PSE Lectures Ep 32 - Identifying the group structure of n-torsion points

Transcript

okay so today I'm going to definitely finish the the classification of torsion points which we we almost did last time we did the major part last time but there are still some loose ends to tie um let me just for yeah

so um let me I mentioned the LMA last time that I I didn't prove so so let me try to prove it quickly ER dilemma is about recurrence relations um it goes like this H let C1 C2 be re numbers um and um suppose we have a a polom which is going

to be characteristic polinomial of some recurrence relation and and C1 [Music] lambus C2 has M only one root uh well is a double root I guess no then um

sequence a is a solution to the currence relation um can do dn- C1 D nus1 minus C2 n 2 equal Z if and only

if it is it is of the form um a m equals um now we have like a linear combination and Alpha 1 RM plus Al 2 m r to the m well uh a double I should say the dou root iote is r and then the r is is appearing in the generic

solution um um where alpha 1 and Alpha 2 or some real number some constants so this LMA is saying I mean you can read it in in different order but suppose you you are given recurrence relation of this form it means that it's um it's a

recurrence relation of degree 2 because I have I mean a n depends on a n minus one or DN depends on DN minus one and on DN minus two and if you remember that's exactly the the relation that we had for the the number of torsion points um then what you need to do in order to to find all the solution all the possible solutions for the recurrence relation you look at the the characteristic polom of the relation H you look for a root assuming that this characteristic polinomial has

one root and in our case it's indeed what is happening um in the number of points of the I the recurrence relation coming from the number of endorion points point the characteristic polom F of Lambda is equal to Lambda minus one square so assuming that this is your setup you can classify all the solutions and all the solutions well one solution is is is looking like R to the m I mean you you

take the the root of the polinomial f as R and and R to the m is if you you substitute R to DM here you can easily see that this is a solution because this is R to the m minus1 r to the m - 2 and you cancel out R to the m minus two out of all the factors and you're left with a exactly the polinomial F for which you know R is a is a root for so so the the the

solution am equals R to the m is always a solution and another thing you can do is you can you can take m r to the m as a as a solution this would also if you check it in the in the orange recurrence relation this would also be a solution for for the relation and what the Lemma says is that any other possible solution has to be a

linear combination of these two so let's prove it yeah I'm sorry I missed the point one solution is R to the power of n and the second solu is M Mr R to the m oh right right right if you want a m Prime is m the m so actually that's part of the proof

is to check that Mr R to the m is a solution um uh let's see m r to the N is a solution um we have F of Lambda is equal to Lambda minus r square right because R is a double root so this

means that R is equal to C1 / 2 and the dis discriminant C1 s + 4 C2 is equal to zero right this is like how you solve a quadratic po omal equation um and this this implies in fact equivalent that now I substitute right so I I get a minus C1 - C1 * C1

/ 2 - 2 C2 which is equal to- C1 r - 2 C2 and this has to be equal to zero right this is just a combination of these two these two equations oh P Nicholas um so just to bring you up to speed I'm I'm proving this Lemma that um if you have a recurrence relation like in the

orange part here and the characteristic polom of the recurrence relation has a double root then you can classify all solutions to the recurrence relation um as as is written in the in the green box or in the green part here okay uh are you are you on board yeah I'll try to follow okay so I'm saying let me show I'm I'm showing that m r to the m is

indeed a solution to the the orange recurrence relation how do I see how do I see it I mean from the assumptions I know that R is equal to C I mean r is a double root which is a solution to a quadratic equation if it's a double root it means that the discriminant Delta H is is zero the discriminant is C1 s + 4 C2 so I know it's zero and then the solution is a just C1 plus minus the discriminant

divided by um by two um which is which is just C1 over the two so I know these two these two equations with the double lines and these two equations imply um what is written here um and now I multiply what is written here by R Sorry by Yeah by r m minus 2 and I get um 0 is equal to minus c 2 sorry minus C1 r

m [Music] -1 this is this and and minus 2 C2 r m - 2 m but I also have our H square is = to C1 R plus C2 and and so I I can substitute well I I can multiply this by

m um R to the nus 2 and what I get is a m r to the N equals C1 m r to the nus1 plus C2 n r to the nus 2 and and now what do I do I

am I have yeah I can I I combined the two the two equations star and D star and I get m m r the m equals um MC1 R to the nus1 plus mc2 R to the n and well actually I this I just

