PSE Lectures Ep 30 - Divisor of g_m-g_n, continued

Transcript

okay so what we did last time is starting to to analyze the divisor of um of GM minus GN let me just remind you um um so we we denote um NP to be some pair of functions

GN GMP GMP G [Music] sorry JP a J&P and and we want um we want to analyze GN and eventually hn um in order to understand n torsion points so um we had a um we had a um a theorem I mean we started to talk about the theorem that says that

um for M greater than n with m m n m + n and M minus n Prime to the characteristic of the field um any elliptic curve e over

K and satisfies the divisor of GM minus GN is the divisor generated by M plus n torsion points namely you take each M plus n torsion point and uh consider the divisor in which this this the sum of all these points each of

them with coefficient one you do the same for the M minus n torsion points and you substract the dev are generated by em well twice the divort and twice the divisor generated by n okay and this is a this is a big Theory I mean basically almost everything we want to know

about torsion points would follow from that and in order to prove the the the there um um several several cases what what is going to happen is you see um one way to one way to prove this theorem is that we can say um I mean I I want to know what is the degree of GM minus GN or rather the not the

degree the um the order of GM minus GN in each of the points if I show that for any point P the order of GM minus GN at p is the same as the the sum of the relevant coefficients then then I've shown um I guess I can I can do write some

so it's enough to show that for any P the the order of GM minus GN at p is the the sum of degrees in uh in the right hand side okay and um in order to do this we

we we will divide um the set of all points on the curve into uh three cases and we show that in each of the cases this this is what happened so several cases um case one we basically did case one is is the um um Point e uh in the intersection of em and

N here what we did um is showing we showed that um of course the the the point at Infinity is one such belongs belongs to this intersection and we showed that um at o GM minus GN has order minus

2 this was I mean essentially it's because um we know that um both GM and GN have um HS of Multiplicity two at o and they have different leading coefficients

and different leing coefficients ER namely one over m² and 1/ n² um and so so er so the PS don't cancel each other

now we had a LMA that says that we can um we can know um so if we if we take two points on the Curve and and a uniformize at one of these points say then if I look at the

translation and TQ from E to e rather from rational points sorry rational functions on E IR rational functions on E which which takes um a function R and sends it to um r of um I can write it like this but maybe it's

better um take Q of R and TQ of R takes a point p and sends it to R of p+ Q and and we so we have this um translation by Q within the argument of the the rational function and we showed that um TQ of U I mean U is a

uniformer and if I apply TQ of U then TQ on you then this is um a uniformis at um P minus Q Now using this LMA and the fact that we know that GM minus GN is um has order minus two at o i can conclude that it has order minus two at all points in the

intersection and this we also talked about it but it's not hard to see um basically um if p is an end toward and point then H TP of g n is is simply GN because I mean you take a TP of

GN well it's it's because n * p + Q for any Q is going to be simply n q because p is p is an N torsion point um so so what what I'm saying is let's see um yes so

so if I want if I want to calculate the order um of GN at P then I might as well ER so I mean I want to express it in terms of the the order of GN at o but this is not so this this I know to be minus two and I claim that by the LMA we have

this equality um with the question mark above it and the reason is if you take some uniformize so GN is equal to um U Theus 2times Time s or S um is finite and non zero at O then I can simply apply a TP on U and this I know to be a uniformer at um at P minus p minus o is

p um hence I can write GN is in TP u - 2 times um TPS s and um TPS evaluated at p is simply s at um should be s o yes simply s o which we

know to be different than than zero and infinity H so I get that I I can express GN also as a um a uniformer to the power of minus 2 at P so we get we get this orange equality um so that that's that's something we talked about I'm just uh sketching um what we did but the conclusion is that um GM minus GN has order minus two at all of these uh green

points any any question about this part thanks a lot for the overview it has been very helpful okay so now we go to the the other case in this case um is points that are in EM but not

in now here it's also um it's also quite easy um these are not in either e m+ n nor in e n minus n h to see this it's very easy like um if p is is not I mean p is in em and not in N it means that NP is different than

zero and MP is zero so obviously m+ NP is different than zero so it's not in M plus n but also M minus NP is different than zero okay so so we have this is um this is this is an easy case because

