure understand why is not because um any point any rational point on E would satisfy fxy equal z Oh yes okay okay sorry if it's a multiple yeah yeah thanks okay so again

ER we need to be careful with this H notion of division h i mean there is a notion of division of of polinomial in two variable it just that we don't have all the all the toolkit of gcd and so on but we can still talk about two pols in two variables that divide each other for example x - y divides x - y * x + y okay so this this is completely a I mean completely legit to uh uh to talk about but we we need to be a bit

careful so if if he divides the the denom denominator G then we um we don't get anything interesting on the on the rational points of e ER and in in the rest of the cases so otherwise if F does not divide G then uh we could F

XY gxy is defined on a all points X node y node such that ones that satisfy GX X node Y no is non and as a remark how many

points can there be such that a g node GX node y node is non zero remember these need to be points in a in E of K so it is really a intersection of two curves and there is a theorem that a um two curves can intersect at most in the the multiplication of their degrees so the number this is like a generalization of the phenomena that we saw that um the tangent line can intersect a cubic at most at at three points this is because

the tangent line is a degree one curve and the cubic is a degree 3 curve so the they can intersect at most in one time three that is three points and in in the the more General case this is called bazo intersection of e with the Curve CG ER is at most degree degree Fe the the defining

polinomial of e times the degree of G okay but we don't rely on it this this is a um it's quite involved to prove it but we we don't need it um but it's it's nice to know that uh we have some uh some kind of bound on how many bad points this rational function can can have so we will

call um rational function f modu g kxy such that F does not divide G um a rational function on E okay and

um we we can Define we Define equivalence relation on on on such functions by fxy modu gxy is equal to F Prime XY modul G Prime

XY if an only if or I should say if now instead of of saying FG Prime minus um f Prime G is equal to zero I want to say that it's equal to zero modu of so um Fe divides this polinomial okay this is basically what I'm doing here is H recall that when I had pols in one

variable and I had a polom fee it didn't it didn't have to be a prime polinomial this this would This was um a mod Fe arithmetic right it means that f is equivalent to G if and only if F minus G instead of equal being equal to zero it's equal to zero modul modu of Fe which is again it's the same as saying that Fe divides f f minus

G and what I'm doing here is this the same in two variables okay so I'm considering the the mod fee arithmetic on pols in two so um so in other um the the field of I mean the this the set of rational functions on E with this equivalence

relation is a in in in notations it is kxy modul f which means right that it's also the same as a um the equivalence relation as I said is is um is this okay so this is all to motivate the the

following definition for an elliptic curve K the polinomial over e or maybe on e um is the set K which which is simply modul um this defining polinomial

Fe okay so this is modul Y 2 - x Cub - a x - b equal Z all right so by definition in A in E of k sorry uh I I have

I Mis wrote this should be K of e okay this is like Po polinomial on E is K of e um and in in K of e we can replace every I mean we can we can look at it as as just ordinary pols in two variables and whenever we have a a term that is y Square we can simply replace it by X Cub + a x + b um so if f is a polinomial in K of

e h we can replace um any term y s with um X Cub plus a X plus b and this means that um um fxy can be written

in what is called a canonical form that is what notations do you that choose say VX plus y WX for some some pairs pair of pols v and and W um in one variable right polom in X because I replace all the terms y

Square H potentially I I I mean I have a term y that I cannot replace I mean if I had square root I would be able to replace y with the square root of X Cub plus a X plus b but in general I don't have it so I I keep it as is and um and I get that fxy is is equal in in K right this equality in K not not an equality of of um General polinomial in two variables um but but in K of E I can write fxy like

so I claim that the canonical form is unique I mean you cannot you cannot write it in two different canonical forms um before the proof are there any questions on no okay so H this this is it's very

straightforward I mean you could have have done it as an exercise but just to to give you a sense I mean fxy say it's equal to vxy VX plus y WX and let's say it's also equal to V X Plus y w Prime X um then I can do um just reorganization and and get

