Newton polygons of higher order in algebraic number theory

We develop a theory of arithmetic Newton polygons of higher order, that provides the factorization of a separable polynomial over a $p$-adic field, together with relevant arithmetic information about the fields generated by the irreducible factors. This carries out a program suggested by \O{}. Ore. As an application, we obtain fast algorithms to compute discriminants, prime ideal decomposition and integral bases of number fields.


Introduction
R. Dedekind based the foundations of algebraic number theory on ideal theory, because the constructive attempts to find a rigorous general definition of the ideal numbers introduced by E. Kummer failed. This failure is due to the existence of inessential discriminant divisors; that is, there are number fields K and prime numbers p, such that p divides the index i(θ) := (Z K : Z[θ]), for any integral generator θ of K, where Z K is the ring of integers. Dedekind gave a criterion to detect when p ∤ i(θ), and a procedure to construct the prime ideals of K dividing p in that case, in terms of the factorization of the minimal polynomial of θ modulo p [Ded78].
M. Bauer introduced an arithmetic version of Newton polygons to construct prime ideals in cases where Dedekind's criterion failed [Bau07]. This theory was developed and extended by Ø. Ore in his 1923 thesis and a series of papers that followed [Ore23,Ore24,Ore25,Ore26,Ore28]. Let f (x) ∈ Z[x] be an irreducible polynomial that generates K. After K. Hensel's work, the prime ideals of K lying above p are in bijection with the irreducible factors of f (x) over Z p [x]. Ore's work determines three successive factorizations of f (x) in Z p [x], known as the three classical dissections [Ber27], [Coh00]. The first dissection is determined by Hensel's lemma: f (x) splits into the product of factors that are congruent to the power of an irreducible polynomial modulo p. The second dissection is a further splitting of each factor, according to the number of sides of certain Newton polygon. The third dissection is a further splitting of each of the late factors, according to the factorization of certain residual polynomial attached to each side of the polygon, which is a polynomial with coefficients in a finite field.
Unfortunately, the factors of f (x) obtained after these three dissections are not always irreducible. Ore defined a polynomial to be p-regular when it satisfies a technical condition that ensures that the factorization of f (x) is complete after the three dissections. Also, he proved the existence of a p-regular defining equation for every number field, but the proof is not constructive: it uses the Chinese remainder theorem with respect to the different prime ideals that one wants to construct. Ore himself suggested that it should be possible to introduce Newton polygons of higher order that continue the factorization process till all irreducible factors of f (x) are achieved [Ore23,Ch.4, §8], [Ore28,§5].
Ore's program was carried out by the second author in his 1999 thesis [Mon99], under the supervision of the third author. For any natural number r ≥ 1, Newton polygons of order r were constructed, the case r = 1 corresponding to the Newton polygons introduced by Ore. Also, analogous to Ore's theorems were proved for polygons of order r, providing two more dissections of the factors of f (x), for each order r. The whole process is controled by an invariant defined in terms of higher order indices, that ensures that the process ends after a finite number of steps. Once an irreducible factor of f (x) is detected, the theory determines the ramification index and residual degree of the p-adic field generated by this factor, and a generator of the maximal ideal. These invariants are expressed in terms of combinatorial data attached to the sides of the higher order polygons and the residual polynomials of higher order attached to each side. The process yields as a by-product a computation of ind(f ) := v p (i(θ)), where θ is a root of f (x). An implementation in Mathematica of this factorization algorithm was worked out by the first author [Gua97].
We present these results for the first time in the form of a publication, after a thorough revision and some simplifications. In section 1 we review Ore's results, with proofs, which otherwise can be found only in the original papers by Ore in the language of "höheren Kongruenzen". In section 2 we develop the theory of Newton polygons of higher order, based in the concept of a type and its representative, which plays the analogous role in order r to that played by an irreducible polynomial modulo p in order one. In section 3 we prove analogous in order r to Ore's Theorems of the polygon and of the residual polynomial (Theorems 3.1 and 3.7), that provide two more dissections for each order. In section 4 we introduce resultants and indices of higher order and we prove the Theorem of the index (Theorem 4.18), that relates ind(f ) with the higher order indices constructed from the higher order polygons. This result guarantees that the factorization process finishes at most in ind(f ) steps.
Although the higher order Newton polygons are apparently involved and highly technical objects, they provide fast factorization algorithms, because all computations are mainly based on two reasonably fast operations: division with remainder of monic polynomials with integer coefficients, and factorization of polynomials over finite fields. Thus, from a modern perspective, the main application of these results is the design of fast algorithms to compute discriminants, prime ideal decomposition and integral bases of number fields. However, we present in this paper only the theoretical background of higher order Newton polygons. We shall describe the concrete design of the algorithms and discuss the relevant computational aspects elsewhere [GMN08a,GMN08b]. Let λ ∈ Q − be a negative rational number, expressed in lower terms as λ = −h/e, with h, e positive coprime integers. We denote by S(λ) the set of segments of the Euclidian plane with slope λ and end points having nonnegative integer coordinates. The points of (Z ≥0 ) 2 are also considered to be segments in S(λ), whose initial and final points coincide. The elements of S(λ) will be called sides of slope λ. For any side S ∈ S(λ), we define its length, ℓ := ℓ(S), and height, H := H(S), to be the length of the respective projections of S to the horizontal and vertical axis. We define the degree of S to be Note that any side S of positive length is divided into d segments by the points of integer coordinates that lie on S. A side S ∈ S(λ) is determined by the initial point (s, u) and the degree d. The final point is (s + ℓ, u − H) = (s + de, u − dh). For instance, the next figure represents a side of slope −1/2, initial point (s, u), and degree three. The set S(λ) has the structure of an abelian semigroup with the following addition rule: given S, T ∈ S(λ), the sum S + T is the side of degree d(S) + d(T ) of S(λ), whose initial point is the sum of the initial points of S and T . Thus, the addition is geometrically represented by the process of joining the two segments and choosing an apropriate initial point. The addition of a segment S with a point P is represented by the translation P + S of S by the vector represented by P . The neutral element is the point (0, 0). The invariants ℓ(S), H(S), d(S) determine semigroup homomorphisms ℓ, H, d : S(λ) −→ Z ≥0 .
For technical reasons we consider also a set of sides of slope −∞, which is formally defined as S(−∞) := Z >0 ×(Z ≥0 ) 2 . If S = (ℓ, (s, u)) is a side of slope minus infinity, we define ℓ(S) := ℓ, H(S) := ∞, d(S) := 1. Also, we take by convention h = ∞, e = ℓ. This set has an obvious structure of an abelian monoid, and the length determines a monoid homomorphism, ℓ : S(−∞) −→ Z >0 . There is a geometric representation of such an S as a side whose end points are (s − ℓ, ∞) and (s, u). Note that the points of (Z ≥0 ) 2 belong to S(λ) for all finite λ, so that it is not possible (even in a formal sense) to attach a slope to them.
We have a natural geometric representation of a side. Let us introduce a geometric representation of a formal sum of sides as an open convex polygon of the plane. Let N = S 1 +· · ·+S t be a formal sum of sides of negative slope. Let S ∞ = (ℓ ∞ , P ∞ ) be the sum of all sides of slope −∞ among the S i , and let P 0 be the sum of all initial points of the S i that don't belong to S(−∞) (in case of an empty sum we consider respectively P ∞ = (0, 0) and/or P 0 = (0, 0)). Let P = P ∞ + P 0 . Then, N is represented as the polygon that starts at P and is obtained by joining all sides of positive length and finite slope, ordered by increasing slopes. If i 1 is the abscissa of P , we have to think that the polygon starts at the abscisa i 0 = i 1 − ℓ ∞ , that formally indicates the starting point (at infinity) of a side of slope −∞. The typical shape of this polygon is By definition, every principal polygon represents a formal sum, N = S 1 +· · ·+S t , of sides S i ∈ S. This expression is unique in any of the two following situations (1) N = S, with S ∈ (Z ≥0 ) 2 , (2) N = S 1 + · · · + S t , with all S i of positive length and pairwise different slopes. It is clear that any N ∈ PP can be expressed in one (and only one) of these canonical forms. Usually, when we speak of the sides of a principal polygon, we mean the sides of this canonical expression. If we need to emphasize this we shall use the term canonical sides of N . The finite end points of the canonical sides are called the vertices of the polygon.