um we we can pin exactly where where the point is it's it's basically only in em and not in the the other components of the divisor um now I claim okay so what we need to show that um GM minus GN has order minus 2 at and this minus two again comes from um

just looking at the divisor right we um take the blue D Resort and in this blue divisor a p p appears only in in em and it does not appear in any of the other components right so the coefficient of p is minus two in the divisor hence we need to show that

the left hand side namely the divisor of GM minus GN has um I mean we need to show that GN minus GN has order minus two at at such Point spe now we saw okay so we saw

that GM has order minus two at at any point in in EM this was a a basic LMA um and sorry I um I need to take a look at something outside e

hey sorry there was a some drama outside um okay so we want to show that GN minus GN has order minus two at at these points p and we already so we know that GM has order minus two so this means that I can write GM as some uniformer time I mean to the minus two times s with s um at P not zero and not infinite but

now I can I can do like simple manipulation for GM minus GN I write GM as I know u - 2 * s and GN I write as um u - 2 * U the 2 * GN and now I take a a uus 2 Factor common factor and I get S minus U2

GN so in order to show that GN minus GN has order minus 2 at P it's enough to show that this guy does not vanish and is not equal to Infinity at P but if we check what it is in s minus u² GN at p is s at P minus um basically minus Z because u u

vanishes at p and GN h n of p is not zero so it's I can write it as a multiplication u sare p times GN of p and the reason I can I can write this multiplication is because I know that this is not zero I mean we assume that P is not an end tent point and I know that this is

zero we had a LMA that if you you want to evaluate a a product of rational functions at a point and you know that both values of the the rational function the point are are finite then you can just ER you could just take the product of the evaluations so all in all this is this is simply s of P which we know is different than zero and infinity yes why do we want to be sure that H it can't be zero multiplied by

zero 0 multipli by Z is okay but we don't want GN P to be Infinity ah right right zero I mean this is the Gen like a general rule of thumb in analysis and other places ER we can do finite non zero times infinity this usually makes sense infinity times infinity also usually make sense but the the hard part is I mean what tends to be chaotic is zero times infinity so this is like a just a way to

remember it but we of course we proved we proved the claim that if both values are finite then then you can do the product but then we should uh write that the second term is not Infinity not that it's not zero yes you're right yes than okay um yes sorry I'm not sure well I'm not sure I mean it's a fact that I'm not following [Music] um what why the the g m minus n has

order minus two at P so sorry so so you say that that the so we you are showing this um by U um like based on the fact that GM has there minus two yes so this this is something we proved in a previous ER okay so okay the the fact about GM I mean it was like uh let me just look for it because I was not finding it but but if we already saw this um I will look for it later yeah it's

essentially um don't worry I mean I can look for it later so we had let me just just remind you where to look for it it's it's it's a bit I mean we we proved something slightly slightly different uh we showed that GN over X at O Okay is is n Square I think one divided by yeah one divided by n s okay yes this is but what this

means um this means that GN at all is like X over n² and X has degree um degree two right I mean as a VAR yes so so it means that when you substitute o i mean o is like Infinity comma infinity and you substitute it instead of the X X has like X is like Infinity Square I see okay so so it means that GN

has order um multiplicity to at at all and and the same the same goes we had another Lama for GM sorry for hn yeah right we had like H and H and one y I think it was 1/ m² so this implies at o right at O So this implies that um that the order of of hn at o is three okay because now y has degree

three okay so yeah so basic I mean this is of course in O but using the LMA the translation LMA this is the what I mentioned earlier now that using the translation LMA we can we can deduce that the the order of GN at at any other relevant point in in for example is also going to be minus two nice okay make a lot of sense now thanks okay so going back to the proof so this basically finishes the um the

case we we considered I mean we wanted to show that GM minus GN has order minus two at at all these points that are in em but not in and that is fine we we are able to express GM minus GN as um as U to the minus 2 for uniformer at P where the times some some function that doesn't vanish and and is not Infinity okay yes okay so the third case is the the

interesting and third case [Music] um these These Are points well actually is like a a case 2 Prime these are points um in EN n but not in

em and this is a a symmetric um to to case number two I mean we don't care that I mean we don't care that m is bigger than n or n i me it doesn't it doesn't matter for the analysis we did in in case number two and then we have case three which are a points um

not in N say m and not in n and this case has H three subcases and the first um p

e e n minus n and not in n+ N um for this then we have we have the following equation we know that MP is equal to n n p um but NP is not equal to minus NP right can you can you see this from

a basically we know that M minus n p is equal to O and we know that m plus NP is not equal to so we we conclude that m is equal MP is equal to NP and NP is not equal to minus NP and since we also know that P is not in EM Union

n um we know that MP is not is not zero is not o and NP is okay so um one thing to notice is that GM minus

