Thank you. My lecture will be to share with you, particularly with young students, all

the pleasure I had to study this question about number field, put anything in question concerning class group, for instance. When preparing this lecture, I was thinking of when I was a student, I had two supervisors, Desiree and Yvaniette. And Yvaniette told me,

when you do not know what to do, do class number. And now I would say, when you do not know what to do, do class group. And it's fantastic. Really, really, we know nothing. Everything

is new. And the challenge is to put, I would say, not strange mathematics, any mathematics and you have pleasure, and you start from Gauss, of course, and so on. It's very exciting. For the moment, I'm much more excited by that than by prime number theory. So now, let's start.

So some basics, facts. So I start from the finite extension of Q. This is the ring of

integer. And we consider ideal. And so I consider i to be the set of non-zero ideal of OK.

And then you say that i and j are equivalent if and only if there exists alpha and alpha prime belonging to OK, such that when you multiply, when you have this equality.

So it's an equivalence relation. And what is very surprising is that you will consider the class group. And this is a theorem 0, which is that this set with this relation

is a finite abelian group for the law induced by the multiplication, the composition,

not multiplication of ideals. Very classical. And what we get is CLD, what I call CLD. So

you must not make confusion. It's what we call ordinary class group. Oh, yes, you are right, CLK. So this CLK, at least three questions. I would like to three types of mathematics

which are interested by that. First of all, it's algebra, people from algebraic number theory. And now we can say people from analytic and also people from computational

point of view. So it's very interesting, these three points of view. And I shall be more

concerned not with the case of quadratic field. What is this? So you impose k over q equals 2. And what is important is a discriminant, fundamental discriminant. So I shall write it

D. So how do you build a fundamental discriminant? You start from D star square free,

and you impose it to be, so if it is square free, it is congruent to 1, 2, or 3 mod 4, and it is positive or negative. And D is equal to fundamental, you obtain 4D star, and this is, you multiply, so D is equal to D star or 4D star. And here, when D is congruent

to 1 mod 4, and D is congruent to 2 or 3 mod 4. So we know exactly, so what is lovely in the case of quadratic field, we can enumerate all the fields, quadratic field,

without problem. So that means you consider all the q of square root of D, and there is a theorem which is very easy. And like, okay, curly D for this set of

discriminant, this is the theorem, it is at 0 of 1 is 3 over pi square x plus capital O of square

root of x. This is not deep at all, and it is the same type of this at counting square free number less than x. And it's important, you can, this formula, so slight modification,

we will use it, you can do it for D negative, you can impose some congruencies, D congruent to 1 mod 4, or D congruent to 4 mod 8, or D congruent to 0 mod 8, you have exactly the same type of formula. This is important because it's surprising for me. Yes, so what is,

what is very, people must be conscious of that, that this formula, it has been, it is very rarely extending. So my question is, I count nk of x to be the cardinality,

of up to i. So you give a meaning to this formula, such that the discriminant of k is less than capital X. So you are counting how many extension of degree k that you fix,

the discriminant less than x. And the question, so for 2 it's known, and of course you can put minus, and there are very few results. I check with, with Belabas, so for k equals 3 is due

to Davenport-Iron. So you see at the beginning of the 70s, for k equals 4, it's more recent, is due to several people, Olivier, Diaz-Diaz, Cohen, and Bhargava, and k equals 5, a couple

of, to Bhargava, and that's all. So I insist about this problem, people are able to count

subfamilies, but this type of, as far as I know, this type of generality, we did not, we do not know how to count. So already there is a wall in some sense. So I continue to,

when k is q of square root of D, so I recall some fundamental facts, that the ring of potential, you have an easy formula, which is z, with z omega, with omega is D plus square

root of D divided by 2. Of course you have the norm of an element, x plus y. And so here we have to put x and y in q, and it already becomes a difference. So of course

we will speak about that. We have the set of units, it's completely different when

D is positive and D positive. Here you know that there is a number w which is equal to 2, 4, or 6, which counts the number of units. For D positive, you have something like plus or minus 1 multiplied by epsilon D, n belonging to z, and this epsilon D is what

we call the fundamental unit, which is linked with a Pele equation, which is a nightmare of

a real quadratic field. So there is another class group which arrives, it is C or LD, will be the ordinary class group of q of square root of D. You have the narrow cross

loop, which is, you consider you have a new relation with an n, and you ask that alpha

e is equal to alpha prime g with norm of alpha, norm of alpha prime positive. So this is a narrow class group, and sometimes you are very puzzled to guess what people have in mind when they speak of the class group. So as you can guess, when D is negative, no problem.

And what happens is that CL of D is a factor group of C of D with index 1 or 2.

And it's a very difficult question to see when we have equality. So yes,

so an easy proposition is that these two class groups coincide. So first, possibility is that D is negative or D positive, and the norm of epsilon D is equal to minus 1.

So we may ask about what is this? Is it frequent this thing? First of all,

to see if we make confusion. So my question is, when is solvable? So it's not so easy, I say it's tricky, to say that n epsilon D, this is solvable, is equivalent to the

Pell equation with parameter D star is solvable. So you have to take a q, because when you think of the integer, you have number 2 here, so you may have Pell equation with plus or minus 4.

And now, yes, so this implies all the prime factor p divided by D star implies p equals

2, or p congruent to 1 by 4. And all the difficulty is that the converse is false. It's not a local global, it's over Z. So I write these things, so it receives the

cardinality of D between 0 and x, such that CLD equals CD is very rare,

it's a severe process. Usually we have the bigger group, the cardinality at least,

is equal to 2 times the cardinality of CLD. So what do we know also? We know the class number. Class number, so when D is negative, no confusion, so usually you will find the

notation H of D, and this is W, which is equal to 2, 4, 6, L, 1, D, square root of D

divided by 2 pi. And for D positive, the formula is only for what we call H of D, which is the ordinary class group. And here you have square root of D, L, 1, and here log of epsilon

of D, which is our nightmare. Yes, another thing which is interesting that, of course,

you are thinking of Gauss, Gauss was our thinking of quadratic form. Of course, CD can be interpreted by a binary quadratic form over Z via the action of SL2 of Z.

So I would like to now give some conjecture. I would like to say yes,

what is it, why is it the nightmare? Because the nightmare is here. Because so many formulas, you have always formulas when H of D log epsilon D are married. This is the case. It's very, when we can't separate them, we are doing, I would not say silly thing, but we are not

approaching the difficulty, the real difficulty of saying the same thing, H of D alone, or log of epsilon D alone. Now, I write some conjecture to make the state of art conjecture

that I would like to stress, and they will concern mainly the D positive. So I suppose D positive, and certainly the most famous one is due to Gauss. There exists, so a question,

infinitely many p such that with, so p congruent to 1 mod 4 such that H of p equal 1. So that

means there are infinitely many real quadratic field with, which are principle. So of course,

when I say that, I mean that p is positive. It's a usual prime. The second one is strange. I remember it was quoted to me by Cohen. I discussed about that. But what happens? We do not know this thing. So you can, I would like to write the name of Cohen. There exists

infinitely many, such that H of p different from 1. We do not know that. This is very strange, at least. That's strange. We have this one. Always prime, you see. We will meet

someday the parity question. A third one, so of course, this one was implicitly, certainly the brain of a lot of people. So I would like to, so just at the same moment, three people, a logical coincidence. I say Sarna, Cooley, and Cohen wrote these two conjectures.

So there exists C1 strictly positive, such that when you sum the class number between 0 and x, it is asymptotic to some C naught x log square x. So you see,

I was speaking to you of marriage, and now I speak of divorce. So let me comment. There

is H of p, so I write p congruent to 1.4 is equivalent to, I think the constant would be 1

over 8, something like that, C2 x. And now the fifth one, so I write it in a very, so here also the convergence of several mathematicians, the three I quoted. So prove that

epsilon d, that most epsilon d concentrate around exponential of square root of d. So I

do not define what means concentrate, and you can guess what is a nightmare for computational

point of view. You take a d reasonable, suppose with 3 digits, and what is epsilon d? We see the difficulty. So in fact, this conjecture, we can, I would not say they are equivalent, but they have the same feeling. As soon as you have that, you may think that in the class

number formula, where is it? Yes, the log epsilon d is almost square root of d. So you see there's a huge consolation. So in fact, they are almost equivalent to not this one. To say that the odd, the number of classes which are the exponent is very vaguely around log

of d. And I shall explain that after. So by prime number theorem, you get that. Here

is something else, appears the two part, the even part, which is something that two power omega of d, something like log of d. And so this one, so they are very challenging question.

