skew

. — This paper constitutes a ﬁrst attempt to do analysis with skew polynomials. Precisely, our main objective is to develop a theory of residues for skew rational functions (which are, by deﬁnition, the quotients of two skew polynomials). We prove in particular a skew analogue of the residue formula and a skew analogue of the classical formula of change of variables for residues. Résumé (Une théorie des résidus pour les fractions rationnelles tordues) Cet article constitue un premier pas en direction du développement de méthodes analytiques pour les polynômes tordus. Précisément, notre principal objectif est de développer une théorie des résidus pour les fractions rationnelles tordues (qui sont, par déﬁnition, les quotients de deux polynômes tordus). Nous démontrons en particulier des analogues tordus de la formule des résidus et de la formule classique de changement de variables.

polynomials also appear in the literature; for instance, one can cite Cohn's book [2] or Jacobson's book [7].
In the classical commutative case, polynomials are quite interesting because they exhibit at the same time algebraic and analytic aspects: typically, the Euclidean structure of polynomials rings has an algebraic flavor, while derivations and Taylor-like expansion formulas are highly inspired by analysis.However, as far as we know, analysis with Ore's polynomials has not been systematically studied yet.This article aims at laying the first stone of this study by extending the theory of residues to the so-called skew polynomials, which are a particular type of Ore polynomials.
Let K be a field equipped with an automorphism θ of finite order r.We consider the ring of skew polynomials K[X; θ] in which the multiplication is governed by the rule Xa = θ(a)X for a ∈ K.The first striking result of this article is the construction of Taylor-like expansions in this framework: we show that any skew polynomial f ∈ K[X; θ] admits expansions of the form: (1) for any given point z in a separable closure of K.Moreover, when r is coprime with the characteristic of K, we equip K[X; θ] with a canonical derivation and interpret the coefficients a n appearing in Equation (1) as the values at z of the successive divided derivatives of f (X).All the previous results extend without difficulty to skew rational functions, that are elements of the fraction field of K[X; θ]; in this generality, Taylor expansions take the form: (2) These results lead naturally to the notion of residue: by definition, the residue of f (X) at z is the coefficient a −1 in the expansion (2).Residues at infinity can also be defined in a similar fashion.
In the classical commutative setting, the theory of residues is very powerful because we have at our disposal many formulas, allowing for a complete toolbox for manipulating them easily and efficiently.In this article, we shall prove that residues of skew rational functions also exhibit interesting formulas, that are: -(cf.Theorems 3.2.1 and 3.2.2) a residue formula, relating all the residues (at all points) of a given skew rational function, -(cf.Theorems 3.3.2and 3.3.4)a formula of change of variables, making explicit how residues behave under an endomorphism of Frac(K[X; θ]).
Our theory of residues has interesting applications to coding theory as it allows for a nice description of duals of linearized Reed-Solomon codes (including Gabidulin codes) which have been recently defined by Martínez-Peñas [12].This application will be discussed in a forthcoming article [1].
J.É.P. -M., 2021, tome 8 This article is structured as follows.In Section 1, we recall several useful algebraic properties of rings of skew polynomials.Special attention is paid to the study of endomorphisms of K[X; θ] and of its fraction fields.In Section 2, we focus on Taylorlike expansions of skew polynomials and establish the formulas (1) and (2).Finally, the theory of residues (including the residue formula and the effect under change of variables) is presented in Section 3.
Convention. -Throughout this article, all the modules over a (possibly noncommutative) ring A will always be left modules, i.e., additive groups equipped with a linear action of A on the left.Similarly, a A-algebra will be a (possibly noncommutative) ring B equipped with a ring homomorphism ϕ : A → B. In this situation, B becomes equipped with a structure of (left) A-module defined by a • b = ϕ(a)b for a ∈ A, b ∈ B.