GN at this point p is is always going to be equal to zero H how how do you see this I mean it's basically because we know that um yeah I mean we know that MP is equal to NP and so we want to determine the the multiplicity of this this zero um we also

know that H NP is equal to H MP sorry is um yeah okay we know that h&p is equal to HP um because of this the green the green equality remember h n Agent P is simply the the y coordinate of um of the point n n * P

um and now we had a Lemma about derivation we calculated the derivation of GM minus GN in terms of of HL this was a two 2 * m- n time this was 2 m

hm minus 2 N hn but we know we know that hn and hm are equal at P so when we evaluate this uh derivation equation at P we get um 2 m minus n Times age U you choose one let's say h

m even just take a the the two so now we know that M minus n is prime to P Prime to the the characteristic of the of the field so um and uh and

because MP is not equal to MP sorry to minus MP P cannot be of order two but if p is not of order two then it's it's y coordinate is different than zero um sorry M MP MP cannot be to and this means that hm

p is different than zero but um okay so this really means this really means that the the derivation of GM minus GN at p is is non zero so we

conclude what we conclude we conclude that GM minus GN has a zero of order one also called Simple zero ATP okay remember that this H this derivation that we defined on rational functions um it has the um the

properties that you you know from a derivations of pols namely that if you have a a root of a polom such that the derivation of the polinomial does not vanish at the root then this root is of order one and if if derivation um vanishes n times then and not n plus one times then this this root is of order n okay so this ER this proves the the first sube any questions about

that now do you do you follow I feel like I follow uh I mean so far uh things make sense but but let's let's pin it completely it's not hard I I can go over it again it's uh maybe it's it's easier I mean I I think it makes sense I mean everything like matches so but but I mean yeah Lisa what do you think H I always for going uh over again please go again okay so we're saying we're um

in case number three these are points in em but not in sorry points that are neither in em or in and uh we take the first subcase which is points that are in e m minus n but not in E M plus n for these points we have the ER the green equations and the green equations translate to the the other green equations with the double n um and we also know of course that um that P does not belong to em Union this

is a this is because we're like in case number three so this this gives the two the two other green equations or inequalities now it's easy to check the GM minus GN evaluated at p is zero basically by the just by the assumptions um and also that hn of p is equal to hm of P this is because of this uh this specific uh specific equation if if MP is equal to

NP then the y coordinate has to has to be equal so far you okay L sir yes okay and now I'm saying we had a so I we clearly have a we know that P is a zero of GM minus GM and we want to to compute the multiplicity of this zero um so it will be enough to show I mean I claim that the multiplicity is one here and and to prove this claim it's enough to show that the derivation of GM

minus GN at p is non zero you remember we had that um uh we had a theorem that says that the the the order of the derivation I mean if okay basically the order of the derivation of r at p is a um the order of r at P minus one right this was like a one of the things we proved about derivations yes but

how uh we will know that the multiplicity is one if we prove that the derivation is not zero because zero it might be like three no I mean the thing is like a a function that doesn't vanish and and doesn't I mean this derivation of GM minus GN is just another rational function right which is a which is always a um which is always finate in fact right um so if we prove that it's non

zero then its order is is zero right because I mean a function that doesn't then if we prove that it's non zero at p and we also know that it's not Infinity at P then it it doesn't have a zero or or a pole at P so the order of this function at p is zero but if the order of this function at p is zero if this is zero then I know got that order of of r at p is one okay okay so so it's enough to um to show that the derivation of of GM minus

G n at p is non zero but that's very easy because I do I I use this calculation about derivations which expresses the derivation of GN minus GN in terms of hn hn and hn but in our case hm is equal to hn so when I evaluate p i get this expression and this expression I know does not is not zero and also not Infinity H at P because M minus n is assumed to be prime to p and hm hm of p is assumed to be non

zero this is this is because m is not a point of order two the reason m is not a point of order two is because MP is not equal to n to minus MP this is one of the assumptions okay so all in all um we know that the derivation of GM minus GN at p is non zero a not Infinity so uh so we get the it has a

simple zero there okay sounds good uh Nicholas are you following yes sorry I was onone yeah um I'm following I think on the high level yeah the details is get some but the last explanation but you already was clear thanks okay great yeah that's that's the main issue I think the the the heart of the matter in this subcase and also it's going to be in the the the other two subcases is we use this H this

properties of derivations in order to calculate the order of a rational function at a point this comes in handy it's kind of a useful so let's take a break and uh we come back to the other two subes

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okay um um let's go back to it and what are the other two subcases um the second second subcase is is that a p is in e m + n but not in e m minus n and and this is symmetric to a case number

one some case H because if you if you go over the argument in CA in subcase number one you see that we didn't use anywhere um the well it's it's it's just symmetric I mean you can you can do the derivation of GM minus GN at thisp and you you get the so we are left with the the third the third

subcase which means that um um p is not in is not in a in m + n wait sorry p is in the intersection e n + n and e n

minus and remember we also assume that P is not in the Union of M and so what what equations uh can we conclude conclude from the these assumptions um we have MP is not equal to sorry is equal to minus

NP this is because p is in in em or e e of M minus n we have MP is equal to NP um and we know also that um MP is not o and NP is not o

um and and from this we conclude that 2 MP is equal to m which is equal to right I just uh I can add these two equations um I guess I need to transfer one of them this if I edit like this oh yeah um yeah I I can uh either Ed these two equations as they are and

then I get 2 MP equal to o or I could um I could swap sides or substract one equation from the other and then I get the 2 N isal ER now we we know from from a the previous case um that if I take the derivation of GM minus

GM at P well here it's going to be zero the reason is I mean I can express it as is let's see 2 m hn minus 2 N hn at P but I know that um 2 MP is equal to 2 NP so they have the same y-coordinate and and therefore this this

expression vanishes so far it's okay can you say again like this last thing like it banishes because uh to MP okay yeah okay sorry I see I mean sorry sorry yeah I mean I'm just saying like 2 MP as as points on the curve 2 MP is equal to 2 NP and if if you multiply by hm it's just it just means you take the

y-coordinate if the points are equal they have the same y-coordinate yeah totally thank you um and Okay so so we know I mean okay we know that for these for the points p in in this third subcase um GM minus GN at p is equal to zero right because um MP is equal to NP so they have also in particular the same H the

same x coordinate um and so so we know that P is is a zero of GM minus GN and it is of order bigger than one because the derivation does not vanish and I want to show that um it has in fact order to um I mean we know that the multip I mean the the multiplicity is at least two because the the derivation also vanish H in order to really show that

it's that the multiplicity is exactly two I would need to take a look at the the derivation of the derivation derivation of derivation of GM minus GN um but that's okay because we have a formula for the derivation of GM minus GN this is this 2 m this is no D of 2 m

hmus 2 n hn and I want to evaluate this at P now we know this this is another calculation we did when we talked about derivations is that if I do the derivation of hn I get n * um 3 G n² +

a and obviously the same goes to M right hm is m g so now when I want to calculate the the double derivation we need to do the derivation of this uh 2 m hm minus 2 N hn but derivation of of a difference is the difference of derivations um so so I get that

[Music] um this is now um 2 m² 3 g m + a minus 2 n² G n² so three 3 G n²

+ okay so now I I I substitute a I substitute p and this means I know that 2 NP is equal to zero oh it means that um that NP is a point of order two which means that H it's um

it's of the form Omega comma 0 for a Omega a root of [Music] um X um [Music] okay so so in particular sorry NP is is a point of

order to it means that NP is is Omega comma Zer for some some omega and now when I substitute NP into this H into this equation I get the the derivation of the derivation of GM minus GN evaluated at p is simply I mean here I have um GM squar and when I do it to P it's like it's doing taking the the x coordinate of U of MP and raising it

the^ two so so I simply get 2 m² 3 omega squ + a um - 2 n² Omega 2 which is the same as saying 2 a 2 N minus n n + n right this this last

part is simply m m s s okay and now this is non zero because um we know we assume that M plus n and M minus n are prime to P this is a so in order to show that the double

derivation of GM minus GN at p is non zero it's enough to um to say why this why this factor is nonzero any idea why why this Factor should be non zero because Omega is non zero Omega is non zero but maybe 3 omega square is equal to minus a but it can't be equal to something with sign minus because Omega Square will be positive and multipli by three

will be positive yes but a a is just the general coefficient it could be minus be three 3 omega Square can't be with s minus okay but then then maybe a is I mean the 3 omega square is a is positive let's say again you need to think of Omega in in a finite field so it's not exactly negative or positive right but even if it's just a real numbers okay three 3 omega Square would be always positive

but it could be that a is is is negative a is just a general coefficient of the the curve right you have um y² = X sorry um right here and it's X CU x + a x b right this this is a a General coefficients but there is something else this we we talked about it so Omega where did I write it I should be written X

Cube Omega is a root of SX right but it we know that it cannot be a double root do do you remember why no I don't remember so elliptic curve generally does not have multiple roots of this polinomial ah right right right right okay so this means that um if I take the derivation of of S Nam 3

x² + a and I substitute Omega I get something that is non zero this is equivalent to say that Omega if I know that Omega is a root if I want to say that it it is not a double root or or of any other Multiplicity then it's enough to say that the derivation of the polom at Omega is is non zero it's equivalent in fact

so okay so I know the 3 omega sare + a is is non zero it's basically because Omega is not a double loot or because Omega is is a simple loot s he this by the way to say that Omega is a simple root of s it was equivalent to say that e is a is

smooth this was kind of the the idea behind this uh this property okay so all in all I get that the double derivation of GM minus GN at p is non zero and this means that the order order P GN minus

GN is exactly P well um yes it's it's exactly P okay and this this fits our case I mean remember our subase and where we yeah our subcase is um p is in E M plus n and also in e m minus n but not in the the other two not in e m and not in so when I take the the coefficients in the the divisor um er in in the main

the main statement of the theorem I get 1 + one which is two so I I showed that um that if I take the order of GM minus GN I also get two and this this basically compc um I'm curious should we somehow take care of infinity or is just somewhere outside of all cases and we don't care about it no we we do care about the point at Infinity this let me

remind you oh we just uh never mind we just have Point pointed Infinity we don't have separately zero and infinity yeah yeah yeah I mean this was the first case points which are both in em and in N they cannot be in M plus n and they cannot be in e m minus n um and O is is one of these points and we showed that the order of of GM minus GN at all is minus two yes yes right okay

uh now how do you feel about the the proof so the this last part was was cool thank you the thing that I'm not I mean I'm I'm I'm just going going away over the notes but so we have the case one concludes that the order is minus two case two concludes that the order is is minus 2 but then case three concludes that the order is two is this last sentence correct um case three has um three subcases ah right I was

mixing yeah so I mean in case three we take points that are not in em and not in so in this in this divisor in blue yes yes yes yes yes yeah in this divisor in blue we do we take points that are um here maybe here yes it's so it's one and and the subcase two is also yeah order one and then the last okay they could they could appear either in both of E M plus n and E M minus n or just in

one of them but they definitely do not appear here this is case this is case three yes and you see you divide it into into subcases in the subcase in which the points are in E M plus n and in e m minus n this is actually the third subcase then the coefficient be you know 1 + one so two we need to show that the order of GM minus GN at these points is is two but if if the point belongs to E M plus n but doesn't belong to e m minus

n then we need to show that the order of GM minus GN is one yeah okay and if it if it's in a and this is what we showed and if it's in e m minus n but not in EM plus n it's it's symmetry yeah okay all the three subes thank you very much okay so Nicholas do you have a questions about the proof okay um so

cor I now I take the ER degrees yes huh okay um I now take the degrees so I I know remember I have divisor of GN minus GN is equal to the divisor generated by M plus n plus the divisor generated by m minus n

