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On the index of an algebraic integer and beyond
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Journal of Pure and Applied Algebra www.elsevier.com/locate/jpaa
On the index of an algebraic integer and beyond Anuj Jakhar a , Sudesh K. Khanduja b,c,∗ a
The Institute of Mathematical Sciences (IMSc), HBNI, CIT Campus, Taramani, Chennai, 600113, Tamil Nadu, India b Indian Institute of Science Education and Research Mohali, Sector 81, Knowledge City, SAS Nagar, Punjab, 140306, India c Department of Mathematics, Panjab University, Chandigarh, 160014, India
a r t i c l e
i n f o
Article history: Received 8 July 2019 Received in revised form 31 October 2019 Available online xxxx Communicated by V. Suresh Dedicated to Professor Raman Parimala on her 70th Birthday MSC: 11R04; 11R29; 12J10 Keywords: Rings of algebraic integers Dedekind domains Valued fields
a b s t r a c t Let K = Q(θ) be an algebraic number field with θ in the ring AK of algebraic integers of K having minimal polynomial f (x) over Q. For a prime number p, let ip (f ) denote the highest power of p dividing the index [AK : Z[θ]]. Let f¯(x) = φ¯1 (x)e1 · · · φ¯r (x)er be the factorization of f (x) modulo p into a product of powers of distinct irreducible polynomials over Z/pZ with φi (x) ∈ Z[x] monic. Let the r integer l ≥ 1 and the polynomial N (x) ∈ Z[x] be defined by f (x) = φi (x)ei + p N (x), N (x) = ¯ 0. In this paper, we prove that ip (f ) ≥ l
r
i=1
ui deg φi (x), where ui is a
i=1
constant defined only in terms of l, ei and the highest power of the polynomial φ¯i (x) dividing N (x). Further a class of irreducible polynomials is described for which the above inequality becomes equality. The results of the paper quickly yield the well known Dedekind criterion which gives a necessary and sufficient condition for ip (f ) to be zero. In fact, these results are proved in a more general set up replacing Z by any Dedekind domain. © 2019 Elsevier B.V. All rights reserved.
1. Introduction Let K = Q(θ) be an algebraic number field with θ in the ring AK of algebraic integers of K and f (x) be the minimal polynomial of θ over the field Q of rational numbers. The explicit determination of the discriminant of K is an important problem in Algebraic Number Theory. The computation of the discriminant dK of K is closely connected with the computation of the index of the subgroup Z[θ] in AK in view of the formula discr(f ) = [AK : Z[θ]]2 dK , where discr(f ) stands for the discriminant of the polynomial f (x). Let p be a prime number and ip (f ) denote the highest power of p dividing the index [AK : Z[θ]] which is clearly independent of the choice of the root θ of f (x) with K = Q(θ). For g(x) belonging to Z[x], let g¯(x) denote * Corresponding author. E-mail addresses: [email protected] (A. Jakhar), [email protected] (S.K. Khanduja). https://doi.org/10.1016/j.jpaa.2019.106281 0022-4049/© 2019 Elsevier B.V. All rights reserved.
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e the polynomial over Z/pZ obtained on replacing each coefficient of g(x) modulo p. Let φ¯1 (x)e1 · · · φ¯r (x) r be the factorization of f¯(x) into a product of powers of distinct irreducible polynomials over Z/pZ with each φi (x) ∈ Z[x] monic. Let the integer l ≥ 1 and the polynomial N (x) ∈ Z[x] be defined by
f (x) =
r
φi (x)ei + pl N (x), N (x) = ¯0.
(1.1)
i=1
Note that when ei = 1 for each i, then clearly p does not divide discr(f ) and hence ip (f ) = 0. In 1878, Dedekind gave a necessary and sufficient condition known as Dedekind criterion to be satisfied by f (x) for ip (f ) to be zero. It asserts that ip (f ) = 0 if and only if either ei = 1 for each i or whenever ej > 1 for some j, then l = 1 and φj (x) does not divide N (x) for such an index j (see [2, Theorem 6.1.4], [3]). In this paper, we give a simple formula for ip (f ) in terms of ei , l and deg φi (x) for all those monic irreducible polynomials f (x) ∈ Z[x] which when expressed as (1.1) satisfy the condition that whenever ej > 1, then gcd(ej , l) = 1 and φ¯j (x) does not divide N (x). We also give a lower bound for ip (f ) involving only the constants ei , l, deg φi (x) and the highest power of φ¯i (x) dividing N (x) for any monic irreducible polynomial f (x) belonging to Z[x] written in the form (1.1). Moreover this lower bound becomes zero for only those polynomials which satisfy the condition of Dedekind criterion as pointed out in Remark 1.2. These results besides yielding Dedekind criterion generate families of irreducible polynomials belonging to Z[x] for which ip (f ) is quickly calculable. Precisely stated, we prove: Theorem 1.1. Let K = Q(θ) be an algebraic number field with θ in the ring AK of algebraic integers of K having minimal polynomial f (x) over Q. Let p be a prime number and ip (f ) denote the highest power of p e dividing the index [AK : Z[θ]]. Suppose that f¯(x) = φ¯1 (x)e1 · · · φ¯r (x) r is the factorization of f¯(x) obtained by replacing each coefficient of f (x) modulo p into a product of powers of distinct irreducible polynomials over Z/pZ with φi (x) ∈ Z[x] monic. Let ti ≥ 0 denote the highest power of φ¯i (x) dividing the polynomial r N (x), where N (x) belonging to Z[x] is defined by f (x) = φi (x)ei + pl N (x), l ≥ 1, N (x) = ¯0. Let ui i=1
stand for the non negative integer1 given by ⎧ ei (e − 1)l + gcd(ei , l + 1) − 1 ⎪ ⎪ i , if ti > ⎨ 2 l+1 ui = ⎪ (e − 1)(l − 1) + gcd(e , l) − 1 ⎪ i ⎩ max{lti , i }, otherwise. 2 Then the following hold: (i) ip (f ) ≥
r
ui deg φi (x).
i=1
(ii) If gcd(ej , l) = 1 and tj = 0 whenever ej > 1, then equality holds in (i), i.e., ip (f ) = r (ei −1)(l−1) deg φi (x). 2 i=1
Remark 1.2. With notations as in the above theorem, ui = 0 implies that either ei = 1 or ei > 1 in which case l = 1 and ti = 0. So if ip (f ) = 0, then by Theorem 1.1(i), ui = 0 for 1 ≤ i ≤ r and hence f (x) satisfies the condition of Dedekind criterion. Conversely if f (x) satisfies this condition, then in view of Theorem 1.1(ii), ip (f ) = 0. We now illustrate the above theorem through examples given below. 1
In view of [6, Problem 435],
(n−1)(t−1)+gcd(n,t)−1 2
is an integer for positive integers n, t.