And I would say some, what is known, I say what I like about one and two, so nothing precise, is known about one and two. But I have a strange result with Karim Belabas, which says that there

exists, so it's a theorem, not a feeling, these are conjecture, there exists a positive proportion of p congruent to one mod four, such that three does not divide h of p. So

it's a very few result with h of p infinitely, and a positive proportion of prime. And

we can escape this here. And you see, you may ask, so with this thing, you say nothing here, nothing here, of course. And so it was proved around this here, and people know this result,

and it's always known, a question to prove that if you take p congruent to one mod four, then such that three does not divide h of minus p. So we have the feeling that the real

quadratic field are much more complicated, and here an example of something we know about real, and it is not known about complex quadratic field. So the idea is about the number of, there is a difference, we will see it with Leventport, about the number of elements

in the three parts. So now about this one, yes, so I have a result about this one, it's a, yes, so it's a, I would say something, it's due to myself, Jouve, and after

him improve by Ross, so which proves that for almost all d, epsilon d is greater than d power three minus delta. It's very, I would say it's ridiculous, because when you compare this

number with this one, it's like the height of my, a tree in my garden is the distance to the sun. Nevertheless, if someone proves with four, it will be accepted, I'm sure, because you see the question is, suppose you improve that, which type of mathematics

you improve? Ross used the determinant method. Okay, you see, anything can arrive in this question. Okay, yes, yes, so now we little by little, we are converging with

Cohen and Landstra. So what about d negative? d negative, I think about this question, we know much more, and our philosophy is that the odd part is something like

square root of d, so it's completely different, the class number, and this one, no, for almost all d, for almost d, so, excuse me, yes,

that's a definition. Yes, but the number three is not so easy, you have to push to get to three.

It's not for all d. Yeah, that's for almost all d, you see, for almost all d, okay, and Enrique, to come to my mind, the question, suppose that you know that infinity prime of the form n squared plus one, many questions here are solved.

I don't know if you'll be talking later, which I'll be done, about the server place form in this collection. So maybe I would like to say something, the separation of h of d from the unit is very well done in the server place form,

right, and because you have asymptotic for the, I mean, the length of your data, and so you can average class number according to the length of your data, that means you separate, etc, etc, and other observations, another recorded difference is that if you look at the contribution in the first formula,

this is of those that exist in the main contribution, it's a very special experience. To comment what you are saying, that I insist about this type of summation,

it's what you have in mind, people, some people may result about the log of separation. So now we go to, so always the quadratic field, what do we know about the

p-rank? So first of all, I must define what is, so R and abelian group, and what is the p-rank of this, it is a dimension of fp of a divided by p a. So I use also sometimes the

notation, the cyclic group of order p pi nu, and if you write, so maybe it's to have a feeling,

so suppose that you write this as a product of all the prime

nu to one to infinity of a cyclic group, and which one appearing with the frequency alpha p nu, then the p-rank is a sum of alpha p nu.

So what is well known is the theorem of Gauss, always M, which says that the two rank of C d is equal to the number of prime divisor of d minus one.

So in some sense, we know any question about the two rank is a question about omega of d,

so it's rather simple, so it's solved. And now what about the literature, what about cl of d? What is the two rank of cl of d? So here we have problem with the literature, because they are false proof, but they are

correct one. So I know that cl of d is a factor group of C d of index one and two, so it affects the two parts, but it may affect the two rank, or maybe a factor C8 may become

a factor C4. This is not sufficient to guess what is the two rank. So I recall that if you have that, so we have always the two class group coincide. And this is a theorem now,

I think we found a correct proof in ASL. And so suppose that d positive and epsilon d

is equal to one, so the two class groups are not the same, then the two statements are equivalent. The first one is C of d is equal to the cyclic group multiplied by cl of d, so that means that here the two rank is lower by one, and this is equivalent to

there exists a p which divides d such that p is congruent to one. So in other words,

this is equivalent. In other words, if x y minus d star is locally solvable, but not

it's you, that's what I say. Yes, so we have to guess, so we know how behave the two rank

of the ordinary class group. Now we arrive at Cohen Landstra heuristic, so it's number two.

So it's published in lecture, notes in mathematics, in some journal arithmetic in this year, and a lot of people have thought about this heuristic.

They have been generalized, some of them have been proven extending the situation. So they were, so I shall not enter in the algebraic interpretation, very superficial thing. So they use a computer first of all, and they count the proportion of d

negative up to the capacity of the computer, the d negative such that the three part of Cd is

C9. So is it true that the proportion is something like, okay, and now is another group of cardinality nine, and here they find that it is only

one. So how can you explain that? So first of all you, but you must be careful in this approach with a copulation because the convergence is very,

very slow. I will give you examples. So the ratio of these two numbers is eight, okay, and eight is correct, and this is nine. Yes, I like it.

Yes, and what is this? This is the cardinality of the automorphism of C3 squared divided by the cardinality of automorphism of C, and the three is the exponent is here.

Okay, so it's what I find to explain you what is behind. So they are the brilliant intuition to affect to some group, some question where you want to meet some group,

not the cardinality because here the cardinalities are the same cardinality, but this weight, the inverse of the cardinality. So you see here, this group has much, much more automorphism than this one, so it appears much oftenly, less oftenly.

And so some easy, so to, so some cardinality I want to,

you must be very careful what you put. This cardinality is p r minus one, p r minus p r minus one. This is easy, and it's natural. This formula is classical.

It will be natural. We'll always find formula with eta function. So this function appears in theory of partition, and so the first one is you take eta k of t is the product. This is a real function. Okay, t power minus i, and you suppose that

of course t is less than one, and eta infinity of t is equal to the product i greater than one,

one minus t minus i. So this product is very quickly convergent. We will benefit from that, and for instance with this notation, this quantity is p power r square eta r of p.

So you see, when I write koenleinz-reich conjecture, we will not be surprised to see this function meet frequently. So what about so I write koenleinz-reich in the way that I like, and I,

yes, so okay, so let p prime at least three. So they were frightened by the prime if p equals

two because Gauss was waiting them here, and so first of all, so we suppose that d is negative. So I use the notation c phi number five because there are other conjecture the probability. So you can guess what I mean by probability that the p rank of c d is equal to

r. Yes, this probability is equal to p eta infinity of p product one minus p minus k

minus two from k. C6 says that the average for alpha integer greater than or equal to zero,

the average value, if you want the expectation, expectation of this crazy product, not crazy, but I say the p rank of c d minus, okay, so when you sum from this expectation is equal to

one. Okay, so when you see that, you can remember, oh yes, that remind me some counting automorphism. Okay, this is the origin of that, and for real quadratic field,

so they call them c9, c10. So here the probability will be slightly modified.

It will be p minus r, r plus one, so it's almost useless that I write it. Here's the probability, no, this one is, the last one is p minus alpha.

So these are the heuristic of Kohen Lenstra in the original paper. So now we must make some comments, what is known, what is not known.

So it's, so what they have to support that their conjecture at that time, only computational data, and this conjecture we approve only in one case. So we are going to,

we approve only in one case, that means we will meet him several time in Davenport, Iron, which gives, so it was, I told it already, it's the same thing as counting

cubic fields, but I told you, they only give this expression here, so they were, so they were only proving,

I think is it alpha equals one or alpha equals two, zero, alpha equals, yes, alpha equals one, yes. Okay, it was only proof for that, only.

It does not much change, I would say, but we have extension which confirm

the intuition of Kohen and Lenstra. So I would like to make some comments. The first one is, if you believe it, okay, so you see this, you must have in mind that this coefficient goes to zero very quickly, very very quickly. So for instance,

I doubt that someone, a computer, is it known a D with the three rank of CD is equal to 10, for instance.

That means you have the first coefficient, the second one is bounded, the first coefficient p power, that means three power minus 100, it's very very small. So I think is it true for the specialists of computational, I think it is not known, and it's not surprising,

it's very difficult. And first of all, with my colleague Kluners, I would like, you may be, I would not criticize of course, but command C6 and C10. So we make some observation

and I call nkp is a cardinality of the total number of, I don't know what is English word,

linear vector spaces, subspaces of any dimension of hp power k, yes,

okay, of any dimension. So the computation is almost done. So you have nkp is equal to the sum from l equals to 0 to k and you have a ratio pk minus 1, pk minus

p minus l minus 1 divided by pl minus pl minus 1. So the eta function is here, the eta k function. So C6, so it's only playing with, I would say, a formula. So C6 is equivalent to,

for all alpha, for all alpha integer, the average values, that means what I wrote, expectation of p power alpha

cd is equal to this number of vector subspaces. You see, so we have the moment for someone from Analytic Number Theory prefer computing moments than this expression which is full of algebraic

meaning in terms of automorphism. And now C10 is equivalent for all alpha integer. So it is a slightly modified n alpha plus 1 p minus n alpha over p over p power alpha. And in some lecture,

I will explain in our proof about the forum with Kuehnert, we see these vector spaces. We do not see the automorphism but we see these linear vector spaces. Okay, so now we made some

remarks with Kuehnert because and I think that Cohen and Lestrade are not very precise. Maybe it was clear, I don't know. So it's not, I would say, a proposition. Proposition is that C6 implies C5

and C10 implies C9. That means it is so and I have this version of C6. So as soon as you

are able to compute all the moment, you know the probability. Here also. It's not surprising but it's not evident, I would say. It's not evident. And the converse, we cannot prove,

of course, we do not know what is the rate of convergence that is probability. From probability, if you write that, you cannot deduce the expectation. So nevertheless, I like this proof and because I must admit that sometimes we meet the paper of Heath Brown concerning

the two-selmer group. We have the same type of solution and we meet an infinite linear system. So I write what is it. So I take only the negative and so when using Lebesgue

