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

By Lhoussain El Fadil and Mohamed Faris

Submitted: April 21st 2021Reviewed: August 20th 2021Published: October 14th 2021

DOI: 10.5772/intechopen.100021

## 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>0for every 0=i,,n1, and gcdνpa0n=1for some prime integer p, then fxis 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 Kνand for a monic polynomial fx=ϕnx+an1xϕn1x++a0xRνx, where Rνis a valuation ring of a discrete rank one valuation and ϕbeing a monic polynomial in Rνxwhose reduction ϕ¯is irreducible over Fν, aixRνx, degai<degϕfor every i=0,,n1, if νai1i/nνa0for every i=0,,n1and gcdνa0n=1, then fxis irreducible over the field K. In this paper, based on absolute value, we give an irreduciblity criterion of monic polynomials. More precisely, let Kbe a discrete rank one valued field, Rits valuation ring, F, its residue field, and Γ=Kits value group, we show that for a monic polynomial fx=ϕnx+an1xϕn1x++a0xRx, where ϕbeing a monic polynomial in Rxwhose reduction ϕ¯is irreducible over F, aixRx, degai<degϕfor every i=0,,n1, if aniγifor every i=0,,n1and nis the smallest integer satisfying γnΓ, where γ=a01/n, then fxis 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 Lextending , where Kis a discrete rank one valued field. Our results are illustrated by some examples.

## 2. Preliminaries

### 2.1 Newton polygons

Let L=Qαbe a number field generated by a complex root αof a monic irreducible polynomial fxZxand ZLthe ring of integers of L. In 1894, K. Hensel developed a powerful approach by showing that the prime ideals of ZLlying above a prime pare in one–one correspondence with monic irreducible factors of fxin 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, Rνits valuation ring, and L=Kαa simple extension generated by αK¯a root of a monic irreducible polynomial fxRνx, the valuations of Lextending νare in one–one correspondence with monic irreducible factors of fxin Khx, where Khis the henselization of Kwill be defined later. So, in order to describe all valuations of Lextending ν, one needs to factorize the polynomial fxinto 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 Kνbe a rank one discrete valued field (νK=Z), Rνits valuation ring, Mνits maximal ideal, Fνits residue field, and Khνhits henselization; the separable closure of Kin K̂, where K̂is the completion of K, and is an associated absolute value of ν. By normalization, we can assume that νK=Z, and so Mνis a principal ideal of Rνgenerated by an element πKsatisfying νπ=1. Let also νbe the Gauss’s extension of νto Khx. For any monic polynomial ϕRνxwhose reduction modulo Mνis irreducible in Fνx, let Fϕbe the field Fνxϕ¯.

Let fxRνxbe a monic polynomial and assume that fx¯is a power of ϕ¯in Fνx, with ϕRνxa monic polynomial, whose reduction is irreducible in Fνx. Upon the Euclidean division by successive powers of ϕ, we can expand fxas follows fx=i=0laixϕxi, where degai<degϕfor 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 Nϕfis the lower boundary of the convex envelope of the set of points iνaii=0lin the Euclidean plane. For every edge Sj, of the polygon Nϕf, let ljbe the length of the projection of Sjto the x-axis and Hjthe length of its projection to the y-axis. ljis called the length of Sjand Hjis its height. Let dj=gcdljHjbe the degree of Sj, ej=ljdjthe ramification degree of Sj, and λj=HjljQthe slope of Sj. Geometrically, we can remark that Nϕfis the process of joining the obtained edges S1,,Srordered by increasing slopes, which can be expressed by Nϕf=S1++Sr. The segments S1,,and Srare 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 Sof the polygon Nϕ+fof slope λand initial point sus, let lbe its length, Hits height and ethe smallest positive integer satisfying Z. Since =HZ, we conclude that edivides l, and so d=l/eZcalled 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 πϕis the maximal ideal of Rνxgenerated by πand ϕ.

Let λ=h/ebe the slope of S, where h=H/dand d=l/e. Notice that, the points with integer coordinates lying in Sare exactly sus,s+eush,,s+deusdh. Thus, if iis not a multiple of e, then s+ius+idoes 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+t0Fϕybe the residual polynomial of fxassociated to the side S, where for every i=0,,d, ti=cie. For every λQ+, the λ-component of Nϕfis the largest segment of Nϕfof slope λ. If Nϕfhas a side Sof slope λ, then T=S. Otherwise, Tis reduced to a single point; the end point of a side Si, which is also the initial point of Si+1if λi+1<λ<λior the initial point of Nϕfif λi<λfor every side Siof Nϕfor the end point of Nϕfif λi<λfor every side Siof Nϕf. In the sequel, we denote by Rλfy, the residual polynomial of fxassociated 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) Letfx=f1xf2xinRνxbe monic polynomials such thatfx¯is a positive power ofϕ¯. Then for everyλQ+, ifTiis theλ-componenet ofNϕfi, thenT=T1+T2is theλ-componenet ofNϕfand

Rλfy=Rλf1yRλf2y

up to multiplication by a nonzero element ofFϕ.