Preliminaries
We consider a field K equipped with an automorphism θ : K → K of finite order r.We let F be the subfield of K consisting of elements a ∈ K with θ(a) = a.The extension K/F has degree r and it is Galois with cyclic Galois group generated by θ.
We denote by K[X; θ] the Ore algebra of skew polynomials over K.By definition elements of K[X; θ] are usual polynomials with coefficients in K, subject to the multiplication driven by the following rule: (3) ∀a ∈ K, X • a = θ(a)X.
Similarly, we define the ring K[X ±1 ; θ]: its elements are Laurent polynomials over K in the variable X and the multiplication on them is given by (3) and its counterpart: In what follows, we will write A = K[X ±1 ; θ].Letting Y = X r , it is easily checked that the center of A is F [Y ±1 ]; we denote it by Z .We also set C = K[Y ±1 ]; it is a maximal commutative subring of A .We shall also use the notations A + , C + and Z + for K[X; θ], K[Y ] and F [Y ] respectively.
In this section, we first review the most important algebraic properties of A + and A , following the classical references [14,8,10,11,2,7].We then study endomorphisms and derivations of A + , A and some of their quotients as they will play an important role in this article.
1.1.Euclidean division and consequences.-As usual polynomials, skew polynomials are endowed with a Euclidean division, which is very useful for elucidating the algebraic structure of the rings A + and A .The Euclidean division relies on the notion of degree whose definition is straightforward.Definition 1.1.1.-The degree of a nonzero skew polynomial f = i a i X i ∈ A + is the largest integer i for which a i = 0.
By definition, the degree of 0 ∈ A + is −∞.(ii) There exist Q left , R left ∈ A + , uniquely determined, such that We underline that, in general, Q right = Q left and R right = R left .For example, in C[X, conj] (where conj is the complex conjugacy), the right and left Euclidean divisions of aX by X − c (for some a, c ∈ C) read: A further consequence is the existence of left and right gcd and lcm on A + .They are defined in term of ideals by: for f, g ∈ A + .A noncommutative version of Euclidean algorithm is also available and allows for an explicit and efficient computation of left and right gcd and lcm.
1.2.Fraction field.-For many applications, it is often convenient to be able to manipulate quotient of polynomials, namely rational functions, as well-defined mathematical objects.In the skew case, defining the field of rational functions is more subtle but can be done: using Ore condition [13] (see also [9, §10]), one proves that there exists a unique field Frac(A ) containing A and satisfying the following universal property: for any noncommutative ring A and any ring homomorphism ϕ : A → A such that ϕ(x) is invertible for all x ∈ A , x = 0, there exists a unique morphism ψ : Frac(A ) → A making the following diagram commutative: (5) Under our assumption that θ has finite order the construction of Frac(A ) can be simplified.Indeed, we have the following theorem.
For the proof, we will need the following lemma.Proof.-If f = 0, the lemma is obvious.Otherwise, the quotient A /f A is a finite dimension vector space over F .Hence, there exists a nontrivial relation of linear dependence of the form: In other words, there exists showing that f has a right multiple in Z .Multiplying the relation f g = N by g on the left, we get gf g = N g = gN .Simplifying now by g on the left, we are left with gf = N , showing that f has a left multiple in Z as well.
Proof of Theorem 1.2.1.-Clearly Frac(Z )⊗ Z A contains A .Let us prove now that it is a field.Reducing to the same denominator, we remark that any element of Frac(Z ) ⊗ Z A can be written as D −1 ⊗ f with D ∈ Z and f ∈ A .We assume that f = 0.By Lemma 1.2.2, there exists g ∈ A such that f g ∈ Z .Letting N = f g, one checks that N −1 ⊗ gD is a multiplicative inverse of D −1 ⊗ f .Consider now a noncommutative ring A together with a ring homomorphism ϕ : This proves that, if such an extension exists, it is unique.On the other hand, using that Z is central in A , one checks that the formula (6) determines a well-defined ring homomorphism Frac(A ) → A making the diagram (5) commutative.
The notion of degree extends without difficulty to skew rational functions: -Given an integer n ∈ Z and a Laurent polynomial C ∈ C written as C = i a i X i , we define θ(C) = i θ(a i )X i .The morphism θ extends to Frac(C ).For n 0 and C ∈ Frac(C ), we set: and, when C = 0, we extend the definition of N n to negative n by: We observe that N 0 (C) = 1 and N 1 (C) = C for all C ∈ C .Moreover, when n = r, the mapping N r is the norm from Frac(C ) to Frac(Z ).In particular N r (C) ∈ Frac(Z ) for all C ∈ Frac(C ).
) be a morphism of K-algebras.Then there exists a uniquely determined element C ∈ C + (resp.invertible (1) Conversely any element of C as above gives rise to a well-defined endomorphism of A + (resp.A , resp.Frac(A )).
Remark 1.3.2.-An endomorphism of Frac(A ) is entirely determined by Equation (7).Indeed, by definition, the datum of γ : Frac(A ) → Frac(A ) is equivalent to the datum of a morphism γ : A → Frac(A ) with the property that γ(f ) = 0 whenever f = 0.Moreover, in the above equivalence, γ appears as the restriction of γ to A .This shows, in particular, that γ is determined by its restriction to A .
Proof of Theorem 1.3.1.-Unicity is obvious since C can be recovered thanks to the formula C = γ(X)X −1 .We first consider the case of an endomorphism of A + .Write γ(X) = i c i X i with c i ∈ K. Applying γ to the relation (3), we obtain: for all a ∈ K. Identifying the coefficients, we end up with c i θ i (a) = c i θ(a).Since this equality must hold for all a, we find that c i must vanish as soon as i ≡ 1 (mod r).Therefore, γ(X) = CX for some element C ∈ C + .An easy induction on i then shows that γ(X i ) = N i (C)X i for all i, implying eventually (7).Conversely, it is easy to check that Equation (7) defines a morphism of K-algebras.
For endomorphisms of A , the proof is exactly the same, except that we have to justify further that C is invertible.This comes from the fact that X γ(X −1 ) has to be an inverse of C. (1) We notice that the invertible elements of C are exactly those of the form aY n with a ∈ K, a = 0 and n ∈ Z.We now come to the case of endomorphisms of Frac(A ).Writing γ(X) = f D −1 with f ∈ A + and D ∈ Z + and repeating the proof above, we find that f X −1 ∈ C .Thus γ(X) = CX with C ∈ Frac(C ).As before, C cannot vanish because it admits X γ(X −1 ) as an inverse.From the fact that γ is an endomorphism of K-algebras, we deduce that γ |A is given by Equation (7).Conversely, we need to justify that the morphism γ defined by Equation ( 7) extends to Frac(A ).After Remark 1.3.2, it is enough to check that γ(f ) = 0 when f = 0, which can be seen by comparing degrees.
For C ∈ Frac(C ), C = 0, we let γ C : Frac(A ) → Frac(A ) denote the endomorphism of Theorem 1.3.1 (X → CX).When C lies in C + (resp.when C is invertible in C ), γ C stabilized A + (resp.A ); when this occurs, we will continue to write γ C for the endomorphism induced on A + (resp.on A ).We observe that γ C takes Y to: and, therefore, maps Frac(Z ) to itself.In other words, any endomorphism of K-algebras of Frac(A ) stabilizes the center.This property holds similarly for endomorphism of A + and endomorphisms of A .
-If γ C is an endomorphism of C -algebras, it must act trivially on Z , implying then (ii).By Hilbert's Theorem 90, if C ∈ Frac(C ) has norm 1, it can be written as θ(U )/U for some U ∈ Frac(C ), U = 0; (iii) follows.Finally it is routine to check that (iii) implies (i).
For endomorphisms of A + and A , Proposition 1.3.3can be made more precise.
Proposition 1.3.4.-For C ∈ C , the following assertions are equivalent: Proof.-The proof is the same as that of Proposition 1.3.2.Morphisms between quotients.-Let N ∈ Z + be a nonconstant polynomial with a nonzero constant term.The principal ideals generated by N in A + and A respectively are two-sided, so that the quotients A + /N A + and A /N A inherit a structure of K-algebra.By our assumptions on N , they are moreover isomorphic.
We consider in addition a commutative algebra Z over Z .We let θ act on Z + ⊗ Z C by id ⊗θ and we extend the definition of γ C to all elements C ∈ Z ⊗ Z C .Namely, for C as above, we define γ C : Moreover, the following assertions are equivalent: Proof.-The proof is entirely similar to that of Theorem 1.3.1 and Proposition 1.3.3.Note that, for the point (iii), Hilbert's Theorem 90 applies because the extension As an example, let us have a look at the case where Z = Z and N 1 and N 2 have Y -degree 1. Write N 1 = Y − z 1 and N 2 = Y − z 2 with z 1 = 0 and z 2 = 0.By Theorem 1.3.6,any morphism γ : A /N 1 A → A /N 2 A has the form X → cX for an element c ∈ K with the property that: Obviously, Equation (8) implies that c does not vanish.Hence, any morphism γ as above is automatically an isomorphism.Moreover, Equation ( 8) again shows that z 1 /z 2 must be a norm in the extension K/F .Conversely, if z 1 /z 2 is the norm of an element c ∈ K, the morphism γ C induces an isomorphism between A /N 1 A to A /N 2 A .We have then proved the following proposition.-For j ∈ Z, we define the section operator σ j : A → C by the formula: J.É.P. -M., 2021, tome 8 For 0 j < r and f ∈ A , we notice that σ j (f ) is the j-th coordinate of f in the canonical basis (1, X, X 2 , . . ., X r−1 ) of A over C .When j 0, we observe that σ j takes A + to C + and then induces a mapping A + → C + that, in a slight abuse of notations, we will continue to call σ j .
Lemma 1.3.8.-For f ∈ A , C ∈ C and j ∈ Z, the following identities hold: Proof.-It is an easy checking.Lemma 1.3.8ensures in particular that σ 0 is C -linear and the σ j 's are Z -linear for all j ∈ Z. Consequently, for any integer j, the operator σ j induces a Frac(C )-linear mapping Frac(A ) → Frac(C ).Similarly, for any N ∈ Z and any integer j, it also induces a (Z /N Z )-linear mapping A /N A → C /N C .Tensoring by a commutative Z -algebra Z , we find that σ j induces also a (Z /N Z )-linear mapping In a slight abuse of notations, we will continue to denote by σ j all the extensions of σ j defined above.
It is worth remarking that the section operators satisfy special commutation relations with the morphisms γ C , namely: Applying now σ j , we end up with . This proves the lemma.
Using Lemma 1.3.9, it is possible to construct some quantities that are invariant under all γ C , that is, after Theorem 1.3.1 or 1.3.6,under all morphisms of K-algebras.Precisely, for a tuple of integers (j 1 , . . ., j m ) ∈ Z m , we define: (i) If γ is a morphism of K-algebras, then γ commutes with σ j1,...,jm as soon as (ii) If γ is a morphism of C -algebras, then γ commutes with σ j1,...,jm as soon as Proof.-By Theorem 1.3.1 or 1.3.6, it is enough to prove the Proposition when γ = γ C for some C. By Lemma 1.3.9,combined with the relation (for j, j ∈ Z), we find: The first assertion follows while the second is a direct consequence of the characterization of morphisms of C -algebras given by Proposition 1.3.4 or Theorem 1.3.6.
1.4.Derivations over Ore polynomials rings.-Given a (possibly noncommutative) ring A and a A-algebra B, we recall that a derivation ∂ : A → B is an additive mapping satisfying the Leibniz rule: Classification.-As we classified endomorphisms of K-algebras in Section 1.3, it is possible to classify K-linear derivations over Ore rings.For C ∈ Frac(C ), and n ∈ Z, we define: We observe that Tr r is the trace from Frac(C ) to Frac(Z ).In particular, it takes its values in Frac(Z ).
) be a K-linear derivation, i.e., a derivation whose ring of constants contains K.Then, there exists a uniquely determined