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Example 1.3. Let f (x) = xn + axn−1 + b belonging to Z[x] be an irreducible2 trinomial. Let p be a prime number not dividing a and dividing b with exact power l ≥ 1. With notations as in Theorem 1.1, set φ1 (x) = x and φ2 (x) = x + a, so that e1 = n − 1, e2 = 1, t1 = 0, t2 = 0. By assertion (i) of Theorem 1.1, we have ip (f ) ≥ (n−2)(l−1)+gcd(n−1,l)−1 . Furthermore if l is coprime to n − 1, then in view of Theorem 1.1(ii) 2 we see that ip (f ) = (n−2)(l−1) . 2 It may be pointed out that the condition stated in assertion (ii) of Theorem 1.1 is not necessary for the equality to hold in assertion (i) as the following example shows. Example 1.4. Let K = Q(θ) with θ satisfying the polynomial f (x) = x4 + 9x3 + 27x2 + 27x + 375. Then f (x) is irreducible over Q by Eisenstein’s irreducibility criterion. It can be easily checked that the absolute value of the discriminant of f (x) is 33 × 56 × 31271. By a basic result (cf. [7, Lemma 2.17]), the prime 3 does not divide the index of the subgroup Z[θ] in AK . Further by virtue of the equality [AK : Z[θ]]2 |dK | = |discr(f )| = 33 × 56 × 31271, the prime 31271 can not divide [AK : Z[θ]]. For the prime p = 5, f (x) ≡ x(x + 3)3 mod 5. Retaining the notations of Theorem 1.1 and taking φ1 (x) = x, φ2 (x) = x + 3, we have e1 = 1, e2 = 3, t1 = 0, t2 = 0, l = 3. So assertion (ii) of Theorem 1.1 is not applicable here as e2 and l are 2 ,l)−1 not coprime. By Theorem 1.1(i), we see that i5 (f ) ≥ (e2 −1)(l−1)+gcd(e = 3. Since the exact power of 5 2 dividing discr(f ) is 6, it now follows that i5 (f ) = 3. Therefore [AK : Z[θ]] = 53 . With notations as in Theorem 1.1, recall that the index [AK : Z[θ]] equals the determinant of the transition matrix from a Z-basis of AK to {1, θ, · · · , θn−1 } (see [1, Chapter 2, §2, Theorem 2]). Let Z(p) denote the localization of Z at pZ. Since a Z-basis of AK is also a Z(p) -basis of the integral closure S(p) (say) of Z(p) in K, it follows that ip (f ) is the exact power of p dividing the determinant of the transition matrix from a Z(p) -basis of S(p) to {1, θ, · · · , θn−1 }. We use this observation to extend the notion of ip (f ) replacing Z by any Dedekind domain R and p by a non-zero prime ideal of R. This is done after introducing some notations. Notation. Let R be a Dedekind domain whose quotient field will be denoted by K. Let K(θ) be a finite separable extension of K with θ integral over R having minimal polynomial f (x) of degree n over K. Let p be a non-zero prime ideal of R and Rp be the localization of R at p. For any ξ in Rp , the image of ξ under ¯ Similarly for a polynomial the canonical homomorphism from Rp onto Rp /pRp R/p will be denoted by ξ. g(x) ∈ Rp [x], g¯(x) will denote the polynomial obtained on replacing each coefficient c of g(x) by c¯. Let Sp denote the integral closure of Rp in K(θ). Then as is well-known, Sp is a free Rp -module of rank n (see [11, Chapter V, §4, Corollary 2]). For any non-zero prime ideal p of R, we define the p-index of θ to be the highest power of p dividing the principal ideal of Rp generated by the determinant of the transition matrix from an Rp -basis of Sp to {1, θ, · · · , θn−1 }. Clearly the p-index of θ is independent of the choice of the root θ of f (x); we shall denote it by ip (f ). With the above notation, we shall prove the following more general version of Theorem 1.1. Theorem 1.5. Let R, K(θ), f (x), p, Rp , ip (f ) be as in the above paragraph. Let π be a prime element e of the discrete valuation ring Rp . Suppose that f¯(x) = φ¯1 (x)e1 · · · φ¯r (x) r is the factorization of f¯(x) into a product of powers of distinct irreducible polynomials over R/p with φi (x) ∈ R[x] monic. Let ti ≥ 0 be the highest power of φ¯i (x) dividing the polynomial N (x), where N (x) belonging to Rp [x] is r i ,l+1)−1 defined by f (x) = φi (x)ei + π l N (x) with N (x) = ¯0, l ≥ 1. Let ui stand for (ei −1)l+gcd(e or 2 i=1
i ,l)−1 max{lti , (ei −1)(l−1)+gcd(e } according as ti > 2
2
ei l+1
or not. Then the following hold:
By Perron’s criterion (cf. [10]), xn + axn−1 + b belonging to Z[x] is irreducible over Q when |a| > 1 + |b|.
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(i) ip (f ) ≥
r
ui deg φi (x).
i=1
(ii) If gcd(ej , l) = 1 and tj = 0 whenever ej > 1, then equality holds in (i), i.e., ip (f ) = r (ei −1)(l−1) deg φi (x). 2 i=1
2. Preliminary results In what follows, v is a discrete valuation of a field K with valuation ring Rv having maximal ideal Mv vˆ) will denote completion of (K, v) and v˜ˆ the unique and value group Z. We fix a prime element π of Rv . (K,
For any ξ in the valuation ring of v˜ˆ, ξ¯ will denote of K. prolongation of vˆ to a fixed algebraic closure K ˜ its vˆ-residue, i.e., the image of ξ under the canonical homomorphism from the valuation ring of v˜ˆ onto its
containing K, L will stand for the residue field of the valuation of L residue field. For a subfield L of K obtained by restricting v˜ˆ. When g(x) ∈ Rv [x], g¯(x) will stand for the polynomial obtained by replacing each coefficient of g(x) by its image in Rv /Mv . We now prove a lemma which plays a significant role in the proof of Theorem 1.5. Lemma 2.1. Let π be a prime element of a discrete valuation ring Rv . Assume that φ1 (x), · · · , φr (x) belonging to Rv [x] are monic polynomials such that φ¯1 (x), · · · , φ¯r (x) are distinct irreducible polynomials over Rv /Mv . For 1 ≤ i ≤ r, let fi (x) = φi (x)ei + π ki Ni (x) be a polynomial with Ni (x) ∈ Rv [x] of degree less than ei deg φi (x), ki ≥ 1 and N i (x) = ¯0. If N (x) ∈ Rv [x] with N (x) = ¯0 is such that f1 (x) · · · fr (x) = φ1 (x)e1 · · · φr (x)er + π l N (x), then l = min{k1 , · · · , kr }. Proof. We shall prove the lemma by induction on r. We first prove the lemma for r = 2. On multiplying, we have f1 (x)f2 (x) =
2
φi (x)ei + π k1 N1 (x)φ2 (x)e2 + π k2 N2 (x)φ1 (x)e1 + π k1 +k2 N1 (x)N2 (x).
(2.1)
i=1
So when we write f1 (x)f2 (x) =
2
φi (x)ei + π u A(x)
(2.2)
i=1
with A(x) ∈ Rv [x], A(x) = ¯0, then u ≥ min{k1 , k2 }; it is to be shown that equality holds here. This is obvious when k1 = k2 in view of (2.1). Suppose to the contrary that k1 = k2 and u > k1 . Comparing equations (2.1), (2.2) and keeping in mind that k1 , k2 are positive, it follows on taking the image of the polynomials in K[x] that N 1 (x)φ¯2 (x)e2 + N 2 (x)φ¯1 (x)e1 = ¯0, which is not possible as φ¯1 (x)(= φ¯2 (x)) is irreducible over Rv /Mv and deg N1 (x) < e1 deg φ1 (x) by hypothesis. This contradiction proves that u = min{k1 , k2 }. Let us assume that the lemma holds for r. We now prove it for r + 1. By induction hypothesis, when r fi (x) is written as i=1 r
φi (x)ei + π s B(x), B(x) ∈ Rv [x], B(x) = ¯0,
i=1
then s = min{k1 , · · · , kr }. Consequently r+1 i=1
fi (x) =
r+1 i=1
φi (x)ei + π s B(x)φr+1 (x)er+1 + π kr+1 Nr+1 (x)
r i=1
φi (x)er + π s+kr+1 Nr+1 (x)B(x).
(2.3)
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So when
r+1
5
fi (x) is written as
i=1 r+1
fi (x) =
i=1
r+1
φi (x)ei + π m D(x)
(2.4)
i=1
with D(x) ∈ Rv [x], D(x) = ¯0, then m ≥ min{s, kr+1 }. It only remains to prove that m = min{s, kr+1 }. This equality is immediate from (2.3) when s = kr+1 ; if s = kr+1 , then a comparison of (2.3) and (2.4) together with the inequality m > kr+1 implies that B(x)φ¯r+1 (x)er+1 + N r+1 (x)
r
φ¯i (x)er = ¯0,
i=1
which is not possible because φ¯1 (x), · · · , φ¯r+1 (x) are distinct monic irreducible polynomials over Rv /Mv and deg Nr+1 (x) < er+1 deg φr+1 (x) in view of the hypothesis. This completes the proof of the lemma. 2 In addition to several simplifications, we will use an extension of Ore’s Index Theorem to discrete valued fields (stated as Theorem 2.A below). For this we need to introduce the notions of Gauss valuation, φ-Newton ¯ polygon (Definition 2.2), φ-index of a polynomial for a monic polynomial φ(x) belonging to Rv [x] with φ(x) irreducible over Rv /Mv (Definition 2.3), and the v-index of a monic irreducible polynomial g(x) ∈ Rv [x] (Definition 2.4). We shall denote by v x the Gauss valuation of the field K(x) of rational functions in an indeterminate x which extends a given valuation v of K and is defined on K[x] by v
x
ci x
i
= min{v(ci )}, ci ∈ K. i
i
¯ Definition 2.2. Let φ(x) ∈ Rv [x] be a monic polynomial with φ(x) irreducible over Rv /Mv and g(x) ∈ Rv [x] n ai (x)φ(x)i with deg ai (x) < deg φ(x), an (x) = 0 be the be a monic polynomial not divisible by φ(x). Let i=0
φ(x)-expansion of g(x) obtained on dividing it by successive powers of φ(x). Let Pi stand for the point in the plane having coordinates (i, v x (an−i (x))) when an−i (x) = 0, 0 ≤ i ≤ n. Let μij denote the slope of the line joining the points Pi and Pj if an−i (x)an−j (x) = 0. Let i1 be the largest positive index not exceeding n such that μ0i1 = min{ μ0j | 0 < j ≤ n, an−j (x) = 0}. If i1 < n, let i2 be the largest index such that i1 < i2 ≤ n and μi1 i2 = min{ μi1 j | i1 < j ≤ n, an−j (x) = 0} and so on. The φ-Newton polygon of g(x) (with respect to the underlying valuation v) is the polygonal path having segments P0 Pi1 , Pi1 Pi2 , . . . , Pik−1 Pik with ik = n. These segments are called the edges of the φ-Newton polygon and their slopes form a strictly increasing sequence; these slopes are non-negative as g(x) is a monic polynomial with coefficients in Rv . The region enclosed by the positive side of x-axis, the φ-Newton polygon of g(x) and the vertical line passing through the last vertex of this polygon will be referred to as the region of the φ-Newton polygon of g(x). Definition 2.3. Let φ(x) and g(x) be as in the above definition. Let ν denote the number of points with positive integer coordinates lying on or below the φ-Newton polygon of g(x) (with respect to the underlying valuation v) away from the vertical line passing through the last vertex of this polygon. As in [5], the φ-index of g is defined to be ν deg φ(x) and will be denoted by iφ (g).