Theorem 2.2.(theorem of the polygon) LetfRνxbe a monic polynomial such thatfx¯is a positive power ofϕ¯. IfNϕf=S1++Sghasgsides of slopeλ1,,λgrespectively, then we can splitfx=f1××fgxinKhx, such thatNϕfi=SiandRλifiy=Rλifyup to multiplication by a nonzero.

Theorem 2.3.(theorem of the residual polynomial) LetfRνxbe a monic polynomial such thatNϕf=Shas a single side of finite slopeλ. IfRλfy=i=1tψiyaiis the factorization inFϕy, thenfxsplits asfx=f1x××ftxinKhxsuch thatNϕfi=Sihas a single side of slopeλandRαfiy=ψiyaiup to multiplication by a nonzero element ofFϕfor everyi=1,,t.

### 2.2 Absolute values

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

1. a=0if 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+bmaxabfor every abK2, then the absolute value is called a non archimidean absolute value and we say that Kis a non archimidean valued field.

Lemma 2.4.LetKbe a valued field. Thenis a non archimidean absolute value if and only if the setn1KnNis bounded inR.

Proof.By induction if is a non archimidean absolute value, then the set n1KnNis bounded by 1.

Conversely, assume that there exists MR+such that n1KMfor every nN. Let abK2, nN, and set m=supab. Then a+bn=k=0nnkakbnk, where nkis 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=supabas desired. □

Exercices 1.Let Kbe a valued field.

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

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

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

4. Show that if is a non archimidean absolute value and abK2such that ab, then a+b=maxab.

5. Let pbe a prime integer and p:QR+, defined by ap=pνpafor every aK, where νpis the p-adic valuation on Q. Show that pis a non archimidean absolute value of Q.

### 2.3 Characteristic elements of an absolute value

Let Kbe a non archimedian valued field.

Let R=aKa1and M=aKa<1. Then Ris a valuation ring, called the valuation ring of , Mits maximal ideal, and so F=R/Mis a field, called the residue field of .

Exercices 2.Let pbe a prime integer and the p-adic absolute value of Q, defined by a=pνpafor every aZ, where νpais the greatest integer satisfying pνpadivides afor a0and νp0=.

1. Show that Qis a non archimidean valued field.

2. Determine the characteristic elements of .

Exercices 3.Let Kbe a non archimidean valued field.

1. Show that Γ=Kis 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/Bfor every ABKx2with B0, define an absolute value on Kxcalled the Gauss’s extension of .

3. For every polynomial P=i=0npixϕiKx, let Pϕ=maxpii=0n.Show that, extended by A/Bϕ=Aϕ/Bϕfor every ABKx2with B0, ϕdefine an absolute value on Kx.

### 2.4 Completion and henselization

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

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

Example 1.

Any convergente sequence of Kis a Cauchy sequence.

The converse is false, indeed, it suffices to consider the valued field Q0with 0is the usual absolute value of Qand un=1+1/1!++1/n!for every natural integer n. Then unis a Cauchy sequence, which is not convergente.

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

Example 2.

1. R0is a complete valued field.

2. Q0is not a complete valued field.

Definition 3.Let Kbe a valued field, L/Kan extension of fields, and Lan absolute value of L.

1. We say that Lextends if Land coincide on K. In this case LL/Kis called a valued field extension.

2. Let LL/Kbe 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:Fis called its residue degree.

Definition 4.Let K11and K22be two valued fields and f:K1K2be an isomorphism of fields. fis said to be an isomorphism of valued fields if it preserves the absolute values.

Exercices 4.Let LL/Kbe a valued field extension.

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

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

3. Let Γ=Kand Δ=LL. Show that if Γis a discrete rank one Abelian group and L/Kis a finite extension, then Δis a discrete rank one Abelian group and L:K=ef, where eis the ramification index of the extension and fis its residue degree.

Theorem 2.5.([16], Theorem 1.1.4)

There exists a complete valued fieldLL, which extendsK.