Conversely, any such C gives rise to a unique derivation of
Let ∂ be a K-linear derivation as in the proposition.Applying ∂ to the equality for all index i, showing that c i has to vanish when i ≡ 1 (mod r).
J.É.P. -M., 2021, tome 8 Thus ∂(X) = CX for some C ∈ C + (resp.C ∈ C ).A direct computation then shows that: and, more generally, an easy induction leads to ∂(X i ) = Tr i (C)X i for all i 0. In the cases of A and Frac(A ), we can also compute ∂(X i ) when i is negative.For this, we write: As before, an easy induction on i then gives ∂(X i ) = Tr i (C) X i for all negative i.We deduce that Equation ( 9) holds.
For the converse, we first check that Equation ( 9) defines a derivation on A .In the case of Frac(A ), we need to justify in addition that ∂ (given by Equation ( 9)) extends uniquely to Frac(A ).This is a consequence of the following formula: which holds true because D is central.
Let ∂ C : Frac(A ) → Frac(A ) denote the derivation of Proposition 1.4.1.We have: We deduce that ∂ C stabilizes Frac(C ) and Frac(Z ) and acts on these rings as the derivation -For C ∈ Frac(C ), the following assertions are equivalent: Proof.-The equivalence between (i) and (ii) is clear by what we have seen before.If (ii) holds, then the additive version of Hilbert's Theorem 90 ensures that C can be written as for all integer i.By K-linearity, we deduce that An interesting feature of the derivation ∂ Y,can is that its p-th power vanishes (as we can check easily by hand).This property will be very pleasant for us in Section 2 when we will define Taylor expansions of skew polynomials.Unfortunately, it seems that there is no simple analogue of ∂ Y,can when p divides r, as shown by the following proposition.
Proof.-Our assumptions ensure that ∂ C induces the derivation d/dY on Frac(C ).For i ∈ {1, 2, . . ., p}, we define A direct computation shows that: (10) In particular, we deduce that C i ∈ Frac(C ) for all i.We claim that C has at most a simple pole at 0. Indeed, if we assume by contradiction that C has a pole of order v 2 at 0, we would deduce that C i has a pole of order vi at 0 for i ∈ {1, . . ., p}, contradicting the fact that C p vanishes.We can then write C = aY −1 + O(1) with a ∈ K.The recurrence relation (10) shows that, for i ∈ {1, . . ., p}, we have ), where the coefficients a i 's satisfy: In order to guarantee that a p vanishes, we then need a ∈ F p ⊂ F .Taking the trace, we obtain -Following Section 1.3, we propose to classify K-linear derivations A /N 1 A → A /N 2 A .However, we need to pay attention in this case that such derivations are only defined when A /N 2 A is an algebra over A /N 1 A , that is when N 1 divides N 2 .As in Section 1.3, we consider in addition a commutative Z -algebra Z .We extend readily the definition of Proposition 1.4.6.-Let N 1 , N 2 ∈ Z + be two nonconstant polynomials with nonzero constant terms.We assume that Moreover, the following assertions are equivalent: Proof.-It is entirely similar to the proofs of Propositions 1.4.1 and 1.4.2.

