PSE Lectures - Ep 21 - Grup Law & Dive into Projective space

Transcript

okay so um what we did last time um is showing that the the condition that the discriminant is is non zero um for a a an algebraic curve of of the the type we are interested is equivalent for the the ER the curve being smooth that is to admit a a tangent line at every point and ER I would like to start today with a filling a gap in fact if you if you remember the

the proof that we did last time um we said that um for algebraic curve um of the form y² = X Cub plus a X+ B and Delta of e which is this 4 A a cub +

27 b² is non zero if and only if um e e um has tangent line at every point and the proof ER this is usually the case I mean pols are admit tangent line for almost all the

points and there there are only a a a small number of points in which a an algebraic curve could possibly be non smoth or or to not admit tangent line and what we said the the the interesting case of the proof um last case maybe is that the tangent line that we are looking for

ER needs to so if if if p is x0 y z then the slope of the tangent line is 3x0 SAR + a over 2 y z this is M and of course um it could be that um I mean the the numerator and the denominator are are zero this is the the non-trivial case um all the other cases we it's easy to show that there is

a tangent line possibly vertical but um if um 2 y z e z and 3x0 squar plus a is also zero um then what do we do I mean what we did is um we substitute um I mean instead of a we we replace minus a because clearly for negative a um

this this the second condition cannot happen so sorry for positive a so in this case necessarily A is negative so I replace it by a positive which means the condition is now um 3x² sorry x0 squ minus a is equal to zero which is equivalent to say that x0 Square sorry 3x0 s is equal to a now what I did last time is I said okay this means that um x0 is the square root or

maybe plus minus the square root of a over3 and then I substitute it in the in the equation and I find out um I find out that this this implies the in fact equivalent that Delta of e is zero so and and this this completes the proof it shows that um H this last case cannot happen in in an elliptic curve but can somebody tell me what is the problem with this argument me remember we are talking about um over

a general field arbitrary um okay this square root might not be defined exactly yeah so square root is not a is not a feature that you can have in every in every field um and you I mean I didn't cook up an example but you can easily take small primes and find a ER

elements in some prime field ER for which a there is no square of this element I mean square root ER but this so this this is a is not koser so to speak I mean h it's not meant to be a religious er er proposition but but just a um just to tell you that you cannot do the usual operations that you are used to in real

numbers over a general field so not everything is passes passes through so so I need a way to circumvent this ER this process so instead I get uh um x0 equals a over3 sorry x0 squ equals a over3 by the way a over3 might also not be defined if you are in characteristic three right

because in characteristic three three is zero and then I don't have a I cannot divide by zero but H our assumption is that the characteristic of the the fields is always different than two or three so I can divide by three and now I do the following I know we have um x0 Cube plus sorry minus a ex0 plus

b equals um y z but y z is zero in in our case so it equals zero and now I plug in this x0 Square I get um x0 time a over3 h minus a x0 uh plus b equal Z which is equivalent to say that x0 is - 2 over3 time B over

a okay and now I take this and I plug it in h the triangle equation and I get that um a over3 is equal to 9 b² ID 4 a cub sorry a square which is equivalent to say that um 4 a cub - 27 b² is equal to

Z and now I substitute back minus a instead of a and I get that this is 4 a squ Sorry 4 a cub plus 27 b² = Z so we get that the condition the condition of a having a a a well defined I mean a tangent line at at the point P for which both the the numerator and the denominator are are zero is equivalent

to say that the discriminant is non zero so so this gives us a smoothness any questions about this uh correction any other questions about what we did last time okay so I I remind you that

um the K rational point so so let's say e over K is an elliptic curve the K rational points which I denote as e with brackets K is the solution set for the polinomial um so it's all the X comma Y in k squ such that Y2 = X Cub + a X plus b Union I I have to add one other point

which I call the point at infinity and this will serve as the as the neutral element for the the group law that we are going to Define and what I said last time is a kind of how to how to H view the group Lo geometrically over the real numbers but and okay let me maybe repeat I mean you take this elliptic take two points

say um p and Q you pass the line between p and Q this line has to intersect a the curve in a third point this is p maybe we can they noted as P star q and then I I reflect along the the xaxis to get um what what I Define as P plus

Q okay so my drain skills again are not uh not very good but H ER intuitively um because the curve is is um is a cubic is of degree three then generically a line um in in I mean a line intersects a cubic in in three points of course there could be a lines that do not intersect the cubic in a in three points for example you take take the tangent line here this tangent line intersects the ER

the the point I mean the curve only at one point but we have this point at Infinity which we think again for now just intuitively is lying um at at the top and at the bottom at the same time so I can say that these are the the K rational points right I can say that if I add this this point at Infinity then even this tangent line intersects the curve at well at at two

points right at the point at infinity and at p H in fact and we will not get into this um rigorously one can Define the multiplicity of a of a intersection and it turns out that this this point P will have a um intersection multiplicity to and and so using a a correct notion of multi icity the I mean one can say that a line intersects a

cubic always at um three points if you count multiplicity correctly and you have the point at Infinity H okay so this this is the general intuition and now we need to make it a at least well defined over a general field so okay so now we have um let's say

um of um this um this operation plus over a general field so I start with um two points p is x0 y z or maybe sorry I do one here in the notes and H

Q is X2 Y2 um and let L be the line between p and Q so L is of the form Yus um well I guess we could say y equal um

MX plus um y1 where m is I mean if it is if it is defined H Y2 - y1 over X2 - X1 so yeah so the bar over p and Q uh or like L is equal to P bar this is the

line line um between oh I mean the pass that passes I was trying to make a shortcut line that passes through a p and Q okay so so we have this line um if

X X1 equals X2 then in fact the line must be vertical I mean the X1 equals X2 it means that I have this uh let's say p is here and Q is here and then the the line that passes through them does not well it has an infinite slope so uh so m is not defined but still the line is defined um so

um vertical um and and we deal with this later so suppose X1 is different than X2 um now we have the the equation of L right y equal m x - X1 h plus y1 I think I had a mistake here this this should

be x - X1 H and I I want to find the intersection of L with E and so I substitute instead of a I mean I sub this uh this thing into the equation of E I get that M xus X1 + y1 squar equals x Cub plus a x + B so this is a

um this is a key I mean this can be rearranged as as a cubic right H this is equivalent after some some calculation to the equation 0 equals x cub and now what's the coefficient of x square the coefficient of x square is simply M Square sorry it's a minus n s and there are other things but I don't care about the other terms

now this is a cubic equation we know that that um X1 and X2 are a roots of um of this equation star um and also I mean over the algebraic closure I can write

um x - X1 x - X2 X - X3 = Z okay it's the same it's it's just another representation of this uh this equation that um at least a priori we only know it is valid over the algebraic closure but if you open this um this I think we had last time or maybe a session before

that this is equivalent to say and this is always true in a cubic that you get a x Cube minus X1 + X2 + X3 x² plus some other terms equal Z okay this this is what happens when you you open up ER you open up this uh this equation and we know that this is an equation it's the same as we have here in the in the

purple one which is a polom over K not not over K bar so we conclude that X1 + X2 + X3 is equal to m² which is an element in K and since already X1 and X2 are are in K it follows that X3 is also in k okay and not not in K bar that is kind of inaccessible okay so now we're almost

done for this case because we have the x coordinate of the intersection right we have um this this line intersect the curve at P star q and we have the x coordinate x coordinate of this is X3 what is the y-coordinate the y-coordinate is easy because I know that y I mean I know that this P star Q must lie on on the line L so Y3 Y3 is equal sorry ah yeah Y3 is equal to to

minus H the the Y3 that uh that we have in P star q that is m x - X3 sorry xus X1 plus y1 so p+ Q is defined as X3 Y3 okay and this is valid in any field um any questions about this uh this case yes Nicholas um I'm I'm still

unclear exactly because the the first star that you see here you said it was defined over the algebraic closure and then you say okay X3 must be on K and not the algebraic clure I'm I'm sure how you made this gag yeah how did you pass from reduce one okay so so I'm saying we have this this cubic here in purple right and like any cubic like any polinomial I mean I can Factor it linearly over the algebraic closure yes so I know that over the algebraic closure I have I have this guy yep and this guy has to be equal to I mean these two things has

have to be equal so now I open up h i open up the brackets in maybe we call it double star okay I open up the brackets in double star I get here we can say Triple Star right I didn't do the the calculation it's a bit Yeah Yeah but I okay but this you can believe I get Triple Star so now I but I know that Triple Star has to be equal to Star

right is a polinomial equation so it means that minus m² has to be equal to minus X1 + X2 plus X3 and since m is in K so all the 3x are in K as well yeah since since m is in k then M square is in K and since X1 and X2 are in K as well then then X3 is simply You Know M Square minus X1 minus X2 makes sense thank you okay um what about the other cases um so this is for so this is a a a definition

of the the binary operation for any um for any two points that are not equal and um that don't have the same ex coordinate um okay ER in

case yes uh can you list all the cases we want to consider um I I will I didn't um write it very organized in the notes so I will I will list it as we go along if okay if that's okay um I mean basically we want um here I should say this is the the generic case is that P is not equal to Q X1 is not equal to X2 and also p and

Q are not the point at Infinity but you know if p and Q if one of them is the point at infinity or even two of them then we know what we have to do because the point at Infinity has to be the I mean we want it to be the zero element of the the group law so this this is easy um in case X1 is equal to X2 but y1 is not equal to Y2 right then the line is vertical we

said it then L is vertical and and then of course L cannot intersect the curve in any ER finite point but it will intersect the the curve in the point at Infinity right this is a we will have this the case is the the purple case this line will intersect um the K rational points at the point of infinity so of course in this case we

Define um p+ Q is equal to a well it will intersect thec rational points at o and then we have to reflect but reflecting o is is just a is just o again so I guess if you want it by steps you could say the the star is is O and hence p+ Q is defined also is O is the reflection of O is O

um and now what is the case we are left with the case we are left with is um if p is equal to q and of course p and Q are not the point at Infinity I mean again the point at Infinity it's we know how to add zero to zero in any anything that needs to be a group uh this this is Trivial but what happens in the the P equals Q case I mean or or maybe let me just remark on a on the last case the

the this is the last case we discussed p and Q and we said that P plus Q is zero or o this means that if we are about to have a group then p is equal to minus Q okay so the ER the the additive inverse in the um I mean the additive inverse if if we have a group then the add additive inverse would have to be simply reflecting along the the x-axis again reflecting along the x-axis is just a a to help you psychologically but over a general field

I simply put a minus in front of the the the y-coordinate okay so this is the last the the case we had uh before and and now maybe I need to choose another point for the drawing now now I want a um to add a point to itself this is called in in cryptography people call it doubling a

point so to double a point I need to draw the line that passes through p and Q this is like the tangent line to e at p and I claim that this tangent line will intersect e exactly at an additional point this would be P star p and then I will reflect and we'll get p+ p okay but I need to show it algebraically yes uh do we say that

double I mean do we split double from addition to itself of an element such that everything holds up together so what I'm say if we sumon element to itself um since one is in the top of the other you can kind of draw the tangent line to intersect different like the line the line that touches both can go anywhere to not some sense so that's why you define double or no no no I call it doubling a point is simply adding a it's just a term that people use in in cryptography in elliptic curve cryptography is like a just adding

a point to itself according to the elliptic curve group law okay I mean maybe this is advanc into doubling and adding have different that I mean this we will later denote it as as 2p okay enough okay and this this is in in accordance to annotation we already had in a billion groups right I mean okay yeah makes sense okay so this this is why it's called doubling um okay so so this is the kind of the

geometric intuition but ER Let's Do It algebraically um this is L right the the slope of L and and P is is say X1 X1 y1 H and so the slope is h 3x1 2 + a over 2 y1 okay now if y1 is equal to zero which is I mean

it means that the slope is not really defined then L is vertical if L is vertical then of course then a p let's say p star P would be the the the only possible intersection of the the vertical line with the the K rational points which is zero and this means that P plus p eer so this tell you that

a the points here and possibly if we had a like a small Lake here then also this these uh these points are points that according to the group law if you add them to thems you get zero so this is called the two tent points okay otherwise so y I know that y1 is not zero um if y1 is not zero then the equation of L is given

by y - y1 or maybe y equal m x - X1 + y1 and and again I substitute it um now I Define G of x to be um this x Cub + a X+ B minus a m x - X1 + y1

squar and clearly G of X1 is zero right because we know that that P lies on the curve and P is x0 y y z sorry X1 y1 um now how do we find the the other root here we we have a bit of a problem because the argument we had before was assuming that we have two

distinct Roots if we if we had two two distinct Roots then we know that the third root over the algebra right closure must be infect in the field K and then we find that we this this will be the x coordinate of the um of of the result we are looking for um but here what is happening is in fact because it's a tangent line ER the the X1 is a double root

claim that X1 is a double root of G and this is very easy ER I take G Prime first I do it generically so for General X it's going to be 3x² + a minus 2 m 2 m m X+