Definition 5.The smallest complete valued field extending Kis called the completion of Kand 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)

LetfRxbe a monic polynomial such thatfx¯=g1xg2xinFxandg1xandg2xare coprime inFx. IfKis a complete valued field valued field, then there exists two monic polynomialsf1xandf2xinRxsuch thatf¯1x=g1xandf¯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 qbe a prime integer which is coprime to p, n2an integer, and fx=xn+qx+pqZx. First f¯x=xxn1+qin Fx. As fxis q-Eisenstein fxis irreducible over Q. Thus, we conclude that Hensel’s lemma is not applicable in Q.

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

Exercices 5.Let Kbe 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 fieldKis Henselian if and only if it extends uniquely toKs, where Ksis the separable closure of K.

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

Theorem 2.8.LetKbe a valued field. ThenKhis the separable closure ofKinKs.

## 3. Main results

Let Kbe a non archimidean valued field, νthe associated valuation to defined by νa=Lnafor every aK, Rits valuation ring, Mits maximal ideal, Fits residue field, and Khνhits 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 fand gin Rx,

1. fand gare said to be -equivalent if fg<f.

2. We say that gis -divides fif there exists qRxsuch that fand gqare -equivalent.

3. We say that a polynomial ϕRxis -irreducible if for every fand gin Rx, ϕ-divides fgimplies that ϕ-divides for ϕ-divides g.

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

1. ϕis monic,

2. ϕis -irreducible,

3. ϕis -minimal; for every nonzero polynomial fRx, ϕ-divides fimplies that degϕdegf.

It is easy to prove the following lemma:

Lemma 3.1.LetϕRxbe a monic polynomial. Ifϕ¯is irreducible overF, thenϕis a MacLane-Vaquié key polynomial of.

### 3.2 Augmented absolute values

Let ϕRxbe a MacLane-Vaquié key polynomial of and γR+with γϕ. Let ω:KxR0, defined by ωP=maxpiγii=0lfor every PKx, with P=i=0lpiϕiand degpi<degϕfor every i=0,,land extended by ωA/B=ωAωBfor every nozero Aand Bof Kx.

Lemma 3.2LetP=i=0nbiϕibe aϕ-expansion ofP, where the conditiondegbi<degϕfor everyi=0,,nis omitted. Ifϕ¯does not dividebix/b¯for everyi=0,,n, thenωP=maxbiγii=0n, wherebRνsuch thatbi=b. Such an expansion is called an admissible expansion.

Theorem 3.3.LetϕRxbe MacLane-Vaquié key polynomial ofandγR+withγϕ. The mapω:KxR0, defined byωP=maxpiγii=0lfor everyPKx, withP=i=0lpiϕianddegpi<degϕfor everyi=0,,l, and extended byωA/B=ωA/ωBfor every nonzero polynomialsABKx2, is an absolute value ofKx.

Proof.It suffices to check that ωsatisfies the three proprieties of an absolute value in Kx. Let ABKxbe tow polynomials, A=i=0kaiϕi, and B=i=0sbiϕithe ϕ-expansions.

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

2. Let A=i=0laiϕiand B=i=0tbiϕibe the ϕ-expansions of Aand B, with degai<degϕand degbi<degϕfor 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=0Lfiϕiis the ϕ-expansion of AB, where fi=ri+qi1with q1=0. Since ϕ=1, we conclude that riciand qicifor every i=0,,L. Let i1and i2be the smallest integers satisfying ωA=ai1γi1and ωB=bi2γi2. Then by the ultra-metric propriety ωciϕiai1γi1bi2γi2for every i=0,,L. For the equality, by definition of i1and i2, ωai1bjϕi1+j<ai1bjγi1+jfor every j<i2and ωajbi2ϕj+i2<ajbi2γj+i2for every j<i1. Thus by using the expression of ci1+i2, we conclude the equality. For i=i1+i2, let abRν2, with ai1=aand 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 ϕis MacLane-Vaquié key polynomial of , ϕdoes not -divide ai1/a, and ϕdoes not -divide bi2)/b. Therefore, ri=ci. Hence ωAB=maxaiγii=0lmaxbiγii=0tand ωAB=ωAωBas desired.

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

Definition 9.The absolute value ωdefined in Theorem 3.3 is denoted by ϕγand called the augmented absolue value of associated to ϕand γ.

