 Welcome to this number tier seminar. Our guest today is Professor Subramanian Uttakrishnan from Chennai in India. He's our associate and his first visit to ICTP. Today, he'll be talking to us today about including our on number few related problems. Thank you, Subramanian. It's a pleasure to have you here. Thank you. Thank you very much. First of all, I thank ICTP for the support and opportunity to receive ICTP. We'll talk about current algorithms on number fields and econ algorithms and some ID classes in number fields and it's a later problem. And in the end, if time permits, I'll talk about divisibility problems on class numbers. First, define what's the including algorithm. So, it gave me an algebraic number field for us and OK, the each ring of integers. That's a total ring of integers. And we know, and we say that a non-negative function from OK to non-negative function, OK, is including if whenever for all elements A and B, for B non-zero, there should be a Q and R so that A can be written as BQ plus R with pi of R is less than pi of B. We can clearly see that this is just a generalization of divisional algorithm, what we used to learn for itself. The question is whether for a in general, instead of OK, whether if you have an integral domain R or in particular for a ring of integers, OK, whether we have an including function or not. So, why do we want to study about integral algorithm? Suppose we have an integral algorithm for an integral domain R, then that will imply that R is a PID and PID implies V of B, therefore, it is interesting for us to study whether given integral domain as an equivalent algorithm or not. So, in this talk, we'll always concern about number fields and its ring of integers. When we go for the first non-trivial number field, real quadratic field, quadratic field in general, we have a natural norm function, which is a non-negative function on Q square root 1 or Ms square free design. And we first ask the question, when will this function is an equidion function, whether we have a quotient and remain that always? So, let's have some examples. We know that Z is equidion with respect to the modulus function. And the ring of all polynomial things is equidion. With respect to the function, P of X is one plus a degree of P of X and zero, whenever P of X is non-zero and zero if P of X is zero. And Gaussian integers are also equidion with respect to the absolute value, square root of the absolute value. So, the question is, whether how do we determine that a given integral domain is equidion or not? So, we briefly discuss non-ecruidion, non-ecruidion, non-ecruidion. So, let's recall that how we prove that, how we prove that Z is equidion. So, if we have, if I consider the real line, so if you want to prove that we have a Diction Algorithm, we have a quotient and remainder for every a and b for being non-zero. We can see that the proof is essentially whether it's essentially is equal and true if I take the interval of length zero of b, whether this covers the entire real line or not. So, this, so, so modulus on this set is equidion is equivalent to saying that the intervals of length b covers the entire real line. So, based on this observation, we can define a equidion minimum of on any number of k. So, let's say that for any k and k, we say that m of k to be the infimum of norm of k minus eta, where the infimum runs over, all eta belongs to okay. And we define m of k, which is weekly minimum of k, which is supremum of m of k and where k runs over k. And by the definition, so let's recall that, what is included in Algorithm? So, if you have, you say that you want to say that for every a and b, you want this to be whole. And when we divide norm of p both sides, we can see that which is norm of a over p plus p is less than 1, which is equivalent to saying that norm of a by p minus p. So, this is equivalent to saying that for every, for every element in k, there should be an element in okay, so that this has size less than 1. So, in generalizing that, we can define including minimum and including minimum, minimum at k and including minimum of k. So, clearly if m of k is less than 1, clearly k is norm included in. If m of k is bigger than 1, k is not norm included in. So, you want to see what is the size of m of k for a given Algorithm number. So, to find m of k, we can naturally, for any given number of k, we can embed it into r to the power n. Where if k over, if k is a number filled of beginning n, then you can embed k in r to the power n. We can naturally extend the definition of m of k and m of k to m of k bar, which is r to the power n. So, we can ask, we can, because it uses a continuous function on m of k bar, if we simply extend the definition of m of k and m of k to k bar. So, we ask the question whether what is the, what is the superman value for m of k bar, especially m of r to the power n. So, if m of r to the power n is less than, by definition of m of k and m of k bar, clearly m of k is does not equal to m of k bar. Therefore, we want to estimate what is m of k. So, there's a conjecture regarding, including minimum amount of time. It says that for a total, if suppose k is totally real, algebraic number, but totally real, I mean all embeddings of k are totally real. There is no complex ability for that. When the discriminant be, and if the degree is at least five, then we can say that, then the conjecture says that m of k is bounded above a root d by 2 to the power n. And still the