The addition of polygons is defined in terms of the expression as a formal sum of sides (not necessarily the canonical ones). That is, if N = S 1 + · · · + S r and N ′ = S ′ 1 +· · ·+S ′ s , then N +N ′ is the geometric representation of S 1 +· · ·+S r +S ′ 1 +· · ·+S ′ s . The reader may check easily that this is well-defined and PP has a structure of semigroup with neutral element (0, 0). Also, it is clear that this addition is compatible with the sum operations that we had on all S(λ). Note that the addition of N ∈ PP with (the polygon represented by) a point P ∈ (Z ≥0 ) 2 is the translation P + N . The fact of adding to N (the polygon represented by) a side of slope −∞ is reflected by a horizontal shift of the finite part of N , without changing the starting abscissa i 0 of N . Definition 1.2. We define the length of a principal polygon N = S 1 + · · · + S r to be ℓ(N ) := ℓ(S 1 ) + · · · + ℓ(S r ). The length determines a semigroup homomorphism, Let N ∈ PP. Let i 0 be the abscissa where the polygon starts and i 1 the abscissa of the point P where the finite part of N starts. For any integer abscissa i 0 ≤ i ≤ i 0 + ℓ(N ) we denote by For i ≥ i 1 these rational numbers form an strictly decreasing sequence. Definition 1.3. Let P = (i, y) be a "point" of the plane, with integer abscissa i 0 ≤ i ≤ i 0 + ℓ(N ), and ordinate y ∈ R ∪ {∞}. We say that P lies on N if y = y i , and in this case we write P ∈ N . We say that P lies above N if y ≥ y i . We say that P lies strictly above N if y > y i .
For any i 1 < i ≤ i 0 + ℓ(N ), let µ i be the slope of the segment joining (i − 1, y i−1 ) and (i, y i ). The sequence µ i1+1 ≤ · · · ≤ µ i0+ℓ(N ) is an increasing sequence of negative rational numbers. We call these elements the unit slopes of N . Consider the multisets of unit slopes: Let N ′ be another principal polygon with starting abscissa j 0 and starting abscissa for the finite part j 1 . Consider analogous multisets U j (N ′ ), for all j 1 ≤ j ≤ j 0 + ℓ(N ′ ). By the definition of the addition law of principal polygons, the multiset U k (N + N ′ ) contains the smallest k − i 1 − j 1 unit slopes of the multiset U i0+ℓ(N ) (N ) ∪ U j0+ℓ(N ′ ) (N ′ ) that contains all unit slopes of both polygons. Thus,

and equality holds if and only if
Lemma 1.4. Let N, N ′ ∈ PP. Let P = (i, u) be a point lying above the finite part of N and P ′ = (j, u ′ ) a point lying above the finite part of N ′ . Then P + P ′ lies above the finite part of N + N ′ and Proof. Clearly, u + u ′ ≥ y i (N ) + y j (N ′ ) ≥ y i+j (N + N ′ ) and P + P ′ ∈ N + N ′ if and only if both inequalities are equalities.
Definition 1.5. Let λ ∈ Q − and N ∈ PP. Consider a line of slope λ far below N and let it shift upwards till it touches N for the first time. Denote by L λ (N ) this line of slope λ having first contact with N . We define the λ-component of N to be S λ (N ) := N ∩ L λ (N ). We obtain in this way a map: If N has a canonical side S of positive length and finite slope λ, we have S λ (N ) = S, otherwise the λ-component S λ (N ) reduces to a point. (1) for all N, N ′ ∈ PP and all λ ∈ Q − . 1.2. φ-Newton polygon of a polynomial. Let p be a prime number and let Q p be a fixed algebraic closure of the field Q p of the p-adic numbers. For any finite extension, Throughout the paper O L will denote the ring of integers of L, m L its maximal ideal, and F L the residue field. The canonical reduction map red L : O L −→ F L will be usually indicated by a bar: α := red L (α). We fix a finite extension K of Q p as a base field, and we denote v := v K , O := O K , m := m K , F := F K , q := |F|. We fix also a prime element π ∈ O.
We extend the valuation v to polynomials with coefficients in O in a natural way: be a monic polynomial of degree m whose reduction modulo m is irreducible. We denote by F φ the finite field O[x]/(π, φ(x)), and by red: O[x] −→ F φ the canonical homomorphism. We denote also by a bar the reduction of polynomials modulo m,¯: admits a unique φ-adic development: which is a point of the plane if u i is finite, and it is thought to be the point at infinity of the vertical line with abscissa i, if u i = ∞.
Definition 1.6. The φ-Newton polygon of a nonzero polynomial f (x) ∈ O[x] is the lower convex envelope of the set of points P i = (i, u i ), u i < ∞, in the euclidian plane. We denote this polygon by N φ (f ).
The length of this polygon is by definition the abscissa of the last vertex. We denote it by ℓ(N φ (f )) := n = ⌊deg(f )/m⌋. Note that deg f (x) = mn + deg a n (x). The typical shape of this polygon is the following e e e e e e e e e e e e Definition 1.8. The principal φ-polygon of f (x) is the element N − φ (f ) ∈ PP determined by the sides of negative slope of N φ (f ), including the side of slope −∞ represented by the length ord φ (f ). It always starts at the abscissa i 0 = 0 and has From now on, we denote N = N − φ (f ) for simplicity. The principal polygon N and the set of points P i = (i, u i ) that lie on N , contain the arithmetic information we are interested in. Note that, by construction, the points P i lie all above N .
We attach to any abscissa ord φ (f ) ≤ i ≤ ℓ(N ) the following residual coefficient Note that c i is always nonzero in the latter case, because deg a i (x) < m. Let λ = −h/e be a negative rational number, with h, e positive coprime integers. Let S = S λ (f ) be the λ-component of N , (s, u) the initial point of S, and d := d(S) the degree of S. The points (i, u i ) that lie on S contain important arithmetic information that is kept in the form of two polynomials that are built with the coefficients of the φ-adic development of f (x) to whom these points are attached. Definition 1.9. We define the virtual factor of f (x) attached to λ (or to S) to be the polynomial We define the residual polynomial attached to λ (or to S) to be the polynomial: Note that only the points (i, u i ) that lie on S yield a nonzero coefficient of R λ (f )(y). In particular, c s and c s+de are always nonzero, so that R λ (f )(y) has degree d and it is never divisible by y.
If π ′ = ρπ is another prime element of O, and c =ρ ∈ F * , the residual coefficients of N − φ (f ) with respect to π ′ satisfy c ′ i = c i c −ui , so that the corresponding residual polynomial R ′ λ (f )(y) is equal to c −u R λ (f )(c h y). We can define in a completely analogous way the residual polynomial of f (x) with respect to a side T , which is not necessarily a λ-component of N − φ (f ). Definition 1.10. Let T ∈ S(λ) be an arbitrary side of slope λ, with abscissas s 0 ≤ s 1 for the end points, and let d ′ = d(T ). We say that the polynomial Thus, if all points of S λ (f ) lie strictly above T we have R λ (f, T )(y) = 0. Note that deg R λ (f, T )(y) ≤ d ′ and equality holds if and only if the final point of T belongs to S λ (f ). Usually, T will be an enlargement of S λ (f ) and then, where s is the abscissa of the initial point of S λ (f ). The motivation for this more general definition lies in the bad behaviour of the residual polynomial R λ (f )(y) with respect to sums. Nevertheless, if T is a fixed side and f (x), g(x) lie both above T , it is clear that f (x) + g(x) lies above T and 1.3. Admissible φ-developments and Theorem of the product. Let , and let N ′ be the principal polygon of the set of points (i, u ′ i ). Let i 1 be the first abscissa with a ′ i1 (x) = 0. To any i 1 ≤ i ≤ ℓ(N ′ ) we attach a residual coefficient as before: For the points (i, u ′ i ) lying on N ′ we can have now c ′ i = 0; for instance, in the case a ′ 0 (x) = f (x), the Newton polygon has only one point (0, v(f )) and Finally, for any negative rational number λ = −h/e as above, we can define the residual polynomial attached to the λ-component S ′ = S λ (N ′ ) to be and s ′ is the abscissa of the initial point of S ′ .
Definition 1.11. We say that the φ-development (4) is admissible if for each abs- By the uniqueness of the φ-adic development we have In particular, all points (i, u i ) lie above N ′ ; in fact, for some 0 ≤ k 0 ≤ i, we have From now on, i will be an integer abscissa of the finite part of N ′ . Clearly, (7) and (5) we have Let us denote this common polygon by N . Finally, let us prove the equality of all residual coefficients. If c i = 0, then u i = y i (N ), and from (6) we get k 0 = 0 and u i = w i,0 = u ′ i . By (7), (5) and (8), , and from (5) and (7) we get w i,0 > y i (N ) too. By (8) we get c ′ i = 0. The construction of the principal part of the φ-Newton polygon of a polynomial can be interpreted as a mapping . Also, for any negative rational number λ, the construction of the residual polynomial attached to λ can be interpreted as a mapping The Theorem of the product says that both mappings are semigroup homomorphisms.
Proof. Consider the respective φ-adic developments Denote by N ′ the principal part of the Newton polygon of f g, determined by this φ-development. We shall show that N ′ = N f + N g , that this φ-development is admissible, and that R ′ λ (f g) = R λ (f )R λ (g) for all λ. The theorem will be then a consequence of Lemma 1.12.
Let w k := v(A k (x)) for all 0 ≤ k. Lemma 1.4 shows that the point (i, u i ) + (j, v j ) lies above N f + N g for any i, j ≥ 0. Since w k ≥ min{u i + v j , i + j = k}, the points (k, w k ) lie all above N f + N g . On the other hand, let P k = (k, y k (N f + N g )) be a vertex of N f + N g ; that is, P k is the end point of S 1 + · · · + S r + T 1 + · · · + T s , for certain sides S i of N f and T j of N g , ordered by increasing slopes among all sides of N f and N g . By Lemma 1.4, for all pairs (i, j) such that i + j = k, the point (i, u i ) + (j, v j ) lies strictly above N f + N g except for the pair i 0 = ℓ(S 1 + · · · + S r ), j 0 = ℓ(T 1 + · · · + T s ) that satisfies (i 0 , u i0 ) + (j 0 , v j0 ) = P k . Thus, (k, w k ) = P k and This shows that N ′ = N f + N g and that the φ-development (9) is admisible.
Finally, by (1), the λ-components ) be a point of integer coordinates lying on S ′ (not necessarily a vertex). Denote by I the set of the pairs (i, j) such that (i, u i ) lies on S f , (j, v j ) lies on S g , and i + j = k. Take P (x) = (i,j)∈I a i (x)b j (x). By Lemma 1.4, for all other pairs (i, j) with i + j = k, the point (i, . Notation. Let F be a field and ϕ(y), ψ(y) ∈ F[y] two polynomials. We write ϕ(y) ∼ ψ(y) to indicate that there exists a constant c ∈ F * such that ϕ(y) = cψ(y).
Let f (x) = F 1 (x) · · · F r (x) be the factorization into a product of monic polynomials of O[x] satisfying F i (X) ≡ φ i (x) ni (mod m), provided by Hensel's lemma. Then, for all 1 ≤ i ≤ r and all λ ∈ Q − .
Proof. For any 1 ≤ i ≤ r, let G i (x) = j =i F j (x). Since φ i (x) does not divide G i (x) modulo m, the principal φ i -Newton polygon of G i (x) reduces to the point (0, 0). By the Theorem of the product, On the other hand, N φi (F i ) = N − φi (F i ) because both polygons have length n i . Now, for any λ ∈ Q − , S λ (G i ) is a point and R λ (G i )(y) is a nonzero constant. By the Theorem of the product, 1.4. Theorems of the polygon and of the residual polynomial.
, with monic polynomials f φ (x), G(x) such that red(G(x)) = 0 and f φ (x) ≡ φ(x) n (mod m). The aim of this section is to obtain a further factorization of f φ (x) and certain arithmetic data about the factors. Thanks to Corollary 1.14, we shall be able to read this information directly on f (x); more precisely, on Proof. By the Theorem of the product and Corollary 1.14, it is sufficient to show that if F (x) := f φ (x) is irreducible, then N φ (F ) = S is one-sided and the roots θ ∈ Q p have all v(φ(θ)) equal to minus the slope of S.
In fact, for all the roots θ ∈ Q p of F (x), the rational number v(φ(θ)) takes the same value because the p-adic valuation is invariant under the Galois action. Since F (x) is congruent to a power of φ(x) modulo m we have λ := −v(φ(θ)) < 0. We have λ = −∞ if and only if F (x) = φ(x), and in this case the theorem is clear.
be the minimal polynomial of φ(θ) and let is an irreducible factor of Q(x) and by the Theorem of the product N φ (F ) is also one-sided with slope λ.
We discuss now how Newton polygons and residual polynomials are affected by an extension of the base field by an unramified extension.
Lemma 1.18. We keep the above notations for f (x), λ = −h/e, θ, L. Let K ′ ⊆ L be the unramified extension of K of degree m, and identify F φ = F K ′ through the embedding (10).
Then, for any nonzero polynomial P (x) ∈ O[x]: where R ′ denotes the residual polynomial with respect to φ ′ (x) over K ′ , ǫ ∈ F * K ′ does not depend on P (x), and s is the initial abscissa of S λ (P ).
Proof. Consider the φ-adic development of P (x): and v(ρ(θ)) = 0. Therefore, the above φ ′ (x)-development of P (x) is admissible and N − φ ′ (P ) = N − φ (P ) by Lemma 1.12. Moreover the residual coefficients of the two polygons are related by

λ by the Theorem of the polygon, admits a factorization
Proof. By the Theorem of the product, we need only to prove that if F ( if and only if R ′ λ (G)(y) has the same property over F K ′ . In conclusion, by extending the base field, we can suppose that deg φ = m = 1.
Consider now the minimal polynomial , and it has the same property by the Theorem of the product.
Corollary 1.20. With the above notations, let θ ∈ Q p be a root of G i (x), and L = K(θ).
Proof. The statement about f (L/K) is a consequence of the embedding of the finite field F φ [y]/(ψ i (y)) into F L determined by red(x) → θ, y → γ(θ). This embedding is well-defined by item 4 of Proposition 1.17. The other statements are a consequence of the Theorem of the product and Corollary 1.16.
1.5. Types of order one. Starting with a monic and separable polynomial f (x) ∈ O[x], the Newton polygon techniques provide partial information on the factoriza- , obtained after three dissections [Ber27]. In the first dissection we obtain as many factors of f (x) as pairwise different irreducible factors modulo m (by Hensel's lemma). In the second dissection, each of these factors splits into the product of as many factors as sides of certain Newton polygon of f (x) (by the Theorem of the polygon). In the third dissection, the factor that corresponds to a side of finite slope splits into the product of as many factors as irreducible factors of the residual polynomial of f (x) attached to this side (by the Theorem of the residual polynomial).
The final list of factors of f (x) obtained by this procedure can be parameterized by certain data, which we call types of order zero and of order one.
. We attach to any type of order zero the map be a monic and separable polynomial. We say that the type t is f -complete if ω t (f ) = 1. In this case, we denote by By truncation of t we obtain the type of order zero t 0 := φ(y) (mod m).
We denote by t 0 (f ) the set of all monic irreducible factors of f (x) modulo m. We denote by t 1 (f ) the set of all types of order one obtained from f (x) along the process of applying the three classical dissections: for any non-f -complete ψ 0 (y) ∈ t 0 (f ), we take a monic lift φ(x) to O[x]; then we consider all finite slopes λ of the sides of positive length of N − φ (f ), and finally, for each of them we take the different monic irreducible factors ψ(y) of the residual polynomial R λ (f )(y) ∈ F φ [y]. These types are not intrinsical objects of f (x). There is a non-canonical choice of the lifts φ(x) ∈ O[x], and the data λ, ψ(y) depend on this choice.
We denote by T 1 (f ) the union of t 1 (f ) and the set of all f -complete types of order zero. By the previous results we have a factorization in , and, if t has order one, f t (x) is the unique monic divisor of f (x) in O[x] satisfying the following properties: The factor f ∞ (x) is already expressed as a product of irreducible polynomials in O[x]. Also, if a = 1, the Theorem of the residual polynomial shows that f t (x) is irreducible too. Thus, the remaining task is to obtain the further factorization of f t (x), for the types t ∈ t 1 (f ) with a > 1. The factors of f t (x) will bear a reminiscence of t as a birth mark (cf. Lemma 2.4).
Once a type of order one t = (φ(x); λ, ψ(y)) is fixed, we change the notation of several objects that depend on the data of the type. We omit the data from the notation but we include the subscript "1" to emphasize that they are objects of the first order. From now on, for any nonzero polynomial P (x) ∈ O[x], any principal polygon N ∈ PP, and any T ∈ S(λ), we shall denote The aim of the next two sections is to introduce Newton polygons of higher order and prove similar theorems, yielding information on a further factorization of each f t (x). As before, we shall obtain arithmetic information about the factors of f t (x) just by a direct manipulation of f (x), without actually computing a p-adic approximation to these factors. This fact is crucial to ensure that the whole process has a low complexity. However, once an irreducible factor of f (x) is "detected", the theory provides a reasonable approximation of this factor as a by-product (cf. Proposition 3.12).

Newton polygons of higher order
Throughout this section, r is an integer, r ≥ 2. We shall construct Newton polygons of order r and prove their basic properties and the Theorem of the product in order r, under the assumption that analogous results have been already obtained in orders 1, . . . , r − 1. We also assume that the theorems of the polygon and of the residual polynomial have been proved in orders 1, . . . , r − 1 (cf. section 3). For r = 1 all these results have been proved in section 1.

Types of order
, λ i are negative rational numbers and ψ r−1 (y) is a polynomial over certain finite field (to be specified below), that satisfy the following recursive properties: (1) φ 1 (x) is irreducible modulo m. We denote by ψ 0 (y) ∈ F[y] the polynomial obtained by reduction of φ 1 (y) modulo m. We define F 1 := F[y]/(ψ 0 (y)).
(2) For all 1 ≤ i < r − 1, the Newton polygon of i-th order, N i (φ i+1 ), is one-sided, with positive length and slope λ i .
. The type determines a tower F =: F 0 ⊆ F 1 ⊆ · · · ⊆ F r of finite fields. The field F i should not be confused with the finite field with i elements.
By the Theorem of the product in orders 1, . . . , r − 1, the polynomials Let us be more precise about the meaning of N i (−), R i (−), used in items 2,3.
Notation. We denote t 0 = ψ 0 (y). For all 1 ≤ i < r, we obtain by truncation of t a type of order i, and a reduced type of order i, defined respectively as: We have semigroup homomorphisms: is the residual polynomial of i-th order with respect to λ i . The polynomial R i (P )(y) has degree d(S i (P )). Both S i and R i depend only on the reduced type t 0 i . Finally, we denote by s i (P ) the initial abscissa of S i (P ). Other data attached to the type t deserve an specific notation. For all 1 ≤ i < r: where, by convention: R 0 (P ) = P (y)/π v(P ) ∈ F[y]. By Lemma 2.17 in order r − 1 (see Definition 1.8 and Remark 1.7 for order one): . Definition 2.1. We say that a monic polynomial P (x) ∈ O[x] has type t when (1) P (x) ≡ φ 1 (x) a0 (mod m), for some positive integer a 0 , (2) For all 1 ≤ i < r, the Newton polygon N i (P ) is one-sided, of slope λ i , and , for some positive integer a i .
(3) If P (x) has type t then all inequalities in item 1 are equalities, and Proof. Item 1 is a consequence of (11); in fact, for all 1 ≤ i < r: . Item 2 is a consequence of (11) and item 1: Finally, if P (x) has type t the two inequalities of (12) are equalities, so that m i ω i (P ) = m i+1 ω i+1 (P ); on the other hand, deg P = m 1 a 0 = m 1 ω 1 (P ).
be a monic polynomial with ω r (P ) > 0. We denote by P t (x) the monic factor of P (x) of greatest degree that has type t. By the Theorems of the polygon and of the residual polynomial in orders 1, . . . , r − 1, this factor exists and it satisfies , then it is of type t if and only if ω r (P ) > 0.
(2) P (x) is of type t if and only if deg P = m r ω r (P ) > 0.
(3) P (x)Q(x) has type t if and only if P (x) and Q(x) have both type t.
Proof. The polynomial P (x) is of type t if and only if ω r (P ) > 0 and P t (x) = P (x); thus, items 1 and 2 are an immediate consequence of (13). Item 3 follows from the Theorem of the product in orders 1, . . . , r − 1.
We fix a type t of order r − 1 for the rest of section 2.
2.2. The p-adic valuation of r-th order. In this paragraph we shall attach to t a discrete valuation v r : . Consider the mapping that assigns to each side S ∈ S(λ r−1 ) the ordinate of the point of intersection of the vertical axis with the line L λr−1 of slope λ r−1 that contains S. If (i, u) is any point of integer coordinates lying on S, then H r−1 (S) = u + |λ r−1 |i; thus, H r−1 is a semigroup homomorphism.
Note that v r depends only on the reduced type t 0 .
All points of N P lie above the line L P and all points of N Q lie above the line L Q . If v r (P ) ≤ v r (Q), all points of both polygons lie above the line L P . Thus, all points of N − r−1 (P + Q) lie above this line too, and this shows that v r (P + Q) ≥ v r (P ). Finally, for any a ∈ O, we have v r (a) = e r−1 v r−1 (a) by definition, since the (r − 1)-th order Newton polygon of a is the single point (0, v r−1 (a)).
This valuation was introduced by S. MacLane without using Newton polygons [McL36a], [McL36b]. In [Mon99, Ch.2, §2], J. Montes computed explicit generators of the residue field of v r as a transcendental extension of a finite field. These results lead to a more conceptual and elegant definition of residual polynomials in higher order, as the reductions modulo v r of the virtual factors. However, we shall not follow this approach, in order not to burden the paper with more technicalities.
The next proposition gathers the basic properties of this discrete valuation.
times the ordinate at the origin of the line L that has slope λ r−1 and passes through Since v r is a valuation, this proves item 3, and item 4 is a particular case.
In a natural way, ω r induces a group homomorphism from K(x) * to Z, but it is not a discrete valuation of this field. For instance, for K = Q p , π = p, t = (x; −1, y + 1) and P (x) = x + p, Q(x) = x + p + p 2 , we have However, we shall say that ω r is a pseudo-valuation with respect to v r ; this is justified by the following properties of ω r .
We can reinterpret the computation of v(P (θ)) given in item 5 of Proposition 3.5 in order r − 1 (cf. Proposition 1.17 for r = 2), in terms of the pair v r , ω r .
and equality holds if and only if ω r (P ) = 0.
2.3. Construction of a representative of t. By Lemma 2.2, a nonconstant polynomial of type t has degree at least m r . In this section we shall show how to construct in a effective (and recursive) way a polynomial φ r (x) of type t and minimal degree m r .
We first show how to construct a polynomial with prescribed residual polynomial.
be a nonzero polynomial of degree less than f r−1 , and let ν = ord y (ϕ). Then, we can construct in an effective way a polynomial P (x) ∈ O[x] satisfying the following properties Proof. Let L be the line of slope λ r−1 with ordinate V /e r−1 at the origin. By item 4 of Proposition 2.7, V /e r−1 ≥ f r−1 v r (φ r−1 ) ≥ f r−1 h r−1 ; thus, the line L cuts the horizontal axis at the abscissa V /h r−1 ≥ e r−1 f r−1 . Let T be the greatest side contained in L, whose end points have nonnegative integer coordinates. Let (s, u) be the initial point of T and denote u j := u − jh r−1 , for all 0 ≤ j < f r−1 , so that (s + je r−1 , u j ) lies on L. Clearly, s < e r−1 and, for all j, We proceed by induction on r ≥ 2. For r = 2 the polynomials c j (y) belong to F[y]; we abuse of language and denote by c j (x) ∈ O[x] the polynomials obtained by choosing arbitrary lifts to O of the nonzero coefficients of c j (y). The polynomial Thus, the coefficient π u−jh1 c j (x) determines a point of N − 1 (P ) lying on T , and v 2 (P ) = V . Finally, it is clear by construction that ν = (s 1 (P ) − s)/e 1 and y ν R 1 (P )(y) = R 1 (P, T )(y) = ϕ(y).
Let now r ≥ 3, and suppose that the proposition has been proved for orders 2, . . . , r − 1.
the last equality by (16) below, in order r − 1. Let L j be the line of slope λ r−2 with ordinate at the origin V j /e r−2 . Let T (j) be the greatest side contained in L j , whose end points have nonnegative integer coordinates. Let s j be the initial abscissa of T (j). Consider the unique polynomial ϕ j (y) ∈ F r−2 [y], of degree less than f r−2 , such that ϕ j (y) ≡ y (ℓr−2uj −sj )/er−2 c j (y) (mod ψ r−2 (y)), and let ν j = ord y (ϕ j ). By induction hypothesis, we are able to construct a polynomial P j (x) of degree less than m r−1 , with v r−1 (P j ) = V j , ν j = (s r−2 (P j )−s j )/e r−2 , and y νj R r−2 (P j )(y) = ϕ j (y) in F r−2 [y].
Definition 2.12. A representative of the type t is a monic polynomial φ r (x) ∈ O[x] of type t such that R r−1 (φ r )(y) ∼ ψ r−1 (y). This object plays the analogous role in order r − 1 to that of an irreducible polynomial modulo m in order one.
From now on, we fix a representative φ r (x) of t, without necessarily assuming that it has been constructed by the method of Propositon 2.10. We denote bỹ t = (φ 1 (x); λ 1 , φ 2 (x); · · · ; λ r−2 , φ r−1 (x); λ r−1 , φ r (x)), the extension of t, which is half-way in the process of extending t to different types of order r.
2.4. Certain rational functions. We introduce in a recursive way several rational functions in K(x). We let h r , e r be arbitrary coprime positive integers, and we fix ℓ r , ℓ ′ r ∈ Z such that ℓ r h r − ℓ ′ r e r = 1.
Definition 2.13. We define π 0 (x) = 1, π 1 (x) = π, and, for all 1 ≤ i ≤ r, Each of these rational functions can be written as π n0 φ 1 (x) n1 · · · φ r (x) nr , for adequate integers n i ∈ Z. Also, where the dots indicate a product of integral powers of π and φ j (x), with 1 ≤ j < i. We want to compute the value of v r on all these functions.
Proof. We proceed by induction on r. For r = 2 all formulas are easily deduced from v 2 (φ 1 ) = h 1 , that was proved in Proposition 2.7. Suppose r ≥ 3 and all statements true for r − 1. Let us start with item 1. By Proposition 2.7 and (16), Hence, the formula for i = r − 1 follows from the induction hypothesis. Suppose from now on i < r − 1. By Lemma 2.14, φ i (x) = φ i (x) is an admissible φ r−1 -adic development of φ i (x), and by Lemma 2.25 in order r − 1 (Lemma 1.12 in order one) we get N − r−1 (φ i ) = (0, v r−1 (φ i )), so that v r (φ i ) = e r−1 v r−1 (φ i ) and the formula follows by induction.
Item 4 is easily deduced from the previous formulas, and item 5 is an immediate consequence of (17) and Lemma 2.14. The last statements follow from (16) and the previous formulas.
2.5. Newton polygon and residual polynomials of r-th order. Let f (x) ∈ O[x] be a nonzero polynomial, and consider its unique φ r -adic development We define the Newton polygon N r (f ) of f (x), with respect to the extensiont of t, to be the lower convex envelope of the set of points (i, u i ), where Note that we consider the v r -value of the whole monomial a i (x)φ r (x) i . Actually, we did the same for the Newton polygons of first order, but in that case ( Proof. The third item is obvious. Let us prove items 1, 2. Let u := min 0≤i≤n {u i }, and consider the polynomial All monomials of g(x) have the same v r -value and a different ω r -value: because ω r (a i ) = 0 by Lemma 2.2. By item 2 of Proposition 2.8, v r (g) = u and ω r (g) = i 0 , the least abscissa with u i0 = u. Since, v r (f − g) > u, we have v r (f ) = v r (g) = u, and this proves item 1. On the other hand, item 1 of Propositon 2.8 shows that ω r (f ) = ω r (g) = i 0 , and this proves item 2.
The following observation is a consequence of Lemmas 2.2 and 2.17.
From now on let N = N − r (f ). As we did in order one, we attach to any integer abscissa i of the finite part of N a residual coefficient c i ∈ F r . The natural idea is to consider c i = R r−1 (a i )(z r−1 ) for the points lying on N . However, this does not lead to the right concept of residual polynomial attached to a side; it is necessary to twist these coefficients by certain powers of z r−1 .  We define the virtual factor of f (x) attached to S (or to λ r ) to be the rational function

the rational functions introduced in Definition 2.13.
We define the residual polynomial attached to S (or to λ r ) to be the polynomial: Only the points (i, u i ) that lie on S yield a non-zero coefficient of R λr (f )(y). In particular, c s and c s+de are always nonzero, so that R λr (f )(y) has degree d and it is never divisible by y. We emphasize that R λr (f )(y) does not depend only on λ r ; as all other objects in Sect.2, it depends on t too.
We define in an analogous way the residual polynomial of f (x) with respect to a side T that is not necessarily a λ r -component of N . Let T ∈ S(λ r ) be an arbitrary side of slope λ r , with abscissas s 0 ≤ s 1 for the end points. Let d ′ = d(T ). We say that f (x) lies above T in order r if all points of N with abscissa s 0 ≤ i ≤ s 1 lie above T . In this case we define Note that deg R λr (f, T )(y) ≤ d ′ and equality holds if and only if the final point of T belongs to S λr (f ). Usually, T will be an enlargement of S λr (f ) and then, where s is the abscissa of the initial point of S λr (f ). For technical reasons, we express c i in terms of a residual polynomial attached to certain auxiliary side.
In particular, Proof. If v r (aφ i r ) = y i (N ), we have v r (a) = V and S r−1 (a) ⊆ T (i). Then, the lemma follows from (19) in order r − 1. If v r (aφ i r ) > y i (N ) then S r−1 (a) lies strictly above T (i) and R r−1 (a i , T (i))(y) = 0. 2.6. Admissible φ r -developments and Theorem of the product in order r. Let be a φ r -development of f (x), not necessarily the φ r -adic one. Let N ′ be the principal polygon of the set of points (i, . Let i 1 be the first abscissa with a ′ i1 (x) = 0. As we did in order one, to each integer abscissa i 1 ≤ i ≤ ℓ(N ′ ) we attach a residual coefficient For the points (i, u ′ i ) lying on N ′ we may have now c ′ i = 0; for instance in the case a ′ 0 (x) = f (x) the Newton polygon has only one point (0, v r (f )) and c ′ 0 = 0 if ω r (f ) > 0. Finally, for any negative rational number λ r = −h r /e r , with h r , e r positive coprime integers, we define the residual polynomial attached to the λ r -component Definition 2.24. We say that the φ r -development (21) By the uniqueness of the φ r -adic development we have . Therefore, by (22) and (23), all points (i, u i ) lie above N ′ ; in fact On the other hand, for any abscissa i of the finite part of N ′ and for any 0 < k ≤ i we have by (23) The following claim ends the proof of the lemma: Claim. Let i be an abscissa of the finite part of N ′ such that (i, u ′ i ) ∈ N ′ . Then, u i = u ′ i if and only if c ′ i = 0; and in this case c ′ i = c i . In fact, suppose c ′ i = 0, or equivalently, ω r (a ′ i ) = 0. We decompose (22) and (25) we have
Proof. Consider the respective φ r -adic developments and denote by N ′ the principal part of the Newton polygon of order r of f g, determined by this φ r -development. We shall show that N ′ = N f + N g , that this φ r -development is admissible, and that R ′ λr (f g) = R λr (f )R λr (g) for all negative λ r . The theorem will be then a consequence of Lemma 2.25.
Let w k := v r (A k φ k r ) for all 0 ≤ k. Lemma 1.4 shows that the point (i, u i )+(j, v j ) lies above N f + N g for any i, j ≥ 0. Since w k ≥ min{u i + v j , i + j = k}, the points (k, w k ) lie all above N f + N g too. On the other hand, let P k = (k, y k (N f + N g )) be a vertex of N f + N g ; that is, P k is the end point of S 1 + · · · + S r + T 1 + · · · + T s , for certain sides S i of N f and T j of N g , ordered by increasing slopes among all sides of N f and N g . By Lemma 1.4, for all pairs (i, j) such that i + j = k, the point (i, u i ) + (j, v j ) lies strictly above N f + N g except for the pair i 0 = ℓ(S r−1 + · · ·+ S r ), j 0 = ℓ(T r−1 + · · · + T s ) that satisfies (i 0 , u i0 ) + (j 0 , v j0 ) = P k . Thus, (k, w k ) = P k . This shows that Finally, by (1), the λ r -components S ′ = S λr (N ′ ), S f = S λr (N f ), S g = S λr (N g ) are related by: S ′ = S f + S g . Let (k, y k (N ′ )) be a point of integer coordinates lying on S ′ (not necessarily a vertex), and let T (k) be the corresponding side of slope λ r−1 given in Lemma 2.22, with starting abscissa s k . Denote by I the set of the pairs (i, j) such that (i, u i ) lies on S f , (j, v j ) lies on S g , and i + j = k. Take P (x) = (i,j)∈I a i (x)b j (x). By Lemma 1.4, for all other pairs (i, j) with i + j = k, the point (i, u i ) + (j, v j ) lies strictly above N ′ . By Lemma 2.23,

Lemma 2.22, (19) and the Theorem of the product in order
This shows that the residual polynomial attached to S ′ with respect to the φ rdevelopment (26) is R λr (f )R λr (g). Proof. Let f (x) = f t (x)g(x). By (13), ω r (g) = 0. By the Theorem of the product, and R λr (f ) = R λr (f t )R λr (g). Since N − r (g) reduces to a point with abscissa 0 (cf. Lemma 2.17), the polygon N − r (f ) is a vertical shift of N − r (f t ) and R λr (g) is a constant.

Dissections in order r
In this section we extend to order r the Theorems of the polygon and of the residual polynomial. We fix throughout a type t of order r − 1 and a representative φ r (x) of t. We proceed by induction and we asume that all results of this section have been proved already in orders 1, . . . , r − 1. The case r = 1 was considered in section 1.

Theorem of the polygon in order r. Let f (x) ∈ O[x]
be a monic polynomial such that ω r (f ) > 0. The aim of this section is to obtain a factorization of f t (x) and certain arithmetic data of the factors. Thanks to Corollary 2.27, we shall be able to read this information directly on N − r (f ), and the different residual polynomials R λr (f )(y).
Theorem 3.1 (Theorem of the polygon in order r). Let f (x) ∈ O[x] be a monic polynomial such that ω r (f ) > 0. Suppose that N − r (f ) = S 1 + · · · + S g has g sides with pairwise different slopes λ r,1 , . . . , λ r,g . Then, f t (x) admits a factorization as a product of g monic polynomials of O[x] satisfying the following properties: (1) N r (F i ) is equal to S i up to a translation, Proof. Let us denote e = e 1 · · · e r−1 . We deal first with the case f t (x) irreducible.
Note that deg f t = m r ω r (f ) > 0, by Lemma 2.2, and N r (f t ) = N − r (f t ) by Corollary 2.18. Since f t (x) is irreducible, ρ := v(φ r (θ)) is constant among all roots θ ∈ Q p of f t (x), and 0 ≤ v r (φ r )/e < ρ, by Proposition 2.9. We have ρ = ∞ if and only if f t (x) = φ r (x), and in this case the theorem is clear. Suppose ρ is finite. Let be the minimal polynomial of φ r (θ), and let Q(x) = P (φ r (x)) = 0≤i≤k b i φ r (x) i . By the Theorem of the polygon in order one, the x-polygon of P has only one side and it has slope −ρ. The end points of N r (Q) are (0, ekρ) and (k, kv r (φ r )). Now, for all 0 This implies that N r (Q) has only one side and it has slope λ r := −(eρ − v r (φ r )).
