 So, I'm going to talk about off-gala extension. I will present this structure. I will also talk about the minimal structure off-gala structure on separable field extension. Then I will introduce normal basis for this field extension and then also application in finite field arithmetic. So, we start with off-gala extension, some background concerning off-algebra, co-commutative of algebra. The important fact is that here I have two structure, a structure of algebra and a structure of co-algebra. So, here I have the multiplication and here the multiplication. Here I have the unit and on the other side I have the co-mortification and the co-unit. I have also an empty pot here and I will say that the off-algebra is, the H here is an off-algebra if this diagram is commutative, which means that which is expressed here. And I also want the off-algebra to be co-commutative, which means that the application of the switch map with the co-mortification is right. We have a simple example. We have the example of off-algebra. Here it is an algebra, as we know. We just want to describe the co-algebra structure. We have the commutification here. Since this algebra here is generated by the element in J. So we just say what is the behavior of our G. G is sent to G times G. And the co-unit is sent every element in G to 1. And it is, the anti-pod sends an element to its inverse. So it's easy to see that we have the diagram which commutes. And since we have this definition for the co-mortification, it is easy to see that we have this variant. The co-mortification is a variant by the switch map. And we have an interesting, which is also basic example of off-algebra. We have the cube root of 2. And we consider the field obtained over Q. We have this linear map and the morphism of this field. They find here. And this is an information which will be easier to clear later. In fact, it is a sort of way to say that Q is the field or a variant of this thing. But we know that this is not Galois over Q. And it is a co-commutative of Galois, because it is, we have, sorry, we have from this under morphism, C and S, we construct H, which is a co-commutative of algebra, because it is a quotient ring, since it is an algebra. And the co-mortification, we just want to define it on the generator. On C, it is C times tensor C minus 3 times S tensor S. And on S, it is like this. The co-unit sends C to 1 and S to 0, the antipode stabilized at C, and sends S to its 2 minus S. So here, the commutativity of the diagram is given by this relation here. In fact, we make the quotient because we want it to satisfy the commutativity of the diagram, the diagram, sorry. And here we have this variant of the co-mortification because remember that we define C in this. And if we switch in C, it's OK. Because if we are C and C, here it's OK. But here we have C, we will get 0 here. But remember that the co-unit sends S to 0, so it's free work. We can talk about of Galois extension. So if I have a finite field extension, L over K, and a group of co-automorphism L, we have a linear representation here. And this linear representation can be extended to a morphism of algebra, which is a morphism of the two states. And we know that the dimension of this group algebra is the order of G. And we have this equivalence. The extension is Galois if and only if. This is an isomorphism. Clearly, we can identify the element as the multiplication by them. So it is included in this. This also includes it. We have this inclusion, but the dimension is the same. So it is an isomorphism. And from this, we generalize by saying that we will have a structure of up Galois here in this field extension. If we have a co-commutative of algebra H, such that the dimension as a K vector space is the same as the dimension of the degree of this field. And we have an up action like this, such that we have an isomorphism here. And it is also important to see that this, in fact, since H will act on L, we have the fact that L is H module, which will be interesting later. So this is what we call an up Galois extension. Sorry, for when you took G in the beginning, it was not like the whole Galois group, right? No. Some group of the Galois group. This one, sorry. You are talking about this G? Yeah. No, it is a group, maybe a short group if you want. But it is just a group of automorphism. This is to do the setting. I just want an extension and a group of automorphism. And what is important is here. Here is here, because this extension is Galois if and only if I say it. In the beginning, it is not the Galois group. So you imagine that I introduce the group algebra and the structure of this because this gives us an example. Here, we want an up Galois structure on this extension. This can be defined by this group algebra because everything is OK, as we said. This is a very easy example. Everything is OK. And here, we have also that H2 is restored that it is an up algebra. And we know the dimension. It is generated by c1, cns, dimension 3. So here, we have this which is a vector square dimension 3. And all of these go in here. We have the isomorphism. And it is co-commutative. Everything is OK. But we can see that, in fact, of Galois, extensions generalize Galois extension. Now, on separable, this is what we want to the topic which interests us. So what do we do? We have a finite separable field extension. So