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# On the Irreducible Factors of a Polynomial and Applications to Extensions of Absolute Values

Written By

Lhoussain El Fadil and Mohamed Faris

Reviewed: 20 August 2021 Published: 14 October 2021

DOI: 10.5772/intechopen.100021

From the Edited Volume

Edited by Kamal Shah

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## Abstract

Polynomial factorization over a field is very useful in algebraic number theory, in extensions of valuations, etc. For valued field extensions, the determination of irreducible polynomials was the focus of interest of many authors. In 1850, Eisenstein gave one of the most popular criterion to decide on irreducibility of a polynomial over Q. A criterion which was generalized in 1906 by Dumas. In 2008, R. Brown gave what is known to be the most general version of Eisenstein-Schönemann irreducibility criterion. Thanks to MacLane theory, key polynomials play a key role to extend absolute values. In this chapter, we give a sufficient condition on any monic plynomial to be a key polynomial of an absolute value, an irreducibly criterion will be given, and for any simple algebraic extension L=Kα, we give a method to describe all absolute values of L extending ∣∣, where K is a discrete rank one valued field.

### Keywords

• Irreducibly criterion
• irreducible factors
• Extensions of absolute values
• Newton polygon’s techniques

## 1. Introduction

Polynomial factorization over a field is very useful in algebraic number theory, for prime ideal factorization. It is also important in extensions of valuations, etc. For valued field extensions, the determination of irreducible polynomials was the focus of interest of many authors (cf. [1, 2, 3, 4, 5, 6, 7]). In 1850, Eisenstein gave one of the most popular criterion to decide on irreducibility of a polynomial over Q [1]. A criterion which was generalized in 1906 by Dumas in [8], who showed that for a polynomial fx=anxn+an1xn1++a0Qx (a00), if νpan=0, nνpainiνpa0>0 for every 0=i,,n1, and gcdνpa0n=1 for some prime integer p, then fx is irreducible over Q. In 2008, R. Brown gave what is known to be the most general version of Eisenstein-Schönemann irreducibility criterion [9]. He showed for a valued field Kν and for a monic polynomial fx=ϕnx+an1xϕn1x++a0xRνx, where Rν is a valuation ring of a discrete rank one valuation and ϕ being a monic polynomial in Rνx whose reduction ϕ¯ is irreducible over Fν, aixRνx, degai<degϕ for every i=0,,n1, if νai1i/nνa0 for every i=0,,n1 and gcdνa0n=1, then fx is irreducible over the field K. In this paper, based on absolute value, we give an irreduciblity criterion of monic polynomials. More precisely, let K be a discrete rank one valued field, R its valuation ring, F, its residue field, and Γ=K its value group, we show that for a monic polynomial fx=ϕnx+an1xϕn1x++a0xRx, where ϕ being a monic polynomial in Rx whose reduction ϕ¯ is irreducible over F, aixRx, degai<degϕ for every i=0,,n1, if aniγi for every i=0,,n1 and n is the smallest integer satisfying γnΓ, where γ=a01/n, then fx is irreducible over K. Similarly for the results of extensions of valuations given in [10, 11], for any simple algebraic extension L=Kα, we give a method to describe all absolute values of L extending , where K is a discrete rank one valued field. Our results are illustrated by some examples.

## 2. Preliminaries

### 2.1 Newton polygons

Let L=Qα be a number field generated by a complex root α of a monic irreducible polynomial fxZx and ZL the ring of integers of L. In 1894, K. Hensel developed a powerful approach by showing that the prime ideals of ZL lying above a prime p are in one–one correspondence with monic irreducible factors of fx in Qpx. For every prime ideal corresponding to any irreducible factor in Qpx, the ramification index and the residue degree together are the same as those of the local field defined by the irreducible factors [6]. These results were generalized in ([12], Proposition 8.2). Namely, for a rank one valued field Kν, Rν its valuation ring, and L=Kα a simple extension generated by αK¯ a root of a monic irreducible polynomial fxRνx, the valuations of L extending ν are in one–one correspondence with monic irreducible factors of fx in Khx, where Kh is the henselization of K will be defined later. So, in order to describe all valuations of L extending ν, one needs to factorize the polynomial fx into monic irreducible factors over Kh. The first step of the factorization was based on Hensel’s lemma. Unfortunately, the factors provided by Hensel’s lemma are not necessarily irreducible over Kh. The Newton polygon techniques could refine the factorization. Namely, theorem of the product, theorem of the polygon, and theorem of residual polynomial say that we can factorize any factor provided by Hensel’s lemma, with as many sides of the polygon and with as many of irreducible factors of the residual polynomial. For more details, we refer to [7, 13] for Newton polygons over p-adic numbers and [14, 15] for Newton polygons over rank one discrete valued fields. As our proofs are based on Newton polygon techniques, we recall some fundamental notations and techniques on Newton polygons. Let Kν be a rank one discrete valued field (νK=Z), Rν its valuation ring, Mν its maximal ideal, Fν its residue field, and Khνh its henselization; the separable closure of K in K̂, where K̂ is the completion of K, and is an associated absolute value of ν. By normalization, we can assume that νK=Z, and so Mν is a principal ideal of Rν generated by an element πK satisfying νπ=1. Let also ν be the Gauss’s extension of ν to Khx. For any monic polynomial ϕRνx whose reduction modulo Mν is irreducible in Fνx, let Fϕ be the field Fνxϕ¯.

Let fxRνx be a monic polynomial and assume that fx¯ is a power of ϕ¯ in Fνx, with ϕRνx a monic polynomial, whose reduction is irreducible in Fνx. Upon the Euclidean division by successive powers of ϕ, we can expand fx as follows fx=i=0laixϕxi, where degai<degϕ for every i=0,,l. Such a ϕ-expansion is unique and called the ϕ-expansion of fx. The ϕ-Newton polygon of f, denoted by Nϕf is the lower boundary of the convex envelope of the set of points iνaii=0l in the Euclidean plane. For every edge Sj, of the polygon Nϕf, let lj be the length of the projection of Sj to the x-axis and Hj the length of its projection to the y-axis. lj is called the length of Sj and Hj is its height. Let dj=gcdljHj be the degree of Sj, ej=ljdj the ramification degree of Sj, and λj=HjljQ the slope of Sj. Geometrically, we can remark that Nϕf is the process of joining the obtained edges S1,,Sr ordered by increasing slopes, which can be expressed by Nϕf=S1++Sr. The segments S1,, and Sr are called the sides of Nϕf. The principal ϕ-Newton polygon of fx, denoted by Nϕ+f, is the part of the polygon Nϕf, which is determined by joining all sides of negative slopes. For every side S of the polygon Nϕ+f of slope λ and initial point sus, let l be its length, H its height and e the smallest positive integer satisfying Z. Since =HZ, we conclude that e divides l, and so d=l/eZ called the degree of S. Remark that d=gcdlH. For every i=0,,l, we attach the following residue coefficient ciFϕ:

ci=0,ifs+ius+iliesstrictlyaboveSas+ixπus+imodπϕ,ifs+ius+iliesonS.E1

where πϕ is the maximal ideal of Rνx generated by π and ϕ.

Let λ=h/e be the slope of S, where h=H/d and d=l/e. Notice that, the points with integer coordinates lying in S are exactly sus,s+eush,,s+deusdh. Thus, if i is not a multiple of e, then s+ius+i does not lie on S, and so ci=0. It follows that the candidate abscissas which yield nonzero residue coefficient are s,s+e,, and s+de. Let Rλfy=tdyd+td1yd1++t1y+t0Fϕy be the residual polynomial of fx associated to the side S, where for every i=0,,d, ti=cie. For every λQ+, the λ-component of Nϕf is the largest segment of Nϕf of slope λ. If Nϕf has a side S of slope λ, then T=S. Otherwise, T is reduced to a single point; the end point of a side Si, which is also the initial point of Si+1 if λi+1<λ<λi or the initial point of Nϕf if λi<λ for every side Si of Nϕf or the end point of Nϕf if λi<λ for every side Si of Nϕf. In the sequel, we denote by Rλfy, the residual polynomial of fx associated to the λ-component of Nϕf.

The following are the relevant theorems from Newton polygon. Namely, theorem of the product and theorem of the polygon. For more details, we refer to [15].

Theorem 2.1. (theorem of the product) Let fx=f1xf2x in Rνx be monic polynomials such that fx¯ is a positive power of ϕ¯. Then for every λQ+, if Ti is the λ-componenet of Nϕfi, then T=T1+T2 is the λ-componenet of Nϕf and

Rλfy=Rλf1yRλf2y

up to multiplication by a nonzero element of Fϕ.

Theorem 2.2. (theorem of the polygon) Let fRνx be a monic polynomial such that fx¯ is a positive power of ϕ¯. If Nϕf=S1++Sg has g sides of slope λ1,,λg respectively, then we can split fx=f1××fgx in Khx, such that Nϕfi=Si and Rλifiy=Rλify up to multiplication by a nonzero.

Theorem 2.3. (theorem of the residual polynomial) Let fRνx be a monic polynomial such that Nϕf=S has a single side of finite slope λ. If Rλfy=i=1tψiyai is the factorization in Fϕy, then fx splits as fx=f1x××ftx in Khx such that Nϕfi=Si has a single side of slope λ and Rαfiy=ψiyai up to multiplication by a nonzero element of Fϕ for every i=1,,t.

### 2.2 Absolute values

Let be an absolute value of K; a map :KR+, which satisfies the following three axioms:

1. a=0 if and only if a0,

2. ab=ab, and

3. a+ba+b. (triangular inequality)

for every abK2.

If the triangular inequality is replaced by an ultra-inequality, namely a+bmaxab for every abK2, then the absolute value is called a non archimidean absolute value and we say that K is a non archimidean valued field.

Lemma 2.4. Let K be a valued field. Then is a non archimidean absolute value if and only if the set n1KnN is bounded in R.

Proof. By induction if is a non archimidean absolute value, then the set n1KnN is bounded by 1.

Conversely, assume that there exists MR+ such that n1KM for every nN. Let abK2, nN, and set m=supab. Then a+bn=k=0nnkakbnk, where nk is the binomial coefficient. As is a non archimidean absolute value, a+bnsup{nk1KakbnkMmn. Thus a+bM1/nm. Go over the limit, we obtain a+bm=supab as desired. □

Exercices 1. Let K be a valued field.

1. Show that if K is a finite field, then is a non archimidean absolute value.

2. More precisely, show that if K is a finite field, then is trivial; x=1 for every xK.

3. Let ν:KR be the map defined by νa=Lna for every aK, where Ln is the Napierian logarithm defined on R+. Show that is a non archimidean absolute value if and only if ν is a valuation of K. ν is called an associated valuation to .

4. Show that if is a non archimidean absolute value and abK2 such that ab, then a+b=maxab.

5. Let p be a prime integer and p:QR+, defined by ap=pνpa for every aK, where νp is the p-adic valuation on Q. Show that p is a non archimidean absolute value of Q.

### 2.3 Characteristic elements of an absolute value

Let K be a non archimedian valued field.

Let R=aKa1 and M=aKa<1. Then R is a valuation ring, called the valuation ring of , M its maximal ideal, and so F=R/M is a field, called the residue field of .

Exercices 2. Let p be a prime integer and the p-adic absolute value of Q, defined by a=pνpa for every aZ, where νpa is the greatest integer satisfying pνpa divides a for a0 and νp0=.

1. Show that Q is a non archimidean valued field.

2. Determine the characteristic elements of .

Exercices 3. Let K be a non archimidean valued field.

1. Show that Γ=K is a sub-group of R., called the value group of .

2. For every polynomial A=i=0naixiKx, let A=maxaii=0n. Show that, extended by A/B=A/B for every ABKx2 with B0, define an absolute value on Kx called the Gauss’s extension of .

3. For every polynomial P=i=0npixϕiKx, let Pϕ=maxpii=0n. Show that, extended by A/Bϕ=Aϕ/Bϕ for every ABKx2 with B0, ϕ define an absolute value on Kx.