I just developed a double star into here it this I write in slightly nicer form and this is C1 M -1 R the nus1 plus C2 n [Music] -1 so nus 2 R to the N minus 2 okay so this is

a um and what we get is so so Mr R to the m um is solution I mean you see this is like a d m then I have um C1 d n minus1 and then I have C2 d n minus this is exactly the

relation in Orange is this clear this a bit technical but H it should be it should be not hard to follow I'm not sure I understood the step when you introduced M minus one and M minus 2 yes so you see I have yeah actually let me sh this it

yeah so it's because um it's because of I think it's because of this you see what I do here is I know that I mean if you if you take the the one out of here you get C1 sorry you get you get from here take the minus one out you get minus C1 R to the m minus one which is exactly this right if you if you take

M because I I I I turned this into into this where I I turned the green into the orange right yes um what's the difference between the green and the orange just it's minus C1 I mean minus C1 R to the m minus one it's like the the the Orange is has additional term C1 I mean C1 ER * -1 * R

nus1 and this is exactly the term here I mean I'm I'm saying orange minus green what is orange minus green is exactly C1 R to the m minus one yeah minus minus C1 yeah minus yeah minus C1 R to the N minus one now what is orange the other orange minus the other green this is a this is simply this this guy yes yes but the sum of these guys is

zero right okay so I'm allowed I'm allowed to to it's like I added zero to the the initial ration I mean it's like I I had this the green part here and I added two terms that sum to zero I added okay one term I added here and one term I added here but the sum of these terms is zero yes yes so I'm allowed to do it so the green is indeed equal to the orange thank you is this clear or

no I think so yes thank you okay so okay so so we have we have one solution we have actually two solutions right we said that's very easy to check that R to the m is indeed a solution to this recurrence for ation so so now um since we know that um the set of solutions ER to

um the relation let's call this relation r is a a a two-dimensional Vector space it is enough to show and I I have two a a a two dimensional Vector space and I have two vectors that live in this ve in in this Vector space um so it is enough to

show that um let's call it a m equal R to the m and a prime m equals m r to the m are linearly independent in the space of solutions right so remember the the vector space of solutions is simply a subset let's call the solution set um I

mean the space of all solution s then s is a subset of the set of all sequences of real numbers sequences of real numbers is like um you have some sequence of real numbers and they have they have no no constraints just H just being real numbers and within this and this is a vector space because you can add two sequences coord coordinate wise and you

can multiply a sequence by a scalar and then it's again a sequence and the solutions to the solution set to R is is a sub Vector space um I mean basically if it's a if you have because R is a homogeneous recurrence relation then if I have a a a Sol two solutions and I I add them two together then it's also going to be a solution because I would simply add the two equations and I would still get

zero and in fact if you you multiply a solution by escalar let's say you multiply R to the m by Alpha that's completely fine I mean you you just I mean it it would also satisfy the the relation R because in the equation you just can can cancel cancel out Alpha so if R if R to the m satisfy the the equation also Alpha R to the m and this is true for any solution so I'm saying the set of all solutions is a sub Vector space of um the the real numbers sorry the

sequences of real numbers and we showed that this is in fact the kernel of um you take the characteristic polom um and you you make a a transformation um out of it f of big Lambda so a big big Lambda that is applied on a sequence is just taking it into the H just shifting it by one this is a n+ one

and you can do F of Lambda this is just Lambda composed with Lambda minus C1 Lambda what is again minus C2 minus C2 * the identity on um I don't know it's called the sequence space called so the the identity of so we have