conjecture is open, except for some special cases. And for non-totally real number fields, we don't know any conjectural bound for that. So, it is only for the real product. You said that degree is at least five or at most five. You said something which was likely, degree is at least three, sorry. For the five, we have m of k is 2 to the power minus n square root of d. And if you go for real product fields, we know m of k. So, it says that if m is 2, 3, modulo 4, then we know the exact value for the equivalent minimum. And for, which is mod m plus 1 over 4, for m is 3, modulo 4, we know that m of k is modulo sub m plus 1 over square divided by 16 m. And with this proportion, we can see the upper and lower bound for the m of k for the imaginary quadratic field. And using this upper bound, we have a, we know all of the imaginary quadratic field, which are 1 square root m for m is equal to 1, 2, 3, 7 and 11. So, there are nine imaginary quadratic fields, which are, which are class number one, one out of nine, there are five of them are non-includient. In fact, if you have a more general question, only these five, you could be a number, only these five imaginary quadratic for equilibrium with respect to any function. They cannot find even in function for the remaining four imaginary quadratic fields. So, how the proof goes with an example. If we take q square root minus 2. So, we know that we want to cover, we want to know that m of k is less than 1. So, m of k is less than 1 is equivalent to saying that, if we have a fundamental parallelogram for a number, fundamental parallelogram for the lattice, if we have unit circles above, if I have, so m of k bar less than 1 is equal, is saying, is implied by the fact that, if I take the unit circles around the lattice points on the parallelogram, if this covers the entire fundamental parallelogram, then we can see that, for this condition holds for every, for every element x in k and we have a q in k. So, the, so this, I mean, this is kind of jump with the proof to say that q square root minus 2 is non-includient. And if you go for q square root minus 5, you can clearly say that, you can clearly see that the circles does not cover the fundamental parallelogram. So, there we have a gap, therefore it is not non-includient. In fact, q square root minus 5 have class number 2. For real quadratic number fields, we have a upper bound for m of k, for we have a lower and upper bound for that. And using that, and with, so we, and also we know all complete classification of non-includient numbers too. There are only 16 of them, 16 of them which are non-includient. But we can ask the questions, we know that there are, we know that, we know that there are 16 of them are non-includient. But the question is, non-includient algorithm for real quadratic fields, what about any other function? Can we have a function which is different from non-includient? So, we, with the Ramuditi and Kshilivas, we noticed that with some mild conjectures on non-includient. I will explain it later. For mild conjectures on real quadratic fields, we can say that every real quadratic field of class number one is etude. So class number one implies etude n for every real quadratic field. But we assume that hard little conjecture and we will, we can just say that if I take the set of four primes of primes x in a aquatic progression so that two p plus one is also primes, two p plus one is also primes, as we are at least exo-orlogics, as is due to infinity. And if rich prime conjecture says that if I have a unit in the, in more case star, and if we look for the set of all primes in x so that epsilon could have got p minus one is one model of this product. And this coordinate is etude n of exo-orlogics. So we assume this, mild, we assume these statements and we could able to see that real number. In fact, we have put algorithm for every class number one to etude n. And also we have certain families, we can simply a family and we do that if at all there is, if we have a class number one, a quadratic field in that family, where d square root of d, where d is equal to a plus one square, d square n square plus two a plus one square n plus 23. If in this family also we could end for every class number one product. When you go for cubic number three, we don't have a bound for m of k bigger, but we have a bound for m of k bar. And m of k bar says that if we have a complex cubic number field, the m of k bar has a bound. And if we consider totally real cubic number field, we have the bound for m of k bar, which is one over eight square root of the treatment of k. And if we consider the Galois cubic number field and we proved that there are certain families, there are certain family of cyclic cubic fields, which are Euclidian, there are certain cyclic cubic fields, which are not Euclidian. So he proved that if a cyclic cubic field with the contact, if it has contact area of equal to the values, 7, 9, 13, and they are Euclidian with respect to now. If we have the conductor is equal to 73, 79, and any conductor between 163 and 10 to the power four, they are not normal. So we took up this problem and we proved that all the fields what Smith listed, all of them are Euclidian with respect to, with respect to some function. We don't know what is the, we proved that the existence of Euclidian algorithm for such fields. Then we moved to a quartic number field. If you go for quartic number field, what is the bound for m of k? And if suppose if k is totally complex quartic field, then we know that we know the upper bound for non-equivalent number fields. If the discriminant is less than, if the discriminant is bigger than that number, then clearly k is not non-equivalent. And using that bound on the explicit computation, we then proved that there are only two such number fields which are non-equivalent. When you go for bicoated field, we also have complete classification of non-equivalent bicoated fields. Usually the classification of non-equivalent number field is most of the time it is complete only for if it has a complex input. So for totally real number fields, there it is still open. Even in the basic case, even in the cubic case, it is still open. Even if we have a complex quartic field, if it contains real quadratic fields, q square root two, we know all non-equivalent number fields. If it contains some other real quadratic fields and two ramified, we know only one which is non-equivalent number field. If two is, if k contains a real quadratic field and also two does not ramified, unless the second point, if two only ramified ramification index two in k and initial degree two, then also we know only four of them are equally number fields. And if k is equal to that particular form i square root of b f s b a, then we know a complete class, we know a list of number fields are non-equivalent. And if in the case of totally complex quartic field, we just not have a real quartic, a quadratic, a real quadratic subfield, we know the bound, at least bound for m of k. So, we took up the problem for bico-radic number fields. And with K-Shearing-Vauss and Ushank and Sandali, we proved that all imaginary bico-radic fields of class number one of the form q square root m and q square root n, they are equally. And in fact, we could extend the result to all cyclic quartic. So, we proved that all cyclic quartic fields, cyclic quartic fields are also equally, they have less number. So, we have a quick literature to be about, what about other number fields? I consider, we consider up to number fields of degree four. What do we know about number fields of IID? Then we concern to normic number. So, most of the theory relies on only, we know, most of the time, we know only for number fields of prime degree. So, for non-prime degrees, we don't, it's not well developed. So, for the, so we have the question, suppose if we have a number field k over q, and it has a degree l, where l is a prime, where l is a prime, then what can we say about whether k is equal or not? And Mekum proved that, he proved that if another l is prime, we can always have a constant cl so that when the discriminant is bigger than that constant, it's not normative. So, we know that, of course, assuming GRH, assuming GRH, we know that there is a, whenever we fix a l, we know that there is a constant, if the discriminant is above the constant, it is not normative. So, using the RSL, they themselves classified cyclic to be non-ecronian number fields of degree five and seven. And they proved that if the degree l is equal to these numbers, 19, 31, 37, 43, then they are not normative. There is no number field of degree 19, which is not normative. So, even though the first term gives a bound, but how do we classify that? The classification of the non-ecronian number fields when the l is fixed, it is still widely open. So, we need to find an upper bone and then we need to know what is cl for doing it. So, then we can use the computation to classify a classified normative. Let's come back to the non-ecronian, not non-ecronian stuff. So, people always look for non-ecronian function, non-ecronian number fields. So, a D.A. Clarke who first showed that the real-quality field Q square root of 69 is Ecrudian, but it is not non-ecrudian, but he gave a explicit function for that. He said that if we have to define a function pi in this field, then this particular pi serves surface parts. And we can clearly see that this pi function is almost a norm function. So, whenever we have pi of a plus b square, one plus pi root of 69 over two, whenever a b is not equal to 10 comma three, the function he defined is a square plus a b minus something this way which is a quadratic form which is given by the norm well, norm map of Q square root of 69. So, only at the point 10 comma three, it is differs from the norm, it is 26. So, he cleverly constructs such a function so that it becomes of Ecrudian. So, there is nothing special about 26. We can take any number bigger than 26 that will serve surface. So, to go for norm Ecrudian, say how do we prove that? How do we prove that existence of such an Ecrudian value? So, mode scheme gave a one-to-one correspondence between the Ecrudian functions. Whenever Ecrudian functions are okay, in fact for any integral domain, increasing sequence of correspondence between the Ecrudian function and the one increasing sequence of sets satisfying certain conditions so that if we take a, you will cover the complete integral term. If we have, he proved that if we have the increasing sequence of sets with certain conditions and which covers okay, then that will imply that we have a function phi which is Ecrudian function. So, the construction says the following. So, he, I'm not going to the detail of the construction. I just explained the construction itself. So, he said take A naught, which is union, zero union, the set of all units in the integral domain, R. And let's define AI by inductively. AI is AI minus one union. All such elements in R, so that when we consider the quotient ring, R over A, and if you consider the natural canonical map, AI minus one to R over A, this should be, this should be a subjective. So, he said that collect all such A's so that when we go from AI minus one to R over A, that should be a subjective map. So, clearly by the construction, by the construction, we can see that this AI is an increasing sequence of sets and if this covers okay, we are done. So, for example, if you take Z, the A naught is zero comma minus one comma one. And then A one, so we need to look for integers so that when you go from A naught to Z over A, that should be a subjective map. And it is clearly, we can see that A naught contains zero minus one, therefore A one should contains integers from minus two, minus one, zero, one. So, in A two is numbers from minus four, minus seven to plus four, and so on. And this says that, you can clearly see that this covers Z, therefore that will guarantee that Z is not, Z is equal, Z is equal. This is another way to show that Z is equal. So, so using ModSkins criteria, remember the proof that if you assume ZRH, then we can always have an increasing sequence of sets which covers okay. So, we covers okay. And in fact, the precise answer says that if we have all primes in this set, if we have all primes in this set, all primes, all prime ideals in this set, prime elements in this set, then there is some n so that n is equal to one. And he assumes, precisely assumes ZRH is to show that all primes are in okay, all primes are in A two. And therefore by ModSkins construction, we can ModSkins says that if this covers okay, then K is equal. So, by ModSkins K is equal. So, to remove ZRH, so all of our work so far is restricted to a number of things of small degree and to remove ZRH from the input, that's it. So, to remove ZRH, you need to show that how do we put all primes in the set here? How do we make sure that all primes are in the set? The first non conditional result says that if we have set a number field K and satisfying, and if we consider the ring of S integers, where he has ring of S integers, let's take S be a set of all primes of K, there are finite primes and infinite primes. And S contains all the infinite primes. And we take set of all X in okay, so that the order of X with respect to the prime P is at least zero. So, for all primes not in this. So, we fix a finite number of places of K and then we look for all integers which is one zero. And then, if we consider such ring of S integers for us, and if the size of S is sufficiently large, if it is at least maximum of phi and two times at equal to minus three, and if K has a real embedding or it can be the G-th root of Nt, the G is the G series of gamma P minus one, or finite prime in S, then such a ring of integers is equal. So, here we can see that there is no assumption on GRH. And to remove for the actual number field K, for the also ring of integers K, Kankan-Ramurthi introduced this concept of admissible primes. So, remember, we want to see that whether A2 contains all prime materials or not, in order to say that A2 contains all prime materials or not, they modify Mohrstein's construction. They modify the Mohrstein construction and they boost the admissible primes. So, that when we consider the sequence here, that A2 contains all primes in all this set, corresponding set here contains all prime materials. So, that is their idea. So, they introduce that set of admissible prime and they studied the problem. So, what is an admissible prime? So, admissible set of primes. So, let's say pi 1, pi 2, pi S be a distinct non-associated prime elements. And we say that they said pi 1, pi 2, pi S is admissible. If we consider a beta, a beta is pi 1 to the power a 1, pi 2 to the power a 2, pi S to the power a S, where a is a non-aggressive. Then, if you go for OK model of beta star, then the natural map from OK star to OK model of beta star should be such it. And this should be true for every integer A1, A2, A3. If this is true, then we can say that A1, A2, A3 is their admissible set. And in fact, using this, they proved that if a totally real model, they proved that if we have a totally real a Galois extension and if it has S number of admissible primes where the size of the gas is modulus of N of k minus 4 plus 1, then OK is equal. So, there are N of k is a degree of k over k. If a real total real Galois extension has sufficient number of admissible primes, then OK is equal to N if it is P a. And also, they proved that every finite Galois extension of unit primes bigger than 3 also holds true. P a implies equal to N. And for abelian extension, they have a sharp addition. They proved that abelian, if we have k over q is abelian extension of degree 1. If we have S number of admissible primes that's how the R plus S is bigger than or equal to 3, then OK is equal to N. So, this leaves us to consider number of smaller degrees. So, smaller hand, smaller hand. So, they themselves considered the real quadratic case and they proved that q square root 14 is equal to N. So, so, so this is the first, this is the long standing conjecture to prove that q square root 14 is equal to N. Special about q square root 14 is the first smallest real quadratic field which is not non-equal N, but it is F a. So, they proved that that q square root 14 is equal to N by exhibiting two admissible primes for that, which is 5 minus square root 14 and 3 