Since Q(θ) = 0, f t (x) divides Q(x) and the Theorem of the product shows that N r (f t ) is one-sided, with the same slope. Also, R λr (f t ) ∼ R λr (f ) by Corollary 2.27. This ends the proof of the theorem when f t (x) is irreducible. If f t (x) is not necessarily irreducible, we consider its decomposition f t (x) = j P j (x) into a product of monic irreducible factors in O[x]. By Lemma 2.4, each P j (x) has type t and by the proof in the irreducible case, each P j (x) has a onesided N r (P j ). The Theorem of the product shows that the slope of N r (P j ) is λ r,i for some 1 ≤ i ≤ s. If we group these factors according to the slope, we get the desired factorization. By the Theorem of the product, 27. The statement about v(φ r (θ)) is obvious because P j (θ) = 0 for some j, and we have already proved the formula for an irreducible polynomial.
We recall that the factor corresponding to a side S i of slope −∞ is necessarily Remark 1.7).
Let λ r = −h r /e r , with h r , e r positive coprime integers, be a negative rational number such that S := S λr (f ) has positive length. Let f t,λr (x) be the factor of f (x), corresponding to the pairt, λ r by the Theorem of the polygon. Choose a root θ ∈ Q p of f t,λr (x), and let L = K(θ). By item 4 of Propositions 1.17 and 3.5, in orders 1, . . . , r − 1, there is a well-defined embedding F r −→ F L , determined by This embedding depends on the choice of θ. After this identification of F r with a subfield of F L we can think that all residual polynomials of r-th order have coefficients in F L .
Proof. Item 1 is a consequence of the Theorem of the polygon and the formula for v r (φ r ) in Proposition 2.15. Item 2 follows from Proposition 2.9, because v r (π r ) = 1, ω r (π r ) = 0 by Proposition 2.15. Item 3 follows from the Theorem of the polygon and item 2 in order r − 1. Item 4 follows from items 2,3. We prove now an identity that plays an essential role in what follows.
where (s, u) are the coordinates of the initial point of S. By Corollary 3.2, On the other hand, by the Theorem of the polygon and Proposition 2.9, for all i: with equality if and only if (i, u i ) ∈ S. This proves item 2. Also, (29) and (30) show that v(P S (θ)) ≥ 0, so that P S (θ) belongs to O L . Denote for simplicity z r = γ r (θ). In order to prove the equality P S (θ) = R λr (P )(z r ), we need to show that for every (i, u i ) ∈ S: Let (s(a i ), u(a i )) be the initial point of S r−1 (a i ). By items 1,2 of the proposition in order r − 1 (Proposition 1.17 if r = 2), applied to the polynomial a i (x), Since v r (a i )e(L/K)/e = v L (a i (θ)), it suffices to check the following identity in L: which is a consequence of Lemma 3.4. This ends the proof of item 1.
Also, (30) shows that v(P (θ)) ≥ H/e, and v(P (θ)) = H/e ⇐⇒ v(P 0 (θ)) = H/e . For simplicity we denote by (s(a i ), u(a i )) the initial point of S r−1 (a i ). By (33), the initial point of )/e r−1 ). Now, by induction, the Theorem of the product, and (34), we have We get an analogous expression for γ ′ r (θ) j γ ′ r−1 (θ) tr−1(i) , just by putting ′ everywhere and by replacing u(a i ) by u(a i ρ i )/e r−1 . By taking the quotient of both expressions and taking classes modulo m K ′ we get into a product of t monic polynomials, with all N r (G i ) one-sided of slope λ r , and Proof. Let us deal first with the case F (x) := f t,λr (x) irreducible. We need only to prove that R λr (F )(y) is the power of an irreducible polynomial of F r [y]. Let θ ∈ Q p be a root of F (x), take L = K(θ), and fix the embedding F r → F L as in (27). Let K ′ be the unramified extension of K of degree f 0 · · · f r−1 , and let G( be the minimal polynomial of θ over K ′ , so that F (x) = σ∈Gal(K ′ /K) G σ (x). Under the embedding F r → F L , the field F r is identified to F K ′ . By Proposition 3.6, we can construct a type t ′ of order r − 1 over K ′ such that R ′ λr (F )(y) ∼ R λr (F )(cy), for some nonzero constant c ∈ F K ′ . By the construction of t ′ , for any σ = 1, the polynomial G σ (x) is not divisible by φ ′ 1 (x) modulo m K ′ ; thus, ω ′ r (G σ ) ≤ ω ′ 1 (G σ ) = 0, and R ′ λr (G σ )(y) is a constant. Therefore, by the Theorem of the product, R ′ λr (G)(y) ∼ R ′ λr (F )(y) ∼ R λr (F )(cy), so that R λr (F )(y) is the power of an irreducible polynomial of F r [y] if and only if R ′ λr (G)(y) has the same property over F K ′ . In conclusion, by extending the base field, we can suppose that By (17), Π(x) admits an expression Π(x) = π n ′ 0 φ 1 (x) n ′ 1 · · · φ r−1 (x) n ′ r−1 for some integers n ′ 1 , . . . , n ′ r . Take Φ(x) := π n0 φ 1 (x) n1 · · · φ r−1 (x) nr−1 with sufficiently large non-negative integers n i so that Π(x) k Φ(x) is a polynomial in O[x]. Then, the following rational function is actually a polynomial in O[x]: Moreover, by item 5 of Proposition 2.15, ω r (B jer ) = 0 for all j such that B jer = 0, so that this φ r -development of g(x) is admissible.
Our aim is to show that N r (Q) is one-sided with slope λ r , and R λr (Q)(y) is equal to P (y) modulo m, up to a nonzero multiplicative constant. Since P (x) is irreducible, R λr (Q)(y) will be the power of an irreducible polynomial of F[y]. Since Q(θ) = 0, F (x) is a divisor of Q(x) and the residual polynomial of F (x) will be the power of an irreducible polynomial too, by the Theorem of the product. This will end the proof of the theorem in the irreducible case.
Let R λr (g)(y) = k j=0 c jer y j . We want to show that c jer = cb j for certain constant c ∈ F * independent of j. If (je r , u jer ) ∈ T , then c jer = 0, and by (35), this is equivalent tob j = 0. Suppose now (je r , u jer ) ∈ T ; by item 1 of Proposition 3.5 (cf. (31)) Hence, we want to check that for all j for some nonzero constant c. Now, if we substitute B jer (x) and Π(x) by its defining values, the left hand side is equal to cb j , for c = red L (Φ(θ)/π r (θ) u ). This ends the proof of the theorem in the irreducible case.
In the general case, consider the decomposition, F (x) = j P j (x), into a product of monic irreducible factors in O[x]. By Lemma 2.4, each P j (x) has type t, so that ω r (P j ) > 0. By the Theorem of the product, N r (P j ) is one-sided, of positive length and slope λ r . By the proof in the irreducible case, the residual polynomial R λr (P j )(y) is the positive power of an irreducible polynomial, and by the Theorem of the product it must be R λr (P j )(y) ∼ ψ r,i (y) bj for some 1 ≤ i ≤ t. If we group these factors according to the irreducible factor of the residual polynomial, we get the desired factorization.
Corollary 3.8. With the above notations, let θ ∈ Q p be a root of G i (x), and L = K(θ). Let f r = deg ψ r,i (y).
Proof. The statement about f (L/K) is a consequence of the extension of the embeding (27) to an embedding which is well-defined by item 4 of Proposition 3.5. The other statements follow from the Theorem of the product. The computation of f (L/K) and e(L/K) follows from and the fact that f (L/K) is divisible by f 0 · · · f r and e(L/K) is divisible by e 1 · · · e r (cf. Corollary 3.3).

Types of order r attached to a separable polynomial. Let f (x) ∈ O[x]
be a monic separable polynomial.
Definition 3.9. Let t be a type of order r − 1. We say that t is f -complete, if ω r (f ) = 1. In this case, f t (x) is irreducible and the ramification index and residual degree of the extension of K determined by f t (x) can be computed in terms of some data of t, by applying Corollary 3.8 in order r − 1 (Corollary 1.16 if r = 2).
The results of section 3 can be interpreted as the addition of two more dissections, for each order 2, . . . , r, to the three classical ones, in the process of factorization of f (x). If t is a type of order r − 1 and ω r (f ) > 1, the factor f t (x) experiments further factorizations at two levels: first f t (x) factorizes into as many factors as the number of sides of N − r (f ), and then, the factor corresponding to each finite slope splits into the product of as many factors as the number of pairwise different irreducible factors of the residual polynomial attached to the slope.