y1 - mx1 and now I substitute X1 and um I hope you can believe me that you get you get zero mean uh it's an easy calculation with a um with substituting ER M for being a for being this this expression okay so so X1 is a double

root it means that this cubic GX which I know over over K Bar it is equal to x - X1 squar time x minus x uh 3 let's say H which means that the coefficient of um of x² which is a minus m s minus m s must be equal to

um or sorry M sare must be equal to the sum of the sum of all the roots that is um X1 + X1 plus X3 this is just as before only that instead of having two distinct Roots X1 and X2 I have a double root X1 and this means that X3 is m² minus 2x1 which is an element in k thus we and that's the only possible

this shows you by the way that there could be no other intersection point of the line with the with the Curve okay also in the previous cases there could be no no more than three intersection points because intersection points would would have each coordinates being a root of a cubic and the could only be three possible roots of a cubic um and this in this case h p+ q will be this X3 and

minus um we have the line L H so y Y3 is going to be m x 3 - X1 + y1 and we need to reflect it so it's going to be minus M x3 - X1 andus y1 okay yes I'm a bit confused how we introduced G Prime

G Prime is the is the derivative it's the formal derivative ah it's not g Prime it's derivative okay yeah maybe clear then it is clear it's a maybe it's a good it's a good H um standard to H to write the derivative like this in in our because we don't do a lot of derivatives the only thing we do is like formal derivatives of pols and then I want to reserve the prime to just denoting another function so maybe I can I will denote it

as D okay with derivative we had a we had a Lema this I don't I mean this appeared in the note I don't remember exactly where but we definitely talked about it that um an element in in a field is a double root of a polom if and only if it it nullifies the polom and the derivative of the polom right

okay so just to show you that kind of the the things are um er are also defined with the point at Infinity I mean how do I add the point at Infinity I according to the the kind of geometric intuition we we have if is I draw the line um let's say how how do I add the point at Infinity to itself I draw the line between itself I mean between the two points and H I look for the intersection

the intersection is is again the point at Infinity you reflect you get again the point at Infinity so this is kind of but this is really not necessary I mean since this is going to be the neutral element H we know that o plus o has to be O Okay so this this defines yes maybe this is stup but can we understand the point that Infinity as a subfield of a single element well no not as a subfield because it doesn't have an but I mean it's like another I will I

will talk about it in the next in the next part of this uh this lecture okay this This concerns the the projective space we talked about a bit about the projective space and one can Define the like the project projectivization or the elliptic curve in in projective coordinates and then you have a precise ER meaning of what is the the point at Infinity okay thank you but um just to say I mean what I what I defined here is also well defined it's not a it's not hand waving as as

mathematicians say I mean the we introduced a new symbol o and I defined the binary operation and this binary operation is is well defined for every every pair of points in the in the K rational points e ek okay yes um this is maybe slightly off topic but uh I think there are also other type of cures where you don't need to have special cases that we call complete addition formulas and F and will we be seeing those or this is a a the projective I think in the projective

coordinates ER you have a cleaner a CER formulation well it depend on the type of the curve like short Warrior stress don't have this but and like this have you don't care about if it's a zero or not zero this that makes it easier to be to have constant time as well implementation yeah there are okay there are a we can I mean there are different types of coordinate systems to Define elliptic curves um I forgot to there are various names and in in different coordinate systems ER you might have a cleaner definition that's true but

um er this is mostly important for H for implementation and H for us we are we are going to stick with this H this definition because we need we need it for kind of theoretical development I could I could later on do a session on I forgot the names um but the different coordinate systems for elliptic curves ER but it's rather Technical and I think I mean it would not help us directly in any any of the things we do so thanks yes there was a another question maybe in the

chat no sorry I just linked there's a paper that der complete formulas for sh by yeah you don't you don't need to use yes so so in general um what I will try to do henceforth is to avoid most of the the computations um because it's not it's not helpful I think for also for the audience I mean this is something you need to either do do yourself or or like let a computer verify

um but but still it's it's important to know it's important to to understand how this is this is defined over a general field all right so sorry mad one more question just to understand you said we will not be going through all of the different coordinate systems uh but that this also implies we will not go through different types of Curves like what we do here is completely General I mean any elliptic curve over a field with characteristic that is not two or three can be

expressed in this form okay okay that's why we build short that's why we so we are not losing any generality apart from this characteristic two and three but of course in terms of implementation there could be I mean there are a better ways to I mean efficiently compute the stuff stuff we talk about okay thank you very much okay so I suggest we take a 10 minutes break and we we come back to the projective version of of elliptic

curves so you know um okay um can we start

again all right so H the next thing I want to talk about is H is called the projective coordinates um we we talked a bit about the the projective space just in terms of a terminology what we did in the ER the

previous hour is called apine so this is the distinction is is aine geometry or aine coordinates and projective coordinates um we saw we saw a definition but let me repeat it let K be a field and consider the aine threedimensional space over K which

is I remind you it's um simply K * K * k um and now I take the aine the aine space and I remove the origin and I want to Define an equivalence relation which I call

Tilda on the the collection of all points in the Aline space the Aline three dimensional space H without the origin um and I do so by H by setting x y z equivalent to Lambda XY Z for any a nonzero element Lambda

um maybe exercise a Tilda is indeed equivalence relation um I can I can give you a sketch I mean h uh the the non-trivial thing in in equivalence relations Is to show transitivity so if you have X1 y1 Z1 equivalent to X2

Y2 Z2 and you have in addition that X2 Y2 Z2 is equivalent to X3 Y3 Z3 then um let's say X1 y1 Z1 is equal to Lambda say Lambda 1 X2 Y2 Z

2 and X2 Y2 Z2 is equal to Lambda 2 X3 Y3 Z3 and then you just multiply the lambdas you get that X1 y1 Z1 is equal to a Lambda 1 Lambda 2 x X3 Y3 Z3 okay reflexivity and symmetries is very easy so we have like this equivalence

relation um the way to view it or okay maybe just a the projective plane is a a this set the Aline space minus the origin this is a set with an equivalence relation and I can take the quotient this is called a p k

2 okay notice the the um the change in ER in Dimensions I mean the projective plane is is like Dimension two object but it is defined as taking a dimension three object and moding out by some equivalence relation and um we do what we we always do to get intuition we look over the real numbers over the real numbers H the Aline space Sorry the Aline um three-dimensional space is represented like so

and we say that two points let's say a P1 and a P2 are equivalent if they belong to the same line that passes through the origin okay so so in fact what we have here and this you can look at the notes

in page uh 72 every every line in the the 3D space that pass um passes through the origin is a a representative of an equivalence class so we denote um X colon Y colon Z the equivalence class

of XY Z in a in the projective plane okay so how do I represent it again over the real numbers H what I can do is I take this sphere okay H each line let's say a blue

line a blue line that passes through the origin is one equence class I can take the unit Vector a um in the positive and in the negative direction of this um of this blue line so um I can choose either um to represent the the whole blue line so what I get is that the the projective H plane over the real numbers is is simply um when you can take the upper half of the sphere and um apart from the the bottom

H the bottom circle of the the upper half of the sphere H every every point in the upper half would represent a line passes through through the origin right this line and in the points on the on the a basis Circle I would need to identify antipodal points because they will belong to the same line okay so H you could visualize the the projective space Sorry the projective plane over the real numbers as has a an upper upper half of the

sphere in which in the basis Circle you identify two h two opposite points yes I'm sorry I'm not sure why do you take only the hper half so why do you don't not need to take the opposite points when it's because Z is not equal to zero because this point is equivalent to this point they they if they lie on the same line that passes through the origin they are equivalent okay right to to say that a a two two triples XY Z are the same up to

a multiplication by lamba is to say that they lie on the same line that passes through the origin good okay nice okay okay so now um this is this is for the the kind of a to help you overcome the psychological barrier but what I defined is valid for any field ER we um we call points [Music] um

with a z coordinate or maybe I can write it like this points of of the form X col y col0 are points at Infinity so in other words we we identify

um and uh points X col y col 1 again this one is just um I can always I guess X col y col Z if Z is non zero then this is going to be equal to X over Zed y over Z 1 okay and these points we call aine

points um yeah so essentially what we are doing here is any line that passes through the origin we treating this as a point in the in the in the P2 and um any line that say a scalar multiple like a Lambda times x is same as X in this sense I mean you could say that the the line itself what is a line a line is a collection of

points satisfying some equation and this this collection of points is the equivalence class okay under the the the equivalence relation yes of course each each particular point on the line is a what we call a representative of an equivalence class right it's is this clear yeah okay yes yeah Carlos yeah I have one question um this are you still excluding order two and order three uh Fields here yes I mean actually not I don't need to

but how do you have a projective space with an order to fi Go I mean in the this is this is interesting this is actually interesting lamb times whatever is going to be whatever like it's not the exact same thing there just one element how can how can you have a projective how can you projecti one single element I because I do have a um okay I mean it would mean that okay so um let me just make a definition a line in a PK to is the solution

set of um an equation a X plus b y+ c z equal zero for some a b and c in the field okay so it's it's going to be all the X colon Y col Z this is L satisfying a x + b y plus CZ equals z and then there is a fascinating

example it's called the h think F plane it is what happens when you do the projective um plane for the field F2 so first of all how many points will H what what is the the number of points of the projective plane over F2 can anybody say I would say four like all of the

combinations that that you can but remember that we take FS2 time FS2 * FS2 oh we remove the origin seven exactly seven remove the origin and then we divide by an equivalence relation but as you said Carlos the equivalence relation is is nothing here because I can only M only Lambda I'm allowed to multiply is a one and multiplying by one is doesn't do anything so it's it's simply the same as as

F2 a cube minus the origin and this is a seven points now I defined what is a what is a line in the projective space in the projective plane so you can fiddle around with this and look what are the what are the possible lines and it turns out that what you get is this I mean one repres representation of it is the

following you have three points here three points here and one point here you draw a circle like this and a line like this a line like this and a line like this and this is a representation of the projective ER I mean the so-called F fu fu sorry the fu plane is seven points and

every straight line or the circle is a line passing passing through the points so in fact it is the the minimal model in which a a every um every line passes through exactly three points okay and two distinct lines intersect exactly at one point w yeah for for your next psychedelic Adventure thank you that um all right now

polinomial H let's say in three variables we didn't Define it but okay I can Define is is a sum say over K is a sum of monomial a i j k x to the i y to the j z to the k h

where a i JK is an element is always an element in the field K okay so this this should come as no surprise um let's call it f fxy z is called homo over degree

n if um all its monomial um a i j k x to the i y to the j z to the K satisfy I + J + k equals n okay so homogeneous homogeneous polom of of some degree it means that um the

sum of the exponents of the X Y and Zed in each of the monomials is some some constant constant number n now if um maybe it's better to to take a capital letter for this so if f is a a homogeneous of degree n

what happens when I plug in instead of X Y and Z I plug in Lambda X Lambda y Lambda z i I simply get I mean Lambda to the n f of XY Z so x y and z is uh a zero I guess we could say fxy Z is zero if and only if

F of Lambda X Lambda y Lambda Z is zer and this means that the set of zeros of a homogeneous polinomial makes sense also in the projective in the projective plane um so so F so let's say the solution set of f is well defined

in um and P K2 now how do I I mean if FX Y is any polom in two variables then I can Define f XY Z to be the polom Z to the n f of x over

z y over Z this is a homogeneous polom of degree n i mean and of course here I need to have that H the the degree of the original polom was n to begin with so if you give me a polom of degree

ordinary polinomial in two variables of degree n then I can I can cook a canonical homogene homogeneous polinomial in three variables um of degree n and of course ER when I plug in into this new polinomial capital F I plug in FX y1 I get simply f of x y

so for example I take this polom that we care a lot about x x Cub + a x + b - Y 2 and I want to make the the homogenization of this polinomial this is a polinomial of degree three in two variables and to make the hom homogenization I need more space what I do is I simply multiply by

Zed to a certain power all the terms that are missing missing Powers so X Cube remains as is then I have a x z s then I have b z cub and then y y Cube sorry y² time Z okay and and of course f of x y1 is is going to be simply the the original polinomial F

fxy okay it's so so far is it is it clear all right yeah yeah Thomas sorry I just wanted to say it's clear okay um all right so so now what is the projective version of a of a line I take I take a line I mean example if

L is a let's say y = mx + a line in the Aline space A2 um namely it's the solution set of a polom in two variables um say Yus MX minus a then I take the homogenization um and I get a y = mx Plus plus a

z is a line in the projective space P2 right this is the the homogenization of er of the previous line and um the points the points of let's call it maybe l l Prime the points of L are or can be identified

identified is the points on L Prime which are which are aine remember aine point in the projective space H are simply the points of the form xy1 and indeed when I when I substitute X y1 in to the the equation of L Prime then I

get I I simply get the equation of L okay and now suppose we have two lines L1 is a y equal MX plus h maybe B1 and L2 is y equals a MX plus B2 and these are parallel lines right now I take the homogenization of

these parallel lines so in P2 I get L1 Prime which is a y = mx + B1 Z and L2 y = mx + B2 Z and points ER in the intersection of L1

Prime and L2 Prime are simply the solution set a our Solutions um H to to the following equation I have Yus mxus B1 zus y - MX - B2 Z equals B2 minus B1

Z and uh if Z equal zero then indeed I have a solution then h y = mx um and so so er X MX zero is is a solution to this point to this uh to this equation of course this

solution I can I can refactor it as as simply one m zero okay and that's that's basically uh the only solution so in the projective in the projective plane two parallel lines intersect exactly at one point which is a point a non aine point right a point with a zero coordinate at a a zero Z coordinate is a pointed Infinity yes yeah but can you like is there in the in the notes that you have some uh

intuition on how two parallel lines can be can have an intersection how can can they intersect if they are parallel uh no I mean this they are parallel lines in the Aline parallel lines in oh okay now I understand okay okay okay fair enough that's that's the it's it's it gives you another kind of angle of what is what is this projective space or plane okay so in the projective plane we have we have a line at Infinity um unlike in the projective uh in the

projective line we have a point at Infinity in the projective plane we have a line at Infinity every point of this line is called a point at infinity and it turns out that if you take any any two parallel lines in the aine space in aine plane then the intersect ER uh are now you have you have a question no no was just for for what was mentioning like that I don't know if that's correct but like intuitively I imagine like in the real world you can't see like the the two parallel lines of the train uh line for example and and

and if you squeeze your eyes you can see how they meet at the horizon line kind of uh so I I I don't know if this is correct but to me to imagine so I'm I'm not going to I'm not going to argue with anyone's intuition I mean this this can be this can vary but the the formal aspect of it is is explained here okay um and what matters to us is this I mean the projective version of the the the polinomial of the elliptic curve right so we now

look at a a yes yes sorry just one more question uh according to this which indeed kind of makes sense with what I kind of was thinking a point which has for coordinate z z is a point that we cannot get back into a find coordinates that it's not possible to be represented in a coordinates so does this kind of uh get back to the fact that two parallel lines uh might have an inter well can have an intersection in the projective uh space but this projection only exist in the projective space and

the proof is that we cannot get back this uh this line into the aine one okay yeah okay this I mean of course the I mean if there are two parallel lines they do not they simply do not intersect in the in the aine space okay okay this this you can I mean you can easily write the equations and see that that there is no solution yeah okay so now in the remaining Thomas yes sorry I have one question so this homogeneization we used to get this into the projective space right where did that come from I mean it comes from the the fact that I want to

think of A2 as embedded in in P in P2 P2 is like is like A3 minus something modul something right and I want to take XY here to xy1 this of course this is a it's a bit of a convention but I choose a convention once and for all and once I choose this then I have a a a canonical way to turn a polinomial say fxy into a into a hom homogeneous

polinomial okay so it actually it actually comes from this H this convention okay thanks all right um so now I look at H the the polinomial we care most about and this is um y sare maybe X Cub + a x + b - y 2 = 0

and I look at the the homogenization of this polom um in um in P2 so this is in the Aline space A2 in P2 what I get is um maybe we call it e Prime it's till X Cub + a x z s + b z Cub minus y² z = z and um so of course

points original points K rational points in E they correspond to the Aline points on E on they correspond to X Y one I mean any point x XY that lies on the original curve H it's Alpha inversion in the in the projective plane would would lie on the projectivization of the of the curve but we have we have something new um so

um what are the Nona points of a e Prime in of course in the projective plane so to do this I set Z equal to zero and I get um x x cube is equal to Zer which means that X is equal to

zero and of course 0 0 0 is not is not a legitimate point in the in the um a projective plane so 0 y zero is a a a solution is a point it's in fact it's the only point I mean of course z y z is simply 0 1 0 is the only

point and none fine point on E Prime okay and this this is the ER the socalled a pointed Infinity ER and now you could go and and Define a tangent lines at so this is this e Prime is called a projective curve it's like a and of course a a projective curve of

degree three is called an elliptic curve if the discriminant so e Prime is called projective elliptic curve if the discriminant is non zero okay and you could go go on and Define what is a tangent line and H you will have an analogous result saying that um a projective curve of this of

this form e Prime is an elliptic curve if and only if it is a it is smooth or admits a tangent line at every point and once you did this you can Define the ER the binary operation in terms of the ER of the projective I mean of all the the the projective solutions for for E Prime and and now there is this this big elephant in the room that we didn't talk about and this is the associativity of

the of the binary operation we we defined so this is the this is the involved I mean uh the elementary proof for many years H was unfeasible ER for humans to to track I mean very few mathematicians did it and in modern times people just plugged it into a computer and verified ER the non-elementary there was non-elementary proofs that this is a an associative um operation but they

involve a a theory that we we will not reach I mean this is a so-call reman ro theorem and and then ER I I was looking around for an elementary and feasible for human proof of of associativity and I found the paper ER from 2021 by some Japanese guy um and this appears in the appendix ER the the proof is is still it's like 10 pages or maybe eight or something like that but it is all done

in elementary ER terms and in projective coordinates H so now that you know projective coordinates H you can if you if you like H you can look at the appendix and see ER that uh you actually get an associative uh associative operation and yeah please share the paper it's it's in the appendix in the notes I mean the uh I can tell you there is only one appendix at the moment yeah just just look over there I

mean I I don't remember the the name of the paper and the author and okay so so basically H this this is a point where we skip we might if you if you really ER want we H might later on go over some parts of the proof maybe some of the cases just to give you to give you um a sense of how how to to prove it but to go to the full proof I think is not is not very not very useful and so what we get is

that for an elliptic curve over a field K the the set of K rational points is in a billion group yes yeah I'm sorry just because we have I just wanted make a question related to Thomas last question about the uh homogeneization I still don't see why where do we use the fact that all the monomals I mean the sum of the pors of the monomals are all equal to n like to me like why why do use this fact to

transform to prity SP is not something that was clear to me well because this actually I think the answer is very easy I mean I'm interested in a pols only because I want to look at the the the solution set of these polinomial and a a a solution set of a non-homogeneous polinomial does not make sense in the in the projective space because I want to say that if um if x y z is a solution of a polinomial f then actually if and only

if Lambda X Lambda y Lambda Z is also a solution of the polinomial because I know that in the projective plane XY sorry X colog or or maybe yeah yes XYZ is equivalent to Lambda X Y and Z so to to to say that a a I mean to to be able to make sense of a talking about zeros of pols I must have a homogeneous

polinomial is this is this clear not really I don't see why do we to find Ro of let me let me give you an example I mean let me give you an example I mean I take this polom F fxyz is a x + 3 okay that's a legitimate or maybe x + one okay it's a legitimate polinomial over any field but it's not a homogeneous

polinomial so what are the zeros of this polom I mean it's all the points so the solution set of F in the Aline in the Aline space is is simply the collection of all XY Z such that um I mean x x is equal to minus one so it's simply minus1 comma X comma y

but now I I look I look in in P2 and I find that a a minus one X why it must be a a a solution for this polinomial right but a now I risk SC it I multiply let's say I multiply everything by one by minus one this is this point is is simply equal to 1 comma - x comma minus y and this point is not is not in the

solution set I mean if I substitute it in the polom it's not so I I have I have like a Paradox I have two points that are equal but if I substitute naively their coordinates into the polom one one time I get that it's a it's it's zero and in another time I get it's non zero okay so I the the point of homogeneous polinomial is it's the only the only sensible way to talk about solution set of a pols in the project plane because I I I have to have so I

mean I have to have a polinomial with the property that if X Y and Z is a zero then I must then then also Lambda XY Lambda Lambda y Lambda Z is zero I how does this solve the problem in this specific example of X+ One X+ one is not a homogeneous polinomial because it has a monomial of degree one and a monomial of degree zero yeah sure if we make it homogeneous make it homogeneous then I multiply it by

Z and and now it's okay right now the solutions I mean solution set of um polinomial X Plus Z this is well defined I mean let's call it g if this is a homogeneous polinomial so so if X Y and Z is is a zero of this namely if x is equal to minus to minus Z then also G of Lambda X Lambda y Lambda Z

okay okay so that's the point it's like because because of this equivalence relation you cannot talk about zeros of pols that are not homogeneous makes sense okay thanks okay um let's stop here unless you have a you have other questions or I I have one more uh I understood homogeneous polinomial as Pol s whose each term has the same degree each nonm each monomial yes each monomial okay so okay so it's this criteria that we're apply for for yeah

okay all right um have a have a great weekend and see you next time thank you very much byebye thank you byebye this