Example 3.Let be the 2-adic absolute value defined on Qby a=eν2a, where for every integer b, ν2bis the largest integer satisfying 2kdivides bin Z. Let ϕ=x2+x+1Zx. By Lemma 3.1, ϕis a MacLane-Vaquié key polynomial of . Since ϕ=1, for every real γ, 0<γ1, the map ω:QxR0, defined by ωPα=maxpiγii=0lfor every PKx, with P=i=0lpiϕiand degpi<2.

### 3.3 Extensions of absolute values

The following Lemma makes a one–one correspondence between the absolute value of Land monic irreducible factors of fxin Khxfor any simple finite extension L=Kαof Kgenerated by a root αK¯of a monic irreducible polynomial fxKx.

Lemma 3.4.([19], Theorem 2.1)

LetL=Kαgenerated by a rootαK¯of a monic irreducible polynomialfxKxandfx=i=1tfieixbe the factorization into powers of monic irreducible factors ofKhx. Thenei=1for everyi=1,,tand there are exactlytdistinct valuations1,,andtofLextending. Furthermore for every absolute valueiofLassociated to the irreducible factorfi,Pαi=Pαi¯, where¯is the unique absolute value ofKh¯extendingandαiK¯is a root offix.

Lemma 3.5.([16], Corollary 3.1.4)

LetL/Kbe a finite extension andRLthe integral closure ofRinL. Then

RL=LRL;

for any elemntαL,αRLif and only ifαL1for every absolute valueLofLextending.

Lemma 3.6.LetfxRxbe a monic irreducible polynomial such thatfx¯is a power ofϕ¯inFxfor some monic polynomialϕRx, whose reduction is irreducible overF. LetL=KαwithαK¯a root offx. Then for every absolute valueLofLextending, for every nonzero polynomialPKx,PαLP.

The equality holds if and only ifϕ¯does not divideP0¯, whereP0=Pa, withaKsuch thatP=a.

In particular,ϕαL<1andPαL=Pfor every polynomialPKxsuchdegP<degϕ.

Proof.Let Lbe an absolute value of Lextending , PKxa nonzero polynomial, and aKwith a=P. Then P0=1. Since αis integral over R, we conclude that P0αL1. Thus, PαLa=P.

Moreover, the inequality PαL<Pmeans that P0αML, which means that P0α0modML. Consider the ring homomorphism φ:FxFML, defined by φP¯=Pα+ML. Then P0α0modMLis equivalent to ϕ¯does not divide P0¯.

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

Let PKxbe a nonzero polynomial of degree less than degree of ϕ. Then P0xRxis a primitive polynomial; P0=1. As degree P¯0is less than degree of ϕ¯, ϕ¯does not divide P0¯. Thus PαL=P. □

Theorem 3.7.LetfxRxbe a monic polynomial. Iffxis irreducible overKh, thenfx¯is a power ofϕ¯inFxfor some monic polynomialϕRx, whose reduction is irreducible overF. Moreover if we setfx=i=0naixϕixtheϕ-expansion offx, thenaniγifor everyi=0,,n, whereγ=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,,αkbe the roots of fxin 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 Khextending 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++b0the minimal polynomial of ϕαover Kh, and Fx=gϕx=ϕxt+bt1ϕxt1++b0. By the previous case, we conclude that btiτifor every i=0,,twith τ=b01/t, which means that NϕF=Shas a single side of slope λ=νb0t. Since Fα=0, we conclude that fxdivides Fx, and so Nϕfhas a single side of the same slope λ. Therefore, aniγifor every i=0,,n, where γ=a01/n.

Exercices 6.Let Kbe a non archimidean valued field and fxKhx. Set fx=i=0naixϕixthe 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.LetKbe a non archimidean valued field,Γ=Kits value group, andfxKxa monic polynomial such thatfx¯is a power ofϕ¯inFx. Letfx=i=0naixϕixbe theϕ-expansion offxand assume thataniγifor everyi=0,,n, whereγ=a01/n. Letebe the smallest positive integer satisfyingγeΓ. Thenfxhas at mostdirreducible monic factors inKhx, whered=n/ewith degree at leastemeach, andm=degϕ.

Proof.By applying the map Ln, the hypothesis aniγifor every i=0,,nmeans that νani, where λ=νa0n, which means that Nϕf=Shas a single side of slope λwith respect to ν. Let fx=i=1tfixbe a non trivial factorization of monic polynomials in Khx. Then by Theorem 2.2, Nϕfi=Sihas a single side of slope λ. Fix i=1,,tand let fix=j=0liaijxϕjbe the ϕ-expansion of fi. Then degfi=limand Lnγ=λis the slope of Si. Since eis the smallest positive integer satisfying γeΓ, we conclude that eis the smallest positive integer satisfying νK. On the other hand, since λ=ai0liis the slope of Si, where liis the length of the side Si, we conclude that edivides li. Thus degfi=diem, where di=lie. It follows that every non trivial factor fixhas 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, ife=n, thenfxis irreducible overKh.