### 2.4 Completion and henselization

Let K be a valued field and consider the map d:K×KR0, defined by dab=ab. Then d is a metric on K.

Definition 1. A sequence unKN is said to be a Cauchy sequence if for every positive real number ε, there exists an integer N such that for every natural numbers m,nN, we have unumε.

Example 1.

Any convergente sequence of K is a Cauchy sequence.

The converse is false, indeed, it suffices to consider the valued field Q0 with 0 is the usual absolute value of Q and un=1+1/1!++1/n! for every natural integer n. Then un is a Cauchy sequence, which is not convergente.

Definition 2. A valued field K is said to be complete if every Cauchy sequence of K is convergente.

Example 2.

1. R0 is a complete valued field.

2. Q0 is not a complete valued field.

Definition 3. Let K be a valued field, L/K an extension of fields, and L an absolute value of L.

1. We say that L extends if L and coincide on K. In this case LL/K is called a valued field extension.

2. Let LL/K be a valued field extension and Δ=LL. Then e=|Δ/Γ| the cardinal order of Δ/Γ, is called the ramification index of the extension and f=FL:F is called its residue degree.

Definition 4. Let K11 and K22 be two valued fields and f:K1K2 be an isomorphism of fields. f is said to be an isomorphism of valued fields if it preserves the absolute values.

Exercices 4. Let LL/K be a valued field extension.

1. Show that if is a non archimidean absolute value, then L is a non archimidean absolute value.

2. Assume that K is a non archimidean valued field. Show that the convergence of a series in K is equivalent to the convergence of its general term to 0.

3. Let Γ=K and Δ=LL. Show that if Γ is a discrete rank one Abelian group and L/K is a finite extension, then Δ is a discrete rank one Abelian group and L:K=ef, where e is the ramification index of the extension and f is its residue degree.

Theorem 2.5. ([16], Theorem 1.1.4)

There exists a complete valued field LL, which extends K.

Definition 5. The smallest complete valued field extending K is called the completion of K and denoted by K̂.

Furtheremore, the completion is unique up to a valued fields isomorphism.

Now we come to an important property of complete fields. This theorem is widely known as Hensel’s Lemma. For the proof, we refer to ([16], Lemma 4.1.3).

Theorem 2.6. (Hensel’s lemma)

Let fRx be a monic polynomial such that fx¯=g1xg2x in Fx and g1x and g2x are coprime in Fx. If K is a complete valued field valued field, then there exists two monic polynomials f1x and f2x in Rx such that f¯1x=g1x and f¯2x=g2x.

The following example shows that for any prime integer p, Hensel’s lemma is not applicable in Q, with is the p-adic absolute value defined by a=pνpa. Indeed, let q be a prime integer which is coprime to p, n2 an integer, and fx=xn+qx+pqZx. First f¯x=xxn1+q in Fx. As fx is q-Eisenstein fx is irreducible over Q. Thus, we conclude that Hensel’s lemma is not applicable in Q.

Definition 6. A valued field K is said to be Henselian if Hensel’s lemma is applicable in K. The smallest Henselian field extending K is called the henselization of (K, and denoted by Kh.

Exercices 5. Let K be a valued field K.

Show that KKhK̂. Furthermore, these three fields have the same value group and same residue fields.

We have the following apparently easier characterization of Henselian fields. For the proof, we refer to ([16], Lemma 4.1.1).

Theorem 2.7. The valued field K is Henselian if and only if it extends uniquely to Ks, where Ks is the separable closure of K.

In particular, we conclude the following characterization of the henselization Kh of K.

Theorem 2.8. Let K be a valued field. Then Kh is the separable closure of K in Ks.

## 3. Main results

Let K be a non archimidean valued field, ν the associated valuation to defined by νa=Lna for every aK, R its valuation ring, M its maximal ideal, F its residue field, and Khνh its henselization.

### 3.1 Key polynomials

The notion of key polynomials was introduced in 1936, by MacLane [17], in the case of discrete rank one absolute values and developed in [18] by Vaquié to any arbitrary rank valuation. The motivation of introducing key polynomials was the problem of describing all extensions of to any finite simple extension Kα. For any simple algebraic extension of K, MacLane introduced the notions of key polynomials and augmented absolute with respect to the gievn key.

Definition 7. Two nonzero polynomials f and g in Rx,

1. f and g are said to be -equivalent if fg<f.

2. We say that g is -divides f if there exists qRx such that f and gq are -equivalent.

3. We say that a polynomial ϕRx is -irreducible if for every f and g in Rx, ϕ -divides fg implies that ϕ -divides f or ϕ -divides g.

Definition 8. A polynomial ϕRx is said to be a MacLane-Vaquié key polynomial of if it satisfies the following three conditions:

1. ϕ is monic,

2. ϕ is -irreducible,

3. ϕ is -minimal; for every nonzero polynomial fRx, ϕ -divides f implies that degϕdegf.

It is easy to prove the following lemma:

Lemma 3.1. Let ϕRx be a monic polynomial. If ϕ¯ is irreducible over F, then ϕ is a MacLane-Vaquié key polynomial of .

### 3.2 Augmented absolute values

Let ϕRx be a MacLane-Vaquié key polynomial of and γR+ with γϕ. Let ω:KxR0, defined by ωP=maxpiγii=0l for every PKx, with P=i=0lpiϕi and degpi<degϕ for every i=0,,l and extended by ωA/B=ωAωB for every nozero A and B of Kx.

Lemma 3.2 Let P=i=0nbiϕi be a ϕ-expansion of P, where the condition degbi<degϕ for every i=0,,n is omitted. If ϕ¯ does not divide bix/b¯ for every i=0,,n, then ωP=maxbiγii=0n, where bRν such that bi=b. Such an expansion is called an admissible expansion.

Theorem 3.3. Let ϕRx be MacLane-Vaquié key polynomial of and γR+ with γϕ. The map ω:KxR0, defined by ωP=maxpiγii=0l for every PKx, with P=i=0lpiϕi and degpi<degϕ for every i=0,,l, and extended by ωA/B=ωA/ωB for every nonzero polynomials ABKx2, is an absolute value of Kx.

Proof. It suffices to check that ω satisfies the three proprieties of an absolute value in Kx. Let ABKx be tow polynomials, A=i=0kaiϕi, and B=i=0sbiϕi the ϕ-expansions.

1. ωA=0 if and only if aiγi=0 for every i=0,,k, which means ai=0 for every i=0,,k (because γR+). Therefore A=0.

2. Let A=i=0laiϕi and B=i=0tbiϕi be the ϕ-expansions of A and B, with degai<degϕ and degbi<degϕ for every i=0,,suplt. For every i=0,,L=l+t, let ci=j=0iajbij. Then AB=i=0Lciϕi. For every i=0,,L, upon the Euclidean division, let ci=qiϕ+ri. Then AB=i=0Lfiϕi is the ϕ-expansion of AB, where fi=ri+qi1 with q1=0. Since ϕ=1, we conclude that rici and qici for every i=0,,L. Let i1 and i2 be the smallest integers satisfying ωA=ai1γi1 and ωB=bi2γi2. Then by the ultra-metric propriety ωciϕiai1γi1bi2γi2 for every i=0,,L. For the equality, by definition of i1 and i2, ωai1bjϕi1+j<ai1bjγi1+j for every j<i2 and ωajbi2ϕj+i2<ajbi2γj+i2 for every j<i1. Thus by using the expression of ci1+i2, we conclude the equality. For i=i1+i2, let abRν2, with ai1=a and bi2=b. If ri<ci=ab, then ri/c<ci/c, and so ri/c<1. By reducing modulo M, we deduce that ϕ¯ divides ci/c¯, which means that ϕ -divides (ai1/a)(bi2/b), which is impossible because ϕ is MacLane-Vaquié key polynomial of , ϕ does not -divide ai1/a, and ϕ does not -divide bi2)/b. Therefore, ri=ci. Hence ωAB=maxaiγii=0lmaxbiγii=0t and ωAB=ωAωB as desired.