we can take the normal closure. And we know g is the Galois group of the top field, the normal closure on the base field. And we have g prime, which is the top, the normal closure to the intermediate L. And so it is easy to see that. g operates faithfully on this set of coset, coset mode g prime for this map. And the kernel, in fact, the kernel is included in g prime. And since we know that, we know the fundamental theorem of Nara theory, which gives us the subgroup from the subgroup we have a subfield, from the subfield we have a subgroup, this cannot be non-trivial. It is, of course, it is necessarily trivial. So the action is the full. And what is important for the following is the regular subgroup of the permutation group here, this permutation group here, the regular subgroup of this permutation group. And I recall the definition. We say that subgroup H is a regular subgroup. If the stabilizer of any element is trivial, and the permeability, here is H. It is not N. It is H. The permeability of H is the index of g prime. So we can see that any normal complement N of g prime in g is a regular subgroup. Because the stabilizer is in this intersection. But since it is a normal complement, this is trivial, which means that this is trivial. And we saw that here. And as the order of N is the index of g. This is an example. Now, so in fact, of Galois extension theory, start with Sweetler and another guy, I forgot the name. But for the case of separable extension, the work start by Kreitzer and Paragis. In fact, since when we have a separable extension, we can take the normal closure. So they prove that the of Galois structure on this extension can be studied only from the of Galois constructed from the normal closure. This is what we see here. Because the result says that we have an of Galois structure if we have a normal subgroup N of the permutation group, which is normalized by g. And the of Galois structure is given by this of Galois, co-commutative of Galois algebra. And you see, in fact, they use distance theory. We want of algebra for L over k. But defining structure on L, what we do? We go to the normal closure. We take the subgroup, which is the group algebra. And we look for the fix of algebra. This is the structure. Now, it is important when in of Galois theory, they need to count the number of subalgebra of a given of algebra. It is important for them. And here we have a result for that. Here, I give you the way they are counting. In fact, when we have a Galois extension, we have only one Galois group, which is normal. But when we have a field extension, you can have on this field extension many of Galois structure. You can have many of Galois structure. And here it is a way to count what we do, since things are encoded with the subgroup, the isomorphism class of subgroup in the way I just described. We are going to just first do a classification of the isomorphism class. And then we are going to count for each isomorphism class the structure coming from them. And then we have all the numbers, the whole number of Galois structure. This is the result. Here, what we want to do, we have a field, we have a group. The sub-of-algebra of this of algebra are the one coming from the subgroup of the group. There is it by Prespo-Rio and Vela. And then there is a projection between the subgroup of this end and the sub-of-algebra of the fix of algebra. And this is the way, this is yes. For the inebrious side, what is the half-algebra of type N? Of type N means that this half-type N means that, in fact, the half-Galois structure is generated by this of algebra. Because the group algebra is generated by N. That's when an half-Galois structure is generated by this of algebra. We say that it is of type N. And we know that we have a Galois correspondence when we have a Galois extension between the subgroup and the subfield. And it is a bijection. But here it is not a bijection. We are just an injection, not a bijection. So when we have a field extension, which is h of Galois, which means we have a h-option, we say from this half-algebra, we have this field fixed by h prime. And we have a correspondence between the sub-of-algebra to the subfield. And this correspondence has the same properties, but it is not bijective. And now, what we did with greater is to consider the minimal structure, the minimal structure idols for which there is only two sub-of-algebra, the top one and the base field. This is what we call the minimal of Galois structure. And we have this result. So as usual, we have the field extension, the normal closure, the Galois group to this field. And the minimal of Galois structure, remember that we are in the case where the off-Galois structure comes from, off-algebra, defined in the top. From the normal closure. So the n is very important. The n gives us the structure. And now, the minimal one will be the one for which the n has no proper non-trivial subgroup, normalized by g. And we take this particular case because it works. We want n to have no proper non-trivial if we take h to be a characteristically simple group. It works. This is what we describe here. And if we take also n characteristics that we have for h, we replace by the group of automorphism, we have at least two minimal of Galois structure. And here we have only one. And this only one comes from the previous result from Bayard. Because Bayard gives us these properties. We know how to count. And here, if we count, we can identify only one in the case. In the case, remember, in the case, not here. We do not have the case. I will show you the point. Yes, in this case, yes. Bayard continues as well by saying that when n is a Bernstein number, which means that n is prime with phi. And phi here is the other phi function. This means that this work for n prime, we have only one. And this only one is minimal. In fact, it uses the degree prime. And here we have an example to show that from the claim for group. We know what this means, automorphism group. And we see that we can write the Galois group as this semi-direct product. And the conditions are OK as what we just described to get a minimal of Galois structure. We know that for Galois extension, every separable Galois extension we degree to here. Every separable extension of degree to is Galois. This is true for the degree until 4 in case of Galois. Every field extension of degree at most 4 is of Galois. Because we have the case where n is prime. And you have the other case. But in the other case, the normal closure as a Galois group, which can be written as a semi-direct product. So we can conclude. But that is not true for 5. We have some degree 5 extension which are not of Galois. This separable field extension of degree less than 4 is Galois. Sorry? Is it true this way? What do you say? The separable field extension of degree less than 4? We? I think you can find like. It's hot Galois. I'll talk about oxythritol here. Can you understand what I'm doing? OK. Now I see choose a second part of the talk. And I want to talk about normal basis. What is a normal basis? It is a basis generated by an element we have. Sorry, we have a field extension, Galois field extension, which is Galois here, Galois field extension. And what we call a normal basis is a basis of L over K, which is generated by an element of L and the action of the Galois group. That is what we call a normal basis. This means that, in fact, our L is a module of rank 1 over the group algebra, sorry. And we know that we have this theorem, which is here you have the story of the theorem. This theorem says that such an element always exists. It was conjectured by Eisenstein, proved by Hansel. And this is for the story of the theorem. And it's also true for infinite extension. Now, some construction. I decided to show you some construction. But these construction are not mine, so by just to know that people are working on this. So we have an extension of local field, which is parable and total reunified. This is different. This is a fractional ideal of the ring or integer of this top field. And I do not see. Yes, I put the name. Biot and Ender construct, in fact, they give a characterization of elements who can generalize a normal basis by using the group of ramifications and the group of ramifications. And the number, the number, the lower D max here is a lower ramification number. But there are also from the group of ramifications, we have the lower ramification number and the upper ramification number. And this is how we characterize the man who can generalize, who can generate a normal basis. Also, yes, here, now we have the case where the extension, our extension for the literature, is of Galois, the structure is given by an H. And in fact, L is a module of rank 1 over H. So we have also a normal basis in this case. And here, Biot describes how characterize the element who can generalize a normal basis using the variation of the different. And now we have the case of normal integral basis. In fact, we have, again, an extension of locality. What is the definition of normal in the non-Galois setting? In the non-Galois, for example. Yes, this means that, in fact, we found an element in L such that the action of H in this element will give us a basis. That's definition of normal. Yeah. Here we have, in fact, a crater in 2022 describe or characterize how we can construct a normal integral basis. This means the normal basis of fractional ideal. Because the difference is the fractional. It is a variable. And this, it describes how to compute out a explicitly construct normal basis of this fractional ideal. Our construction, in fact, what we do here is to consider one-dimensional algebraic group. One-dimensional algebraic group over a perfect field with positive characteristics. We have t, a rational point on the algebraic group of order n. We have the quotient isogenic. If the fiber above this a is irreducible, then the field obtained from k by adjoining the fiber is a cyclic extension of degree n. And this fiber above a is a divisor which is invariant by translation. In fact, this translation is the action of a Galois group of an element, an element in this fiber. And in fact, since we have a group here, which is an algebraic group, which is a variety, we have some equation. So using the equation of the variety, we can construct a function which will generate this linear space, meaning the function which dominates minus d. We work divisor, we dominate minus d. And this space is n-dimensional vector space. And if the fiber above a does not meet the n torsion, then the evaluation of this basis at any b in the fiber will give us a normal basis of n over k. So I want to show you the argument of the proof. So the extension is Galois because we assume that the fiber above a is reducible, which means that the Galois group stabilize this. This means that this extension is Galois. And remember that our field is perfect, which means that any extension is separable. So it is Galois. It is normal because this fiber is assumed to be reducible. And it is normal because the base field is a perfect field. Now, using these things here, we prove that the Galois group is obedient because the Galois group is defined by a joint element which is in the n torsion. And when we do it for each element, we can switch by composition. We have that it is Galois. And we show also that the exponent sorry divide n. But in fact, what we do here is a kind of commercial theory for one dimensional algebra group. We are doing commercial theory. That's what we do. So what we have here, we have a pairing from this product to this. That this pairing is non-degenerate. That means that we have this isomorphism here. But this group is a cycle of order n, which is the standard algebra group of order n. And cycle, because here it is cycle. Now, we take a one dimensional algebra group. Why? Because we know the classification. We know that over perfect field, we have only three for a fine algebra group, the additive, the multiplicative, and the lucastoride. And in projective, we have just elliptic curves. So we do the work for all of this. That's the result I show you. What do we do? We know the fiber above A. It is this divisor. It is invariant over this, because the support of the divisor is stabilized by this map. And using the equation, as I said before, we can construct the function. But what is the way? For example, if I assume that the algebra group is an elliptic curve, we have this equation. From the equation, we have this function, which is normal, natural, small x and small y. We have the divisor, which shows us that we have this function will have a pole at the unit element. But we have also that x is a pair function, but y is not. So we know that we see that we have the function as divisor. I do put it here. The divisor of this function is the polar divisor is the unit element minus t. And if we consider the action of the algebra group or the action of this translation, we have a system of n function. In fact, this n function are linear independent, because if we assume that we have this 0 linear combination, this means that the function, this function here, we know it's divisor. But it is a divisor of a function on an elliptic curve, which means that the sum of the points in the support, which gives us 0. But if we sum, we have that nB, it is the unit point, which means that B is in the n torsion. Remember that here, we assume that the fiber does not meet the n torsion. So this gives us the fact that it is a basis. And we show we have shown that it is a normal basis. So now I want to show you the properties coming from a normal basis to do finite field arithmetic. But why are normal basis useful? Because if you have an arbitrary basis and we want to do the arithmetic here, we have to consider addition and multiplication addition is easy because it is constant otherwise. But multiplication can be more difficult. But in fact, what is difficult here is that we have to consider the product of the element in the basis. We have to consider this. And to do that, we have to introduce some scalar and see that we have some bi-linear form, CK, who help us to compute the product. Compute the product means that we compute the coordinate of this product in the base from the coordinate of the two components. But the fact is that when the base is a normal basis, in fact, all this bi-linear form comes only from 1, which is C0. And that is why it is important to use normal basis to do arithmetic in a finite field extension. So now I want to share with you some things about multiplication, finite field extension. First, I consider the general setting, which is I have a field, I have a two vector space, finite dimensional vector space, and I have a k-bi-linear map here. And what I want to do is to compute the product. At the end, it will be the product. But here, I start from general setting, just a bi-linear map. And from this model, bi-linear map, I have what I call a pure bi-linear map. It is a map in this field, in this way, to help for which we can compute the image of V1, V2 by using some linear form in this way. And when the phi 1 and phi 2 are the same, I say it is a pure symmetric k-bi-linear map. The k-bi-linear map, the space of k-bi-linear map at this dimension, which means that I do not define. Yes, I define here the algebraic complexity. In fact, I want to compute this. To compute this, I want to use pure bi-linear map. And what I call the complexity is the expression and the composition of this map into a sum of this map. If I have this decomposition, then it's really easy for me to compute. So it's very important to know how to express. And I call the complexity of this map the smallest integer such that this map is rkT. Time, it can be written as a sum of rkT pure map. Now, the symmetry. What I did for the general bi-linear map, I can do the same for the symmetry. But for a symmetric, we have an interpretation. So the algebraic symmetry complexity, it is a decomposition. But if we have this decomposition, this is equivalent to have to map a dash up and dash down, who will go from the vector space to this power of the base field and the power to the base field to this? Such that to compute the product, we just need to apply this map. And here we have some bound coming from what we know concerning pure bi-linear map and algebraic bi-linear map. Now, remember that we are interested in multiplication in finite field extension. We know that every finite field extension is a Galois extension. And the Galois group generated by the Frobenius automorphism. So what we are doing now is to consider a group. We have the group algebra. And we are inside the convolutional product who can allow us to define the component-wise product. And we are here when this will be the complexity. Remember, complexity is the number of pure map in the decomposition. And the diamond sigma is this map here, component-wise, for each part of the vector. And we see that a bi-linear map, where V is a KC, a KC module, T will be said to be C equivalence if we have this relation here. This means that when we place the same element of the group by the first component and the second, we can give a dissolution. And now, equivalent complexity, we define what is pure equivalent, pure C equivalent K bi-linear map. It is a bi-linear map given by the, remember that V is a KC module. So it is also a K vector space. But what we call a pure C equivalent is the case when we have W a vector in W. And we have 5, 1, and 5, 2, which are linear form, but as a KC linear map. We have pure symmetric also. And if we have here the equivalent complexity, in the same way as before, it is a decomposition, the number of elements is a decomposition. We have also the symmetric equivalent complexity. It is a number element in the decomposition, pure equivalent K-linear map. And for the case where L is, for the case when we have a finite extension, we denote by this the linear C equivalent complexity. It is the new QC of N. N is a degree. Now we have some bound. In fact, we have our group C. We have K. And we take L and M to finitely generate a KC module, such that N is free. Since N is free, we can write M as this direct sum with this vector space, W. And we have a KC form, KC linear form, 5, is defined by a family of K form. Why? Because when we apply 5 to N, we have 2, 2, 2, to get this. And in this way, we have also that for this family, in fact, all this family is described by just one element. I remember that E is the unit element in the group. In this way, we have an isomorphism of vector space between the space of KC form and the space of K form. This means that every KC form is described only by one K form. Now for the C equivalent binary element, we have this map here. It is given by a family of map here. But as before, we can generate all this by one. And in fact, we have an isomorphism between the C equivalent binary element and the binary element. And the result is that since this is an isomorphism, and we have this means that to get the complexity of this, of the map T here, in fact, this complexity comes from this, or is bounded by the complexity of this. This is very important for the following. I recall that a C equivalent K binary element here, its complexity is bounded by the one coming from the K binary element associated to. And here, I just to convince you that in fact, symmetric equivalent complexity controls the multiplication. If I know the symmetric complexity and I know a basis of L over K, which means a normal basis, in fact, I can compute, remember that the complexity, giving the complexity means that we have dashed down and dashed this to map, dashed down and dashed up, who help us to compute the product. And since we have the base of the departure space and the variable space, we can use metrics and we can compute the product. The reason we just publish is the following. We have K, a finite field, and prime to the characteristic. We have this Galois-Cover, unit 5, between curve, as we integrate over the fifth K. We assume that the Galois group C of E cyclic, the Galois group E cyclic of order n, we have, you can remind what we did before. We have a point here, a rational point so that the fiber is reducible. And if you have R, an effective user on the, R is an effective user, it is on X. R is an effective divisor, sorry. No, R is an effective divisor on Y, such that R and the action of the Galois group on R are different for every element in the Galois group except the unit. And we have D, a divisor on the curve X. We write R, capital R for this system. It is the divisor generated from R by the action of the Galois group. And this is the inverse image of the divisor D. If we assume that we have this condition and this condition here, then the complexity of multiplication in the degree, because you know that we have the degree and extension of K. We have all the conditions to get this extension and the complexity of this extension, multiplication in this extension, is less than or equal to sigma, where sigma is, the multiplication is this small ring. So some idea for the proof. So we have, remember we are working with curve. With curve, we have the Riemann-Rohr divisor. And in fact, from the definition of the Riemann-Rohr divisor, we can see that this map is well defined because this function is divisor in the minute minus E. But what is E? E is a divisor and here is a local equation of the divisor. So this map