minus 2 times square root 14. And in fact, they proved they taken all the real quadratic fields of the screen and up to 500 and they proved that class number one implicitly. So, recall that we considered the same along the line of thought, we considered the problem and we proved the results which I already mentioned here. And for the cubic case, we proved that the external smooth result using the same line of thought. So, recently we thought about bi-quadratic fields, in fact, multi-quadratic fields. So, we can take a multi-quadratic field, the problem, multi-quadratic fields, but by the result of Ramurthy and Rangurthy and Harper, it says that if R plus is bigger than or equal to 3, then we have a, at least if R is sufficiently large, we don't need to think about it. Therefore, when we go for multi-quadratic field, that reservistics ourselves to the bi-quadratic field. So, we consider the bi-quadratic fields of class number one. These are the list of bi-quadratic fields which are class number one. And the pink ones are known that they are non-equal. And others are not known whether they are non-equal or not. So, we proved that they are in fact equal. So, to prove, so this is a precise result which I already mentioned earlier. So, to prove that result, we want to know the, we want to know the structure of the unit group. And in all of this, surprisingly, in all of the non-equal, not non-equal bi-quadratic fields, the, they have, they always have rank one. And with the torsion part, they do, the one with the torsion part differs. The torsion part is either C4, C2, C6 and C10. So, based on, so to exhibit a invisible prime, so we want to know, we want to construct, we proved the following proportions. So, that these proportions helps us to prove that individually each bi-quadratic fields, what we are considering is a, has a visible set. And that will imply that considering the quadratic number field is equal. So, the proportion goes as follows. If I take a number field here, and if I take a YL bi-chink of integers, suppose if I have two prime ideals, Q1 and Q2, and they are under-amplified over Q, and with a initial degree one, and with prime number Q1 and Q2. If we have these prime ideals, Q1 and Q2, and rational times Q1 and Q2 satisfies all these four conditions, then we can say that Q1 and Q2 is an invisible, admissible fact. So if, of course, and the proportion two assumes that YL has the torsion group C2 cross Z. And proportion four for the number field K with the torsion part C4 cross Z. And with the same, we look for Q1, Q2 satisfying the three conditions mentioned here in proportion three. If we have the Q1 and Q2 satisfying proportions, we can say that Q1, Q2 is an admissible fact. For number fields with the torsion part C6, we have a slightly modified conditions for such number fields. For C10, we can see that only the conditions two and four is changing in all the proportions, two, three, four, five. So basically, we want to show that if we have an admissible prime, we want to look for prime ideals, Q1 and Q2, of course, with restricted conditions. They're not general prime ideals, they should be under-amplified. And with the initial degree one, with the initial degree one, the prime numbers should satisfy certain directions. So let's recall, let's, I'll just outline the proof of proportion three. So star is subject to the outline is, so what the strategy of the proof is to consider, because P1 and P2 are distinct prime ideals, they are appropriate to each other by Chinese remains as there, and we can consider P1 to the power of A1, which is isometric to okay by P2, so what we wanted, what we proved is that we have taken, so the moment we are looking for prime ideal, which are odd prime. So when we have a prime ideal with odd prime, this is a cyclic group, both of them are cyclic group. So we are looking for an element in here, so that it maps to an element alpha, comma one, where alpha is a generator for this group, and one, this group, and we are looking for some other element, which maps to one, comma two, where beta is a generator for this group. So if we could be able to find such an element in okay star, which maps to element here, then we can clearly say that alpha comes to the power A plus B maps to alpha, comma one. So the strategy is to produce strategy, strategy is to produce whether we have such a unit in okay star, which maps elements here. So these conditions follows by what we are assuming. So we assume such conditions. Assume conditions one, two, three, four, so that this, I mean, so that by computation, we can have these two equations. So that will make sure that we have these two conditions that will imply that we have map. So if you go for the proof of proving that my quadratic field is Ecrudian, so let's exhibit with an example that the first of my quadratic fields, what we are considering. So we consider that the quadratic field is given square root minus one and square root minus 11. And we will need to look for times which satisfies the four conditions what we have listed in proportion three. So here in this case, we can take P1 is 157 and P2 is five and look for its prime decomposition. And if we take any prime ideal in the prime decomposition for P1 or okay, we take the first prime ideal in the prime ideal decomposition. And for P2, okay, we take the first prime ideal in the prime ideal decomposition. And with computation, we have checked that, we simply checked that whether this P1 and P2 satisfies the condition or not. So that's what we did in the proof. So we checked that they all satisfies the conditions and that will say that this particular number will be the first square root minus one and square root 11 is equal. For all the by quadratic fields, what we studied earlier. And these are the admissible prime. So for cyclic quoted fields, we need to deal with the same or all the line, but with the different proportions and with a different computation technique. And recently with ushag sandali, we considered the last smallest number field, what we are interested in. Sixty number fields with a unit range less than three. And we proved that all such a 60 number fields with unit range less than three are equal. So this in fact, this complaints kind of all a billion number fields of class number one with everything with time less than three are equal. So this completes the program. So we go to the second part of the talk. The second part, we consider the generalization of equal algorithm to the ideal class. So algorithm means that the norm case it's equivalent to saying that for every x in k we look for a element alpha in both case so that better x minus alpha is less than one. So that is equivalent to saying that whether we have for every x, we have a better we have an alpha so the end of x minus alpha. So let's generalize this notion, this particular inequality and saying that instead of we can think of one as norm of okay. Norm of okay is one. So he's a simple observation. I mean, a clever observation says that he will replace norm of okay by norm of right. And of course we are looking for alpha and c. Here we are looking for alpha in both case. And here we look for alpha and c. It says that we said that an integral domain c is said to be non-inclusion if for every x in k whether we have an alpha in c so that norm of x minus alpha is less than norm of c. So why do we need to have such a function? What is the advantage of considering this thing to alpha c? And he himself proved that if we have such a non-function then the ideal class c is a generator for the class. And it proves two such things. It's true it generates a, it generates okay star and it proves that okay star is cyclic and also it gives the generator for the cycle. Even if you find a more general definition he said that it's take any dedicated domain r in fact any dedicated domain r and the i will be the set of four integral domains. Unlike the Euclidean algorithm case here the definition works for the dedicated domain not for any integral domain. So let's take r be a dedicated domain and i will be the set of four integral domains. And let's take a ideal c in r and he said that we say that c is Euclidean if we have a function psi from i to w where w is in a well-ordered set. So that whenever for any ideal i in i and for every element x in i inverse c not in c there should be a y in c. So that psi of x minus y i c inverse less than psi of w. So if we have such a function so whatever we have in the definition key pi to generalize is this particular part. So when we replace psi by norm and all these things in that it is very clear that we can see that what the definition here is generalizing the notion. So if we have such a function on c if we have such a function psi then we say that your ideal c is Euclidean. So the question is if I have an ideal c in an integral domain r and if I consider any ideal in ideal class c if I consider the ideal class and if I consider any ideal is ideal class c b in c then d is also Euclidean. So it is we can see by tracing the definition we can see that not only c is Euclidean any ideal class c is also Euclidean c also Euclidean. And he himself proved that if you have a Euclidean function then every element in the ideal class is Euclidean and in fact the ideal class generates h k in in particular c class group is it. The question is here. So similarly for Euclidean number three he has the question whether the converse is true whether we have of course if we have an Euclidean function if we have an Euclidean function psi psi implies that class group is exactly so he himself asked the question what about the converse whether we have a converse or not and he get counter example for that he proved that the one leg of an example exists. So he proved that if we take the imaginary recorded it will use q squared minus d where d is equal to these values and they are not they don't have any Euclidean id class at all even though the class group is exactly. And with assuming g rh he proved that they are the one leg case. If we take any other number field other than imaginary recorded it will class psi implies we have such a function we have such a function and an ideal class so that such a function is an is an Euclidean function on the ideal class. So recently Graves proved that he wants to remove she wants to remove the assumption of g rh and she proved that suppose if I have ideal class c and it generates the class group and the ideal class c contains sufficient number of prime ideals we can see that it's kind of analyzing the mode screen so mode screen side here. So he proved that the ideal class c contains sufficient number of prime ideals more precisely if I have the number of prime ideals up to x mama p up to x so that p belongs to the ideal class and so that the canonical map from ok star to ok by p star ok star to ok by p star is onto if we have the canonical map is onto if the cardinality of such set is at least x over log x square and she proved that c is equal. And later Graves and Ramurthi proved that if I have using above results they proved that if I have a Galois extension with the Hilbert class field Galois extension with the escape with the Hilbert class field ok and the Hilbert class field is a billion over cube and with the rank of ok star is at least 4 then cyclic liberalize the ideal. So they proved at least I want to say that there are some subclass of they proved that if I have a case satisfying this condition then we have the converse point and there are refinement of these results recently. We are able to prove that explicitly giving a number field so that the amount transfer ideal classes has equal. So you want to our aim was to prove that we have explicit as I said earlier we have we have refinement of these results removing Hilbert class field conditions and specific there are results on specific classes. And so we consider bicoated fields of this form q square q and q square root k r where q and k, q, k are prime numbers and satisfying such condition they are magnetic provision. And we of course we assume that it has class number two but our aim was to prove that a non-principal ideal classes has including ideal classes. I want to say that non-principal ideal classes. And also we consider bicoated fields of this form q square root two, q square root p, q where p and q are one model of both. Here also we want to say that assuming h k two is two we want to say that non-principal ideal classes. So the proof needs a following lemma. We want to know the conductor of these clearly the bicoated field is an abelian extension the Kronaker-Wiebert theorem we know that we think it contains in a cyclotomic field we want to know the what is that in a cyclotomic field we want to know that the degree of that. So here by following following this map tower of fields you can clearly see that q square root q, q square root p are contained in K what we are considering. And because q is by assumption q is three model of four therefore it has conducted four q. K and R are both one model of four therefore it has conducted K R and therefore if we take the K, the number field K it should be contained in the composition of q square root four q and q square root K R. That means that K contains in q zeta, q square root four q and q square root K R. And in the converse part we can clearly see that by placing the formulae we can see that the composition is contained in K aspect. So the idea is K contained in the composition of q square root four q, zeta four q and q square root q, zeta K R and they are equal. So that will implies that the number field we are considering has conducted four q, K R. And the second one we wanted to prove that the explicit Hilbert glass field for the bi-quadratic field. And we consider the K prime, the bi-quadratic field q square root q square root K R square root K square root R we wanted to prove that K prime is a Hilbert glass field of K and because it has a K prime is a quadratic extension of K and we are assuming that it has class number two and we wanted to say that, we simply need to say that K prime over K is an undramified extension. And also we can see that both K and K prime are undrammed or infinite primes. So we only need to say that that ramification doesn't happen with the finite primes. So they clearly, they are ramified at the infinite primes. So to prove that they are ramified at the infinite primes we consider, we note that K is contained in K prime and K prime is contained in two zeta of four q, K R. And the only prime in the cyclotomy q is zeta of four q, K R where ramified is the prime radius lies above two q, K R. Therefore, if at all there is a prime in K prime which ramifies the prime value in K prime which ramified that should be a prime value lies in above two q, K R. Therefore, it is sufficient to show that the primes in K lies above two q, K R. They are all undramified prime. So here we consider the, I outlined the proof for the case for the prime two. We consider the tower of which, what we just considered earlier. Q zeta of four q are contained in K prime, K prime contains K, K contains q. And we consider the two in two p, two p, three p for respect to the primes in the respect to the square. And we want to show that p three over p two has ramification index one. And to show that we consider another bicoardic p q is L one, q square of K and square of R. And by the condition of K and R, we can see that this has ramification index one. And this, therefore the ramification index of p three over p two is, this is maximum two because the degree of that extension is two. Therefore, by multiplicity of the ramification index, we can see that the ramification index of p three over p two is maximum by two. And, but that means that p three over two is two, but p two over two is at least two because K contains K and R are one model of four. Therefore, this ramification index is at least two, that will implies that p three over p two has ramification index one. So that will prove that any prime ideal in K rise above two should be a, should be a, should be a ramification index. But this will prove that K prime over K is ramified at the prime ideal rise above two. And we can prove similarly for other primes q, K, R as well. So with this, with this we want to use this result to show our field is the field what we're considering is in fact, the non-threaded class is equal to L. So we have proved that if we have an orderly L number field with unit rank at least three, then if I have an integer U satisfying the conditions, then for any, then the set of all prime ideals of prime degree one with the norm of p is U modulo L. So that the canonical map from minus one comma EI to okay, V star is subjective is at least H whole of x square. So this condition is true for at least one of these even you do. So she proved that this is true for at least one of these even EI to EI three, if we choose cleverly even here. So to use this result, remember we want to say that we want to say that we want to know that non-prince-prided class is included. That means that we want to make sure that getting that this is true. We want to make sure that we choose U so that whenever I have a prime ideal P satisfying the conditions in the set, that should that prime ideal should be in the non-prince-prided class. So how do you make sure that? So to make sure that we consider we want to find a U so that whenever we have a prime ideal in U model of 4 QKR and if it's for the notation here, the initial degree of the initial if we have conditions. So I want to know whenever I have my client, our client is so I have a U so that if I have a prime ideal in U model of R and if I have a, so if I have a prime P in P and P prime in P bar, the initial degree of this is two, the initial degree of this is one and the initial degree of this is two. So we want to find such a U so that this is true for every prime in that model progression. So if we have such a P, why do we need such a conditions? If we have such a P, P prime satisfying the initial degrees of one and two, by class field theory we can make sure that this prime ideal P lies in the non-prince-prided class. So that is our model. So now we need to find a U such a U. So how do we find that? To do that, let's consider all prime ideals in K which speaks completely, I call this XK1, all prime ideals in K prime which speaks completely, K prime is HK1, the aspect of that. All prime ideals of all primes here which speaks completely in K prime. And we know that by again class field theory we can see that XK1, XK and HK prime the set XK1 as density one over four and the XHK1 as density one over eight. Therefore the difference has density one over eight. One over eight. So remember you want to know, you want to know, you want to find a prime integer U satisfying these conditions. So we translated the conditions what we record into this. So this is equal to saying that we look for primes so that we have that Q what we consider in the quadratic field is a residue model of P and then K and R are non-residue model of P. So here that's how the non-residue residue class plays a role here. So we want a P so that when we reduce Q model of P that should be a residue and K and R should be non-residue model of P. And of course we need the conditions on U, the conditions on the hypothesis of the clear series. That should satisfy the GCD of U and 60 QPR should be one and VCD of U minus one over two 16 KQR should be that. So these two conditions we can translate into nine and 10. And then we want to see whether we have an integer U satisfying nine and 10. So here we make use of Pollock's result. Pollock said that whenever we have a prime whenever we have prime B1 or equal to five we have a non-residue less than that fact. Whenever we, which is contrary to three model of four we make use of that result to find primes P1, P2, P3 all of them are non-residues model of QKR respectively. So that the above equation nine and 10 can be solved by solving the lab. So if we solve 11 and if you find the X naught it satisfies all the concurrent conditions in 11 and that say if it takes an integer which satisfies the final data sequence 11 that will serve the purpose of it. Then we can show that then we can show that any P in the arithmetic promotion U model of four QKR which satisfies all the conditions what we supposed to know to satisfy that a non-principle ideal class is Eglurian. So that will satisfy. We can check the computation. That means says that the non-principle ideal class what we are considering is in fact have sufficient number of prime ideals and that will prove that non-principle that is Eglurian. I stop here. One has just shown the generalized Riemann hypothesis how exactly does it enter into the. So here it proves that we want to know that that said A2 contains sufficient number of prime. That means we want to if it can go for A2 by definition we want to look for prime primitive roots. So we want to know how many times are there which looks like a primitive roots in that. So then we assume GRH to say that we have sufficient number of prime ideals so that it has a print. So if we look for that condition A2 when we discuss A2 is set of all elements so that when we go from A1 to A2 all elements okay over here that should be a certificate. So we have a, this says that we have a primitive ideal. So that means we want to know we want to know the set of all elements set of all prime ideal which have a primitive ideal. Want to know counting the all such prime ideal which are so it's basically the GRH comes into the picture to count the number of primitive roots. So it gives you a better quantitative estimates for these things. Yes. And also with the better order of magnitude. Yes, yes. Better order of magnitude so that the prime ideal itself belongs to you. Thanks, Subramanian. Thanks, Subramanian.