minus to the divisor generated by e n minus two the divisor generated by n so taking degrees let D DK be the number of points in N taking the degrees of um this

equation we get zero I mean the degree of a of a principal divisor is is always zero and what is the degree of the the divisor on the the right hand side you simply this is d m+ n i mean this the degree of this divisor is just you sum M plus n i mean sorry you sum ER once um the number of of points in in M plus 10 so this is d m + n plus d nus

n- 2 D M minus 2 D okay this this is an important formula and now we um okay so we can we can already see if I take d k to B K squ and

well this satisfies um the formula start why I mean um I want to show that m + n² + m - n² - 2 m² - 2 n² is equal to zero but that's that's indeed the case right here this is like m²

MN m² and here it's m² minus 2 and plus s and uh this cancels with this and and these guys cancel this guys okay so this so a DK equals k s satisfy the recursive formula and what I claim is that we can show that that's the only the only solution for the recursing for

so this is this is kind of another h another piece of theory that we're are going to use in how to solve recursive relations namely linear recursive relations yes I'm not sure I understand why do you call it recursive relations well I mean ER we we know what is D1 D1 is one right this has to be D1 is the number of points that that are in E1 there's only one such Point namely o yes

D2 is four ah you mean that okay okay and then you move you could you could basic hypothetically you could calculate all the the other d right you have a recursion now here I want to simplify this is a recursion relation with like two [Music] um er two indices this m and n and I want to simplify it so I can take I can take n equals one and I get um

d n +1 plus D nus1 minus 2 d n minus 2 right because D1 is two D1 is one and and now this is maybe more familiar as a recursing ReliOn right right so it's enough to show that um I only have one solution to the the second

um let's call it double St or actually the notes I call it R RM this is some relation between the various this and uh enough to show that DK equals k s is the only solution for now the issue here is I don't want

to get into the whole theory of solving linear recursive relations it's not hard and in the notes you have a reference for for a book that does the complete treatment but basically we in the theory there you distinguish between two cases one in which it's called a homogeneous homogeneous linear recurrence relation it means that you don't have a free um a free coefficient and the other one is is the in in homogeneous case in which you have a

free coefficient but the the homogeneous case is easier to solve and so usually you you you reduce the the non-homogeneous case to the the homogeneous case and the way the way you do it is is as follows um so let's call it a a a AK is a solution and

if um BK and is also a solution then um k s minus BK as a sequence is a solution for the homogeneous relation I call it

rhn which is and a 0 equals d n + 1+ d nus1 - 2 DM and without the the minus two right can you all see this I mean if I have I have two solutions for the the the non-homogeneous uh recurrence relation I substract the two solutions and since both of them are

solutions to these these equations when I substract these two equations the the free coefficient Falls is this clear should I uh I mean just imagine um d n + 1 + d nus1 - 2 DN minus 2 and also for another guide e Prime D Prime m+ 1 + DN -1 f - 2

d nus 2 and both of these are equal to zero then I substract the equations and I get that d m +1 minus D Prime n +1 plus D nus1 minus D Prime nus1 minus M to D M minus D Prim n is equal to zero okay so so the difference

so the difference D DM minus D Prime m is a solution to the the homogeneous linear relation I mean the the one that I that I obtain when I erase the free coefficient is this clear this is clear but I don't understand how it relates to the sequence why we don't take just two solutions but we want to take a sequence well because I I want to show that eventually I want to show that DK equals k square is the unique solution

to RM okay it's it's about so I I really need to consider sequences as Solutions there I mean possibly there could be other sequences I know that one sequence DK is satisfies this relation I I know that this relation has to hold right whatever yeah and and I'm saying okay so H how do I prove uniqueness I will I will eventually I will prove uniqueness of solution to the the homogeneous

relation the homogeneous is is easier to to handle basically you will soon see why um so okay so so I'm saying let's suppose that I have two solutions to the okay H thank you Ru and sorry if it's hard to catch up yeah no is thank you by um okay so so I'm saying let's suppose that we have two solutions for the the non-homogeneous relation one solution is