We can think that the type t has sprouted to produce several types of order r, t ′ = (t; λ r , ψ r (y)), each of them distinguished by the choice of a finite slope λ r of a side of N − r (f ), and a monic irreducible factor ψ r (y) of R λr (f )(y) in F r [y].
Definition 3.10. In Sect. 1.5, we defined two sets t 0 (f ), t 1 (f ). We recursively define t r (f ) to be the set of all types of order r constructed as above, t ′ = (t; λ r , ψ r (y)), from those t ∈ t r−1 (f ) that are not f -complete. This set is not an intrinsic invariant of f (x) because it depends on the choices of the representatives φ 1 (x), . . . , φ r (x) of the truncations of t. We denote by t s (f ) compl the subset of the f -complete types of t s (f ). We define Hensel's lemma and the theorems of the polygon and of the residual polynomial in orders 1, . . . , r determine a factorization The following remark is an immediate consequence of the definitions.
Lemma 3.11. The following conditions are equivalent: (3) For all t ∈ t r−1 (f ) and all λ r ∈ Q − , the residual polynomial of r-th order, R λr (f )(y) is separable.
If these conditions are satisfied, then (37) is a factorization of f (x) into the product of monic irreducible polynomials in O[x], and we get arithmetic information about each factor by Corollary 3.8. As long as there is some t ∈ t r (f ) which is not f -complete, we must apply the results of this section in order r + 1 to get further factorizations of f t (x), or to detect that it is irreducible. We need some invariant to control the whole process and ensure that after a finite number of steps we shall have t r (f ) compl = t r (f ). This is the aim of the next section.
We end with a remark about p-adic approximations to the irreducible factors of f (x), that is an immediate consequence of Lemma 2.2, the Theorem of the polygon and Proposition 2.15.
Proposition 3.12. Let t be an f -complete type of order r, with representative φ r+1 (x). Let θ ∈ Q p be a root of f t (x), and L = K(θ). Then, deg φ r+1 = deg f t , and φ r+1 (x) is an approximation where −h r+1 is the slope of the unique side of N − r+1 (f ), and e r+1 = 1.

Indices and resultants of higher order
We fix throughout this section a natural number r ≥ 1.
are the lengths and heights of the sides S i of N − r (P ), and E ′ j = ℓ(S ′ j ), H ′ j = H(S ′ j ) are the lengths and heights of the sides S ′ j of N − r (Q). We recall that for a side S of slope −∞ we took H(S) = ∞ by convention. Thus, the part of Res t (P, Q) that involves sides of slope −∞ is always (38) f 0 · · · f r−1 (ord φr (P )H(Q) + ord φr (Q)H(P )), where H(P ), H(Q) are the total heights respectively of N − r (P ), N − r (Q).
Proof. The three first items are an immediate consequence of the definition. Item 4 follows from N − r (P P ′ ) = N − r (P ) + N − r (P ′ ).
In the simplest case when N − r (P ) and N − r (Q) are both one-sided, Res t (P, Q) represents the area of the rectangle joining the two triangles determined by the sides, if they are ordered by increasing slope. The reader may figure out a similar geometrical interpretation of Res t (P, Q) in the general case, as the area of a union of rectangles below the Newton polygon Our aim is to compute v(Res(P, Q)) as a sum of several Res t (P, Q) for an adequate choice of types t. To this end, we want to compare types attached to P and Q, and this is uneasy because in the definition of the sets t r (P ), t r (Q), we had freedom in the choices of the different representatives φ i (x). For commodity in the exposition, we shall assume in this section that these polynomials are universally fixed.
Convention. We fix from now on a monic lift φ 1 (x) ∈ O[x] of every monic irreducible polynomial ψ 0 (y) ∈ F[y]. We proceed now recursively: for any 1 ≤ i < r and any type of order i with φ 1 (x), . . . , φ i (x) belonging to the infinite family of previously chosen polynomials, we fix a representative φ i+1 (x) of t. Also, we assume from now on that all types are made up only with our chosen polynomials φ i (x).
Once these choices are made, the set t r (P ) is uniquely determined by r and P (x). More precisely, t r (P ) is the set of all types t of order r such that ω t r+1 (P ) > 0 and the truncation t r−1 is not P -complete; in other words, t r (P ) = {t type of order r such that ω t r+1 (P ) > 0, ω t r (P ) > 1}. However, in view of the computation of resultants, we need a broader concept of "type attached to a polynomial".  Note that the analogous statement for the sets t r (P ) is false. For instance, let P (x), Q(x) be two monic polynomials congruent to the same irreducible polynomial ψ(y) modulo m. We have t 0 (P ) = t 0 (Q) = {ψ(y)} = t 0 (P Q), and the type of order zero ψ(y) is P -complete and Q-complete; thus, t 1 (P ) = ∅ = t 1 (Q). However, ψ(y) is not P Q-complete, and t 1 (P Q) = ∅.
We could also build the sett r (P ) in a constructive way analogous to that used in the last section to construct t r (P ). The only difference is that the P -complete types of order r − 1 are expanded to produce types of order r as well. Thanks to our above convention about fixing a universal family of representatives of the types, these expansions are unique.
. Then, t can be extended to a unique type t ′ ∈t r (P ) such that t ′ r−1 = t. The type t ′ is P -complete too. Proof. By Lemma 2.17, N − r (P ) has length one and finite slope λ r ∈ Q − ; hence, deg R λr (P )(y) = d(N − r (P )) = 1. Let ψ r (y) be the monic polynomial of degree one determined by R λr (P )(y) ∼ ψ r (y). The type t ′ = (t; λ r , ψ r (y)) is P -complete, and it is the unique type of order r such that t ′ r−1 = t and ω t ′ r+1 (P ) > 0. In fact, let us check that ω t ′′ r+1 (P ) = 0 for any t ′′ = (t; λ ′ r , ψ ′ r (y)) = t ′ . If λ ′ r = λ r , then R λ ′ r (P ) is a constant; if λ ′ r = λ r , but ψ r (y) = ψ ′ r (y) then ψ ′ r (y) cannot divide R λr (P )(y). Proof. By Lemma 4.4, we can assume that P (x) is irreducible. If P (x) = φ s (x) is the representative of some type of order s − 1 ≤ r − 1, then N s (φ s ) is one-sided of slope −∞; hence, R λs (φ s ) is a constant for every λ s ∈ Q − , and ω t ′ s+1 (φ s ) = 0, for every type t ′ of order ≥ s. Thus,t r (φ s ) = ∅, for all r ≥ s. Otherwise, the theorems of the polygon and of the residual polynomial show that the unique element of t 0 (P ) can be successively extended to a unique element oft 1 (P ), . . . ,t r (P ). Res t (P, Q).
The following observation is an immediate consequence of Lemma 4.2.
Res s (P, Q), and equality holds if and only if ω ′ r+1 (P ) = 0, i.e. if and only if R λ ′ r (P )(y) is not divisible by ψ ′ r (y). This condition is equivalent to (3) of Lemma 4.8 because R λr (P )(y) ∼ ψ r (y) a for some monic irreducible polynomial ψ r (y) ∈ F r [y], and R λ ′′ r (P )(y) is a constant for any negative rational number λ ′′ r = λ r . This ends the proof of the theorem in this case.
Let now P (x) = P 1 (x) · · · P g (x), Q(x) = Q 1 (x) · · · Q g ′ (x) be the factorizations of P (x), Q(x) into a product of monic irreducible polynomials in O[x]. We have proved above that v(Res(P i , Q j )) ≥ Res 1 (P i , Q j )+· · ·+Res r (P i , Q j ) for all i, j; hence, item 1 follows from Lemma 4.9 and the bilinearity of resultants. Also, equality in item 1 holds for the pair P, Q if and only if it holds for each pair P i , Q j ; that is, if and only if Res r+1 (P i , Q j ) = 0, for all i, j. This is equivalent to Res r+1 (P, Q) = 0, again by Lemma 4.9.
We end this section with an example that illustrates the necessity to introduce the setst r (P ). Let O = Z p , P (x) = x+p, Q(x) = x+p+p 100 , and let t 0 = y ∈ F[y]. Clearly, t 0 (P ) = {t 0 } = t 0 (Q), and t 0 is both P -complete and Q-complete, so that t 1 (P ) = ∅ = t 1 (Q). If we take φ 1 (x) = x, we get Res 1 (P, Q) = Res t0 (P, Q) = 1, whereas v(Res(P, Q)) = 100. Thus, we need to consider the expansions of t 0 to types of higher order in order to reach the right value of v(Res(P, Q)). The number of expansions to consider depends on the choices of the representatives φ i (x); for instance, if we take t = (x; −1, y + 1), with representative φ 2 (x) = x + p, we have already Res 2 (P, Q) = 99.
Nevertheless, the setst r (P ) were introduced only as an auxiliary tool to prove Theorem 4.10. In practice, the factorization algorithm computes only the sets t r (P ), as we shall show in the next section.