is well defined. Here also it is well defined for the same reason. Here we put 2 because we have E to E squared. And this map, EB, the first one, E is subjective. Why is it subjective? Because the dimension of its image is equal to the dimension of L. And this comes from the requirement, the hypothesis. We assume that the degree of D is greater than this. And using Riemann-Rohr's theorem, this means that we have this. But since if we have this using Riemann-Rohr's theorem, this will give us what we want, which says that the dimension of the corner is equal to this. Now we can evaluate at R also. And we have these two maps which are well defined. And this map here is injective for the same reason. We assume in the theorem that we have this condition. But this condition means that we have this. This is some interpretation of the definition of R, D, capital R and E. But when you have a degree less than 0, which means that the linear space associated is trillium, which is just 0, but this is the corner of this map. Since the corner of 0, it means that it is injective. And now what do we do? What we want to do is to compute this product. We have the first fact is that since the order of C is prime to the characteristic of the field, this group, the group algebra, is a semi-simple ring. This means that any sub-module is a direct factor. This means that, in fact, here the corner of E b is a direct factor here. And remember that we saw that E b is sub-jective, which means that we have, in fact, a right inverse. Using the same kind of argument, we have that E r squared has also a left inverse. And when we have these two inverses, we can compute what we want. Remember that we have sigma, the complexity of multiplication here. We have two elements. We have to compute the product. What do we do? We apply, since we can go from here to this, we apply E b star times E b. We are here. Then we apply E squared. We come to this. And that E r squared is a left inverse. And we come back here. And we apply E b squared. We go here. So, in fact, we do nothing. But doing this, we come to this ring here. And in this ring, remember that we have this fact. Yes. That if I have a module, a kc module, which is a direct sum in this way. In fact, the complexity of multiplication in L will be given by the complexity of multiplication in w. And this is what we do. Because what we have is that r is this sum, which means that k r here is a direct sum of small k r. And this means that, in fact, multiplication in here has a vector space r. Sorry. So what we have, yes. This is it. So multiplication according to this comes from multiplication according to this. But when multiplication here bound, multiplication here. Because we have this sum. And it is the same thing we have here. Because capital R is obtained by the action of a group of small r. This means that the group algebra here, capital K of small r, gives us, in fact, capital R is a direct sum of the group algebra generated by small r. And when we have this in mind, we are done. Because when we come here, we want to do the multiplication. But here, we have the two elements here in this ring. And do multiplication in this ring. The complexity is bounded by multiplication is K of small r. We can do it using our hypothesis, which we know this sigma. And we are done. And we come back. What we do is to come back here and then apply eb. This means that, in fact, when we have two elements, we apply these two. And then we apply this. We apply this map here, the dash up and dash down. And then we come back to L to compute the product. But in fact, we can see that we have two maps, two linear maps, one to L, two kc to the sigma. And another map will come, will go from kc to the sigma to L. And these are linear maps. And these two linear maps help us to compute the product. This means that, in fact, the complexity of multiplication in L is less than this sigma. So the question is please. It was tonight. If you take, like, a philic session of finite pieces, say, fq n over fq. So we have sine q in group. So the Garouk group is sine x. In this case, how many of Garouk have in this case? Is it fair? Is it formula for that? Yes. In fact, we have to count. What you are saying is that it depends. If the, what I show here is that if the degree is prime, we have only one. See if the degree is prime. But if the degree is not prime, it can be Galois that have many of Garouk's structure. So for every sub extension, there is one, because it's targeting groups. It is a cyclic group. That's why I say, when the degree is prime, this means that we are a simple group. So no sub, no kind of thing. Normal sub. No normal subgroup, which means that no normal sub, no Galois sub extension. And yeah. I mean, there is an analog of finite fields in the global world, in that cyclotomic fields. Do I understand how to do similar things, multiplications, lessons in cyclotomic fields efficiently, because we know how to do them in finite fields? The question is from the? Is there a passage from finite fields to cyclotomic fields? Do I understand how things, how, I mean, are there similar questions and answers for cyclotomic fields? In fact, I do not see the work in this way. But you are talking about arithmetic in cyclotomic. I cannot answer to this question. Thank you. OK. There are no more questions. Let's thank the speaker again.