VX minus V Prime X Plus y w xus w Prime X is equal to zero so I I call this this one ER new I call this one Omega um it is enough to show

um new x + y Omega x equal Z implies new x = z and Omega x equal Z right this would H this would equate V and V Prime and also W and W Prim so how do I I show that a zero is of course 0 times

um new X Plus y Omega X ER and now I do a trick that ER we will see I mean you you seen in many such situations like in complex numbers I simply multiply by the um by the conjugate so um and wait hang on

ah I want a I want this one right zero times anything is is zero so I I I take it to be the the conjugate uh but now I can replace this zero with a new x + y Omega X and I leave the the second one as before this is new x minus X and

uh this is is new Square x uh minus y² Omega X Omega Square x uh which again is like I can replace the Y squar so this is new sare um let's say plus minus s x Omega sare X where s

x is h x Cub + a x + b um but the degree of of S odd and the degree of of newu and new square and W Square Omega Square H are

even so the only way this could happen is is that Omega needs to be the zero polinomial and hence also new is the zero Pol okay and since we used H this this conjugate let me just make it a definition let F in

um be given in canonical form as a XY = VX + y WX the conjugate of ER is f bar which is simply VX minus w y WX

okay and uh the norm of f is um simply F times the conjugate this this is analogous to H what you have in complex numbers I mean C you have um you take you take a a complex number you multiply it by the conjugate you get what is called the the L2

Norm of the of the the complex numbers okay and so F ftimes F Bar H this is going to be a v² x - y² w² x and as we did in the proof I can replace the Y Square by by by S of X so this is a v v² x minus uh X Cub + a x +

b w Square X okay and this this is in K of all right um we were're getting close to a break but the any questions or remarks maybe a question about um theg GD check uh where you told that uh the degree of uh SX is odd but the degree of VX of mu and Omega even but H in fact if you multiply the

degree of the odd degree of SX and the even degree of Omega X you will get a even degree and mo also have even degree ER all right let me think um so we have yeah I would get this is odd yeah there is an issue

here I mean one thing is that the the the degree of V would be H um of new and let me check

yeah uh let me I would need to think about it actually it's it's a good uh it's a good catch um yeah other questions okay so uh let's uh let's take a 10 minutes break sir see you okay um can we start start

again yeah there also there has been a discussion in the comments about the the topic of the degrees so maybe want to revisit it okay yeah ER yeah Thomas that's exactly right ER that uh that that is what I uh I missed so so the proof is valid right just let me repeat I mean this is

even this is odd and this is even then the degree of the the multiplication of polinomial is the sum and the sum of odd and even is odd so this this is a contradiction um is this clear just to yeah one more question Madame uh just to like have a broader picture where like where is this like going towards or or like how does this fit in with uh like the

previous classes that we've had like this topic specifically maybe I missed it because I was not in the first five minutes uh yeah well um we want to to to this generally speaking to understand a algebra geometric object like an elliptic curve we want to understand functions on it okay um H so the the natural notion of of H I mean functions of course for us this is all Pol like algebraic functions pols but pols on on E is this is this

uh this this notion that we defined K of e um that is a bit more involved than just ordinary polinomial and what we will do eventually I mean I will soon Define functions on H on E again in in the same uh the same vein that we defined rational functions on the the aine line right as kind of a imitation of the the construction of the rational numbers and rational functions on E will be crucial to H to construct the the

pairings okay okay so this this would be but if you want a a a a simpler motivation then remember that when you double a point I mean the the the general function of doubling a point ER forgot what where it but yeah here the general function of doubling a point on on e is a pair of rational functions right yep and so if we want to understand doubling of a pointing from a

more profound way whatever we we want to understand rational functions fair enough thank you thank you very much Madame I'm I'm sorry can I double on that question you mentioned that we're going to look at rational function on E but I thought it was already R of X and Y the rational function on E didn't we already Define this I thought R was we we didn't H we didn't Define um explicitly we it's true that okay or depending on how you ER you treat it but I I want to Define it er er formulas okay okay H

so so maybe this is actually the the next the next thing course okay remark so we have a k of E is this modul this polom Fe and it can be shown

that um Fe is irreducible IE um we cannot write or it cannot be written is um a Fe B = A * B for non constant polinomial A and B in in two

variables so this is what we called um in one variable we called it Prime but uh the proper term for a multivariable polinomial is is and remember that when we took the the Ring of pols in one variable and we divided by a prime polom we got a field okay but the proof that this this was a field relied on the fact that we could do gcd and in two variable we cannot we

don't have a well defined gcd um K of e is not a fi all right and you can find example I think I wrote in the in the notes um there are examples that show you that H you have a I mean it cannot be it cannot be a field because you have zero divisor or you don't have multiplicative inverse for

a let's say EG X has no multiplicative inverse okay so the two variable case is more involved erh but nevertheless we want ER to H to turn K of e to a field um in kind of a minimal way right this is the the whole theme of this rational I mean field of rational functions is is to to extend K with some additional

variables ER to the minimal field ER that contains both H so so the way to do it is again to imitate the construction of the rational numbers um and this is the following definition for E over K an elliptic curve set rational