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Definition 2.4. Let g(x) belonging to Rv [x] be a monic irreducible separable polynomial of degree n having a root θ and S be the integral closure of Rv in K(θ). Then S is a free Rv -module of rank n. If d denotes the determinant of the transition matrix from an Rv -basis of S to {1, θ, · · · , θn−1 }, then v(d) is independent of the choice of the Rv -basis of S. As in [5], v(d) will be called the v-index of g(x) and will be denoted by iv (g). Observe that when p is a non-zero prime ideal of a Dedekind domain R and vp is the p-adic valuation of the quotient field of R defined for any non-zero α ∈ R by vp (α) = the highest power of p dividing the ideal αR,
(2.5)
then ivp (g) has the same meaning as ip (g) defined in the Notation preceding Theorem 1.5. The following theorem to be used in the sequel was originally proved by Ore for p-adic valuations [9] and extended to discrete valued fields in [5]. Theorem 2.A. Let v be a discrete valuation of a field K with valuation ring Rv , residue field Rv /Mv and e value group Z. Let g(x) ∈ Rv [x] be a monic irreducible separable polynomial and φ¯1 (x)e1 · · · φ¯r (x) r be the factorization of g¯(x) into a product of powers of distinct irreducible polynomials over Rv /Mv with φi (x) ∈ Rv [x] monic. Then iv (g) ≥
r
iφj (g).
(2.6)
j=1
We shall state a theorem which gives a sufficient condition for the inequality (2.6) to become equality. For this, we need the following definitions.
¯ with φ(x) Definition 2.5. Let φ(x) ∈ Rv [x] be a monic polynomial having a root α in K irreducible over Rv /Mv and π be a prime element of Rv . Let g(x) ∈ Rv [x] be a monic polynomial not divisible by φ(x) having degree a multiple of deg φ(x) with φ(x)-expansion φ(x)n + an−1 (x)φ(x)n−1 + · · · + a0 (x). Suppose that the φ-Newton polygon of g(x) (with respect to the underlying valuation v) consists of a single edge x x (x)) having positive slope λ = de with d, e coprime, i.e., min{ v (an−i | 1 ≤ i ≤ n} = v (an0 (x)) = de so that n is i a (x) divisible by e, say n = et and v x (an−ej (x)) ≥ dj for 1 ≤ j ≤ t. Thus the polynomial n−ej has coefficients π dj an−ej (α) ˜ in Rv and hence is an element of the valuation ring of vˆ. Corresponding to the φ-Newton polygon π dj of g(x), we associate with g(x) a polynomial T (Y ) not divisible by Y with coefficients in K(α) defined by t an−ej (α) t t−j T (Y ) = Y + Y . π dj j=1
Definition 2.6. Let φ(x), α be as in the above definition. Let G(x) belonging to Rv [x] be a monic polynomial ¯ not divisible by φ(x) such that G(x) is a power of φ(x). Let λ1 < · · · < λk be the slopes of the edges of the φ-Newton polygon of G(x). By Theorem 1.1 of [4], we can write G(x) = G1 (x) · · · Gk (x), where the φ-Newton polygon of Gi (x) ∈ Rvˆ [x] has a single edge with slope λi . Let Ti (Y ) belonging to K(α)[Y ] denote the polynomial associated with Gi (x) described as in the above definition. Then G(x) is said to be v-regular with respect to φ if each Ti (Y ) is a product of distinct monic irreducible polynomials over K(α) for e e 1 ≤ i ≤ k. In general, if g(x) belonging to Rv [x] is a monic polynomial and g¯(x) = φ¯1 (x) 1 · · · φ¯r (x) r is its factorization into irreducible polynomials over Rv /Mv with each φi (x) belonging to Rv [x] monic and ei > 0, then by Hensel’s Lemma there exist g1 (x), · · · , gr (x) belonging to Rvˆ [x] such that g(x) = g1 (x) · · · gr (x) and g¯i (x) = φ¯i (x)ei . The polynomial g(x) is said to be v-regular (with respect to φ1 , · · · , φr ) if each gi (x) is vˆ-regular with respect to φi .
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The result stated below is proved in [5, Theorem 1.4]. Theorem 2.B. Let v, g(x), φ1 (x), · · · , φr (x) be as in Theorem 2.A. If g(x) is v-regular with respect to φ1 , · · · , φr , then iv (g) =
r
iφj (g).
j=1
The following elementary lemma is already known (cf. [6, Problem 435]). We omit its proof. As usual for a real number λ, λ stands for the largest integer not exceeding λ. Lemma 2.C. Let t, n be positive integers. Let S1 denote the set of points in the plane with positive integer coordinates lying inside or on the triangle with vertices (0, 0), (n, 0), (n, t) which do not lie on the line x = n. Then #S1 =
n−1 i=1
1 it = [(n − 1)(t − 1) + gcd(t, n) − 1]. n 2
3. Proof of Theorem 1.5 vˆ) We denote the p-adic valuation vp of K defined in (2.5) by v and the completion of (K, v) by (K, having valuation ring Rvˆ . Fix a prime element π of Rv = Rp . By Theorem 2.A iv (f ) ≥
r
iφj (f ).
(3.1)
j=1
To calculate iφj (f ), we carry out some simplifications in terms of factors of f (x) over Rvˆ . ¯ = φ¯1 (x)e1 · · · φ¯r (x)er leads to a In view of Hensel’s Lemma [8, Chapter 2], the given factorization f(x) factorization of f (x) as f1 (x) · · · fr (x), where fi (x) ∈ Rvˆ [x] is monic and f¯i (x) = φ¯i (x)ei , 1 ≤ i ≤ r. Write fi (x) = φi (x)ei + π ki Ni (x), Ni (x) ∈ Rvˆ [x], N i (x) = ¯0, ki ≥ 1, deg Ni (x) < ei deg φi (x).
(3.2)
Keeping in mind that f (x) =
r
φi (x)ei + π l N (x), N (x) = ¯0,
(3.3)
i=1
and applying Lemma 2.1, we see that l = min{k1 , · · · , kr }.
(3.4)
Since f (x) = f1 (x) · · · fr (x), using (3.2), (3.3), (3.4), it can be easily seen that N (x) =
r j=1
(π kj −l )N
j (x)
r
ei ¯ φi (x) .
(3.5)
i=1,i=j
¯ For any polynomial h(x) ∈ Rvˆ [x], let vφ¯j (h(x)) denote the highest power of the polynomial φ¯j (x) dividing ¯ Recall that vφ¯j (N (x)) = tj and l = min{k1 , · · · , kr }. Keeping in mind that deg N j (x) < ej deg φ¯j (x) h(x). in view of (3.2), it immediately follows from (3.5) that for any j, 1 ≤ j ≤ r, we have
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tj = vφ¯j (Nj (x)) if kj = l,
(3.6)
tj ≥ ej if kj > l.
(3.7)
By Corollary 2.5 of [4] the φj -Newton polygon of f (x) is a translate in the positive horizontal direction of the φj -Newton polygon of fj (x), so the number of points with positive integer coordinates lying on or below the φj -Newton polygon of f (x) and fj (x) are the same. Therefore iφj (f ) = iφj (fj ), 1 ≤ j ≤ r.
(3.8)
Thus (3.1) becomes iv (f ) ≥
r
iφj (fj ) =
j=1
r
nj deg φj (x),
(3.9)
j=1
where nj is the number of points with positive integer coordinates which lie on or below the φj -Newton polygon of fj (x) away from the vertical line passing through the last vertex of the polygon. Therefore assertion (i) of the theorem will follow from (3.9) once we prove that nj ≥ uj for 1 ≤ j ≤ r. Let mj stand for vφ¯j (Nj (x)) and Nj (x) =
sj
(3.10) Aju (x)φj (x)u be the φj (x)-expansion of Nj (x),
u=0
Ajsj (x) = 0. By virtue of (3.2), the φj (x)-expansion of fj (x) is given by
fj (x) = φj (x)
ej
+π
kj
sj u=mj
mj −1 u
Aju (x)φj (x) +
Aju (x)φj (x)
u
.