Lambda is a a linear map on the space of all sequences in of Lambda is also a a linear map on the space of all sequences and the solution set this is this is what we showed last time I mean I can repeat it if you want you can see that the solution set is the kernel of F lambra and it is we showed that um the kernel of f Lambda is in fact or the

solution set is in fact a isomorphic as a vector space to R to the two because what is the solution set it's it's a the collection of all sequences that satisfy the relation R but if you give me such a sequence it is completely determined by the two by the first two elements so I can I can construct a linear map from the space of all sequences that takes a a solution and sends it to the the first two elements A1 and A2 and this is a

bjective linear map because I can I mean I can also do the reverse if you give me two elements a I don't know B1 and B2 I can inductively construct a solution to the relation r i just uh B3 is going to be whatever um whatever is dictated by the by the relation R given that I know what is B1 and B2 so I'm saying um the kernel the solution set [Music] um of a of the relation R is a

two-dimensional Vector space okay actually you you don't even need um you don't even need this business of f of Lambda and all that it's enough to see that the solution set is completely is isomorphic as a vector space to um er R to the 2 by by this uh by this correspondence with this A1 A2 and B1 B2 so it is a the solution space is a two-dimensional Vector space and I have found two

vectors in this space namely the sequence R to the m and the sequence R to the m sorry m r to the m and if I show that these two vectors are linearly independent then I'm done then I mean in a two-dimensional Vector space all vectors can can be described as a linear combination of two linearly independent vectors is this clear Lisa now Nicholas yes yes um now you follow I mean this is a bit

like a a vector space theory that we didn't discuss in detail so I can try to fill in some gaps if you I think it makes sense okay so let's see why these two vectors are linearly independent um so suppose um Omega 1 and Omega 2 are two real

numbers such that Omega 1 time m r to the M plus plus Omega 2 R to the N is equal to the Zer sequence that is to say that for any integer M Omega 1 * m r n plus Omega 2 R to the N is equal to

Z but this is infinitely many equations the only solution for infinitely many equations is Omega one equals Omega 2 equals to Z this infinite set infinite system and equations is Omega 1 = Omega 2al Z okay and that's the that's the

definition of what it means to be linearly independent vectors it means if I can express a um the zero element in this case the zero sequence is a linear combination of these two vectors then this linear combination has to be a trivial IE with zero coefficients okay and this this completes the proof uh we know that we have two linearly independent vectors thus the span of these two

vectors m r to the M and M R to the m that we know these two vectors are a solution um and this has to be equal to the solution space of the relation R um IE every er um solution a m to R is of the form

a m equals alpha 1 R to the M plus Alpha 2 m r to the M for all m i mean so this is equivalent to say that for all m a m is Alpha 1 R to the M plus Alpha 2 m r to the m okay that's that's exactly the statement of

the um any question about the the Lemma uh yeah I have a question um we did a conclusion that this will be two dimensional Vector space because H it is isomorphic to H R2 yes right yes okay yeah I mean the isomorphism I didn't do the details but in One Direction you define you take a solution and you send it to the first first two

elements right you take a solution am and you send it to the first two elements of the of the sequence Lisa of sequence of real numbers right yeah this is all sequences of real numbers I mean we we even do a slightly more general setting than what we need because in our case of the the end torsion ER points we the number of points is always an integer but I'm saying even as a sequence of of real numbers you cannot um yeah so the LMA is even slightly more General it's a just easier this

way okay but how do you know that you can always send a first two elements uh to uh solution I mean give me two elements I mean it's like a think about the Fibonacci relation this this is a very good example I mean the Fibonacci relation is is a relation very similar to R the only thing is that when you check the characteristic polinomial of the Fibonacci relation it doesn't have um one one double root but it has two different roots that are by the way complex numbers and not real numbers

but think about the Fibonacci relation it is something like this relation in Orange if you if you give me two different starting conditions than the ordinary starting condition of the Fibonacci relation I can still get a solution right right I I mean I just I mean you give me the first two elements B1 and B2 I calculate what is B3 B3 has to be a C1 all right B B2 plus C2 B B1 okay so and now I can calculate B B4

and so on so this this is a bjective correspondence I I can either take a solution and map it to the two the first two elements and if I have different solutions they have to have different two first two elements because a solution is completely determined by the first two elements yeah okay okay good okay so the the the the vector space of solutions is two dimensional and we demonstrated that we have a a pair of linearly independent vectors in

this SP okay and and this concludes the the details um that showed that the number of points of n for is is n Square for of course um n Prime to P Prime to the the characteris of the the ground field and now we still need to do the the

remaining part of the way that shows that um as a group is integers mod n plus integers mod n and we we kind of um I mean just to to do like a first exploration we obviously en cannot be isomorphic to integers mod n s

because all elements in is a group is an AB bilion group and have a order that divides n so there is no element I mean in in integers module n Square um you have a at least one element ER of order n square right this

is you take the the the element one you add it to itself n Square Times only after you add it to to itself n Square Times you get zero so the order of one in integers modu n square is is n square but in N I remind you n is all the points that if you you add it you add them to themselves n times you get zero so their order is is always smaller or equal to n in fact always divides n so it cannot be I mean these two groups

in Orange and and green cannot be isomorphic is this clear just just this basic reasoning that the okay so so now but I want to completely pin it down I want to say the only the only thing that en can be is a integers mod n plus integers mod n and this will be a very very crucial result for um for the the pairing and and other

things um and to do this we're going to um make a little bit of a detour into combinatorics but I think maybe it's a it's a good time to take a break and uh we come back to the this detour of the oiler tsion function and we complete the proof all right so um see you in 10 minutes see

e e e

e e e

e e e

e e e

e e e

e e e

e e e e

okay um so um as we saw before the break in order to pin down the the group structure of um of ENT torsion points we need to start doing um arguments that rely on on order and um orders of in in cly groups for example which are the the components of what what we want to be we want to show that

is call um in ZN uh an element okay is a generator I if I take all the the the sums of K I get exactly the I mean I get the whole

group The N if and on if gcd of K and N is one that is K is um a prime to n H so obviously one is prime to n and one is going to be a generator of of z n but you know if if if n is even then an even number inside a a ZN cannot be a generator because you you take Z4 for example and you take two then 2 + two is

four which is zero and that's it so you get all the sums of two only give you two and zero um and in order to generate the the group ZN we need an element that is prime to n and and now we can ask how many generators does ZN has uh so um definition the or the poent

function is a m Fe of n is the number of elements K that are between one and n and a are PRI to end okay so um the number of generators of of ZN is

precisely FN now FN is a is a hard function to compute um but we don't necessarily need to compute a function explicitly to to know something about it and one of the things I want to well there is actually one only one thing that we we need from the oiler tsion function and this is the following theorem for any natural number

n if I take the sum of all of D where D is a divisor of n then I get precisely n okay so for example you take nals six then the only divisors D1 is going to be two

D2 is going to be three and you have um D3 is going to be um six maybe I need the zero I forgot but let's see ER yeah I guess D zero equals one and now what is a a fe of one is just one Fe of two is just a one again well sorry few of two is two a fe of

three is a two again yes so but they are all primed to each other no yeah they are primed to each other I'm just H trying to see if the the The Edge Edge case I probably don't include um D3 as the either we don't include well because D3

Fe let's see ah yeah I'm sorry we we need to do something else we need only the the

um we need to yeah the G sorry so Fe F2 is not two and F3 is also not two F2 I mean the number of um of integers that are prime to Prime to two and are smaller or equal to two is just one I mean and F3 is also one and and then F ph6 the number of er of n I mean of integers that are prime to six and

smaller or equal to six is just you have five and you have um four is not prime to six so five and um five one sorry P P6 I have uh one and five and that's it this is two ah F3 is not one F3 is two because one and two are trying to and are to

three okay so you see in this example and the sum of of all the all the oil numbers for all the divisors of of six is equal to six okay I'm sorry for the The Detour it's like a um combinatorial issue that you need to carefully h um is the statement clear the can we go

to the proof okay so um so how do we prove it let a d be a the set of integers just a is the set of integers in the interval one up to n such

that and the gcd of this integer with n is precisely d and this is some kind of Twisted definition of a set I mean I'm looking for all the all the integers in the interval one up to n whose gcd is is some prescribed number D now if um okay is an element in a of

B then I claim that K is equal to D * L for for some l in the interval 1 up to n /d with a a gcd of L um and N / D equals to one and the reason is think about it you can think about it

this way I mean suppose I have a number K that is whose gcd with n is D mean gcd I remind you is that the Great this common divisor so it means that first that D divides n and D divides K so I can write K as is equal to D time something and this something does not have any common factor with n because otherwise it would not be the greatest common divisor h but in fact since since D

already divides n it's like a a a i i extract the maximal factor out of out of n such that after I extract this factor from n k has and and I extract a a um a d from K the remaining part L has no common factor with with what I'm left from from n i mean so so L has to be a a number between one and n/ d and it cannot have any common factor with with n because otherwise you

see just just negated I mean if if L had a common factor with n/ d let's say it would be some D Prime and then I could write L is equal to um n over I could I could write n/ D well sorry then okay I I I don't want to write all the details in that I think

it's a it will be clear this way if um are you following Lisa is this statement that is written here clear to you yeah all good okay and now I think so but I maybe when you continue I will not but so far so good yeah yeah yeah that's that's all I'm asking is about what is so far then you will see the argument is not is not that complicated from here but I'm I'm saying like uh it's a bit of

um bookkeeping usually happens in Comin TOS um and now what is the number of elements l so the number of such L is where it's precisely Oiler function of n / d and each L gives me a different K so I I

I would like to estimate well to count how many elements appear in a a of D each elements in in K of D correspond to a unique a unique L that that has this property that has a a gcd one with n/ d and I can instead of counting the the the numbers k i can simply count the numbers L that satisfy this condition and the number of L's that satisfy this condition is precisely the oiler function of n/d that

plus the number of elements in a of d a of D is the order function of of n of D now a a clearly um the collection of all a all the sets a D is um a partition of a a of the interval one up

to n because well what what does it mean to be a partition it means that if I take the unions then I get everything well obviously every number in the interval one up to n has some gcd with with n so when I I look at all the ad some of the ads um are obviously empty um but um I look at over all the ads then then I get everything I get all the interval one and and also

obviously if I have a two elements that belong one to a d and one to ad Prime they well sorry if I have I mean the two two sets ad and AD Prime they are disjointed because it cannot be that an element has a gcd d with n and also gcd D Prime which is a different than D with n i mean one element has one gcd with n so the the sets are disjoint and this means a it's a partition of the interval one up to n

IE the the union of all a is equal to the interval one up to n um and the sum of all Fe n/ D which is simply the number of elements in a d is going to be equal to n this is D such that D divides n but running over um running over

um p n/ d is the same I mean I do this it's the same as doing this P of d d divides n right because in the sum whenever D divides n um I I mean either I take the the oiler function of n of divided by D or I take the oiler function of of D and the two sums contain the same the same set of

expressions and and that completes the argument so the sum of all the fun all the functions of D running over all the divisors of n is simply n is this clear the the last part or now I think so uh I mean do you understand I mean if do you understand why the number of

elements of a is a Oiler of n/ D no okay so I'm saying what does it mean to be an element K in a of D it means if if you look at it carefully it means that you have to be D time L for some L that sa that that is prime to n/ D that has gcd1 with n this you this you understand yes sorry yeah now I'm saying okay each each element in a of D corresponds to a

unique unique such L right I mean each element K is K time k equals D * n for Unique l i mean d is fixed in this in this in this setup we we're trying trying to count the number of elements in a of D so we fix D and we look at the element K in D there is a unique L for this K such that K is equal to D * L and L is um is prime to n/ D you see this yes thanks yes right it's

like the the I take the I mean I know that taking out Factor taking out the D factor out of K will result to with with some some remainder Factor L that it cannot be cannot have any common factor with n/ d otherwise a d would not be the gcd yes yes yes yes now so so such so every element in a correspond every element K corresponds to a unique l so to count the number of elements in a d I

might as well count the number of possible L's okay now I see thank you so and the number of possible L's is precisely the definition of the oiler to tsion function okay so and now it's it's very easy to see that the the ads are disjoint sets ER I mean you you cannot have a number that appears in ad and in ad Prime because this number would would then have a gcd d with n and a gcd d Prime with n a number can only have one gcd with n right uh

yeah um so so the ads they cover the interval one up to n and and they are disjointed ER and so so I can I can take the the I mean I can I can say that the sum of the number of elements in ad when I run over all the divisors of n is simply the number of elements in the the interval one up to n which is n right yeah and and I get this formula here with the

arrow the number the the sum of all Oiler toan functions of n/ D where D runs over all divisors of n is is n that that's clear yes thank you okay and now I'm saying the last last step of the proof is to observe that these two sums in a a orange and green are the same just to examplify let's say you take n equals 5 sorry

six then you can do 2 Fe of 6 over um 6 h plus P of 6 over 3 plus Fe of 6 over 2 plus Fe of 6 over one and but you could also do okay this is and this is the orange right and you could also do Fe of um 1 plus P of 2 plus P of three plus P of

six it's just a different ordering of the same thing yeah okay so I get that the sum of f of D is is equal to the sum of P of n/ D and and we we show that this is has to be n okay thank you very much thanks okay so now in the remaining part let's let's classify the uh the endorion points on

ethereum uh let's e k and be in curve and and and and integer Prime to and P which is the characteristic okay then en which I remind you again n is the n the the elements in E of K

Bar the set of all p in E of K Bar such that NP is equal to O this is this is a a a an a billian group it's an a billion subgroup of the the a billian group of all um of all points in in E over K bar so this the the group of endion

points um uh and has isomorphism to ZN plus Z okay so the first a kind of big Hummer that we use for this theum is the classification of finite a bilion groups so by um I

guess it's the second classification theing and of finite obedient groups n it is in suab bilan group so it has to be isomorphic to a sum of cyclic groups zn1 plus

zn2 plus so many things up to Z and K some for N1 up to NK such that and this this is why I said it's the second version of the classification theorem um N1 divides N2 that divides N3 and so on so each each of the numbers that appear in

the the index of the cly groups divides the the next one this was one of the I mean we we actually proved that you can go back to the notes and and and see the proof again H but this this was one one of the man calulations of um how to classify um finite a bilan groups Now by this classification uh decomposition let's say um

for any p in NK of p is equal to O right because and and I should say NK divides n um because if you look at this this group in green ER what is this group in green I mean it's like the collection of all Tel

a one up to a k such that AI belongs to z n Ki and um and the the group operation is just a addition coordinate wise and since since all the N the the the orders of all the cly groups ni I divide NK it means that when you take an element in this green group um NK of this element has to be zero right it's like

a ER it it has some kind of um I mean the the order of the of of the first coordinate has to divide N1 the order of the second coordinate has to divide N2 but n N1 and N2 in themselves have to divide NK so the order of each coordinate here has to divide NK so when I do NK times the whole Tel I have to I have to get zero this is the structure of the grouping in green but the group in green is isomorphic to the group in

Orange so I know that this this property has to also be satisfied in the in the orange group and to to say that it's satisfied in the orange group is is simply to say that for any n torsion Point p n k * P has to be equal to O is this clear all right so from this what do I what do I conclude I conclude that whatever NK is NK is some number that is smaller than n

but n is contained in e n k by definition of what is e n k en K is all the points such that nkp is equal to O um but how can E I mean NK is smaller than n so the smaller or equal than n so we must have en K is equal

n but this implies that the number of points namely NK square has to be equal to the number of points in the other side which is exactly to say is equivalent to say that NK is equal to n right because NK and N are positive integers so now we know that in this decomposition the last Factor has to be ZN yes I am not seeing why e n k is equal to

M well because NK since NK is smaller equal to n right NK divides divides n then then this is all always I mean if NK is dividing n then e enk is always contained in N yes okay so you have ah okay you have like a double containment it's it's an equality of sets okay okay okay thank you okay so this this pins down the last factor of the factorization but now we

can we can do um a recursive procedure so continuing um en enk minus one has to be contained in e n right because NK minus one it's here it divides NK and NK divides n so so NK minus one also divides n and um if if some number divides I mean so if I have a a p here

that I know that n k minus one of p is equal to zero then clearly n n of p is is going to be equal to zero because n k minus one divides divides n so um and and observe that the elements in e n k minus1 are

precisely the elements in the nend um such that well that vanish after vanish after multiplication by NK minus one so yes

no now what is what is n n we said that it's a um some it's isomorphic to some Z N1 plus Z N2 plus z k Sorry NK minus one plus ZN K but ZN K is ZN now what are the elements so if you look at this

part how do I say it and I I divide n into two h two parts this in this and and I want to model what is what is the elements of e en K minus one en K minus one is the elements in N that vanish after multiplication by NK minus one let's call this this this component one and this component two in component one all

elements vanish after um multiplication with um with n k minus one right because um every element I mean every element in component number one has order that has to has to divide and minus one NK minus one is it in the top of the food chain um in component number two namely

Z ZN how many elements um there are that um whose order divides NK minus one and it I claim there are there are exactly n k minus one elements who ordered

divides MK One and why I mean the the number of elements um okay so this is this is a coroller of um the oiler totient s it's like um well actually here there is a little bit of a gap um but in Z end what are the

the I mean the number of ele of elements um f d where D for D that divides n k minus one the number of elements in ZN of order D is FD so elements um of

order order precisely D in ZN is a f ofd so the the number of elements in ZN whose order simply divides divides n k minus one is the sum of all

P of D where D divides NK min so what I get is this is by the the formula I will prove this is NK minus one so all in all what do I get I know that um e n k minus one I mean the number of elements in or elements in e n k minus one are precisely the ones whose order divides NK minus one um I look at the two components in in

Orange and I see who's what are the the the how many elements in the first component in the second component have order that divides NK minus one in the first component the number of elements that who whose order divides NK minus one is simply everything because NK minus one is in the the top of the devising chain um and in in the component in the second component ZN I'm using this sum of Oiler torent functions to conclude that in ZN the

number of elements whose order divides n K minus one is n k minus one it's like simply the the the sum of all this thus the number of elements in e n k minus one has to be equal to um the number of elements in the first component namely N1 * ns2 * up to n - one times the number of elements in the

second component whose order is NK minus one but we said that there are NK minus one such elements H and and this formula is like N1 * N2 * n k - 2 * n kus1 s but we know that the number of elements in e n k minus

one is n K -1 s so N1 N2 and n k minus 2 all have to be equal to one in order for this this formula star to hold thus e is isomorph to z n k minus1 some ZN and

since the number of points in N is known to be known to us to be n² we must have NK minus one equals to n i e n and has to be isomorphic to ZN plus ZN okay this is this is the argument how do you feel about it Lisa

now Nicholas I feel bad about it could could you just repeat it yes um yeah it's it's a bit of like a bookkeeping I'm saying um let's see if you you understood how I got into this point into the the to conclude that NK is equal to n yes okay so so now I if I if I have this point I know that um n is isomorphic to this summing with two orange components right right okay and now I'm saying the

following I look at the next the next guy NK minus one and I look at at the at the the the group of torsion points for NK minus one what is the group of torsion points for NK minus one it's first of all because NK minus one divides n it is a subgroup of right right so it's a subset at least um and H what is what is it it's the collection of all points in such that their order divides NK minus one because if NK minus one p is

is equal to O it's not necessarily that the order of p is NK minus one it could be smaller right the the order is the minimal number the order of p is the minimal number s such that SP is equal to to O right right so if I know that NK minus one of P is equal to O then I know that the order of P whatever it is has to divide NK minus one okay right so so the the NK minus one torsion

points are all can be characterized as points in N whose whose order divides NK minus one but but now I look at the decomposition in Orange I know I I know in advance that N is a um is isomorphic to this decomposition so and and or order of elements or number of elements of a certain order is is a a a property that is preserved by isomorphism right if I have an element of Order X I I apply

isomorphism it's still an element of the same order EX in the other the other group right yes so it's good enough for me to analyze how many how many elements in this green um group do have order that divides NK minus one and what I'm going to do I mean since element in this group are just sequences you know like A1 up to a a um K where each of the AI belongs to ZN zni

I then it's enough for me to to count how many elements of order that divides n k minus one have in the first component and how many elements in that with order that divides NK minus one are in the second component so far good yes and now are you following the the retake yes thanks um okay so now in the first component every element has order that divides NK minus one you have no no choice because all the N I divide n k minus

one right okay so the number of elements I'm just counting the number of elements in e n k minus one according to this H to this description so it it is N1 * N2 * N3 up to I mean simply the the the number of elements the first orange component it's just the the product of all the orders of each of the cly groups right so this is this is this yeah now I want to analyze how many elements in

ZN have order that divides NK minus one I claim that if you want to count the number of elements in ZN whose order is precisely D this is going to be FD okay this is a this I didn't prove completely but you can H you can reason about it not so not so hard afterwards but let's say you accept it then I I say the number of elements in ZN of order D is

precisely FD so how many elements in z n have order that divides NK minus one well I mean I just take the sum of all 3D for D that divides and minus one right yes and and this we show to be NK minus one okay okay it makes sense okay so then I get this NK minus one here and I do some algebraic manipulation and I conclude that since I already know in advance that the number

of elements in EN K minus one is NK minus one squ um I conclude that all the other ends have to be one okay and then and then I conclude that NK minus one is a has has to be has to be n as well because I I mean I have this n is is the sum but I know that the number of points in N has ER is is n squ so n k minus one has to be has to be

n okay again How would how did we arrive at the green statement plus green statement because if I know that the orange statement then I know that all the other nks are equal to one right so it's like just saying yeah back from to this I only have I mean ZN K minus one and ZN I mean this yes when you when you say Z1 this is just the trivial group zero and when you do a um direct

sum with the trivial group it's like doing nothing yes yes yes yes okay so so all in all I get this um all the other ends are are one and I get that e n is isomorphic to the sum and and I conclude H AR now sounds good yeah yeah okay so we're precisely out of time but let me just just summarize what what happened here um we had a lot of detours so it may may

be confusing but on the other hand um I think there is a certain Beauty into it because for for this um this final result in the Box we we used a a a plura of tools and we developed a a lot of a lot of side issues that ended up converging into this proof so it's like a it's an interesting mix um but that's it I I think each of the the components that we used for for

the proof namely the divisors and the the classification of ailan groups and the oiler toan function each of them has independent interest and you you may run into it in other contexts so it's it's a worth worth learning about even even independen anyway next time we start the wh pairing yeah thank you okay have a good weekend you too bye one quick question yeah I mean well um so it's about these

kind of things I am I assume that this teach in universities is like do people do exams where they need to prove this kind of stuff uh like reasoning about this or because I mean I I cannot imagine myself being able to like uh create this kind of pro by myself or or memorizing the whole thing and and so on um look this is a this is this is a o an obvious material that appears in master degree in pure mathematics okay um it's not a surprise that you can't imagine that you could prove it in your

current situation but but to to compare what is happening in universities you would go through a a whole bachelor degree in pure mathematics and then half a master degree in pure mathematics it's like um in terms of like hours even if I take the hours that we spend here um and just count the hours for this I mean for someone with I mean that that takes this course in a university he had previous education that is equivalent to I don't know like 10 times the hours

that we spent even more maybe 20 times okay so after 20 times the the hours that we spent your your and exercises and exams and and so on you you feel much more comfortable to to remember the argument or even even come up with a small small proofs that are similar to this okay okay cool good to know I mean I'm always curious on on this yeah that's that's um that's that's what what is going on in this course I mean I'm taking you to very Advanced parts or

very I mean somewhat Advanced a ER by just going through the minimal Path of Knowledge that is required to to see the full details but to feel very com to feel comfortable in like proving it is a is a different issue okay so it's a different level of understanding cool thank you cool all right have a good weekend goodbye thanks same