E is um I take um polom on E times polom on E that are non zero non zero again non zero as pols in right and I divide by an equivalence relation um which is as you expect I mean f f comma G is equivalent to F Prime comma G Prime if and only f g Prime

minus FR Prime G is equal to zero in K e right so basically in K of e it means that H if this expression is is equal to zero if and only if a fe divides let's say this is equal to T * V for

something okay so this a we are um um you have layers of equivalence relations but this is I hope that I could convince you that this is a natural chain of um definitions to make H now to check uh to check if um this of course this f f comma G we will denote it as as F / G so to check if if f modu g is equal to

F Prime modul G Prime um we can put f * G Prime and F Prime Time G in canonical form and and then it's just H just pols uh in one variable um I mean plus this y times polinomial in one variable and we can simply compare coefficients

all right so this is a this is a practical way uh to see if two rational functions on E are are equal yes yeah sorry mad I got a bit lost in this last uh thing so you started giving a definition that you can if I understood correctly count the exact amount of rational functions on e say again yeah I can like what you stating that you can count the exact amount of rational functions on E no okay so sorry can you then repeat what the definition is stating because I

got lost so the definition is um this one k k um round brackets e this is this is analogous to the process we had right that we had in in the the kind of aine aine case sure we we first we Define the notion of polom and then we we Define the the the field of rational functions as a quotient of of polinomial such that the denominator is is non zero right or alternatively it's like a a set of pairs

modulus some equivalence relation this specifically this equivalence relation right so now that I have the notion of pols on E right polom on E is is what I would naturally consider as as pols restricted to the to the elliptic curve e right it's like pols in two variables mod the defining polinomial of e sure okay so now I just continue with a kind of the same recipe that I have I have um rational function so is simply um I

mean alternatively you could you could simply Define it as um the set of all f modu g such that f and g are pols in E on E and G is non zero but again non zero means non zero as a polom in K of e right okay yeah remember that in K of e for example the polinomial Fe is is zero right so what I need is to say that g is non zero is a essentially saying that um

that Fe does not divide G okay okay I I could even write it like this F modul G such that FG or Pol nomials over E and F Fe does not divide G okay okay but this again there are there are kind of layers of a equivalence relations ER because of the the um I mean because of how how we defined it we we want a a notion of pols over I mean

pols on e h this is not just ordinary pols it's like pols with the the if you want pols on E is like pols in in two variables in which you can replace every term y s by the term X Cub + a X plus b that's polom does this imply any sort of uh linear dependence relation or something like that or H between between what between the polom that you define in this set and the Primitive polinomial that defines the that that would be the

the curve equation to put it on some sensor that defines the curve I'm not sure I understand the question entirely but it it does not need to be linear relation in any case Okay um we we simply have I mean the defining defining equation of of the elliptic curve give you a relation sure right y² equals x Cub plus a X plus b yeah you say that a a polinomial in two variable is is a polinomial on E once you you can replace all the all the terms y squar by by their equivalent ER I mean by x x

Cube plus ax plus b you and and this is called polom on E because you can only do it for points on in E points on E satisfy this relation fair enough fair enough fair enough fair enough I see now I see I see this is the bit that I was missing okay fair enough that makes sense and once I once I have a notion of pols over e i I naturally Define rational functions over e as as this this Q construction basically okay yep and er of course K of e is field with the the the natural addition

and multiplication that you want and and the inverse ER I mean the multiplicative inverse is again as we had in the the previous cases okay yeah now yeah sorry Madan one more question uh have you proven that this is actually a field in the sense of how do we know that every polinomial I mean I don't even know if you can say that they have an inverse but a field implies that every element inside of it has a valid inverse so yeah is is it already like implicit or proven somewhere or we are

just going to skip it and that's I I can prove it if you want but it's I think it's actually a very good exercise in definitions okay to to prove it but but how do you do it I mean you have addition and multiplication H defined for example how do you define H multiplication H you simply say that this is the F time F Prime modu of G * G Prime right yeah in addition you define similarly just like you know from a addition of fractions sure and what will be the