(3.11)
u=0
In view of the definition of mj , we have Aju (x) = ¯0 for 0 ≤ u ≤ mj − 1 and Ajmj (x) = ¯0; further by (3.4), kj ≥ l. We split the proof of (3.10) into two cases. ¯ It can be easily seen that the φj -Newton polygon of fj (x) Case I. mj = 0. In this case we have Aj0 (x) = 0. has a single edge which is the straight line segment joining the points (0, 0) and (ej , kj ). This case is divided into two subcases. First consider the subcase when kj = l. Then by (3.6), we have tj = mj = 0 and hence (e −1)(l−1)+gcd(ej ,l)−1 (e −1)(l−1)+gcd(ej ,l)−1 in this situation uj = max{ltj , j }= j . Keeping in mind that the 2 2 φj -Newton polygon of fj (x) has a single edge joining points (0, 0), (ej , l), it follows from Lemma 2.C that (e −1)(l−1)+gcd(ej ,l)−1 nj = j and (3.10) is verified in this subcase. Now consider the subcase when kj > l. 2 Then by (3.7), we have tj ≥ ej . Since kj ≥ l + 1 in this subcase, it is clear that the triangle with vertices (0, 0), (ej , 0) and (ej , l+1) is contained in the region of the φj -Newton polygon of fj (x). So using Lemma 2.C, (e −1)l+gcd(ej ,l+1)−1 we see that nj ≥ j = uj in this subcase, which completes the proof of assertion (i) of the 2 theorem in Case I. Case II. mj > 0. Denote v x (Aj0 (x)) by zj . Note that zj > 0 in this case. Before proceeding with the proof, we make two observations. Keeping in mind that l + 1 ≤ kj + zj and the fact that the last vertex of the φj -Newton polygon of fj (x) is (ej , kj + zj ), we see that the point (ej , l + 1) will lie on the vertical line ej passing through the last vertex of this polygon. Observe that when mj ≤ kj +1 , then the point (ej − mj , kj ) will be contained in the region of the φj -Newton polygon of fj (x). This case is split into two subcases. ej Subcase (i). kj = l; consequently by (3.6) we have tj = mj > 0. In this subcase, when tj ≤ l+1 , then in view of what has been said in the above paragraph, the points with coordinates (0, 0), (ej − tj , l), (ej , l + 1) and
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(ej , 0) will be contained in the region of the φj -Newton polygon of fj (x). Note that (0, 0), (ej −tj , l), (ej , l+1) ej . So the quadrilateral with vertices (0, 0), (ej −tj , l), (ej , l+1), (ej , 0) lie on the same line if and only if tj = l+1 or the triangle with vertices (0, 0), (ej , l + 1), (ej , 0) will be contained in the region of the φj -Newton polygon ej of fj (x) when tj ≤ l+1 . By virtue of Lemma 2.C applied to the triangle with vertices (0, 0), (ej − tj , l), (ej − tj , 0) and keeping in mind that the number of points with positive integer coordinates lying on or inside the quadrilateral with vertices (ej − tj , 0), (ej − tj , l), (ej , l + 1), (ej , 0) away from the line x = ej is ltj , we conclude that nj ≥
(ej − tj − 1)(l − 1) + gcd(ej − tj , l) − 1 + ltj = C (say). 2
(3.12)
Since tj > 0 in the present situation, it can be easily seen that (ej − 1)(l − 1) + l − 1 (ej − 1)(l − 1) + (l + 1)tj + gcd(ej − tj , l) − 1 > 2 2 (ej − 1)(l − 1) + gcd(ej , l) − 1 . ≥ 2
C=
Since C ≥ ltj in view of (3.12), it now follows from the above inequality that nj ≥ uj . Further in the ej present subcase when tj > l+1 , then keeping in mind that the triangle with vertices (0, 0), (ej , 0), (ej , l + 1) is contained in the region of the φj -Newton polygon of fj (x) and applying Lemma 2.C, we see that nj ≥ (ej −1)l+gcd(ej ,l+1)−1 = uj in this situation. 2 e
e
j j Subcase (ii). kj > l. By (3.7), we have tj ≥ ej > l+1 . When mj < l+2 , then arguing as in the first paragraph of Case II, it can be seen that the quadrilateral with vertices (0, 0), (ej − mj , l + 1), (ej , l + 2) and (ej , 0) is contained in the region of the φj -Newton polygon of fj (x). Applying Lemma 2.C to the triangle with vertices (0, 0), (ej − mj , l + 1), (ej − mj , 0), we see that
(ej − mj − 1)l + gcd(ej − mj , l + 1) − 1 + (l + 1)mj 2 (ej − 1)l + mj l + 2mj − 1 + gcd(ej − mj , l + 1) = 2 (ej − 1)l + gcd(ej , l + 1) − 1 (ej − 1)l + (l + 1) − 1 ≥ = uj . > 2 2
nj ≥
e
j When mj ≥ l+2 , then it can be easily seen that the triangle with vertices (0, 0), (ej , 0), and (ej , l + 2) is contained in the region of the φj -Newton polygon of fj (x). Using again Lemma 2.C, we have
(ej − 1)l + (ej − 1) + gcd(ej , l + 2) − 1 (ej − 1)(l + 1) + gcd(ej , l + 2) − 1 = 2 2 (ej − 1)l + gcd(ej , l + 1) − 1 (ej − 1)l + ej − 1 ≥ = uj . ≥ 2 2
nj ≥
This completes the proof of the first assertion of the theorem. Now to prove the second assertion of the theorem, we first show that with hypothesis as in this assertion, fj (x) is vˆ-regular with respect to φj for each j, 1 ≤ j ≤ r and we also calculate iφj (fj ). For this, two cases are distinguished. Consider first the case when ej = 1. Then by (3.2), fj (x) = φj (x) +π kj Nj (x) with kj ≥ 1 and deg Nj (x) < deg φj (x). The φj -Newton polygon of fj (x) has a single edge with slope kj . So by Definition 2.3, we have iφj (fj ) = 0
(3.13)
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and the polynomial associated with fj (x) for this Newton polygon is a linear polynomial not equal to Y in view of Definition 2.5. Hence fj (x) is vˆ-regular with respect to φj in case ej = 1. Consider now the case when ej > 1. Then by hypothesis of assertion (ii), tj = 0 and gcd(ej , l) = 1. When tj = 0, we have mj = 0 and kj = l by (3.6), (3.7). So in the φj (x)-expansion of fj (x) given by (3.11), Aj0 (x) = ¯0; consequently the φj -Newton polygon of fj (x) has a single edge joining the points (0, 0), (ej , l). As ej and l are coprime, the polynomial associated with fj (x) for this Newton polygon is again a linear polynomial. Thus fj (x) is vˆ-regular with respect to φj . Further, applying Lemma 2.C to the triangle with vertices (0, 0), (ej , l), (ej , 0) and keeping in mind that gcd(ej , l) = 1, we see that in the present case iφj (fj (x)) =
(ej − 1)(l − 1) deg φj (x). 2
(3.14)
Using the fact that fj (x) is vˆ-regular with respect to φj for each j proved above, we see that f (x) is v-regular with respect to φ1 , · · · , φr in view of Definition 2.6. It now follows from Theorem 2.B that iv (f ) =
r
iφj (f ).
j=1
Using (3.8), (3.13) and (3.14), the above equality can be rewritten as iv (f ) =
r
iφj (fj ) =
j=1
r (ej − 1)(l − 1) deg φj (x). 2 j=1
This completes the proof of the theorem. Acknowledgements The authors are highly grateful to the referee for his/her several valuable suggestions which have greatly improved the exposition of the paper. The financial support from the respective institutions is gratefully acknowledged by the authors. The second author is also thankful to Council of Scientific and Industrial Research, New Delhi for financial assistance through Emeritus Scientistship. References [1] Z.I. Borevich, I.R. Shafarevich, Number Theory, Academic Press, 1966. [2] H. Cohen, A Course in Computational Algebraic Number Theory, Springer-Verlag, Berlin-Heidelberg, 1993. [3] R. Dedekind, Über den Zusammenhang zwischen der Theorie der ideale und der Theorie der höheren Kongruenzen, Götttingen Abh. 23 (1878) 1–23. [4] S.K. Khanduja, S. Kumar, On prolongations of valuations via Newton polygons and liftings of polynomials, J. Pure Appl. Algebra 216 (2012) 2648–2656. [5] S.K. Khanduja, S. Kumar, A generalization of a theorem of Ore, J. Pure Appl. Algebra 218 (2014) 1206–1218. [6] J.M. de Koninck, A. Mercier, 1001 Problems in Classical Number Theory, Amer. Math. Soc., Providence Rhode Island, 2007. [7] W. Narkewicz, Elementary and Analytic Theory of Algebraic Numbers, Springer-Verlag, Berlin, Heidelberg, 2004. [8] J. Neukirch, Algebraic Number Theory, Springer-Verlag, Berlin, Heidelberg, 1999. [9] Ø. Ore, Newtonsche Polygone in der Theorie der algebraischen Körper, Math. Ann. 99 (1928) 84–117. [10] E.S. Selmer, On the irreducibility of certain trinomials, Math. Scand. 4 (1956) 287–302. [11] O. Zariski, P. Samuel, Commutative Algebra, Vol. I, D. van Nostrand Company, 1958.