3. Completing by zeros; ai=0 if i>k and bi=0 if i>s, we have ωA+B=maxai+biγii=0supksmaxaiγii=0k+maxbiγii=0s=ωA+ωB. Thus, ωA+BωA+ωB.

Definition 9. The absolute value ω defined in Theorem 3.3 is denoted by ϕγ and called the augmented absolue value of associated to ϕ and γ.

Example 3. Let be the 2-adic absolute value defined on Q by a=eν2a, where for every integer b, ν2b is the largest integer satisfying 2k divides b in Z. Let ϕ=x2+x+1Zx. By Lemma 3.1, ϕ is a MacLane-Vaquié key polynomial of . Since ϕ=1, for every real γ, 0<γ1, the map ω:QxR0, defined by ωPα=maxpiγii=0l for every PKx, with P=i=0lpiϕi and degpi<2.

### 3.3 Extensions of absolute values

The following Lemma makes a one–one correspondence between the absolute value of L and monic irreducible factors of fx in Khx for any simple finite extension L=Kα of K generated by a root αK¯ of a monic irreducible polynomial fxKx.

Lemma 3.4. ([19], Theorem 2.1)

Let L=Kα generated by a root αK¯ of a monic irreducible polynomial fxKx and fx=i=1tfieix be the factorization into powers of monic irreducible factors of Khx. Then ei=1 for every i=1,,t and there are exactly t distinct valuations 1,, and t of L extending . Furthermore for every absolute value i of L associated to the irreducible factor fi, Pαi=Pαi¯, where ¯ is the unique absolute value of Kh¯ extending and αiK¯ is a root of fix.

Lemma 3.5. ([16], Corollary 3.1.4)

Let L/K be a finite extension and RL the integral closure of R in L. Then

RL=LRL;

for any elemnt αL, αRL if and only if αL1 for every absolute value L of L extending .

Lemma 3.6. Let fxRx be a monic irreducible polynomial such that fx¯ is a power of ϕ¯ in Fx for some monic polynomial ϕRx, whose reduction is irreducible over F. Let L=Kα with αK¯ a root of fx. Then for every absolute value L of L extending , for every nonzero polynomial PKx, PαLP.

The equality holds if and only if ϕ¯ does not divide P0¯, where P0=Pa, with aK such that P=a.

In particular, ϕαL<1 and PαL=P for every polynomial PKx such degP<degϕ.

Proof. Let L be an absolute value of L extending , PKx a nonzero polynomial, and aK with a=P. Then P0=1. Since α is integral over R, we conclude that P0αL1. Thus, PαLa=P.

Moreover, the inequality PαL<P means that P0αML, which means that P0α0modML. Consider the ring homomorphism φ:FxFML, defined by φP¯=Pα+ML. Then P0α0modML is equivalent to ϕ¯ does not divide P0¯.

In particular, since ϕRx, ϕ1. Furthermore as ϕ¯ divide ϕ¯, we conclude that ϕ<1.

Let PKx be a nonzero polynomial of degree less than degree of ϕ. Then P0xRx is a primitive polynomial; P0=1. As degree P¯0 is less than degree of ϕ¯, ϕ¯ does not divide P0¯. Thus PαL=P. □

Theorem 3.7. Let fxRx be a monic polynomial. If fx is irreducible over Kh, then fx¯ is a power of ϕ¯ in Fx for some monic polynomial ϕRx, whose reduction is irreducible over F. Moreover if we set fx=i=0naixϕix the ϕ-expansion of fx, then aniγi for every i=0,,n, where γ=a01/n.

Proof. The first point of the theorem is an immediate consequence of Theorem 2.6.

For the second point, let m=degϕ.

1. For m=1, let fx=i=1kxαi, where α1,,αk be the roots of fx in K¯, the algebraic closure of K. Then the formula linking roots and coefficients of fx, we conclude that fx=i=0ksixi, where sk=1, si=j1<<jiαj1αji. Keep the notation for the valuation of Kh extending and let ¯ be the unique extension of to Kh¯=K¯. Then α1¯==αk¯=τ, ski¯τi, and τ=γ.

2. For m2, let Ł=Khα, where αK¯ is a root of fx, gx=xt+bt1xt1++b0 the minimal polynomial of ϕα over Kh, and Fx=gϕx=ϕxt+bt1ϕxt1++b0. By the previous case, we conclude that btiτi for every i=0,,t with τ=b01/t, which means that NϕF=S has a single side of slope λ=νb0t. Since Fα=0, we conclude that fx divides Fx, and so Nϕf has a single side of the same slope λ. Therefore, aniγi for every i=0,,n, where γ=a01/n.

Exercices 6. Let K be a non archimidean valued field and fxKhx. Set fx=i=0naixϕix the ϕ-expansion of fx.

Show that fx=maxana0.

Based on absolute value, the following theorem gives an hyper bound of the number of monic irreducible factors of monic polynomials. In particular, Corollary 3.9 gives a criterion to test the irreducibility of monic polynomials.

Theorem 3.8. Let K be a non archimidean valued field, Γ=K its value group, and fxKx a monic polynomial such that fx¯ is a power of ϕ¯ in Fx. Let fx=i=0naixϕix be the ϕ-expansion of fx and assume that aniγi for every i=0,,n, where γ=a01/n. Let e be the smallest positive integer satisfying γeΓ. Then fx has at most d irreducible monic factors in Khx, where d=n/e with degree at least em each, and m=degϕ.

Proof. By applying the map Ln, the hypothesis aniγi for every i=0,,n means that νani, where λ=νa0n, which means that Nϕf=S has a single side of slope λ with respect to ν. Let fx=i=1tfix be a non trivial factorization of monic polynomials in Khx. Then by Theorem 2.2, Nϕfi=Si has a single side of slope λ. Fix i=1,,t and let fix=j=0liaijxϕj be the ϕ-expansion of fi. Then degfi=lim and Lnγ=λ is the slope of Si. Since e is the smallest positive integer satisfying γeΓ, we conclude that e is the smallest positive integer satisfying νK. On the other hand, since λ=ai0li is the slope of Si, where li is the length of the side Si, we conclude that e divides li. Thus degfi=diem, where di=lie. It follows that every non trivial factor fix has degree at least em. Since degf=i=1tdegfitem, we conclude that tne=d. □

Corollary 3.9. Under the hypothesis and notations of Theorem 3.8, if e=n, then fx is irreducible over Kh.

Proof. If n=e, then d=1, and so there is a unique monique polynomial of Kx which divides fx and this factor has the degree at least mn. As degf=nm, we conclude that fx is this unique monic factor. □

Theorem 3.10. Let L=Kα be a simple extension generated by αK¯ a root of a monic irreducible polynomial fxRx such that fx¯=ϕ¯n in Fx. Let fx=i=0naixϕix be the ϕ-expansion of fx. Assume that aniγi for every i=0,,n, where γ=a01/n. Then for every absolute value L of L extending , PαLmaxpiγii=0l for every PKx, with P=i=0lpiϕi and l<n.

Proof. Let L be an absolute value of L extending and let us show that ϕαL=γ. For this reason, let τ=ϕαL. By Lemma 3.6, 0<τ<1. By hypotheses and Lemma 3.6, aiαϕαiLγniτi for every i=0,,n. Thus, if τγ, maxaiαϕαiLi=0n=maxτnγn. Since is a non archimidean absolute value, we conclude that L is a non archimidean absolute value, and so by the ultra-metric propriety, fαL=maxτnγn>0, which is impossible because fα=0. Therefore ϕαL=γ.

Now, let P=i=0lpiϕi be a polynomial in Kx. By the ultra-metric propriety, PαLpiαLγi. □

Theorem 3.11. Let L=Kα be a simple extension generated by αK¯ a root of a monic irreducible polynomial fxRx such that fx¯=ϕ¯n in Fx. Let fx=i=0naixϕix be the ϕ-expansion of fx. Assume that aniγi for every i=0,,n, where γ=a01/n. If n is the smallest positive integer satisfying γeΓ, then there is a unique absolute value L of L extending . Moreover this absolute value is defined by PαL=maxpiγii=0l for every PKx, with P=i=0lpiϕi and l<n.

Furthermore, its ramification index is n and its residue degree is m=degϕ.

Proof. By Corollary 3.9, if n=e, then fx is irreducible over Kh. Thus by Hensel’s Lemma, there is a unique absolute value L of L extending . By Theorem 3.10, we conclude that ϕαL=γ and PαLpiαLγi=pi for every polynomial P=i=0lpiϕi in Kx. Let us show the equality. Let s be the smallest integer which satisfies ωP=psγs. Let i be an integer satisfying ωP=piγi. Then γsi=pi/psΓ. Thus n divides is because n is the smallest positive integer satisfying γeΓ. Since l<n, then is=0. Therefore, PαL=psγs=ωP.

For the residue degree and ramification index, since ϕαL=γ and n=Γγ:Γ, we conclude that n divides the ramification index e of L. On the other hand, since FFϕFL, with Fϕ=Fxϕ¯, we have m=Fϕ:F divides FL:F. As mn=degf, we conclude the equality. □

Exercices 7. For every positive integer n2 and p a positive prime integer, let fx=xnp.

1. How that f is irreducible over Q.

2. Conclude a new proof for R:Q=.

3. Let L=Qα with α a complex root of fx. Show that there is a unique absolute value of L extending the absolute p, defined on Q by ap=eνpa for every aQ, where νp is the p-adic valuation on Q. Calculate its residue degree and its ramification index.

Combining Lemma 3.4 and Theorem 3.8, we conclude the following result:

Corollary 3.12. Let L=Kα be a simple extension generated by αK¯ a root of a monic irreducible polynomial fxRx. Let fx¯=i=1rϕi¯nix be the factorization of fx¯ in Fx, with every ϕiRx is a monic polynomial. For every i=1,,r, let Nϕi+f=Si1++Sigi be the principal ϕi-Newton polygon of fx. Then L has t absolue value extending with rti=1rj=1gidij, where dij=lijeij is the degree of Sij, lij is the length of Sij, and eij=lijdij for every i=1,,r and j=1,,gi.

## 4. Applications

1. Let be the p-adic absolute value defined on Q by a=pνpa and fx=xnpZx. Show that fx is irreducible over Q. Let L=Qα with α a complex root of fx. Determine all absolute value of L extending .

Answer. First Γ=pkkZ is the value group of . Since p=p1, γ=a01/n=p1/n, we conclude that the smallest integer satisfying γeΓ is n. Thus, by Corollary 3.9, fx is irreducible over Qh,and so is over Q. Since fx¯=xn in Fpx, by Theorem 3.11, there is a unique absolute value of L extending and it is defined by PαL=max{piγi,i=o,,l} for every polynomial P=i=0lxi with l<n.

2. Let be the p-adic absolute value and fx=xnaZx such that p does not divide νpa. Show that fx is irreducible over Q. Let L=Qα with α a complex root of fx. Determine all absolute value of L extending .

Answer. First Γ=pkkZ is the value group of . Since p=p1, γ=a01/n=p1/n, we conclude that the smallest integer satisfying γeΓ is n. Thus, by Corollary 3.9, fx is irreducible over Qh,and so is over Q. Since fx¯=xn in Fpx, by Theorem 3.11, there is a unique absolute value of L extending and it is defined by PαL=max{piγi,i=o,,l} for every polynomial P=i=0lxi with l<n.

3. Let fx=ϕ6+24xϕ4+24ϕ3+1516x+32ϕ+48 with ϕZx a monic polynomial whose reduction is irreducible in F2x. In Q2x, how many monic irreducible factors fx gets?, where Q2 is the completion of Q and is the 2-adic absolute value.

Answer. It is easy to check that fx satisfies the conditions of Theorem 3.8; a6iγi with γ=24/6=21/32. Thus e=3 and d=2. By Theorem 3.8, fx has at most 2 monic irreducible factors in Q2x.

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Written By

Lhoussain El Fadil and Mohamed Faris

Reviewed: 20 August 2021 Published: 14 October 2021