dominated convergence theorem and so on and so on. So I call dr the probability, the proportion

that the p-rank of C d is equal to r. So we use, it's not evident but it is not deep. So you meet this infinite linear system which is d0 plus d1 plus d2 is equal to is equal to n0p which is 1,

of course. So the sum of probability is equal to 1. So and after it's more d0 plus p,

d1 plus p, d2 plus p squared, d2 plus pq and so on. And d0 plus p4. So it's like, I would say,

an infinite Van Der Monde determinant. Heath Brown considers that like Van Der Monde with

we find another technique. So I would give some to prove that it has to check that it has the solution and in fact that the solution is unique. So first of all check that dr is equal to

the probability I wrote which is here is a solution. So I was very happy we had

to open first of all R. D. N. Wright book in the partition function. And so we find this, of course, it's classical. So you know if this is a partition and not the n primes,

you know that this is 1 over and it is exactly the function eta infinity 1 over x.

And now there is a classical extension always in R. D. N. Wright which is a lemma so certainly they say it was known to Euler. So it was known. So this is equal to

1 plus x divided by 1 minus x squared plus x4. Such a formula. And in fact we required another

one which is we found it in yes excuse me we can write this formula as eta infinity of 1 over x is equal to the sum x k squared divided by eta. So we are in the I would say in the kingdom

of partition function. So to check this is not sufficient this function we have to equality so we have to generalize it and it was proved already and the generalization is this one.

So you take so suppose D positive integer so we will we will slightly here you have the the sequence of square here we will slightly modify this

it will not be r square but r r minus t but it gives something different which is that always the partition function p of n xn so is the sum from r equals t to infinity

x a r minus t and then i write it because and so you finish one minus x x

so it is not in rd little wood not at the end right so this formula you can use it to check

that it satisfies this infinite linear equation system and i like the proof of that because you are you are exhausted so we make drawing okay to explain with what we call derfy rectangle so we found it in a book of conte so i take 1929 and i decompose it like 8 plus 7

plus 4 plus 4 plus 3 plus 2 plus 1 and i make the following drawing corresponding to 1 2 3 4 5 6 7 8

so what is the so this is the usual description of a partition so we go we go maybe in the 19th century so nothing deep so what is the derfy rectangle with defect t equals one so you

take the largest rectangle you can you can put inside this drawing with size size r r minus one

so you and the drawing is the following it's four three you see the largest rectangle is here four three okay you cannot you see there is one different between the longer and larger i don't know english so when you are right what happens is that you are

going to play to each partition of that type you will you will associate first of all a derfy rectangle and partition here what happens here you have three okay so you have a certain number of here the r r minus t points and then suppose that l suppose that here you have l

integer l points and here m points what is here you you see the drawing here you have

a partition of this number m with a cement less than this number less than three you see so you you you analyze this drawing like that

you see you have the partition of the number l with someone less than here it is four here it is less than three even if it is two and so you have a bijection and this gives this very say integrate formula which is the key of the fact that the dr

r solution the dr and now about the unicity so the unicity for me is is is strange because you you are so i claim that this this system i call s1 so it's the first part the second one

s1 s s minus as at most one solution in positive di and the key what is important that n k p

it's it's something like is less less capital o of p k square over p k square over two something

like that yes it's only n what happened is that the second what we call a second member the second part of here which is n 1 p and so on goes to infinity but not too quickly

and here we will use so it's i shall not explain jensen formula to create we we built a complex variable function and there are system of that type which has which have not a unique solution for instance so suppose here

replace suppose here I replace here but some ck so it will be c0 c1 so it may have more than one solution may have more than a unique positive solution so

I give you so I shall take ck equal the sine hyperbolic sine of pi p power k and the end of

course you will check excuse me so you will find so to find solution you use a Taylor expansion and you use their expansion of the function of sine hyperbolic sine of PA k at the X k and you

will find another solution if you consider the Taylor expansion of sine hyperbolic sine of pi

X k plus a usual the usual sine function use a Taylor expansion of these both function you will see that the ck what I call the ck yes it has this value and the Taylor coefficient are

not the same and positive and you see this function goes to infinity it are absolutely not the same side it's something that exponential p k power k so it's go to infinity much much

quicker so it's so it was a pleasure with cleaners to make a Jensen formula okay but he's brought another technique so thank you no you do not know the rate of convergence of

the probability this is stereo of moment and we are in the edge sees does the theory of

amount implies that we know the probabilities it's work here if I have a rate of convergence if I say more I do not say probability but the number is equal to pump pump pump plus a very good

error term yes I can do but see what I mean