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Contents lists available at ScienceDirect
Journal of Pure and Applied Algebra www.elsevier.com/locate/jpaa
On the index of an algebraic integer and beyond Anuj Jakhar a , Sudesh K. Khanduja b,c,∗ a
The Institute of Mathematical Sciences (IMSc), HBNI, CIT Campus, Taramani, Chennai, 600113, Tamil Nadu, India b Indian Institute of Science Education and Research Mohali, Sector 81, Knowledge City, SAS Nagar, Punjab, 140306, India c Department of Mathematics, Panjab University, Chandigarh, 160014, India
a r t i c l e
i n f o
Article history: Received 8 July 2019 Received in revised form 31 October 2019 Available online xxxx Communicated by V. Suresh Dedicated to Professor Raman Parimala on her 70th Birthday MSC: 11R04; 11R29; 12J10 Keywords: Rings of algebraic integers Dedekind domains Valued fields
a b s t r a c t Let K = Q(θ) be an algebraic number field with θ in the ring AK of algebraic integers of K having minimal polynomial f (x) over Q. For a prime number p, let ip (f ) denote the highest power of p dividing the index [AK : Z[θ]]. Let f¯(x) = φ¯1 (x)e1 · · · φ¯r (x)er be the factorization of f (x) modulo p into a product of powers of distinct irreducible polynomials over Z/pZ with φi (x) ∈ Z[x] monic. Let the r integer l ≥ 1 and the polynomial N (x) ∈ Z[x] be defined by f (x) = φi (x)ei + p N (x), N (x) = ¯ 0. In this paper, we prove that ip (f ) ≥ l
r
i=1
ui deg φi (x), where ui is a
i=1
constant defined only in terms of l, ei and the highest power of the polynomial φ¯i (x) dividing N (x). Further a class of irreducible polynomials is described for which the above inequality becomes equality. The results of the paper quickly yield the well known Dedekind criterion which gives a necessary and sufficient condition for ip (f ) to be zero. In fact, these results are proved in a more general set up replacing Z by any Dedekind domain. © 2019 Elsevier B.V. All rights reserved.
1. Introduction Let K = Q(θ) be an algebraic number field with θ in the ring AK of algebraic integers of K and f (x) be the minimal polynomial of θ over the field Q of rational numbers. The explicit determination of the discriminant of K is an important problem in Algebraic Number Theory. The computation of the discriminant dK of K is closely connected with the computation of the index of the subgroup Z[θ] in AK in view of the formula discr(f ) = [AK : Z[θ]]2 dK , where discr(f ) stands for the discriminant of the polynomial f (x). Let p be a prime number and ip (f ) denote the highest power of p dividing the index [AK : Z[θ]] which is clearly independent of the choice of the root θ of f (x) with K = Q(θ). For g(x) belonging to Z[x], let g¯(x) denote * Corresponding author. E-mail addresses: [email protected] (A. Jakhar), [email protected] (S.K. Khanduja). https://doi.org/10.1016/j.jpaa.2019.106281 0022-4049/© 2019 Elsevier B.V. All rights reserved.
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e the polynomial over Z/pZ obtained on replacing each coefficient of g(x) modulo p. Let φ¯1 (x)e1 · · · φ¯r (x) r be the factorization of f¯(x) into a product of powers of distinct irreducible polynomials over Z/pZ with each φi (x) ∈ Z[x] monic. Let the integer l ≥ 1 and the polynomial N (x) ∈ Z[x] be defined by
f (x) =
r
φi (x)ei + pl N (x), N (x) = ¯0.
(1.1)
i=1
Note that when ei = 1 for each i, then clearly p does not divide discr(f ) and hence ip (f ) = 0. In 1878, Dedekind gave a necessary and sufficient condition known as Dedekind criterion to be satisfied by f (x) for ip (f ) to be zero. It asserts that ip (f ) = 0 if and only if either ei = 1 for each i or whenever ej > 1 for some j, then l = 1 and φj (x) does not divide N (x) for such an index j (see [2, Theorem 6.1.4], [3]). In this paper, we give a simple formula for ip (f ) in terms of ei , l and deg φi (x) for all those monic irreducible polynomials f (x) ∈ Z[x] which when expressed as (1.1) satisfy the condition that whenever ej > 1, then gcd(ej , l) = 1 and φ¯j (x) does not divide N (x). We also give a lower bound for ip (f ) involving only the constants ei , l, deg φi (x) and the highest power of φ¯i (x) dividing N (x) for any monic irreducible polynomial f (x) belonging to Z[x] written in the form (1.1). Moreover this lower bound becomes zero for only those polynomials which satisfy the condition of Dedekind criterion as pointed out in Remark 1.2. These results besides yielding Dedekind criterion generate families of irreducible polynomials belonging to Z[x] for which ip (f ) is quickly calculable. Precisely stated, we prove: Theorem 1.1. Let K = Q(θ) be an algebraic number field with θ in the ring AK of algebraic integers of K having minimal polynomial f (x) over Q. Let p be a prime number and ip (f ) denote the highest power of p e dividing the index [AK : Z[θ]]. Suppose that f¯(x) = φ¯1 (x)e1 · · · φ¯r (x) r is the factorization of f¯(x) obtained by replacing each coefficient of f (x) modulo p into a product of powers of distinct irreducible polynomials over Z/pZ with φi (x) ∈ Z[x] monic. Let ti ≥ 0 denote the highest power of φ¯i (x) dividing the polynomial r N (x), where N (x) belonging to Z[x] is defined by f (x) = φi (x)ei + pl N (x), l ≥ 1, N (x) = ¯0. Let ui i=1
stand for the non negative integer1 given by ⎧ ei (e − 1)l + gcd(ei , l + 1) − 1 ⎪ ⎪ i , if ti > ⎨ 2 l+1 ui = ⎪ (e − 1)(l − 1) + gcd(e , l) − 1 ⎪ i ⎩ max{lti , i }, otherwise. 2 Then the following hold: (i) ip (f ) ≥
r
ui deg φi (x).
i=1
(ii) If gcd(ej , l) = 1 and tj = 0 whenever ej > 1, then equality holds in (i), i.e., ip (f ) = r (ei −1)(l−1) deg φi (x). 2 i=1
Remark 1.2. With notations as in the above theorem, ui = 0 implies that either ei = 1 or ei > 1 in which case l = 1 and ti = 0. So if ip (f ) = 0, then by Theorem 1.1(i), ui = 0 for 1 ≤ i ≤ r and hence f (x) satisfies the condition of Dedekind criterion. Conversely if f (x) satisfies this condition, then in view of Theorem 1.1(ii), ip (f ) = 0. We now illustrate the above theorem through examples given below. 1
In view of [6, Problem 435],
(n−1)(t−1)+gcd(n,t)−1 2
is an integer for positive integers n, t.