inverse of f modul g i mean the inverse remember in a field we only require that nonzero elements have have multiplicative inverse okay I see for so this this is if this is non a nonzero element it means that not only G is non zero but also f is non zero yeah so the inverse is simply G module G okay fair enough um so we we naturally get a field this this is the field of rational functions e thank you thank

you as another remark remember we had canonical forms for pols in pols on E but we also we can do canonic forms for a rational functions only so for um let's say Ral F over G in rational functions over E I can write F ided by G and now I do the the conjugate trick so I multiply and divide by by G

conjugate and this is equal to um something in in the numerator divided by the norm of G right G * G conjugate is the norm of write the the numerator f a g conjugate in canonical form uh so let's say f g conjugate is XY is a VX + y

and remember the the the norm of G is a polinomial only in X right we had it you see the the norm of G is I mean I have y Square over there and I I can simply plug in this SX instead of Y Square so this is a V Prime x minus s of x w

Prime sorry this is V let's say new new square of x and Omega squar of X um so this is a polinomial in X and just a minute uh it means that I can write F over G as um VX + divided by some polinomial in X so this is new squ X

x- s x Omega s of x h and this is simply some some rational function in in X Plus y times some other rational in okay so this is a in this is in the canonical form of

yes yeah you you use the term Norm as the definition of a size of a vector space or something like this like as in Geometry or H it's it's a it's by alogy I mean the term Norm we defined it as just you take G you take take it in canonical form so this is some VX plus plus ywx and then you have the conjugate of G this is VX minus y WX and the norm of G is simply defined as G * G conjugate okay fair enough okay and we we saw that you can I mean you

can manipulate the symbols to get this formula does there's any like underlying uh similarity between how we understand the norm here and how you understand the norm for like vectors and planes and stuff like that um it's it's just by analogy okay fair enough yeah Fair yeah it's not it's not tend to be like a very deep um so okay it's just a term for for us ER but if you if you open the formula you you find out that the norm of G I mean G is a is a rational function on

E sorry G is a is a polinomial on E H so typically it is a it is something in two variables you you have X and Y but what we did the manipulation of the formula showed that when you take the norm of G that is simply gtimes G conjugate you get you can turn it into a polinomial in only one variable okay fair enough right because this this SX yeah replaces y s h so using the using this this fact that the norm of G is a polinomial in one

variable I can turn a rational function f over G this I guess I I mean to be more precise I should write fxy ided gxy a I can turn it into a rational function only in X Plus y * another rational function only in x and this is this is analogous to the canonical form of a pol of a polom on E okay that makes sense thank you very much and maybe okay

exercise the canonical form of a rational function R in the field of rational functions over e is unique okay this H this again it will be a good exercise to Simply revise the the proof of of uniqueness in the in the case of polinomial over e and and do something

similar yes um sorry I'm going to ask a question that you something you did on the first hours how can we get only a y term in the canonical form if the eliptic equation contains the Y Square because we we take a polinomial I mean let's say you take fxy a polom in two variable over K right and now I if you want you can Define the the modum map from kxy to K of e pols over e this K of e are simply defined as

modul f is is the defining polom of e so f x y is Y 2 - x cubus a x minus B okay so this this map is simply taking F into a or we can call it I don't know F I mean this is f f modu of Fe basically okay but what does it mean to be to do the the quotient ring I mean kxy modu it means that in this quotient ring Y sare

is equal to I mean if V is equal to zero this means that y² is equal to X Cub + a X plus b in this K of e yep okay so the the kind of the simple way to to do the modu I mean operation is I I take all the terms in fxy that are y square or powers of of Y Square so all the all the even oh I see yeah and I replace them by by X Cub plus a X plus b

okay so yeah and then I I am left with I am left with a some polom in X Plus y times some other polinomial in X makes sense thanks so we have this we have this um this perspective that a rational function can be defined on all the the

finite rational points of e but we actually want it to be defined over all the all the rational points of e that is we we want to make sense of what what is r evaluated at the point at uh now to do this we um we draw an analogy from calculus so let me remind you ER

let's say so we think of um um for for a rational function r on on the elliptic curve um what is r a value valuated at the point so I mean we will we will do rigorous definitions but as an intuition you can think of O as as like