Taylor expansions
The aim of this subsection is to show that skew polynomials admit Taylor expansion around any closed point of F and to study its properties.Besides, when r is coprime to p, we will prove that the Taylor expansion is canonical and given by a Taylor-like series involving the successive divided powers of the derivation ∂ can .
2.1.The commutative case: reminders.-By definition, we recall that the Taylor expansion of a Laurent polynomial f ∈ C around a point c ∈ K, c = 0 is the series: where T is a formal variable playing the role of Y +c and the notation f [n] stands for the n-th divided derivative of f defined by: We recall also that the n-th divided derivative satisfies the following Leibniz-type relation: from what we deduce that the mapping C → K[[T ]] taking a Laurent polynomial to its Taylor expansion is a homomorphism of K-algebras.Even better, it induces an isomorphism: More generally, let us consider an irreducible separable polynomial N ∈ C .Let also c ∈ C /N C be the image of X, which is a root of N by construction.In this generality, the Taylor expansion around c is well-defined and induces a homomorphism of K-algebras C → (C /N C )[[T ]], inducing itself an isomorphism: The image of N under this isomorphism is a series of valuation 1.As a consequence, twisting by an automorphism of there exists an isomorphism of K-algebras: mapping N to T and inducing the identity map C /N C → C /N C after reduction modulo N on the left and modulo T on the right.Moreover τ C N is uniquely determined by these properties.In addition, we observe that when N = Y −c is a polynomial of degree 1, the isomorphisms τ C Y −c and τ C c agree.It turns out that the existence of the unicity of τ C N continues to hold under the sole assumption that N is separable; this can be proved by noticing that N factors as a product of distinct irreducible factors N 1 • • • N m and, then, by gluing the corresponding τ C Ni using the Chinese Remainder Theorem.In this general setting, the inverse of τ C N can be easily described: it maps T to N and X ∈ C /N C to the unique root of N in lim ← −m>0 C /N m C which is congruent to X modulo N .The existence and the unicity of this root follows from Hensel's Lemma thanks to our assumption that N is separable: it can be obtained as the limit of the Newton iterative sequence: Of course, the above discussion is still valid when C is replaced by Z (and K is replaced by F accordingly).For any separable polynomial F ∈ Z , we then have constructed a well-defined isomorphism: We note that N remains separable in C , implying that τ C N is also defined.The unicity property ensures moreover that the following diagram is commutative: where the vertical arrows are the canonical inclusions.12) by adding a top row at the level of Ore rings.For now on, we fix a separable polynomial N ∈ Z .To simplify notations, we set: Here is our first theorem.
Theorem 2.2.1 (i) There exists an isomorphism of K-algebras τ N : ] mapping N to T and inducing the identity of A /N A after quotienting out by N of the left and T and the right.
(ii) Any isomorphism τ N satisfying the conditions of (i) sits in the following commutative diagram: In what follows, we shall say that a Laurent polynomial N ∈ Z is normalized if N ∈ Z + , N is monic and N has a nonzero constant coefficient.With this definition, any ideal of Z has a unique normalized generator.
Proof of Theorem 2.2.1.-The general strategy of the proof is inspired by the characterization of the inverse of τ N we gave earlier: we are going to construct the inverse of τ N by finding a root of N in A N .Without loss of generality, we may assume that N is normalized.
In addition, we choose and fix an element a ∈ K with Tr K/F (a) = 1.
As in Hensel's Lemma, we proceed by successive approximations in order to find a root of N in A N .Precisely, we shall construct by induction a sequence In what follows, we will often write C m for 1 + aZ m ∈ C + .We assume that Z m has been constructed for some m 1.The second condition we need to fulfill implies that Z m+1 must take the form Z m+1 = Z m + aN m Z for some Z ∈ Z + .The third condition then reads N (C m+1 X) ∈ N m+1 Z + .
Let us first prove that N (C m+1 X) lies in Z + .For this, we observe that The latter is obviously a polynomial in X r with coefficients in K. Since it is moreover stable by the action of θ, its coefficients must lie in F and we have proved that (C m+1 X) r ∈ Z + .The fact that N (C m+1 X) ∈ Z + follows directly.
It remains now to ensure that N (C m+1 X) is divisible by N m+1 for a suitable choice of Z.For any positive integer n, we have the following sequence of congruences modulo N m+1 : We are then reduced to prove that there exists a polynomial Z ∈ Z + such that S + X r N Z ≡ 0 (mod N ), which follows from the fact that X r N is coprime with N .The sequence (Z m ) m 1 we have just constructed defines an element Z ∈ Z N .We set C = 1 + aZ; it is an element of C N .Besides, by construction, CX is a root of N , in the sense that N (CX) = 0.This property together with the fact that C is invertible in C N ensure that the map ι : A /N A → A N , X → CX is a well-defined morphism of K-algebras (see also Section 1.3).Moreover, since C ≡ 1 (mod N ), ι reduces to the identity modulo N .Mapping T to N , one extends ι to a second morphism of K-algebras: τ The latter induces the identity after reduction modulo T on the left and N on the right.Since the source and the target are both separated are complete (for the T -adic and the N -adic topology respectively), we conclude that τ has to be an isomorphism.We finally define τ N = τ −1 and observe that it satisfies all the requirements of the theorem.
it is an endomorphism of K-algebras of A N .Besides, thanks to the unicity result in the commutative case, τ N,1 and τ N,2 have to coincide on C N .This means that γ is in fact a morphism of C N -algebras.Applying Theorem 1.3.6 and passing to the limit, this implies the existence of an invertible element this equality, we find that the proposition holds with V = τ N,2 (U ).
When p does not divide r, the situation is even better because one can select a canonical representative for τ N .Precisely, we have the following theorem.
Theorem 2.2.5.-We assume that p does not divide r.
(i) The homomorphism of K-algebras: N (Y )/Y is congruent to 1 modulo T and raising it to the power 1/r makes sense in (Z /N Z )[[T ]] thanks to the formula: Observe that all the coefficients c n 's lie in Z[1/r] and so can be safely reduced modulo p if p does not divide r.
Proof of Theorem 2.2.5.-The first part of the theorem is easily checked.We now assume that we are given two isomorphisms of K-algebras τ N,1 , τ N,2 : ] satisfying the conditions of the theorem.For i ∈ {1, 2}, we write τ Taking the trace from K to F , we end up with

2.3.
Taylor expansions of skew rational functions.-Recall that we have defined in Section 1.2 the fraction field Frac(A ) of A and we have proved that Frac(A ) = Frac(Z ) ⊗ Z A (see Theorem 1.2.1).