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Example 1.3. Let f (x) = xn + axn−1 + b belonging to Z[x] be an irreducible2 trinomial. Let p be a prime number not dividing a and dividing b with exact power l ≥ 1. With notations as in Theorem 1.1, set φ1 (x) = x and φ2 (x) = x + a, so that e1 = n − 1, e2 = 1, t1 = 0, t2 = 0. By assertion (i) of Theorem 1.1, we have ip (f ) ≥ (n−2)(l−1)+gcd(n−1,l)−1 . Furthermore if l is coprime to n − 1, then in view of Theorem 1.1(ii) 2 we see that ip (f ) = (n−2)(l−1) . 2 It may be pointed out that the condition stated in assertion (ii) of Theorem 1.1 is not necessary for the equality to hold in assertion (i) as the following example shows. Example 1.4. Let K = Q(θ) with θ satisfying the polynomial f (x) = x4 + 9x3 + 27x2 + 27x + 375. Then f (x) is irreducible over Q by Eisenstein’s irreducibility criterion. It can be easily checked that the absolute value of the discriminant of f (x) is 33 × 56 × 31271. By a basic result (cf. [7, Lemma 2.17]), the prime 3 does not divide the index of the subgroup Z[θ] in AK . Further by virtue of the equality [AK : Z[θ]]2 |dK | = |discr(f )| = 33 × 56 × 31271, the prime 31271 can not divide [AK : Z[θ]]. For the prime p = 5, f (x) ≡ x(x + 3)3 mod 5. Retaining the notations of Theorem 1.1 and taking φ1 (x) = x, φ2 (x) = x + 3, we have e1 = 1, e2 = 3, t1 = 0, t2 = 0, l = 3. So assertion (ii) of Theorem 1.1 is not applicable here as e2 and l are 2 ,l)−1 not coprime. By Theorem 1.1(i), we see that i5 (f ) ≥ (e2 −1)(l−1)+gcd(e = 3. Since the exact power of 5 2 dividing discr(f ) is 6, it now follows that i5 (f ) = 3. Therefore [AK : Z[θ]] = 53 . With notations as in Theorem 1.1, recall that the index [AK : Z[θ]] equals the determinant of the transition matrix from a Z-basis of AK to {1, θ, · · · , θn−1 } (see [1, Chapter 2, §2, Theorem 2]). Let Z(p) denote the localization of Z at pZ. Since a Z-basis of AK is also a Z(p) -basis of the integral closure S(p) (say) of Z(p) in K, it follows that ip (f ) is the exact power of p dividing the determinant of the transition matrix from a Z(p) -basis of S(p) to {1, θ, · · · , θn−1 }. We use this observation to extend the notion of ip (f ) replacing Z by any Dedekind domain R and p by a non-zero prime ideal of R. This is done after introducing some notations. Notation. Let R be a Dedekind domain whose quotient field will be denoted by K. Let K(θ) be a finite separable extension of K with θ integral over R having minimal polynomial f (x) of degree n over K. Let p be a non-zero prime ideal of R and Rp be the localization of R at p. For any ξ in Rp , the image of ξ under ¯ Similarly for a polynomial the canonical homomorphism from Rp onto Rp /pRp R/p will be denoted by ξ. g(x) ∈ Rp [x], g¯(x) will denote the polynomial obtained on replacing each coefficient c of g(x) by c¯. Let Sp denote the integral closure of Rp in K(θ). Then as is well-known, Sp is a free Rp -module of rank n (see [11, Chapter V, §4, Corollary 2]). For any non-zero prime ideal p of R, we define the p-index of θ to be the highest power of p dividing the principal ideal of Rp generated by the determinant of the transition matrix from an Rp -basis of Sp to {1, θ, · · · , θn−1 }. Clearly the p-index of θ is independent of the choice of the root θ of f (x); we shall denote it by ip (f ). With the above notation, we shall prove the following more general version of Theorem 1.1. Theorem 1.5. Let R, K(θ), f (x), p, Rp , ip (f ) be as in the above paragraph. Let π be a prime element e of the discrete valuation ring Rp . Suppose that f¯(x) = φ¯1 (x)e1 · · · φ¯r (x) r is the factorization of f¯(x) into a product of powers of distinct irreducible polynomials over R/p with φi (x) ∈ R[x] monic. Let ti ≥ 0 be the highest power of φ¯i (x) dividing the polynomial N (x), where N (x) belonging to Rp [x] is r i ,l+1)−1 defined by f (x) = φi (x)ei + π l N (x) with N (x) = ¯0, l ≥ 1. Let ui stand for (ei −1)l+gcd(e or 2 i=1
i ,l)−1 max{lti , (ei −1)(l−1)+gcd(e } according as ti > 2
2
ei l+1
or not. Then the following hold:
By Perron’s criterion (cf. [10]), xn + axn−1 + b belonging to Z[x] is irreducible over Q when |a| > 1 + |b|.
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(i) ip (f ) ≥
r
ui deg φi (x).
i=1
(ii) If gcd(ej , l) = 1 and tj = 0 whenever ej > 1, then equality holds in (i), i.e., ip (f ) = r (ei −1)(l−1) deg φi (x). 2 i=1
2. Preliminary results In what follows, v is a discrete valuation of a field K with valuation ring Rv having maximal ideal Mv vˆ) will denote completion of (K, v) and v˜ˆ the unique and value group Z. We fix a prime element π of Rv . (K,
For any ξ in the valuation ring of v˜ˆ, ξ¯ will denote of K. prolongation of vˆ to a fixed algebraic closure K ˜ its vˆ-residue, i.e., the image of ξ under the canonical homomorphism from the valuation ring of v˜ˆ onto its
containing K, L will stand for the residue field of the valuation of L residue field. For a subfield L of K obtained by restricting v˜ˆ. When g(x) ∈ Rv [x], g¯(x) will stand for the polynomial obtained by replacing each coefficient of g(x) by its image in Rv /Mv . We now prove a lemma which plays a significant role in the proof of Theorem 1.5. Lemma 2.1. Let π be a prime element of a discrete valuation ring Rv . Assume that φ1 (x), · · · , φr (x) belonging to Rv [x] are monic polynomials such that φ¯1 (x), · · · , φ¯r (x) are distinct irreducible polynomials over Rv /Mv . For 1 ≤ i ≤ r, let fi (x) = φi (x)ei + π ki Ni (x) be a polynomial with Ni (x) ∈ Rv [x] of degree less than ei deg φi (x), ki ≥ 1 and N i (x) = ¯0. If N (x) ∈ Rv [x] with N (x) = ¯0 is such that f1 (x) · · · fr (x) = φ1 (x)e1 · · · φr (x)er + π l N (x), then l = min{k1 , · · · , kr }. Proof. We shall prove the lemma by induction on r. We first prove the lemma for r = 2. On multiplying, we have f1 (x)f2 (x) =
2
φi (x)ei + π k1 N1 (x)φ2 (x)e2 + π k2 N2 (x)φ1 (x)e1 + π k1 +k2 N1 (x)N2 (x).
(2.1)
i=1
So when we write f1 (x)f2 (x) =
2
φi (x)ei + π u A(x)
(2.2)
i=1
with A(x) ∈ Rv [x], A(x) = ¯0, then u ≥ min{k1 , k2 }; it is to be shown that equality holds here. This is obvious when k1 = k2 in view of (2.1). Suppose to the contrary that k1 = k2 and u > k1 . Comparing equations (2.1), (2.2) and keeping in mind that k1 , k2 are positive, it follows on taking the image of the polynomials in K[x] that N 1 (x)φ¯2 (x)e2 + N 2 (x)φ¯1 (x)e1 = ¯0, which is not possible as φ¯1 (x)(= φ¯2 (x)) is irreducible over Rv /Mv and deg N1 (x) < e1 deg φ1 (x) by hypothesis. This contradiction proves that u = min{k1 , k2 }. Let us assume that the lemma holds for r. We now prove it for r + 1. By induction hypothesis, when r fi (x) is written as i=1 r
φi (x)ei + π s B(x), B(x) ∈ Rv [x], B(x) = ¯0,
i=1
then s = min{k1 , · · · , kr }. Consequently r+1 i=1
fi (x) =
r+1 i=1
φi (x)ei + π s B(x)φr+1 (x)er+1 + π kr+1 Nr+1 (x)
r i=1
φi (x)er + π s+kr+1 Nr+1 (x)B(x).
(2.3)
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So when
r+1
5
fi (x) is written as
i=1 r+1
fi (x) =
i=1
r+1
φi (x)ei + π m D(x)
(2.4)
i=1
with D(x) ∈ Rv [x], D(x) = ¯0, then m ≥ min{s, kr+1 }. It only remains to prove that m = min{s, kr+1 }. This equality is immediate from (2.3) when s = kr+1 ; if s = kr+1 , then a comparison of (2.3) and (2.4) together with the inequality m > kr+1 implies that B(x)φ¯r+1 (x)er+1 + N r+1 (x)
r
φ¯i (x)er = ¯0,
i=1
which is not possible because φ¯1 (x), · · · , φ¯r+1 (x) are distinct monic irreducible polynomials over Rv /Mv and deg Nr+1 (x) < er+1 deg φr+1 (x) in view of the hypothesis. This completes the proof of the lemma. 2 In addition to several simplifications, we will use an extension of Ore’s Index Theorem to discrete valued fields (stated as Theorem 2.A below). For this we need to introduce the notions of Gauss valuation, φ-Newton ¯ polygon (Definition 2.2), φ-index of a polynomial for a monic polynomial φ(x) belonging to Rv [x] with φ(x) irreducible over Rv /Mv (Definition 2.3), and the v-index of a monic irreducible polynomial g(x) ∈ Rv [x] (Definition 2.4). We shall denote by v x the Gauss valuation of the field K(x) of rational functions in an indeterminate x which extends a given valuation v of K and is defined on K[x] by v
x
ci x
i
= min{v(ci )}, ci ∈ K. i
i
¯ Definition 2.2. Let φ(x) ∈ Rv [x] be a monic polynomial with φ(x) irreducible over Rv /Mv and g(x) ∈ Rv [x] n ai (x)φ(x)i with deg ai (x) < deg φ(x), an (x) = 0 be the be a monic polynomial not divisible by φ(x). Let i=0
φ(x)-expansion of g(x) obtained on dividing it by successive powers of φ(x). Let Pi stand for the point in the plane having coordinates (i, v x (an−i (x))) when an−i (x) = 0, 0 ≤ i ≤ n. Let μij denote the slope of the line joining the points Pi and Pj if an−i (x)an−j (x) = 0. Let i1 be the largest positive index not exceeding n such that μ0i1 = min{ μ0j | 0 < j ≤ n, an−j (x) = 0}. If i1 < n, let i2 be the largest index such that i1 < i2 ≤ n and μi1 i2 = min{ μi1 j | i1 < j ≤ n, an−j (x) = 0} and so on. The φ-Newton polygon of g(x) (with respect to the underlying valuation v) is the polygonal path having segments P0 Pi1 , Pi1 Pi2 , . . . , Pik−1 Pik with ik = n. These segments are called the edges of the φ-Newton polygon and their slopes form a strictly increasing sequence; these slopes are non-negative as g(x) is a monic polynomial with coefficients in Rv . The region enclosed by the positive side of x-axis, the φ-Newton polygon of g(x) and the vertical line passing through the last vertex of this polygon will be referred to as the region of the φ-Newton polygon of g(x). Definition 2.3. Let φ(x) and g(x) be as in the above definition. Let ν denote the number of points with positive integer coordinates lying on or below the φ-Newton polygon of g(x) (with respect to the underlying valuation v) away from the vertical line passing through the last vertex of this polygon. As in [5], the φ-index of g is defined to be ν deg φ(x) and will be denoted by iφ (g).
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Definition 2.4. Let g(x) belonging to Rv [x] be a monic irreducible separable polynomial of degree n having a root θ and S be the integral closure of Rv in K(θ). Then S is a free Rv -module of rank n. If d denotes the determinant of the transition matrix from an Rv -basis of S to {1, θ, · · · , θn−1 }, then v(d) is independent of the choice of the Rv -basis of S. As in [5], v(d) will be called the v-index of g(x) and will be denoted by iv (g). Observe that when p is a non-zero prime ideal of a Dedekind domain R and vp is the p-adic valuation of the quotient field of R defined for any non-zero α ∈ R by vp (α) = the highest power of p dividing the ideal αR,
(2.5)
then ivp (g) has the same meaning as ip (g) defined in the Notation preceding Theorem 1.5. The following theorem to be used in the sequel was originally proved by Ore for p-adic valuations [9] and extended to discrete valued fields in [5]. Theorem 2.A. Let v be a discrete valuation of a field K with valuation ring Rv , residue field Rv /Mv and e value group Z. Let g(x) ∈ Rv [x] be a monic irreducible separable polynomial and φ¯1 (x)e1 · · · φ¯r (x) r be the factorization of g¯(x) into a product of powers of distinct irreducible polynomials over Rv /Mv with φi (x) ∈ Rv [x] monic. Then iv (g) ≥
r
iφj (g).
(2.6)
j=1
We shall state a theorem which gives a sufficient condition for the inequality (2.6) to become equality. For this, we need the following definitions.
¯ with φ(x) Definition 2.5. Let φ(x) ∈ Rv [x] be a monic polynomial having a root α in K irreducible over Rv /Mv and π be a prime element of Rv . Let g(x) ∈ Rv [x] be a monic polynomial not divisible by φ(x) having degree a multiple of deg φ(x) with φ(x)-expansion φ(x)n + an−1 (x)φ(x)n−1 + · · · + a0 (x). Suppose that the φ-Newton polygon of g(x) (with respect to the underlying valuation v) consists of a single edge x x (x)) having positive slope λ = de with d, e coprime, i.e., min{ v (an−i | 1 ≤ i ≤ n} = v (an0 (x)) = de so that n is i a (x) divisible by e, say n = et and v x (an−ej (x)) ≥ dj for 1 ≤ j ≤ t. Thus the polynomial n−ej has coefficients π dj an−ej (α) ˜ in Rv and hence is an element of the valuation ring of vˆ. Corresponding to the φ-Newton polygon π dj of g(x), we associate with g(x) a polynomial T (Y ) not divisible by Y with coefficients in K(α) defined by t an−ej (α) t t−j T (Y ) = Y + Y . π dj j=1
Definition 2.6. Let φ(x), α be as in the above definition. Let G(x) belonging to Rv [x] be a monic polynomial ¯ not divisible by φ(x) such that G(x) is a power of φ(x). Let λ1 < · · · < λk be the slopes of the edges of the φ-Newton polygon of G(x). By Theorem 1.1 of [4], we can write G(x) = G1 (x) · · · Gk (x), where the φ-Newton polygon of Gi (x) ∈ Rvˆ [x] has a single edge with slope λi . Let Ti (Y ) belonging to K(α)[Y ] denote the polynomial associated with Gi (x) described as in the above definition. Then G(x) is said to be v-regular with respect to φ if each Ti (Y ) is a product of distinct monic irreducible polynomials over K(α) for e e 1 ≤ i ≤ k. In general, if g(x) belonging to Rv [x] is a monic polynomial and g¯(x) = φ¯1 (x) 1 · · · φ¯r (x) r is its factorization into irreducible polynomials over Rv /Mv with each φi (x) belonging to Rv [x] monic and ei > 0, then by Hensel’s Lemma there exist g1 (x), · · · , gr (x) belonging to Rvˆ [x] such that g(x) = g1 (x) · · · gr (x) and g¯i (x) = φ¯i (x)ei . The polynomial g(x) is said to be v-regular (with respect to φ1 , · · · , φr ) if each gi (x) is vˆ-regular with respect to φi .
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A. Jakhar, S.K. Khanduja / Journal of Pure and Applied Algebra ••• (••••) ••••••
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The result stated below is proved in [5, Theorem 1.4]. Theorem 2.B. Let v, g(x), φ1 (x), · · · , φr (x) be as in Theorem 2.A. If g(x) is v-regular with respect to φ1 , · · · , φr , then iv (g) =
r
iφj (g).
j=1
The following elementary lemma is already known (cf. [6, Problem 435]). We omit its proof. As usual for a real number λ, λ stands for the largest integer not exceeding λ. Lemma 2.C. Let t, n be positive integers. Let S1 denote the set of points in the plane with positive integer coordinates lying inside or on the triangle with vertices (0, 0), (n, 0), (n, t) which do not lie on the line x = n. Then #S1 =
n−1 i=1
1 it = [(n − 1)(t − 1) + gcd(t, n) − 1]. n 2
3. Proof of Theorem 1.5 vˆ) We denote the p-adic valuation vp of K defined in (2.5) by v and the completion of (K, v) by (K, having valuation ring Rvˆ . Fix a prime element π of Rv = Rp . By Theorem 2.A iv (f ) ≥
r
iφj (f ).
(3.1)
j=1
To calculate iφj (f ), we carry out some simplifications in terms of factors of f (x) over Rvˆ . ¯ = φ¯1 (x)e1 · · · φ¯r (x)er leads to a In view of Hensel’s Lemma [8, Chapter 2], the given factorization f(x) factorization of f (x) as f1 (x) · · · fr (x), where fi (x) ∈ Rvˆ [x] is monic and f¯i (x) = φ¯i (x)ei , 1 ≤ i ≤ r. Write fi (x) = φi (x)ei + π ki Ni (x), Ni (x) ∈ Rvˆ [x], N i (x) = ¯0, ki ≥ 1, deg Ni (x) < ei deg φi (x).
(3.2)
Keeping in mind that f (x) =
r
φi (x)ei + π l N (x), N (x) = ¯0,
(3.3)
i=1
and applying Lemma 2.1, we see that l = min{k1 , · · · , kr }.
(3.4)
Since f (x) = f1 (x) · · · fr (x), using (3.2), (3.3), (3.4), it can be easily seen that N (x) =
r j=1
(π kj −l )N
j (x)
r
ei ¯ φi (x) .
(3.5)
i=1,i=j
¯ For any polynomial h(x) ∈ Rvˆ [x], let vφ¯j (h(x)) denote the highest power of the polynomial φ¯j (x) dividing ¯ Recall that vφ¯j (N (x)) = tj and l = min{k1 , · · · , kr }. Keeping in mind that deg N j (x) < ej deg φ¯j (x) h(x). in view of (3.2), it immediately follows from (3.5) that for any j, 1 ≤ j ≤ r, we have
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A. Jakhar, S.K. Khanduja / Journal of Pure and Applied Algebra ••• (••••) ••••••
tj = vφ¯j (Nj (x)) if kj = l,
(3.6)
tj ≥ ej if kj > l.
(3.7)
By Corollary 2.5 of [4] the φj -Newton polygon of f (x) is a translate in the positive horizontal direction of the φj -Newton polygon of fj (x), so the number of points with positive integer coordinates lying on or below the φj -Newton polygon of f (x) and fj (x) are the same. Therefore iφj (f ) = iφj (fj ), 1 ≤ j ≤ r.
(3.8)
Thus (3.1) becomes iv (f ) ≥
r
iφj (fj ) =
j=1
r
nj deg φj (x),
(3.9)
j=1
where nj is the number of points with positive integer coordinates which lie on or below the φj -Newton polygon of fj (x) away from the vertical line passing through the last vertex of the polygon. Therefore assertion (i) of the theorem will follow from (3.9) once we prove that nj ≥ uj for 1 ≤ j ≤ r. Let mj stand for vφ¯j (Nj (x)) and Nj (x) =
sj
(3.10) Aju (x)φj (x)u be the φj (x)-expansion of Nj (x),
u=0
Ajsj (x) = 0. By virtue of (3.2), the φj (x)-expansion of fj (x) is given by
fj (x) = φj (x)
ej
+π
kj
sj u=mj
mj −1 u
Aju (x)φj (x) +
Aju (x)φj (x)
u
.