4.2.
Index of a polynomial and index of a polygon. All types that we consider are still assumed to be made up with polynomials φ i (x) belonging to a universally fixed family, as indicated in the last section.
Let Recall the well-known relationship, v(disc(F )) = 2 ind(F ) + v(disc(L/K)), linking ind(F ) with the discriminant of F (x) and the discriminant of L/K. Let N = S 1 + · · · + S g be a principal polygon, with sides ordered by increasing slopes λ 1 < · · · < λ g . We define If N has a side S 1 of slope −∞ and length E ∞ := E 1 , it contributes with E ∞ H fin (N ) to ind(N ), where H fin (N ) is the total height of the finite part of N .  For instance, the polygon below has index 25, the infinite side contributes with 18 (the area of the rectangle 3 × 6) and the finite part has index 7, corresponding to the marked seven points of integers coordinates, distributed into ind(S 1 ) = 2, ind(S 2 ) = 1, E 1 H 2 = 4. integer coordinates below N fin , above L and beyond L ′ , as the sum of the points with given abscissa: (40) ind(N fin ) = ⌊y i1+1 ⌋ + · · · + ⌊y i2−1 ⌋.
Since the Newton polygon N − r (P ) depends on the choice of φ r (x), the value of ind t (P ) depends on this choice too, although this is not reflected in the notation. (1) Let t be a type of order r, and suppose that t ∈ t r (P ) or t is P -complete. Then, ind t (P ) = 0.
If ind r (P ) = 0, then ind t (P ) = 0 for all t ∈ t r−1 (P ). For any such t we have ω r (P ) > 0, so that N − r (P ) is not reduced to a point. By Remark 4.13, N − r (P ) is one-sided with either slope −∞, or length one, or height one. In the first case P (x) is divisible by the representative φ r (x) of t and ω r (P ) = ℓ(N − r (P )) = ord φr (P ) = 1, because P (x) is separable; thus, t is P -complete and t is not extended to any type in t r (P ). If N − r (P ) is one-sided with finite slope λ r and the side has degree one, then the residual polynomial R λr (P )(y) has degree one. Thus, t is either P -complete or it can be extended in a unique way to a type t ′ ∈ t r (P ); in the latter case, necessarily ω t ′ r+1 (P ) = 1 and t ′ is P -complete. This proves item 3. Proof. For commodity, in the discussion we omit the weight f 0 · · · f r−1 that multiplies all terms in the identities.
All terms involved in the first identity are the sum of a finite part and an infinite part. If P (x)Q(x) is not divisible by φ r (x), all infinite parts are zero. If φ r (x) divides (say) P (x), then the infinite part of ind t (P Q) is ord φr (P )(H fin (P )+ H fin (Q)), the infinite part of ind t (P ) is ord φr (P )H fin (P ), the infinite part of ind t (Q) is zero, and the infinite part of Res t (P, Q) is ord φr (P )H(Q), by (38). Thus, the first identity is correct, as far as the infinite parts are concerned.
The finite part of the first identity follows from N − r (P Q) = N − r (P ) + N − r (Q) and Remark 4.14. Let N = N − r (P Q) and let R be the region of the plane that lies below N , above the line L and beyond the line L ′ , as indicated in Remark 4.14. The number ind t (P Q) counts the total number of points of integer coordinates in R, the number ind t (P ) + ind t (Q) counts the number of points of integer cordinates in the regions determined by the right triangles whose hypotenuses are the sides of N − r (P ) and N − r (Q). The region of R not covered by these triangles is a union of rectangles and Res t (P, Q) is precisely the number of points of integer coordinates of this region.
In order to prove the second identity, we note first that for any monic separable by (1) of Lemma 4.16. Now, if we apply this to R = P, Q, P Q, the identity follows from the first one and Lemma 4.4, having in mind that ind t (Q) = 0 = Res t (P, Q) if t ∈t r−1 (Q), because N − r (Q) reduces to a point.
We are ready to state the Theorem of the index, which is a crucial ingredient of the factorization process. It ensures that an algorithm based on the computation of the sets t r (f ) and the higher indices ind r (f ) obtains the factorization of f (x), and relevant aritmetic information on the irreducible factors, after a finite number of steps. Also, this algorithm yields a computation of ind(f ) as a by-product. Note that Lemma 4.16 and this theorem guarantee the equality in (1) whenever all types of t r (f ) are f -complete. Also, Theorem 4.18 shows that this latter condition will be reached at some order r. Proof. By the Theorem of the index, there exists r ≥ 1 such that ind r (f ) = 0, and by (2) of Lemma 4.16 this implies t r (f ) = t r (f ) compl .
In the next section we exhibit an example where the factorization is achieved in order three. More examples, and a more accurate discussion of the computational aspects can be found in [GMN08a]. 4.3. An example. Take p = 2, and f (x) = x 4 + ax 2 + bx + c ∈ Z[x], with v(a) ≥ 2, v(b) = 3, v(c) = 2. This polynomial has v(disc(f )) = 12 for all a, b, c with these restrictions. Since f (x) ≡ x 4 (mod 2), all types we are going to consider will start with φ 1 (x) = x. The Newton polygon N 1 (f ) has slope λ 1 = −1/2 • • r r r r r r r r r r r r 0 1 2 3 4 1 2 × × and the residual polynomial of f (x) with respect to λ 1 is R 1 (f )(y) = y 2 + 1 = (y + 1) 2 ∈ F, where F is the field ot two elements. Hence, t 1 (f ) = {t}, where t := (x; −1/2, y + 1). We have e 1 = 2, f 0 = f 1 = 1 and ω 2 (f ) = 2, so that t is not f -complete. The partial information we get in order one is ind 1 (f ) = 2, and the fact that all irreducible factors of f (x) will generate extensions L/Q 2 with even ramification number, because e 1 = 2.
We need still some auxiliary lemmas. The first one is an easy remark about integral parts.
Lemma 4.24. For all x ∈ R and e ∈ Z >0 , we have 0≤k<e x + k e = ⌊x⌋.
Proof. The identity is obvious when x is an integer, 0 ≤ x < e, because x + k e = 1 for the x values of k such that e − x ≤ k < e, and it is zero otherwise. Write x = n + ǫ, with n = ⌊x⌋ and 0 ≤ ǫ < 1; clearly, ⌊(x + k)/e⌋ = ⌊(n + k)/e⌋, because ǫ/e < 1/e. Consider the division with remainder, n = Qe + r, with 0 ≤ r < e. Then, 0≤k<e n + k e = 0≤k<e Q + r + k e = eQ + r = n.
Proof. For any j ∈ N r+1 denote by λ j the positive integer λ j := e 1 · · · e r ν j = r s=1 j s s i=1 e i f i · · · e s−1 f s−1 e i+1 · · · e r h i = r i=1 r t=i j t e i f i · · · e t−1 f t−1 e i+1 · · · e r h i .
Thus, the class of j s modulo e s is uniquely determined, and it depends only on j s+1 , . . . , j r .
By Corollary 4.26, Φ(j) = z k0 0 z k1+i1 1 · · · (z r−1 ) kr−1+ir−1 z kr r = z k0 0 z k1 1 Γ 2 (k 2 , . . . , k r ) · · · Γ r (k r ), where Γ s (k s , . . . , k r ) := z ks s (z s−1 ) is−1 , for s ≥ 2. Now, the family of all Φ(j) for j ∈ J 0 is an F K -basis of F L . In fact, the set of all Γ r (k r ) for 0 ≤ k r < f r , is a F r -basis of F L = F r+1 , because they are obtained from the basis z kr r , just by multiplying every element by the nonzero scalar z ir−1 r−1 ∈ F r , which depends only on k r . Then, the set of all Γ r−1 (k r−1 , k r )Γ r (k r ) for 0 ≤ k r−1 < f r−1 , 0 ≤ k r < f r , is a F r−1 -basis of F L , because they are obtained from the basis (z r−1 ) kr−1 Γ r (k r ), just by multiplying every element by the nonzero scalar z ir−2 r−2 ∈ F r−1 , which depends only on k r−1 , k r , etc. ⌊ν j ⌋ = ind 1 (f ) + · · · + ind r (f ).
This proves both statements of the theorem and it ends the proof of the theorem when f (x) is irreducible.
In the general case, if f (x) = F 1 (x) · · · F k (x) is the factorization of f (x) into a product of monic irreducible polynomials, we have by definition By Lemma 4.17, an analogous relationship holds for every ind s (f ), 1 ≤ s ≤ r. Hence, item 1 of the theorem holds by the theorem applied to each ind(F i ), and by Theorem 4.10. Let us prove now item 2. By Lemma 4.17, ind r+1 (f ) = 0 if and only if ind r+1 (F i ) = 0 and Res r+1 (F i , F j ) = 0, for all i and all j = i. By the theorem in the irreducible case and Theorem 4.10, this is equivalent to ind(f ) = ind 1 (f ) + · · · + ind r (f ).