is definitely K squ the other solution I just call it BK but I have to to consider the sequence you see Lisa yes but why want to consider a sequence from one to K why you want to consider it to Infinity because this relation has to hold for any M including Infinity no I mean when when when I write infinity it just means you have BK for every k for every natural number K you have BK okay okay makes sense okay are now do you follow do you this

is kind of a new new area yeah completely new for me like this these concepts of like sequences and and the thing that you are going it's it's fully new so I'm I'm not fully following Okay so let me it's very easy I mean when I write a b BK like this I just mean B1 B2 ER B3 and so on up to BK and then you continue it's like an infinite it's like an infinite sequence okay it could be I mean the the

sequence a a k AAL 1 to Infinity this is equal to 1 H 4 9 and so on right this is one particular sequence that we know that satisfies the the the non-homogeneous okay relation right thanks thanks for the example with numbers okay and and and now I I so I'm saying if if I had another sequence B BK this is

BK then I know that the difference of the sequences namely K Square it's the the difference of the sequences is simply the sequence in which you take the difference in each uh in each coordinate right so this this is like a a minus DK H the difference of the sequence is is another sequence and this other sequence would have to satisfy the the homogeneous sorry the the nonhomogeneous relation in

sorry it would have to satisfy the the homogeneous relation right this is what I did here um here in this part I showed that if you take a sequence let's say DM that satisfied the the non-homogeneous relation and you take another sequence the D Prime M that also satisfy the non-homogeneous relation and you subtract the two sequences then the the resulting sequence satisfy the homogeneous

relation is this clear or no yes I think yeah thanks I mean what what I'm saying is so k I'm saying that k s minus BK whatever BK is satisfy the the relation r r hm okay and now I'm going to show that the relation RM rhm

ER I I actually know how to um how to solve it so um the the characteristic or you know what let me just because we're running out of time I'll State a Lama that we can prove next time um all solutions m r hm out of the

form um let's put it a k a is equal to Alpha 1 plus Alpha 2 okay and and let's just assume this Lemma it's it's very easy to uh to prove um if I assume this LMA then then I know that k s minus

BK is needs to be equal to Alpha 1 plus Alpha 2 K for all k for some alpha alpha 1 and Alpha 2 that I I I don't know but I don't need to know there are some some real numbers but I have two instances I know that for k equals 1 BK has to be one and for k equals 2 BK has to be four so I get that um 1 minus BK is equal to um Alpha 1 plus Al h no I I get that sorry that one minus -

one is equal to Alpha 1 + Alpha 2 and I also get that a 4 - 4 is equal to Alpha 1+ 2 Alpha but this this system of equations has only unique solution which is alpha 1al alpha 2al Z so I get that a k s minus BK is equal to zero

for all K IE BK is equal to K K is equal to k s okay so modu Delma we proved that um dkal k s is a unique solution okay is uh is this more or less clear I mean we will have to uh save the the proof of

dilemma and and and another small small issue for next time yeah it looks good okay so you see just take a look at the the variety of tools we we did here I mean we eventually we want to to we want to show that en n is is going to be isomorphic to ZN times ZN and all right we started here this was the beginning of the whole thing what we're we basically did now is showing that

n the number of point in N is is n squ in order to prove this part we did a theory of derivations using the derivations ER we we were able to calculate the divisor of GM minus GN and then we got a recurrence relation this is kind of linear algebra stuff we solve we we showed let's say modu dma that the recurrence relation has a unique solution and this is n s so we used at least a derivations and um and

recurrence relations and then for the last part this we do next time I'm simply going to use the um the classification of finite ailan groups so this is like a a very wide set of tools all all meant to prove this uh this result all right uh questions before we end up uh it's not a question but I met someone who told me that they were watching all the lectures that they currently published on YouTube and they were really sad that the lectur stopped

being uploading I thought that maybe as far as we are still recing them maybe there is some way to upload them if someone was upset about that yeah definitely it's a um I will I will do it it's just that I was waiting to collect some other recordings and then upload all of it together yeah it makes sense so yeah you can you can tell this person that it's going to be up sometime the next couple of month I think yeah that's what they said that I think they will be uploaded but just a bit later

yeah all right anything else then uh have have a nice weekend thank you very much have a nice weekend bye