(3.11)
u=0
In view of the definition of mj , we have Aju (x) = ¯0 for 0 ≤ u ≤ mj − 1 and Ajmj (x) = ¯0; further by (3.4), kj ≥ l. We split the proof of (3.10) into two cases. ¯ It can be easily seen that the φj -Newton polygon of fj (x) Case I. mj = 0. In this case we have Aj0 (x) = 0. has a single edge which is the straight line segment joining the points (0, 0) and (ej , kj ). This case is divided into two subcases. First consider the subcase when kj = l. Then by (3.6), we have tj = mj = 0 and hence (e −1)(l−1)+gcd(ej ,l)−1 (e −1)(l−1)+gcd(ej ,l)−1 in this situation uj = max{ltj , j }= j . Keeping in mind that the 2 2 φj -Newton polygon of fj (x) has a single edge joining points (0, 0), (ej , l), it follows from Lemma 2.C that (e −1)(l−1)+gcd(ej ,l)−1 nj = j and (3.10) is verified in this subcase. Now consider the subcase when kj > l. 2 Then by (3.7), we have tj ≥ ej . Since kj ≥ l + 1 in this subcase, it is clear that the triangle with vertices (0, 0), (ej , 0) and (ej , l+1) is contained in the region of the φj -Newton polygon of fj (x). So using Lemma 2.C, (e −1)l+gcd(ej ,l+1)−1 we see that nj ≥ j = uj in this subcase, which completes the proof of assertion (i) of the 2 theorem in Case I. Case II. mj > 0. Denote v x (Aj0 (x)) by zj . Note that zj > 0 in this case. Before proceeding with the proof, we make two observations. Keeping in mind that l + 1 ≤ kj + zj and the fact that the last vertex of the φj -Newton polygon of fj (x) is (ej , kj + zj ), we see that the point (ej , l + 1) will lie on the vertical line ej passing through the last vertex of this polygon. Observe that when mj ≤ kj +1 , then the point (ej − mj , kj ) will be contained in the region of the φj -Newton polygon of fj (x). This case is split into two subcases. ej Subcase (i). kj = l; consequently by (3.6) we have tj = mj > 0. In this subcase, when tj ≤ l+1 , then in view of what has been said in the above paragraph, the points with coordinates (0, 0), (ej − tj , l), (ej , l + 1) and
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(ej , 0) will be contained in the region of the φj -Newton polygon of fj (x). Note that (0, 0), (ej −tj , l), (ej , l+1) ej . So the quadrilateral with vertices (0, 0), (ej −tj , l), (ej , l+1), (ej , 0) lie on the same line if and only if tj = l+1 or the triangle with vertices (0, 0), (ej , l + 1), (ej , 0) will be contained in the region of the φj -Newton polygon ej of fj (x) when tj ≤ l+1 . By virtue of Lemma 2.C applied to the triangle with vertices (0, 0), (ej − tj , l), (ej − tj , 0) and keeping in mind that the number of points with positive integer coordinates lying on or inside the quadrilateral with vertices (ej − tj , 0), (ej − tj , l), (ej , l + 1), (ej , 0) away from the line x = ej is ltj , we conclude that nj ≥
(ej − tj − 1)(l − 1) + gcd(ej − tj , l) − 1 + ltj = C (say). 2
(3.12)
Since tj > 0 in the present situation, it can be easily seen that (ej − 1)(l − 1) + l − 1 (ej − 1)(l − 1) + (l + 1)tj + gcd(ej − tj , l) − 1 > 2 2 (ej − 1)(l − 1) + gcd(ej , l) − 1 . ≥ 2
C=
Since C ≥ ltj in view of (3.12), it now follows from the above inequality that nj ≥ uj . Further in the ej present subcase when tj > l+1 , then keeping in mind that the triangle with vertices (0, 0), (ej , 0), (ej , l + 1) is contained in the region of the φj -Newton polygon of fj (x) and applying Lemma 2.C, we see that nj ≥ (ej −1)l+gcd(ej ,l+1)−1 = uj in this situation. 2 e
e
j j Subcase (ii). kj > l. By (3.7), we have tj ≥ ej > l+1 . When mj < l+2 , then arguing as in the first paragraph of Case II, it can be seen that the quadrilateral with vertices (0, 0), (ej − mj , l + 1), (ej , l + 2) and (ej , 0) is contained in the region of the φj -Newton polygon of fj (x). Applying Lemma 2.C to the triangle with vertices (0, 0), (ej − mj , l + 1), (ej − mj , 0), we see that
(ej − mj − 1)l + gcd(ej − mj , l + 1) − 1 + (l + 1)mj 2 (ej − 1)l + mj l + 2mj − 1 + gcd(ej − mj , l + 1) = 2 (ej − 1)l + gcd(ej , l + 1) − 1 (ej − 1)l + (l + 1) − 1 ≥ = uj . > 2 2
nj ≥
e
j When mj ≥ l+2 , then it can be easily seen that the triangle with vertices (0, 0), (ej , 0), and (ej , l + 2) is contained in the region of the φj -Newton polygon of fj (x). Using again Lemma 2.C, we have
(ej − 1)l + (ej − 1) + gcd(ej , l + 2) − 1 (ej − 1)(l + 1) + gcd(ej , l + 2) − 1 = 2 2 (ej − 1)l + gcd(ej , l + 1) − 1 (ej − 1)l + ej − 1 ≥ = uj . ≥ 2 2
nj ≥
This completes the proof of the first assertion of the theorem. Now to prove the second assertion of the theorem, we first show that with hypothesis as in this assertion, fj (x) is vˆ-regular with respect to φj for each j, 1 ≤ j ≤ r and we also calculate iφj (fj ). For this, two cases are distinguished. Consider first the case when ej = 1. Then by (3.2), fj (x) = φj (x) +π kj Nj (x) with kj ≥ 1 and deg Nj (x) < deg φj (x). The φj -Newton polygon of fj (x) has a single edge with slope kj . So by Definition 2.3, we have iφj (fj ) = 0
(3.13)
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and the polynomial associated with fj (x) for this Newton polygon is a linear polynomial not equal to Y in view of Definition 2.5. Hence fj (x) is vˆ-regular with respect to φj in case ej = 1. Consider now the case when ej > 1. Then by hypothesis of assertion (ii), tj = 0 and gcd(ej , l) = 1. When tj = 0, we have mj = 0 and kj = l by (3.6), (3.7). So in the φj (x)-expansion of fj (x) given by (3.11), Aj0 (x) = ¯0; consequently the φj -Newton polygon of fj (x) has a single edge joining the points (0, 0), (ej , l). As ej and l are coprime, the polynomial associated with fj (x) for this Newton polygon is again a linear polynomial. Thus fj (x) is vˆ-regular with respect to φj . Further, applying Lemma 2.C to the triangle with vertices (0, 0), (ej , l), (ej , 0) and keeping in mind that gcd(ej , l) = 1, we see that in the present case iφj (fj (x)) =
(ej − 1)(l − 1) deg φj (x). 2
(3.14)
Using the fact that fj (x) is vˆ-regular with respect to φj for each j proved above, we see that f (x) is v-regular with respect to φ1 , · · · , φr in view of Definition 2.6. It now follows from Theorem 2.B that iv (f ) =
r
iφj (f ).
j=1
Using (3.8), (3.13) and (3.14), the above equality can be rewritten as iv (f ) =
r
iφj (fj ) =
j=1
r (ej − 1)(l − 1) deg φj (x). 2 j=1
This completes the proof of the theorem. Acknowledgements The authors are highly grateful to the referee for his/her several valuable suggestions which have greatly improved the exposition of the paper. The financial support from the respective institutions is gratefully acknowledged by the authors. The second author is also thankful to Council of Scientific and Industrial Research, New Delhi for financial assistance through Emeritus Scientistship. References [1] Z.I. Borevich, I.R. Shafarevich, Number Theory, Academic Press, 1966. [2] H. Cohen, A Course in Computational Algebraic Number Theory, Springer-Verlag, Berlin-Heidelberg, 1993. [3] R. Dedekind, Über den Zusammenhang zwischen der Theorie der ideale und der Theorie der höheren Kongruenzen, Götttingen Abh. 23 (1878) 1–23. [4] S.K. Khanduja, S. Kumar, On prolongations of valuations via Newton polygons and liftings of polynomials, J. Pure Appl. Algebra 216 (2012) 2648–2656. [5] S.K. Khanduja, S. Kumar, A generalization of a theorem of Ore, J. Pure Appl. Algebra 218 (2014) 1206–1218. [6] J.M. de Koninck, A. Mercier, 1001 Problems in Classical Number Theory, Amer. Math. Soc., Providence Rhode Island, 2007. [7] W. Narkewicz, Elementary and Analytic Theory of Algebraic Numbers, Springer-Verlag, Berlin, Heidelberg, 2004. [8] J. Neukirch, Algebraic Number Theory, Springer-Verlag, Berlin, Heidelberg, 1999. [9] Ø. Ore, Newtonsche Polygone in der Theorie der algebraischen Körper, Math. Ann. 99 (1928) 84–117. [10] E.S. Selmer, On the irreducibility of certain trinomials, Math. Scand. 4 (1956) 287–302. [11] O. Zariski, P. Samuel, Commutative Algebra, Vol. I, D. van Nostrand Company, 1958.