Taylor expansion at central separable polynomials.
-For a given separable polynomial N ∈ Z , the isomorphism τ N of Theorem 2.2.1 extends to an isomorphism Frac(Z ) ⊗ Z A N → (A /N A )((T )) and we can consider the composite: where the first map is induced by the natural inclusion A → A N .By definition TS N (f ) is called the Taylor expansion of f around N .We notice that it does depend on a choice of the isomorphism τ N .However, one can form several quantities that are independent of any choice and then are canonically attached to f ∈ Frac(A ) and N as before.Many of them are actually given by Corollary 2.2.4; here are they: (i) the order of vanishing of f at N , denoted by ord N (f ); it is defined as the T -adic valuation of TS N (f ), (i') for j ∈ Z, the j-th partial order of vanishing of f at N , denoted by ord N,j (f ); it is defined as the T -adic valuation of σ j (TS N (f )), (ii) the principal part of f at N , denoted by P N (f ); it is the element of A /N A defined as the coefficient of T ord N (f ) in the series TS N (f ), (ii') for j ∈ Z, the j-th partial principal part of f at N , denoted by P N,j (f ); it is the element of C /N C defined as the coefficient of T ord N,j (f ) in the series σ j (TS N (f )), J.É.P. -M., 2021, tome 8 (iii) the 0-th section of TS N (f ), namely σ 0 (TS N (f )), (iii') more generally, any quantity of the form σ j1,...,jm (TS N (f )) with j 1 + • • • + j m ≡ 0 (mod r).
The previous invariants are related by many relations, e.g.: ord N (f ) = min ord N,0 (f ), . . ., ord N,r−1 (f ) , ord N,j+r (f ) = ord N,j (f ), -P N (f ) = j P N,j (f ) X j , where the sum runs over the indices j ∈ {0, 1, . . ., r−1} for which ord N,j (f ) = ord N (f ), -P N,j+r (f ) = X r P N,j (f ), ord N (f g) ord N (f )+ord N (g) and equality holds as soon as A /N A is a division algebra, (2)  - We say that f has no pole at N when ord N (f ) 0. It has a simple pole at N when ord N (f ) = −1.Generally, we define the order of the pole of f at N as the opposite of ord N (f ).

Taylor expansion at nonzero closed points.
-In a similar fashion, one can define the Taylor expansion of a skew rational function at a nonzero closed point z of F .When z is rational, i.e., z ∈ F , z = 0, we simply set TS z = TS Y −z .
Otherwise, the construction is a bit more subtle.Let F s be a fixed separable closure of F and let z ∈ F s , z = 0. Let also N ∈ Z + be the minimal polynomial of z.We have recalled in Section 2.1 that the Taylor expansion around z defines an isomorphism: which is characterized by the fact that it sends Y to z + T .In general, τ C z does not agree with τ C N but there exists a series ] taking T to S z (and acting trivially on the coefficients).The latter extends to an endomorphism of (A /N A )[[T ]], that we continue to call ϕ z .By construction, the following diagram is commutative: ] is an isomorphism satisfying the conditions of Theorem 2.2.1.(i).It is worth noticing that the morphisms of the form ϕ z • τ N can be characterized without any reference of τ N .
Proposition 2.3.1.-Given z ∈ F s , z = 0, we have the following equivalence: a mapping τ z : ] is of the form ϕ z • τ N (where τ N satisfies the condition of Theorem 2.2.1) if and only if τ z is a morphism of K-algebras, τ z (X) ≡ X (mod T ) and τ z (Y ) = z + T . Proof.
-If τ z = ϕ z • τ N , it follows from the conditions of Theorem 2.2.1 that τ z is morphism of K-algebras which induces the identity modulo T .Hence τ z (X) ≡ X (mod T ).Moreover, by the second part of Theorem 2.2.1, we know that τ N coincides with τ C N on C n .Therefore τ z has to agree with Conversely, let us assume that τ z satisfies the condition of the proposition.We have to check that τ N = ϕ −1 z •τ z satisfies the conditions of Theorem 2.2.1.The fact that τ N is a morphism of K-algebras is obvious.The assumption τ z (X) ≡ X (mod T ) ensures that τ N acts as the identity modulo T .Finally, the hypothesis τ z (Y ) = z + T implies that τ z coincides with τ C z on C N .Hence: = T and we are done.Definition 2.3.2.-Given z ∈ F s , z = 0 as before, we say that a morphism τ : From now on, we fix a choice of an z-admissible morphism τ z .Accordingly, we define TS z as the composite: Like TS N , the morphism TS z depends upon some choices but some quantities attached to it are canonical, as the order of vanishing at z, the principal part at z, etc.For f ∈ Frac(A ) and j ∈ Z, we use the transparent notations ord z (f ), ord z,j (f ), P z (f ) and P z,j (f ) to refer to them.
Proposition 2.3.4.-Let z ∈ F s , z = 0 and let N ∈ Z + be its minimal polynomial.Then: ') P z,j (f ) = P N,j (f ) for all j ∈ Z.
J.É.P. -M., 2021, tome 8 for all nonnegative integer n.With this definition, it is formal to check that: However, this result does not give much information because C is hard to compute (and the ∂ [n] C 's are even harder) and depends heavily on z.Typically, Proposition 1.4.4 shows that C cannot be rational unless r is coprime with p.Nevertheless, when p does not divide r and τ z is well chosen, we shall see that the computation of C and ∂

[n]
C can be carried out and yields eventually a simple interpretation of the Taylor coefficients.
From now on, we assume that p does not divide r.By Theorem 2.2.5, we know that there is a canonical choice for τ z , called τ z,can .The corresponding element C is: . Therefore: In particular, we observe that C can is rational and, even better, ∂ Ccan is equal to the canonical derivative ∂ can we introduced in Definition 1.4.3.Its divided powers (defined by Equation ( 16)) also have a simple expression: where the coefficients c n,i 's all lie in Z[ 1 r ] and, consequently, can be reduced modulo p without trouble.With these inputs, Equation (17) reads: which can be considered as a satisfying skew analogue of the classical Taylor expansion formula.

A theory of residues
The results of the previous section lay the foundations of a theory of residues for skew polynomials.The aim of the present section is to develop it: we define a notion of residue at a closed point of F for skew rational functions and then prove the residue formula and study how residues behave under change of variables.
Throughout this subsection, we fix a separable closure F s of F , together with an embedding K → F s .For z ∈ F s and C ∈ Frac(C ), we will write res z (C•dY ) for the (classical) residue at z of the differential form C•dY .
J.É.P. -M., 2021, tome 8 3.1.Definition and first properties.-We recall that, for z ∈ F s , z = 0, we have defined in Section 2.3 a (non canonical) morphism of K-algebras: where N ∈ Z + is the minimal polynomial of z.On the other hand, there is a natural embedding Z /N Z → F s obtained by mapping Y to z. Extending scalars from F to K, it extends to a second embedding We observe that the codomain of ι z , namely K ⊗ F F s , is naturally isomorphic to a product of r copies of F s via the mapping: -For z ∈ F s , z = 0, and f ∈ Frac(A ), we define: -the skew residue of f at z, denoted by sres z (f ), as the coefficient of T −1 in the series TS z (f ); it is an element of A /N A , -for j ∈ {0, . . ., r−1}, the j-th partial skew residue of f at z, denoted by sres z,j (f ), as: Here are two important remarks concerning residues.First, we insist on the fact that both sres z (f ) and sres z,j (f ) do depend on the choice of the z-admissible morphism τ z (used in the definition of TS z ) in general.However, Corollary 2.2.4 shows that sres z (f ) and sres z,j (f ) are defined without ambiguity when f has (at most) a simple pole at z. Besides, when p does not divide r, there is a distinguished choice for TS z (see Theorem 2.2.5), leading to distinguished choices for sres z and sres z,j .In the sequel, we will denote them by sres z,can and sres z,j,can .
Second, we observe that, the collection of all the partial skew residues sres z,j (f )'s captures as much information as sres z (f ), given that sres z (f ) is determined by its sections σ j (sres z (f ))'s with 0 j < r thanks to the formula:

3.1.1.
Residues at special points.-It will be convenient to define residues at 0 and ∞ as well.For residues at 0, we recall that we have defined in Section 2.3.3 a canonical Taylor expansion map around 0: TS 0 : Frac(A ) −→ K((X; θ)).Definition 3.1.2.-For f ∈ Frac(A ) and j ∈ {0, 1, . . ., r−1}, we define the j-th partial skew residue of f at 0, denoted by sres 0,j (f ), as the coefficient of X j−r in the series TS 0 (f ).
Residues at infinity are defined in a similar fashion.Let X be a new variable and form the skew algebra A = K[ X ±1 ; θ −1 ].Clearly A is isomorphic to A by letting X correspond to X −1 .We then get a map: where the second map is the morphism TS 0 for A .Definition 3.1.3.-For f ∈ Frac(A ) and j ∈ {0, 1, . . ., r−1}, we define the j-th partial skew residue of f at ∞, denoted by sres ∞,j (f ), as the opposite of the coefficient of X r−j in the series TS ∞ (f ).
Unlike sres z,j (f ), the partial skew residues sres 0,j (f ) and sres ∞,j (f ) do not depend on any choice and so are canonically attached to f .

Commutative residues.
-The skew residues we just defined are closely related, in many cases, to classical residues of rational differential forms.In order to state precise results in this direction, we need extra notations.We observe that the map res z defines by restriction an F -linear mapping Z dY → F s .Tensoring it by K over F , we obtain a K-linear map ρ z : C dY → K ⊗ F F s .Letting res : (C /N C )((T )) → C /N C be the map selecting the coefficient in T −1 , one checks the two following formulas: Proof.-By definition, sres z,0 (f ) = ι z •σ 0 •sres z (f ).Applying Lemma 1.3.9 and passing to the limit, we find that the isomorphism τ z commutes with σ 0 .Hence σ 0 • sres z is equal to the compositum: Composing further by ι z on the left, we get the proposition.Proposition 3.1.4implies in particular that sres z,0 (f ) does not depend on any choice and thus is canonically attached to f and z.According to Corollary 2.2.4,there are other invariants which are canonically attached to sres z (f ).A family of them consists of the σ j1,...,jm (sres z (f ))'s for j 1 , . . ., j m ∈ Z with j 1 + • • • + j m ≡ 0 (mod r).However, these invariants seem less interesting; for example, they do not define additive functions on Frac(A ).
Under some additional assumptions, other partial skew residues are also related to residues of rational differential forms.If z ∈ {0, ∞} or ord z,j (f ) −1, then: Proof.-When z ∈ {0, ∞}, the proposition can be easily checked by hand.Let us now assume that ord z,j (f ) −1.By Lemma 1.3.9,we know that Moreover, from the fact that τ z induces the identity modulo N , we deduce that C ≡ 1 (mod T ).Consequently τ z commutes with σ j modulo T .The end of the proof is now similar to that of Proposition 3.1.4.

3.2.
The residue formula.-In the classical commutative setting, the theory of residues is very powerful because we have at our disposal many formulas, allowing for a complete toolbox for manipulating them easily and efficiently.We now strive to establish analogues of these formulas in our noncommutative setting.We start by the "commutative" residue formula.
Theorem 3.2.1.-For f ∈ Frac(A ), we have: Proof.-Since β is an isomorphism, it is enough to prove that Writing C = σ 0 (f ) ∈ C , Proposition 3.1.4asserts that: The theorem them follows from the classical residue formula applied to the θ j (C)'s for j varying between 0 and r−1.
The reader might be a bit disappointed by the previous theorem as it only concerns 0-th partial skew residues and it reduces immediately to the classical setting.Unfortunately, in general, it seems difficult to obtain a vanishing result involving skew residues since the latter might be not canonically defined.There is however an important special case for which such a formula exists and can be proved.Theorem 3.2.2.-Let f ∈ Frac(A ).We assume that f has at most a simple pole at all points z ∈ F s , z = 0. Then: z∈F s {∞} sres z,j (f ) = 0 for all j ∈ {0, 1, . . ., r−1}.
The case of canonical residues also deserves some attention.As before, the main input is a formula relating the partial skew residues sres z,j,can (f ) to classical residues.We consider a new variable y and form the commutative polynomial ring K[y] and its field of fractions K(y).We embed Frac(C ) into K(y) by taking Y into y r .We insist on the fact that y is not X or, equivalently, K(y) is not Frac(A ): our new variable y commutes with the scalars.Since K(y) is a genuine field of rational functions, it carries a well-defined notion of residue.For f ∈ K(y) and z ∈ F s , we will denote by res z (f •dy) the residue at f of the differential form f •dy.Similarly the map ρ z extends to K(y)dy.Performing the change of variable y → Y = y r , we obtain the relations: which hold true for any C ∈ C and any z ∈ F s .Proposition 3.2.3.-We assume that p does not divide r.For f ∈ Frac(A ), j ∈ {0, 1, . . ., r−1} and z ∈ F s , z = 0, we have: where ζ is any r-th root of z.
Proof.-Set C can = τ z,can (X) X −1 .From Lemma 1.3.9,we know that: On the other hand, it follows from Theorem 2. Plugging this in (19), we obtain: (20) The main observation is that the twisting function (1 + T /z) j/r which is a priori only defined on a formal neighborhood of T = 0 (or, equivalently of Y = z) is closely related to a function of the variable y which is globally defined.Precisely, consider the local parameter t = y − ζ on a formal neighborhood of ζ.The relation y r = Y translates to (ζ + t) r = z + T .Dividing by z on both sides and raising to the power j/r, we obtain: , showing that our multiplier (1 + T /z) j/r is the Taylor expansion of the function ζ −j y j .
Equation (20) then becomes σ j τ z,can (f ) = τ z,can ζ −j y j σ j (f ) .Taking the coefficient in T −1 , we get: which is exactly the formula in the statement of the proposition.
J.É.P. -M., 2021, tome 8 Unfortunately, Proposition 3.2.3does not give an interesting vanishing result for canonical partial skew residues.Indeed, if we apply the residue formula to the differential form y j+r−1 σ j (f )•dy, we end up with: (21) Actually, this formula does not give any information because the sum on the left hand side can be refactored as follows: and each internal sum vanishes simply because ζ r =z ζ j = 0.In other words, the formula (21) holds equally true when sres ζ r ,j,can (f ) is replaced by any quantity depending only on ζ r .However, Proposition 3.2.3remains interesting for itself and can even be used to derive relations on partial skew residues of a skew rational function f .One way to achieve this goes as follows.Let f ∈ Frac(A ) and j ∈ {1, . . ., r−1}.We assume that we know a finite set Π = {z 1 , . . ., z n } containing the points z ∈ F s , z = 0 for which ord z,j (f ) < 0. We assume further, for each index i, we are given an integer n i with the guarantee that ord z,j (f ) −n i .For each i, we choose a r-th root ζ i of z i .Let P ∈ F s [y] be a polynomial such that, for all i, P (ζ i ) = ζ −j i and the derivative P (y) has a zero of order at least (n i − 1) at ζ i .This choice of P ensures that: for all index i.Thanks to Proposition 3.2.3,we obtain: sres zi,j,can (f ) = ρ ζi P (y) y j+r−1 σ j (f ) • dy .Now applying the residue formula with the function P (y) y j+r−1 σ j (f ), we end up with: The right hand side of the last formula can be computed explicitly on concrete examples (though determining a suitable polynomial P (y) might be painful if the order of the poles are large).For example, when ord z,j (f ) 0, the first summand ρ 0 P (y) y j+r−1 σ j (f ) • dy vanishes.
where, by definition, -We say that z is γ-regular if Z has no zero and no pole at Y = z.
-When z is γ-regular, we define γ z as the value taken by Z at the point Y = z.
For f ∈ Frac(C ) and z ∈ F s , we have the formula The aim of this subsection is to extend this relation to any f ∈ Frac(A ), replacing classical commutative residues by skew residues.

3.3.1.
A general formula.-Comparing skew residues at γ z and z is not straightforward because they do not live in the same space: the former lies in A /N 1 A , where N 1 is the minimal polynomial of γ z while the latter sits in A /N 2 A , where N 2 is the minimal polynomial of z.We then first need to relate A /N 1 A and A /N 2 A .For this, we remark that, as γ acts through the change of variables Y → Z on Z , it maps N 1 to a multiple of N 2 .Therefore it induces a morphism of K-algebras A /N 1 A → A /N 2 A .
Let z ∈ F s , z = 0 be a γ-regular point.
(i) For any admissible choice of τ γ z (see Definition 2.3.2) there exists an admissible choice of τ z such that: (ii) A skew rational function f ∈ Frac(A ) has a single pole at γ z if and only if γ(f ) has a single pole at f .When this occurs, Equation (22) holds for any admissible choices of τ γ z and τ z .
The following lemma will be used in the proof of Theorem 3.3.2.
J.É.P. -M., 2021, tome 8 Proof.-When f ∈ (A /N A )[[T ]], both sides of Equation ( 23) vanish and the conclusion of the lemma holds.Moreover, since ψ and res are both K-linear, it is enough to establish the lemma when f = T i with i < 0. Equation (23) then reads res S i dS/dT = res T i and is a direct consequence of the classical formula of change of variables for residues.
Proof of Theorem 3.3.2.-We begin by some preliminaries.As before, we define and consider the mapping: We extend it by K-linearity to a map ϕ We have: We deduce from this equality that the diagram Let us now consider a z 1 -admissible choice of τ z1 and call it τ 1 for simplicity.It is a prolongation of τ C 1 .Besides, by Theorem 1.3.6,there exists ] such that τ 1 (X) = C 1 X.The properties of τ 1 ensure in addition that C 1 ≡ 1 (mod T ) and that: (see also Remark 2.3.3).Applying ϕ C to this relation, we find: (24) (26) Combining Equations.( 24), ( 25) and (26), we obtain: Set C 2 = C • ϕ C (C 1 )/τ 2 (C) and let τ 2 : A N2 → (A /N 2 A )[[T ]] be the morphism mapping X to C 2 X.The above computations show that τ 2 is well-defined and coincide with τ C 2 on C N2 .On the other hand, one checks immediately that C 2 ≡ 1 (mod N 2 ), proving then that τ 2 induces the identity modulo N 2 .As a consequence, τ 2 is an isomorphism and it is a z-admissible choice for τ z .Moreover, it sits in the following commutative diagram: where ϕ is the extension of ϕ C defined by ϕ i a i T i = i γ(a i )S i .The first assertion now follows from Lemma 3.3.3together with the fact that dS/dT = τ Z 2 dZ/dY .The equivalence in assertion (ii) follows from what we have done before after noticing that S has T -valuation 1 by the regularity assumption on z.The fact that Equation (22) holds for any γ z-admissible choices of τ γ z and τ z in this case is a direct consequence of the fact that skew residues do not depend on the choice of the Taylor isomorphisms when poles are simple.

3.3.2.
The case of canonical residues.-We recall that, when p does not divide r, there is a distinguished choice for τ z leading to a notion of canonical skew residues, denoted by sres z,can .After Theorem 3.3.2,one could hope that Equation (22) always holds with canonical residues, as the latter are canonical.Unfortunately, it is not that simple in general.However, there is an important case where our first naive hope is correct.Proof.-After Theorem 3.3.2, it is enough to check that the admissible choice τ γ z,can leads to the admissible choice τ z,can .By Theorem 2.2.5, this reduces further to check that C 2 lies in (Z /N 2 Z )[[T ]] as soon as C 1 is in (Z /N 1 Z )[[T ]] (with the notations of the proof of Theorem 3.3.2).This is obvious from the definition of C 2 .
We now consider the general case.Proposition 2.2.3 tells us that different choices of τ z are conjugated.As a consequence, sres γ z (f ) and sres z,can γ(f )dγ(Y )/dY should be eventually related up to some conjugacy.In the present situation, it turns out that the conjugating function can be made explicit.From now on, we assume that p does not divide r.As before, we consider an endomorphism of K-algebras γ : Frac(A ) → Frac(A ) and we define C = γ(X) X −1 ∈ Frac(C ).We introduce the extension Z of Frac(Z ) obtained by adding a r-th root of N Frac(C )/ Frac(Z ) (C) and form the tensor products C = Z ⊗ Z C and A = Z ⊗ Z A .We emphasize that C is not a field in general but a product of fields.However, the extension C /Z is a cyclic Galois covering of degree r whose Galois group is generated by the automorphism id ⊗θ.Similarly, A could be not isomorphic to an algebra of skew rational functions.Nevertheless, we have the following lemma.

Theorem 1 . 1 . 2 .
-Let A, B ∈ A + with B = 0. (i) There exist Q right , R right ∈ A + , uniquely determined, such that A = Q right • B + R right and deg R right < deg B.

Remark 1 .
1.3.-Without the assumption that θ has finite order, right Euclidean division always exists but left Euclidean division may fail to exist.The mere existence of Euclidean divisions has the following classical consequence.Corollary 1.1.4.-The ring A + is left and right principal.

Lemma 1 .
2.2.-Any skew polynomial f ∈ A has a left multiple and a right multiple in Z .

1 . 3 .
we define deg f = deg g − deg D. This definition is not ambiguous because an equality of the form g/D = g /D implies gD = g D (since D and D are central) and then deg g + deg D = deg g + deg D, that is deg g − deg D = deg g − deg D .Endomorphisms of Ore polynomials rings.-The aim of this subsection is to classify and derive interesting structural properties of the endomorphisms of various rings of skew polynomials.1.3.1.Classification.

1. 4 . 2 .
Extensions of the canonical derivation d/dY .-An important case of interest occurs when Tr Frac(C )/ Frac(Z ) (C) = Y −1 , as ∂ C then induces the standard derivation d/dY on Frac(C ).When p does not divide r, a distinguished element C satisfying this condition is C = r −1 Y −1 .Definition 1.4.3.-When p does not divide r, we set ∂ Y,can = ∂ r −1 Y −1 .Explicitly:

Remark 1 .1. 4 . 3 .
4.5.-With the notation of the proof above, C p is the function by which the p-curvature of the linear differential equation y = Cy acts.With this reinterpretation, one can use Jacobson identity (see Lemma 1.4.2 of [15]) to get a closed formula for C p , which reads: C p = d p−1 C dY p−1 + C p .J.É.P. -M., 2021, tome 8 Derivations over quotients of Ore rings.
J.É.P. -M., 2021, tome 8 2.2.A Taylor-like isomorphism for skew polynomials.-We now aim at completing the diagram ( 2.2.-If N is an irreducible polynomial in Z , the polynomials aX nr N (with a ∈ F and n ∈ Z) are also irreducible in Z and they all generate the same ideal.If τ N satisfies the conditions of Theorem 2.2.1, then a suitable choice for τ aX nr N is ι • τ N , where ι is the automorphism of (A /N A )[[T ]] taking T to aX nr T .

3 . 6 ,
given a positive integer m, any morphism of K-algebras A /N A → A /N m A takes C /N C to C /N m C .Passing to the limit, we find that any morphism of K-algebras A /N A → A N must send C /N C to C N .Therefore, any isomorphism τ N satisfying the conditions of (i) induces a morphism of K-algebras (C /N C )[[T ]] → C N which continues to map T to N and induces the identity modulo T .By the unicity result in the commutative case, we deduce that τ N coincides with τ C N on (C /N C )[[T ]], hence (ii).

3. 3 .
Change of variables.-In this final subsection, we analyze the effect of an endomorphism γ of Frac(A ) on the residues.According to Theorem 1.3.1,γ(X) = CX for some C ∈ Frac(C ) and we have: and γ acts on Frac(C ) through the change of variables Y → Z. Definition 3.3.1.-Let γ as above and let z ∈ F s .

2
be the reduction of C modulo N 2 .We shall often view C as a constant series in(A /N 2 A )[[T ]].Since the norm of C in the extension C /Z is by definition Z Y −1 , we find: (25) N K⊗ F E2[[T ]]/E2[[T ]] C = N K⊗ F E2/E2 C = z 1z 2 and:

Lemma 3 .
3.5.-Given a γ-regular point z ∈ F s and its minimal polynomial N ∈ Z + , any admissible isomorphism τ z :A N ∼ −→ (A /N A )[[T ]] extends uniquely to an isomorphism:τ A z : Z ⊗ Z A N ∼ −→ (A /N A )((T ))inducing the identity after reduction modulo N on the left and modulo T on the right.Proof.-Let us first prove an analogous statement for τ Z z : Z N → (Z /N Z )[[T ]].For simplicity, set Z 0 = N Frac(C )/ Frac(Z ) (C) ∈ Z and let Z 0 be the reduction of Z 0 modulo N .By the regularity assumption, Z 0 = 0. Hence τ Z z (Z 0 ) has a unique r-th root in (Z /N Z )[[T ]] whose constant term is the image of r √ Z 0 in Z /N Z .This basically proves the existence and the unicity of a prolongation τ Z z of τ Z z .Now, a prolongation of τ z is given by τ A z = τ Z z ⊗ τ z , which proves the existence.For unicity, we remark that, by unicity of τ Z z , any isomorphism τ A z satisfying the conditions of the lemma has to coincide with τ Z z on Z ⊗ Z Z N .Since τ A z is a ring homomorphism, we deduce that it necessarily agrees with τ Z z ⊗ τ z on its domain of definition.Unicity follows.Lemma 3.3.5 shows that the function sres z,can : Frac(A ) → A /N A admits a canonical extension to C .We will continue to call it sres z,can in the sequel.We now consider the element:C = C r N Frac(C )/ Frac(Z ) C ∈ C .By construction, it has norm 1 in the extension C /Z .Hilbert's Theorem 90 then guarantees the existence of an invertible element U ∈ C (uniquely determined up to About unicity.-Unfortunately,unlike the commutative case, the isomorphism τ N is not uniquely determined by the conditions of Theorem 2.2.1.We nevertheless have several results in this direction.Proposition 2.2.3.-Letτ N,1 , τ N,2 : A N → (A /N A )[[T ]] be two isomorphisms of K-algebras satisfying the conditions of Theorem 2.2.1.Then, there exists J.É.P. -M., 2021, tome 8 2.2.1.
The morphism τ N,can is the unique isomorphism τ N :A N → (A /N A )[[T ]]which satisfies the conditions of Theorem 2.2.1 and the extra property τ N 2}, we define N i as the minimal polynomial of z i .The quotient ring Z /N i Z is an algebraic separable extension of F ; we will denote it by E i in the rest of the proof.By construction, E i admits a natural embedding into F s (obtained by mapping Y to z i ).The fact that γ acts on Z by right composition by Z shows that γ C induces a field inclusion E 1 → E 2 , which is compatible with the embeddings in F s .In what follows, we shall always view E 1 and E 2 as subfields of F s with E 1 ⊂ E 2 .For i ∈ {1, 2}, we recall that the Taylor expansion around z i provides us with a canonical isomorphism τ Z