um Infinity comma Infinity okay so it's it's a it's as if o in the x coordinate has Infinity and in the y coordinate also Infinity now to make sense of the definition I mean to motivate it really um what do we do let's say in one variable um what is um I have let's say

a rational function say uh x divided x² + 1 and I want to evaluate it at infinity mean of course if I like on the nose substitute Infinity I get I get nothing I mean I get infinity divided by Infinity I cannot make sense of it but in what what what you do is you H you look at the the the value of the function at a sequence of num finite numbers that gets gets close to Infinity I mean what

what is called a t to infinity or a sequence of number whose limit is infinity and and then you define r at Infinity as the limit of of the results I mean of the value of the function on this sequence of finite numbers now what you can you can do in in in simple terms is basically divide ER the numerator and the denominator by the highest H degree monomial so

to see what is are we doing opal or something similar it's a it's it's kind of lupal yes L what is r at Infinity is like the limit where X tends to Infinity of this this expression x / x² + 1 ER and what we do is is we divide by x² both the H um numerator and denominator so we have x² /

x² plus 1 / x² and this is a this is one over X divided by 1 plus 1 / X squ I should put limit everywhere

and now in the the last expression I can I can I can actually calculate it um one divided by by Infinity is zero and also 1 divided by Infinity square is zero so I get zero I get 0 ided 1 + 0 zero um however if I want to do the same thing H for let's say R Prime is x² plus sorry divided by X2 +

1 then r r Prime at Infinity would be again the limit X tends to Infinity um x² / x² + 1 and this is a x² / x² modul x² / x² plus 1 / x² and this is actually H 1 ided by 1 + 0 so it's and the what I want what I want you to

take from these two examples is that Bas I mean we don't need the whole theory of calculus for limits and all that for all functions I mean s and cosine and um natural logarithm and stuff like this we only care about polinomial and for pols you you have a very simple rule I mean if the the degree of the numerator is bigger than okay if the degree of the numerator is smaller than the degree of the H denominator then H

you get zero like in the first example right the degree here is one the degree here is two H and if you you tend to infinity and you you divide polinomial of degree one by polinomial of degree two you always get zero I mean this is not a proof but but I think you can understand it intuitively right once I will divide by highest the highest degre G monomial I will I will always get zero at at the limit on the other hand if well of course if the degree of the the

numerator was bigger than the degree of denominator then when I 10 when I take 10 x to Infinity I get infinity right it's like it's kind of a stronger a a faster a a tendency to Infinity of the the numerator compared to the denominator so I get I get infinity and if the the degree of the the the the numerator and the denominator are equal then I should expect that H that at that when I tend to Infinity I get some finite

number okay okay is this uh this intuition H ER somewhat clear all right Carlos you is it the yeah oh sorry you're not seeing the screen yeah everyone thr thumbs up so we all we all said yes yeah so yeah so I mean I'm I just want you to feel to feel comfortable in I mean eventually we we just make a formal definition it's it's a not a formal matter but rather just to to motivate

you um okay so what what will we do in the so in K of E we have a a bit of a a a subtle situation first we want to define the degree I notion of a I say degree of for f in k

K of e h it needs to be adjusted um since we have a relation y^2 = X Cub + a X+ B so you know why y Square in ordinary polinomial the monomial y Square would

be of degree two and the right hand side would be a polinomial of degree three so um how do we how do we make sense of it ER we set um fxy in canonical form and um again we have this

relation y s so Y2 = X cub and so on H so I want to say that the the degree of the variable y will be I mean the the relation between the degree of the variable Y and the degree of the variable X will be two 2/3 right so uh the between the degree of X and the degree of Y should 2/3 but h of course it's more convenient

to H to have H degrees being integers uh so uh we Define the degree of Y is equal to 2 and sorry the degree of y equals to 3 and the degree of X is equal to two H and ER and hence the degree of

f is going to be the maximum between 2 times the degree of V remember V is just a polinomial in X so it has a well defined notion of degree H and the second term is y * WX but Y is of degree 3 this is 3 times 3 + 2 times the degree of w okay so this is this is our notion of

H the degree of a polinomial I mean a polinomial over over e ER all right we we ran out of time so I suggest we we stop here and I mean you can ask questions but ER where we heading what we'll do next time is we we use this H um this analogy from calculus and a our new notion of degree to Define what is r the point at Infinity cool I'll stop the

recording