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

Theorem 3.10.LetL=Kαbe a simple extension generated byαK¯a root of a monic irreducible polynomialfxRxsuch thatfx¯=ϕ¯ninFx. Letfx=i=0naixϕixbe theϕ-expansion offx. Assume thataniγifor everyi=0,,n, whereγ=a01/n. Then for every absolute valueLofLextending,PαLmaxpiγii=0lfor everyPKx, withP=i=0lpiϕiandl<n.

Proof.Let Lbe an absolute value of Lextending 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, aiαϕαiLγniτifor every i=0,,n. Thus, if τγ, maxaiαϕαiLi=0n=maxτnγn. Since is a non archimidean absolute value, we conclude that Lis 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=0lpiϕibe a polynomial in Kx. By the ultra-metric propriety, PαLpiαLγi. □

Theorem 3.11.LetL=Kαbe a simple extension generated byαK¯a root of a monic irreducible polynomialfxRxsuch thatfx¯=ϕ¯ninFx. Letfx=i=0naixϕixbe theϕ-expansion offx. Assume thataniγifor everyi=0,,n, whereγ=a01/n. Ifnis the smallest positive integer satisfyingγeΓ, then there is a unique absolute valueLofLextending. Moreover this absolute value is defined byPαL=maxpiγii=0lfor everyPKx, withP=i=0lpiϕiandl<n.

Furthermore, its ramification index isnand its residue degree ism=degϕ.

Proof.By Corollary 3.9, if n=e, then fxis irreducible over Kh. Thus by Hensel’s Lemma, there is a unique absolute value Lof Lextending . By Theorem 3.10, we conclude that ϕαL=γand PαLpiαLγi=pifor every polynomial P=i=0lpiϕiin Kx. Let us show the equality. Let sbe the smallest integer which satisfies ωP=psγs. Let ibe an integer satisfying ωP=piγi. Then γsi=pi/psΓ. Thus ndivides isbecause nis 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=γand n=Γγ:Γ, we conclude that ndivides the ramification index eof L. On the other hand, since FFϕFL, with Fϕ=Fxϕ¯, we have m=Fϕ:Fdivides FL:F. As mn=degf, we conclude the equality. □

Exercices 7.For every positive integer n2and pa positive prime integer, let fx=xnp.

1. How that fis irreducible over Q.

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

3. Let L=Qαwith αa complex root of fx. Show that there is a unique absolute value of Lextending the absolute p, defined on Qby ap=eνpafor every aQ, where νpis 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.LetL=Kαbe a simple extension generated byαK¯a root of a monic irreducible polynomialfxRx. Letfx¯=i=1rϕi¯nixbe the factorization offx¯inFx, with everyϕiRxis a monic polynomial. For everyi=1,,r, letNϕi+f=Si1++Sigibe the principalϕi-Newton polygon offx. ThenLhastabsolue value extendingwithrti=1rj=1gidij, wheredij=lijeijis the degree ofSij,lijis the length ofSij, andeij=lijdijfor everyi=1,,randj=1,,gi.

## 4. Applications

1. Let be the p-adic absolute value defined on Qby a=pνpaand fx=xnpZx. Show that fxis irreducible over Q. Let L=Qαwith αa complex root of fx. Determine all absolute value of Lextending .

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

2. Let be the p-adic absolute value and fx=xnaZxsuch that pdoes not divide νpa. Show that fxis irreducible over Q. Let L=Qαwith αa complex root of fx. Determine all absolute value of Lextending .

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

3. Let fx=ϕ6+24xϕ4+24ϕ3+1516x+32ϕ+48with ϕZxa monic polynomial whose reduction is irreducible in F2x. In Q2x, how many monic irreducible factors fxgets?, where Q2is the completion of Qand is the 2-adic absolute value.

Answer.It is easy to check that fxsatisfies the conditions of Theorem 3.8; a6iγiwith γ=24/6=21/32. Thus e=3and d=2. By Theorem 3.8, fxhas at most 2monic irreducible factors in Q2x.

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Lhoussain El Fadil and Mohamed Faris (October 14th 2021). On the Irreducible Factors of a Polynomial and Applications to Extensions of Absolute Values [Online First], IntechOpen, DOI: 10.5772/intechopen.100021. Available from: