1. Introduction and statements of results
Let $K={\mathbf {Q}}(\theta )$
be an algebraic number field with $\theta$
in the ring $A_K$
of algebraic integers of $K$
. Let $f(x)\in {\mathbf {Z}}[x]$
be the minimal polynomial of $\theta$
having degree $n$
over the field ${\mathbf {Q}}$
of rational numbers. It is well known that $A_K$
is a free ${\mathbf {Z}}$
-module of rank $n$
. An algebraic number field $K$
is said to be monogenic if there exists some $\alpha \in A_K$
such that $A_K={\mathbf {Z}}[\alpha ]$
. In this case, $\{1,\, \alpha,\,\cdots,\,\alpha ^{n-1}\}$
is an integral basis of $K$
; such an integral basis of $K$
is called a power integral basis or briefly a power basis of $K$
. If $K$
does not possess any power basis, we say that $K$
is non-monogenic. It is well known that every quadratic and cyclotomic field is monogenic. In algebraic number theory, it is important to know whether a number field is monogenic or not. The first non-monogenic number field was given by Dedekind in $1878$
. He proved that the cubic field ${\mathbf {Q}}(\xi )$
is not monogenic when $\xi$
is a root of the polynomial $x^3-x^2-2x-8$
(cf. [Reference Narkiewicz14, p. 64]). The problem of testing the monogenity of number fields and constructing power integral bases have been intensively studied (cf. [Reference Ahmad, Nakahara and Husnine1,Reference Ahmad, Nakahara and Hameed2,Reference Gaál and Remete7,Reference Jakhar and Kaur9,Reference Jakhar and Kumar10,Reference MacKenzie and Scheuneman13,Reference Soullami and Sahmoudi17]). In $1984$
, Funakura [Reference Funakura5] gave necessary and sufficient condition on those integers $m$
for which the quartic field ${\mathbf {Q}}(m^{1/4})$
is monogenic. Ahmad et al. [Reference Ahmad, Nakahara and Husnine1,Reference Ahmad, Nakahara and Hameed2] proved that for a square-free integer, $m$
is not congruent to $\pm 1\mod 9$
, a pure field ${\mathbf {Q}}(m^{1/6})$
having degree six over ${\mathbf {Q}}$
is monogenic when $m\equiv 2$
or $3\mod 4$
and it is non-monogenic when $m\equiv 1\mod 4$
. In $2017$
, Gaál and Remete [Reference Gaál and Remete7] studied monogenity of algebraic number fields of the type ${\mathbf {Q}}(m^{1/n})$
where $3\leq n\leq 9$
and $m$
is square free. In [Reference Gaál6], Gaál studied monogenity of number fields defined by some sextic irreducible trinomials. El. Fadil characterized when a prime $p$
is a common index divisor of the number field defined by $x^6+ax^3+b\in {\mathbf {Z}}[x]$
in [Reference Fadil4].
Recall that for an algebraic number field $L={\mathbf {Q}}(\xi )$
with $\xi$
an algebraic integer satisfying a monic irreducible polynomial $g(x)$
over ${\mathbf {Q}}$
, the discriminant $D$
of $g(x)$
and the discriminant $d_L$
of $L$
are related by the formula
In what follows, let $K={\mathbf {Q}}(\theta )$
be an algebraic number field with $\theta$
a root of an irreducible trinomial $f(x)=x^5+ax+b\in {\mathbf {Z}}[x]$
. Then for every rational prime $p$
, we provide necessary and sufficient conditions on $a,\,~b$
, so that $p$
is a common index divisor of $K$
. In particular, under these conditions $K$
is non-monogenic. As an application of our results, we provide some classes of algebraic number fields which are non-monogenic.
Throughout this paper, $\mathop {\mathrm {ind}} \theta$
denotes the index of the subgroup ${\mathbf {Z}}[\theta ]$
in $A_K$
and $i(K)$
will stand for the index of the field $K$
defined by
A prime number $p$
dividing $i(K)$
is called a prime common index divisor of $K$
. Note that if $K$
is monogenic then $i(K)=1$
. Therefore, a number field having a prime common index divisor is non-monogenic. There exist non-monogenic number fields having $i(K)=1$
, e.g., $K={\mathbf {Q}}(\sqrt [3]{175})$
is not monogenic with $i(K)=1$
. In what follows, for a prime $p$
and a non-zero $m$
belonging to the ring ${\mathbf {Z}}_p$
of $p$
-adic integers, $v_p(m)$
will denote the highest power of $p$
dividing $m$
. For a non-zero integer $l$
, let $l_p$
denote $\frac {l}{p^{v_p(l)}}$
. If a rational prime $p$
is such that $p^4$
divides $a$
and $p^5$
divides $b$
, then $\theta /p$
is a root of the polynomial $x^5+(a/p^4)x+(b/p^5)$
having integer coefficients. So we may assume that for each prime $p$
In this paper, $D$
will stand for the discriminant of $f(x)$
. One can check that
We abbreviate $a\equiv b\mod n$
by $a\equiv b(n)$
. For integers $u,\,v,\,w,\,z$
and prime number $p$
, by notation $(u,\,v)\equiv (w,\,z)(p)$
, we mean $u\equiv w(p)$
and $v\equiv z(p)$
. Also, $(u,\,v)\in I(p)$
means that $(u,\,v)\equiv (w,\,z)(p)$
for some $(w,\,z) \in I$
.
With the above notation and assumption (1.2), we prove
Theorem 1.1 Let $K={\mathbf {Q}}(\theta )$
with $\theta$
a root of an irreducible polynomial $f(x) = x^5+ax+b \in {\mathbf {Z}}[x]$
. Then for any odd rational prime $p$
, $p\nmid i(K)$
.
Theorem 1.2 Let $K={\mathbf {Q}}(\theta )$
be an algebraic number field with $\theta$
a root of an irreducible polynomial $f(x)=x^5+ax+b$
. If $u=\frac {v_2(f(-\frac {5b}{4a}))-1}{2}$
and $\delta =2^u-\frac {5b}{4a}$
, then the prime ideal factorization of $2A_K$
is given in Table 1. Furthermore, $2\mid i(K)$
if and only if one of the conditions A$7$
, A$9$
, A$13$
, A$15$
, A$18$
, A$19$
, A$21$
, A$22$
, A$24$
, A$29$
hold.
Factorization of $2A_K$
.
Factorization of $3A_K$
.
In particular, if any one of the conditions A$7$
, A$9$
, A$13$
, A$15$
, A$18$
, A$19$
, A$21$
, A$22$
, A$24$
, A$29$
hold, then $K$
is non-monogenic, otherwise $i(K)=1$
.
The following corollaries follow immediately from the above theorem.
Corollary 1.3 Let $K={\mathbf {Q}}(\theta )$
where $\theta$
satisfies the polynomial $x^5+ax+b$
. If $a,\,~b$
are such that $a=1088r+68$
and $b=18496s+ 1088$
with $r,\,~s\in {\mathbf {Z}}$
, then $K$
is non-monogenic in view of Case A$13$
of Table 1 of Theorem 1.2.
Corollary 1.4 Let $a,\,~b$
be such that $a=20r+5$
and $b=100s+10$
with $r,\,~s\in {\mathbf {Z}}$
. Let $K={\mathbf {Q}}(\theta )$
where $\theta$
is a root of a polynomial $x^5+ax+b$
. Then $K$
is non-monogenic by Case A$15$
of Table 1 of Theorem 1.2.
We now provide some examples of non-monogenic number fields.
Example 1.5 Let $K={\mathbf {Q}}(\theta )$
where $\theta$
is a root of $f(x)=x^5+68x+1088$
. Note that $f(x)$
satisfies Eisenstein criterion with respect to $17$
, hence, it is irreducible over ${\mathbf {Q}}$
. So $K$
is non-monogenic by Corollary 1.3.
Example 1.6 Let $K={\mathbf {Q}}(\theta )$
with $\theta$
satisfying $f(x)=x^5+5x+10$
. It can be easily seen that $f(x)$
is $5$
-Eisenstein. Hence, $K$
is non-monogenic by Corollary 1.4.
2. Preliminary results
Let $K={\mathbf {Q}}(\theta )$
be an algebraic number field with $\theta$
a root of a monic irreducible polynomial $f(x)$
belonging to ${\mathbf {Z}}[x]$
. In what follows, $A_K$
will stand for the ring of algebraic integers of $K$
. For a rational prime $p$
, let $\mathbb {F}_p$
denote the finite field with $p$
elements. The following lemma (cf. [Reference Smith16, Theorem 2.2]) will play an important role in the proof of Theorems 1.1, 1.2.
Lemma 2.1 Let $K$
be an algebraic number field and $p$
be a rational prime. Then $p$
is a common index divisor of $K$
if and only if for some positive integer $h$
, the number of distinct prime ideals of $A_K$
lying above $p$
having residual degree $h$
is greater than the number of monic irreducible polynomials of degree $h$
in $\mathbb {F}_p[x]$
.
We shall first introduce the notion of Gauss valuation, $\phi$
-Newton polygon and Newton polygon of second order, where $\phi (x)$
belonging to ${\mathbf {Z}}_p[x]$
is a monic polynomial with $\overline {\phi }(x)$
irreducible over $\mathbb {F}_p$
.
Definition 2.2 The Gauss valuation of the field ${\mathbf {Q}}_p(x)$
of rational functions in an indeterminate $x$
which extends the valuation $v_p$
of ${\mathbf {Q}}_p$
and is defined on ${\mathbf {Q}}_p[x]$
by
Definition 2.3 Let $p$
be a rational prime. Let $\phi (x)\in {\mathbf {Z}}_p[x]$
be a monic polynomial which is irreducible modulo $p$
and $f(x)\in {\mathbf {Z}}_p[x]$
be a monic polynomial not divisible by $\phi (x)$
. Let $\displaystyle \sum \nolimits _{i=0}^{n}a_i(x)\phi (x)^i$
with $\deg a_i(x)<\deg \phi (x)$
, $a_n(x)\neq 0$
be the $\phi (x)$
-expansion of $f(x)$
obtained on dividing it by the successive powers of $\phi (x)$
. Let $P_i$
stand for the point in the plane having coordinates $(i,\,v_{p,x}(a_{n-i}(x)))$
when $a_{n-i}(x)\neq 0$
, $0\leq i\leq n$
. Let $\mu _{ij}$
denote the slope of the line joining the point $P_i$
with $P_j$
if $a_{n-i}(x)a_{n-j}(x)\neq 0$
. Let $i_1$
be the largest positive index not exceeding $n$
such that
If $i_1< n,$
let $i_2$
be the largest index such that $i_1< i_2\leq n$
with
and so on. The $\phi$
-Newton polygon of $f(x)$
with respect to $p$
is the polygonal path having segments $P_{0}P_{i_1},\,P_{i_1}P_{i_2},\,\ldots,\,P_{i_{k-1}}P_{i_k}$
with $i_k=n$
. These segments are called the edges of the $\phi$
-Newton polygon and their slopes form a strictly increasing sequence; these slopes are non-negative as $f(x)$
is a monic polynomial with coefficients in ${\mathbf {Z}}_p$
.
Definition 2.4 Let $\phi (x) \in {\mathbf {Z}}_p[x]$
be a monic polynomial which is irreducible modulo a rational prime $p$
having a root $\alpha$
in the algebraic closure $\widetilde {{\mathbf {Q}}}_{p}$
of ${\mathbf {Q}}_p$
. Let $f(x) \in {\mathbf {Z}}_p[x]$
be a monic polynomial not divisible by $\phi (x)$
with $\phi (x)$
-expansion $\phi (x)^n + a_{n-1}(x)\phi (x)^{n-1} + \cdots + a_0(x)$
such that $\overline {f}(x)$
is a power of $\overline {\phi }(x)$
. Suppose that the $\phi$
-Newton polygon of $f(x)$
consists of a single edge, say $S$
, having a positive slope denoted by $\frac {l}{e}$
with $l,\, e$
coprime, i.e.,
so that $n$
is divisible by $e$
, say $n=et$
and $v_{p,x}(a_{n-ej}(x)) \geq lj$
for $1\leq j\leq t$
. Thus, the polynomial $b_j(x):=\frac {a_{n-ej}(x)}{p^{lj}}$
has coefficients in ${\mathbf {Z}}_p$
and hence $b_j(\alpha )\in {\mathbf {Z}}_p[\alpha ]$
for $1\leq j \leq t$
. The polynomial $T(Y)$
in an indeterminate $Y$
defined by $T(Y) = Y^t + \sum \nolimits \limits _{j=1}^{t} \overline {b_j}(\overline {\alpha })Y^{t-j}$
having coefficients in $\mathbb {F}_p[\overline {\alpha }]\cong \frac {\mathbb {F}_p[x]}{\langle \phi (x)\rangle }$
is called residual polynomial of $f (x)$
with respect to $(\phi,\,S)$
.
The following definition gives the notion of a residual polynomial when $f(x)$
is more general.
Definition 2.5 Let $\phi (x),\, \alpha$
be as in Definition 2.4. Let $g(x)\in {\mathbf {Z}}_p[x]$
be a monic polynomial not divisible by $\phi (x)$
such that $\overline {g}(x)$
is a power of $\overline {\phi }(x)$
. Let $\lambda _1 < \cdots < \lambda _k$
be the slopes of the edges of the $\phi$
-Newton polygon of $g(x)$
and $S_i$
denote the edge with slope $\lambda _i$
. In view of a classical result proved by Ore (cf. [Reference Cohen, Movahhedi and Salinier3, Theorem 1.5], [Reference Khanduja and Kumar12, Theorem 1.1]), we can write $g(x) = g_1(x)\cdots g_k(x)$
, where the $\phi$
-Newton polygon of $g_i(x) \in {\mathbf {Z}}_{p}[x]$
has a single edge, say $S_i'$
, which is a translate of $S_i$
. Let $T_i(Y)$
belonging to ${\mathbb {F}}_{p}[\overline {\alpha }][Y]$
denote the residual polynomial of $g_i(x)$
with respect to ($\phi,\,~S_i'$
) described as in Definition 2.4. For convenience, the polynomial $T_i(Y)$
will be referred to as the residual polynomial of $g(x)$
with respect to $(\phi,\,S_i)$
. The polynomial $g(x)$
is said to be $p$
-regular with respect to $\phi$
if none of the polynomials $T_i(Y)$
has a repeated root in the algebraic closure of $\mathbb {F}_p$
, $1\leq i\leq k$
. In general, if $F(x)$
belonging to ${\mathbf {Z}}_p[x]$
is a monic polynomial and $\overline {f}(x) = \overline {\phi }_{1}(x)^{e_1}\cdots \overline {\phi }_r{(x)}^{e_r}$
is its factorization modulo $p$
into irreducible polynomials with each $\phi _i(x)$
belonging to ${\mathbf {Z}}_p[x]$
monic and $e_i > 0$
, then by Hensel's Lemma there exist monic polynomials $f_1(x),\, \cdots,\, f_r(x)$
belonging to ${\mathbf {Z}}_{p}[x]$
such that $f(x) = f_1(x)\cdots f_r(x)$
and $\overline {f}_i(x) = \overline {\phi }_i(x)^{e_i}$
for each $i$
. The polynomial $f(x)$
is said to be $p$
-regular (with respect to $\phi _1,\, \cdots,\, \phi _r$
) if each $f_i(x)$
is ${p}$
-regular with respect to $\phi _i$
.
To determine the number of distinct prime ideals of $A_K$
lying above a rational prime $p$
, we will use the Newton polygon of second order and the following theorem which is a weaker version of Theorem 1.2 of [Reference Khanduja and Kumar12].
Theorem 2.6 Let $L={\mathbf {Q}}(\xi )$
be an algebraic number field with $\xi$
satisfying an irreducible polynomial $g(x)\in {\mathbf {Z}}[x]$
and $p$
be a rational prime. Let $\overline {\phi }_{1}(x)^{e_1}\cdots \overline {\phi }_r{(x)}^{e_r}$
be the factorization of $g(x)$
modulo $p$
into powers of distinct irreducible polynomials over $\mathbb {F}_p$
with each $\phi _i(x)\neq g(x)$
belonging to ${\mathbf {Z}}[x]$
monic. Suppose that the $\phi _i$
-Newton polygon of $g(x)$
has $k_i$
edges, say $S_{ij}$
having slopes $\lambda _{ij}=\frac {l_{ij}}{e_{ij}}$
with $\gcd (l_{ij},\,~e_{ij})=1$
for $1\leq j\leq k_i$
. If $T_{ij}(Y) = \prod \limits _{s=1}^{s_{ij}}U_{ijs}(Y)$
is the factorization of the residual polynomial $T_{ij}(Y)$
into distinct irreducible factors over $\mathbb {F}_p$
with respect to $(\phi _i,\,~S_{ij})$
for $1\leq j\leq k_i,$
then
where $\mathfrak p_{ijs}$
are distinct prime ideals of $A_L$
having residual degree $\deg \phi _i(x)\times \deg U_{ijs}(Y).$
Let $L={\mathbf {Q}}(\gamma )$
where $\gamma$
is a root of a monic polynomial $g(x)=a_nx^n+\cdots +a_0\in {\mathbf {Z}}[x],\,~a_0\neq 0$
. Let $p$
be a prime number such that $g(x)\equiv x^n(p)$
. Suppose that the $p$
-Newton polygon of $g(x)$
consists of a single edge with positive slope $\lambda =\frac {l}{e},$
where $\gcd (l,\,~e)=1$
. Let the residual polynomial $T_g(Y)\in \mathbb {F}_p[Y]$
of $g(x)$
is a power of monic irreducible polynomial $\psi (Y)$
over $\mathbb {F}_p$
, i.e., $T_g(Y)=\psi (Y)^s$
in $\mathbb {F}_p[Y]$
, where $s\geq 2$
. In this case, we construct a key polynomial $\varPhi (x)$
attached with the slope $\lambda$
such that the following hold:
(i) $\varPhi (x)$
is congruent to a power of $x$ modulo $p$.(ii) The $p$
-Newton polygon of $\varPhi (x)$ of first order is one-sided with slope $\lambda$.(iii) The residual polynomial of $\varPhi (x)$
with respect to $p$ is $\psi (Y)$ in $\mathbb {F}_p[Y]$.(iv) $\deg \varPhi (x)= e\deg \psi (Y)$
.
As described in [Reference Guárdia, Montes and Nart8, Section 2.2], the data $(x;~\lambda,\,~\psi (Y))$
determine a $p$
-adic valuation $V$
of the field ${\mathbf {Q}}_p(x)$
which satisfies the following properties:
(i) $V(x)=l$
where $\lambda =\frac {l}{e}~$ with $\gcd (l,\,~e)=1$.(ii) If $p(x)=\displaystyle \sum \nolimits _{0\leq i}^{}b_ix^i\in {\mathbf {Z}}_p[x]$
is any polynomial, then
(2.1)\begin{equation} V(p(x))=e\displaystyle\min_{ i\geq 0}^{}\{v_p(b_i)+i \lambda\}. \end{equation}
We define the above valuation $V$
to the valuation of the second order.
If $g(x)=\displaystyle \sum \nolimits _{i=0}^{u} a_i(x)\varPhi (x)^i\in {\mathbf {Z}}_p[x]$
is a $\varPhi$
-adic expansion of $g(x)$
, then the Newton polygon of $g(x)$
with respect to $V$
(also called $V$
-Newton polygon of $g(x)$
of second order) is the lower convex hull of the set of the points $(i,\, V(a_{u-i}(x)\varPhi (x)^{i}))$
of the Euclidean plane.
Let the $V$
-Newton polygon of $g(x)$
of second order has $k$
-sides $E_1,\,\cdots,\, E_k$
with positive slopes $\lambda _1,\,\cdots,\,\lambda _k$
. Let $\lambda _t=\frac {l_t}{e_t}$
with $\gcd (l_t,\,~e_t)=1$
and $[a_t,\, b_t]$
denote the projection to the horizontal axis of the side of slope $\lambda _t$
for $1\leq t \leq k.$
Then, there is a natural residual polynomial $\psi _t(Y)$
of second order attached to each side $E_t,$
whose degree coincides with the degree of the side (i.e., $\frac {b_t-a_t}{e_t}$
) [Reference Guárdia, Montes and Nart8, Section 2.5]. Only those integral points of the $V$
-Newton polygon of $g(x)$
which lie on the side, determine a non-zero coefficient of this second-order residual polynomial. We define $g(x)$
to be $\psi _t$
-regular when the second order residual polynomial $\psi _t(Y)$
attached to the side $E_t$
of the $V$
-Newton polygon of $g(x)$
of second order is separable in $\frac {\mathbb {F}_p[Y ]}{\langle \psi (Y)\rangle }$
. We define $g(x)$
to be $V$
-regular if $g(x)$
is $\psi _t$
-regular for each $t$
, $1\leq t\leq k$
. Further, if each residual polynomial $\psi _t(Y)$
, $t\in \{1,\,~2,\,\cdots,\,k\}$
, is irreducible in $\frac {\mathbb {F}_p[Y ]}{\langle \psi (Y)\rangle }$
, then each $\psi _t(Y)$
provides a prime ideal having residual degree $\deg \psi \cdot \deg \psi _t$
and ramification index $e\cdot e_t$
.
3. Proof of Theorems 1.1 and 1.2
Proof of Theorem 1.1. Since the degree of the extension $K/{\mathbf {Q}}$
is $5$
, to prove this theorem it is sufficient to show that $3\nmid i(K)$
and $5\nmid i(K)$
. If $3\mid i(K)$
, then by (1.3), $a^5\equiv b^4(3)$
. Therefore, $(a,\,~b)\in \{(0,\,~0),\, (1,\,~-1),\,~(1,\,~1)\}(3)$
. Note that if $3$
divides both $a,\,b$
and $5v_3(a)<4v_3(b)$
, then $v_3(a)\in \{1,\,2,\,3\}$
by (1.2).
Case B1: $a\equiv 0(3),\,b\equiv 0(3)$
, $5v_3(a)<4v_3(b)$
, $v_3(a)\in \{1,\,3\}$
. Here $f(x)\equiv x ^5(3)$
and the $x$
-Newton polygon of $f(x)$
has two edges. The first edge, say $S_1$
, is the line segment joining the points $(0,\,0)$
and $(4,\,v_3(a))$
. The second edge, say $S_2$
, is the line segment that joins $(4,\,v_3(a))$
to $(5,\,v_3(b))$
. For each $i=1,\,2$
, the residual polynomial of $f(x)$
with respect to $(x,\,S_i)$
is linear. So Theorem 2.6 is applicable. Using this theorem, $3A_K=\mathfrak { p_1^4}\mathfrak { p_2}$
, where the residual degree of each prime ideal $\mathfrak {p_i}$
for $i=1,\,2,$
is $1$
. Hence, in view of Lemma 2.1, $3\nmid i(K)$
.
Case B2: $a\equiv 0(3),\,b\equiv 0(3)$
, $5v_3(a)<4v_3(b)$
, $v_3(a)=2$
, $a_3\equiv 1(3)$
. In this case, $f(x)\equiv x ^5(3)$
. The $x$
-Newton polygon of $f(x)$
has two edges. The first edge, say $S_1$
, is the line segment joining the points $(0,\,0)$
and $(4,\,2)$
. The second edge, say $S_2$
, is the line segment that join $(4,\,2)$
to $(5,\,v_3(b))$
. The residual polynomial of $f(x)$
with respect to $(x,\,S_1)$
is $Y^2+\bar {1}\in \mathbb {F}_3[Y]$
and with respect to $(x,\,S_2)$
is linear. So by Theorem 2.6, $3A_K=\mathfrak { p_1^2}\mathfrak { p_2}$
, where the residual degree of the prime ideals $\mathfrak {p_1}$
and $\mathfrak { p_2}$
is $2$
and $1$
, respectively. Therefore, $3\nmid i(K)$
.
Case B3: $a\equiv 0(3),\, b\equiv 0(3)$
, $5v_3(a)<4v_3(b)$
, $v_3(a)=2$
, $a_3\equiv -1(3)$
. Here $f(x)\equiv x ^5(3)$
. The $x$
-Newton polygon of $f(x)$
has two edges. The first edge, say $S_1$
, is the line segment joining the points $(0,\,0)$
and $(4,\,2)$
. The second edge, say $S_2$
, is the line segment joining the points $(4,\,2)$
and $(5,\,v_3(b))$
. The residual polynomial of $f(x)$
with respect to $(x,\,S_1)$
is $(Y-\bar {1})(Y+\bar {1})$
and with respect to $(x,\,S_2)$
is linear. Therefore, we have $3A_K=\mathfrak { p_1^2}\mathfrak { p_2^2}\mathfrak { p_3}$
, where the residual degree of each prime ideal $\mathfrak { p_i}$
for $i=1,\,2,\,3$
, is $1$
. Hence, $3\nmid i(K)$
.
Keeping in mind (1.2), we see that the case $5v_3(a)=4v_3(b)$
is not possible.
Case B4: $a\equiv 0(3),\, b\equiv 0(3)$
, $5v_3(a)>4v_3(b)$
. Here $v_3(b)\in \{1,\,2,\,3,\,4\}$
. The $x$
-Newton polygon of $f(x)$
has a single edge, say $S$
, which join the points $(0,\,0)$
and $(5,\,v_3(b))$
. The residual polynomial of $f(x)$
with respect to $(x,\,S)$
is linear. So $3A_K=\mathfrak { p^5}$
, where the residual degree of $\mathfrak { p}$
is $1$
. Thus, $3\nmid i(K)$
.
Case B5: $a\equiv 1(3),\,~b\equiv \tau (3)$
, where $\tau \in \{-1,\,1\}$
. Note that $f(x)\equiv (x^3- x^2\tau +\tau ) (x- \tau )^2(3)$
. In this case, there is only one prime ideal, say $\mathfrak { p_1}$
, of residual degree $3$
. Hence, in view of the Fundamental Equality, there can be either atmost two prime ideals of residual degree $1$
each or one prime ideal of residual degree $2$
. Hence, $3\nmid i(K)$
.
It remains to show that $5\nmid i(K)$
. Suppose there exist distinct non-zero prime ideals $\mathfrak {p}_1$
, $\cdots$
, $\mathfrak {p}_r$
of $A_K$
such that $5A_K=\mathfrak {p}_1^{e_1}\cdots \mathfrak {p}_r^{e_r}$
, where $e_i \geq 1$
, then by the Fundamental Equality, $e_1f_1+\cdots +e_rf_r=5$
where $f_i$
is the residual degree of $\mathfrak p_i$
for $i=1,\,2,\,\cdots,\,r$
. Since $e_i\ge 1$
for all $i=1,\,2,\,\cdots,\,r$
, therefore, there can be at most $5$
prime ideals lying above $5$
, but for every positive integer $h$
the number of monic irreducible polynomials of degree $h$
in $\mathbb {F}_5[x]$
is greater than or equal to $5$
. So by Lemma 2.1, $5\nmid i(K)$
. This completes the proof.
Proof of Theorem 1.2. Case A1: $a\equiv 0(2)$
, $b\equiv 1(2)$
. In this case, $f(x)\equiv (x+1) (x^4+x^3+x^2+x+1)$
(2). Using hypothesis and Equations (1.1) and (1.3), we see that $2\nmid \mathop {\mathrm {ind}}(\theta )$
. So, by Dedekind's Theorem on splitting of primes [Reference Khanduja11, Theorem 4.3], we have $2A_K=\mathfrak { p_1}\mathfrak { p_2}$
, where residual degree of $\mathfrak { p_1}$
is $1$
and residual degree of $\mathfrak { p_2}$
is $4$
. Hence, in view of Lemma 2.1, $2\nmid i(K)$
.
Case A2: $a\equiv 1(2)$
, $b\equiv 1(2)$
. Here $f(x)\equiv (x^2+x+1)(x^3+x^2+1)$
(2). Arguing as in Case A1, one can easily check that $2A_K=\mathfrak { p_1}\mathfrak { p_2}$
, where the residual degree of $\mathfrak { p_1}$
is $2$
and residual degree of $\mathfrak { p_2}$
is $3$
. Thus, by Lemma 2.1, $2\nmid i(K)$
.
Case A3: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)\geq 4v_2(b)$
. By (1.2), we have $v_2(b)\leq 4$
. The $x$
-Newton polygon of $f(x)$
has a single edge, say $S$
, having slope $\frac {v_2(b)}{5}$
. The residual polynomial of $f(x)$
with respect to $(x,\,S)$
is linear. So $2A_K=\mathfrak {p}^5$
, where the residual degree of $\mathfrak { p}$
is $1$
. Hence, $2\nmid i(K)$
.
Note that when $a\equiv 0(2)$
, $b\equiv 0(2)$
and $5v_2(a)<4v_2(b)$
, then $v_2(a)\in \{1,\,2,\,3\}$
and $v_2(b)\ge 2$
by (1.2).
Case A4: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)\in \{1,\,3\}$
. In this case, $f(x)\equiv x^5(2)$
. Set $\phi (x)=x$
. The $\phi$
-Newton polygon of $f(x)$
has two edges. The first edge, say $S_1$
, is the line segment joining the point $(0,\,0)$
and $(4,\,v_2(a))$
with slope $\frac {v_2(a)}{4}$
and the second edge, say $S_2$
, join the point $(4,\,v_2(a))$
to $(5,\, v_2(b))$
. So for each $i=1,\,2$
, the residual polynomial of $f(x)$
with respect to $(\phi,\,S_i)$
is linear. Therefore, $f(x)$
is $2$
-regular. Thus, Theorem 2.6 is applicable. Using this theorem, we have $2A_K=\mathfrak { p_1^4}\mathfrak { p_2}$
, where the residual degree of prime ideals $\mathfrak { p_1}$
and $\mathfrak { p_2}$
is $1$
. Since there are only two monic irreducible polynomials of degree $1$
over $\mathbb {F}_2$
, by Lemma 2.1 we see that $2\nmid i(K)$
.
One can easily check that when $v_2(a)=2$
, then $5v_2(a)<4v_2(b)$
implies $v_2(b)\ge 3$
.
Case A5: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)\in \{3,\,4\}$
, $a_2\equiv 3(4)$
. In this case, $f(x)\equiv x^5(2)$
. Set $\phi (x)=x$
. The $\phi$
-Newton polygon of $f(x)$
has two edges. The first edge, say $S_1$
, is the line segment joining the points $(0,\,0)$
and $(4,\,2)$
with slope $\lambda _1=\frac {1}{2}$
. The second edge, say $S_2$
, join the point $(4,\,2)$
to $(5,\,v_2(b))$
and has slope $\lambda _2=v_2(b)-2$
. The residual polynomial of $f(x)$
with respect to $(\phi,\,S_1)$
is $(Y+\bar {1})^2$
and with respect to $(\phi,\,S_2)$
is $Y+\bar {1}$
. The edge $S_2$
provides one prime ideal of $A_K$
, say $\mathfrak { p_1}$
, with residual degree $1$
. Set $\psi (Y)=Y+\bar {1}$
.
Consider the numerical invariants $l=1,\,~e=2$
and $f=\deg \psi =1$
, where $\lambda _1 = \frac {l}{e}$
with $\gcd (l,\,e) = 1$
. In this case, $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak a$
is an ideal of $A_K$
. All prime ideals dividing $\mathfrak a$
has ramification index a multiple of $e=2$
and residual degree a multiple of $f=1$
. Since $e>1$
, so we analyse second-order Newton polygons. Keeping in mind Lemma 2.1, $2$
divides $i(K)$
if and only if the key polynomial, say $\varPhi$
, attached with the slope $\lambda _1$
provides two prime ideals of residual degree $1$
each. Recall that the second-order valuation on ${\mathbf {Q}}_2(x)$
is defined as $V(\displaystyle \sum \nolimits _{i\geq 0}^{}a_ix^i)=e\cdot \min \{v_2(a_i)+i\lambda _1\}$
. In particular, $V(x)=l=1$
, $V(2)=e=2$
. Since $v_2(a)=2$
and $v_2(b)\in \{3,\,4\}$
, so we can write
where $m\in \{0,\,1\}$
. Take $\varPhi (x)=x^2-2$
. One can easily check that $\varPhi (x)$
is a key polynomial attached with the slope $\lambda _1$
. Note that $V(\varPhi (x))=2$
. The $\varPhi$
-expansion of $f(x)$
is
The $V$
-Newton polygon of $f(x)$
of second order is the lower convex hull of the points of the set $T=\{(0,\,5),\,(1,\,7),\,(2,\, \min \{2v_2(a_2+1)+5,\,2m+6\})\}$
. The $V$
-Newton polygon of $f(x)$
of second order has a single edge of a positive slope $\frac {2m+1}{2}(=\frac {l_1}{e_1})$
joining the points $(0,\,5)$
and $(2,\,2m+6)$
. The residual polynomial, say $\psi _1(Y)$
, attached to this edge is linear. At this stage the numerical invariants are $l_1=2m+1$
, $e_1=2$
and $f_1=\deg \psi _1=1$
. Hence, $2A_K=\mathfrak {p}_1\mathfrak {p_2^4}$
, where the residual degree of the prime ideal $\mathfrak p_2$
is $ff_1=1$
. So $2\nmid i(K)$
.
Case A6: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)\ge 5$
, $a_2\equiv 3(8)$
. With the notation as in Case A5, it can be seen that $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak p_1$
is a prime ideal of residual degree $1$
and $\mathfrak a$
is an ideal of $A_K$
. Let key polynomial $\varPhi (x)$
and valuation $V$
be the same as in the Case A5. The $\varPhi$
-expansion of $f(x)$
is
The $V$
-Newton polygon of $f(x)$
of second order being the lower convex hull of the points $(0,\,5)$
, $(1,\,7)$
, $(2,\,9)$
has a single edge joining $(0,\,~5)$
and $(2,\,~9)$
with a lattice point $(1,\,~7)$
lying on it. The residual polynomial corresponding to this edge is $Y^2+Y+\bar 1\in \mathbb {F}_2[Y]$
. So $2A_K=\mathfrak {p}_1\mathfrak {p_2^2}$
, where the residual degree of $\mathfrak p_2$
is $2$
. Hence, $2\nmid i(K)$
.
Case A7: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)\ge 5$
, $a_2\equiv 7(8)$
. Proceeding as in Case A5, one can check that $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak p_1$
is a prime ideal of residual degree $1$
and $\mathfrak a$
is an ideal of $A_K$
. Take $\varPhi (x)$
and $V$
same as in Case A5. Keeping in mind (3.3), we observe that the $V$
-Newton polygon of $f(x)$
of second order being the lower convex hull of the points of the set $T=\{(0,\,5),\,(1,\,7),\,(2,\, \min \{2v_2(a_2+1)+5,\,2v_2(b)\})\}$
has two edges of positive slope. The residual polynomial attached to each edge is linear. So $2A_K=\mathfrak { p_1}\mathfrak {p_2^2}\mathfrak {p_3^2}$
, where the residual degree of $\mathfrak p_i$
is $1$
, for each $i=2,\,3$
. Hence, in this case, $2\mid i(K)$
.
Note that if $a_2\equiv 1(4)$
and $v_2(b)\ge 4$
, then the $V$
-Newton polygon of $f(x)$
is the lower convex hull of the points $(0,\,5),\,(1,\,7)$
and $(2,\,7)$
. In this case, the residual polynomial with respect to $x^2-2$
is not square-free, so we use another key polynomial $\varPhi (x)=x^2-2x-2$
.
Case A8: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)\in \{3,\,4\}$
, $a_2\equiv 1(4)$
. Let $\phi (x)$
, $\lambda _1$
, $\psi (Y)$
be same as in Case A5. Arguing as in Case A5, it is easy to verify that $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak p_1$
is a prime ideal with residual degree $1$
and $\mathfrak a$
is an ideal of $A_K$
. Let $\varPhi (x)=x^2-2x-2$
. Then $\varPhi (x)$
works as a key polynomial attached with slope $\lambda _1$
. The data $(x,\,\lambda _1,\, \psi (Y))$
define a valuation $V$
of second order given in (2.1), such that $V(x)=1$
, $V(2)=2$
and $V(\varPhi (x))=2$
. The $\varPhi$
-expansion of $f(x)$
is given by
where $m\in \{0,\,1\}$
. The $V$
-Newton polygon of $f(x)$
of second order being the lower convex hull of the points $(0,\,5)$
, $(1,\,8)$
, $(2,\,2m+6)$
has a single edge of positive slope. The residual polynomial attached to this edge is linear. Hence, $2A_K=\mathfrak {p_1}\mathfrak {p_2^4}$
, where the residual degree of $\mathfrak {p_2}$
is $1$
. So $2\nmid i(K)$
.
Case A9: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)=5$
, $a_2\equiv 5(16)$
. Proceeding as in Case A5, we see that, $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak p_1$
is a prime ideal of residual degree $1$
and $\mathfrak a$
is an ideal of $A_K$
. Let $\varPhi (x)$
and $V$
be same as in Case A8. The $\varPhi$
-expansion of $f(x)$
is given by
The $V$
-Newton polygon of $f(x)$
of second order is the lower convex hull of the points of the set $T=\{(0,\,5),\,(1,\,8),\,(2,\,\min \{5+2v_2(a_2+11),\,10+2v_2(1+b_2)\})\}$
. The $V$
-Newton polygon of $f(x)$
of second order has two edges of positive slope and the residual polynomial attached to each edge is linear. Hence, $2A_K=\mathfrak {p_1}\mathfrak {p_2^2}\mathfrak {p_3^2}$
, where the residual degree of $\mathfrak {p}_i$
is $1$
, for each $i=2,\,3$
. Thus, $2\mid i(K)$
.
Case A10: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)=5$
, $a_2\equiv 13(16)$
. Arguing as in Case A5, we see that $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak p_1$
is a prime ideal of residual degree $1$
and $\mathfrak a$
is an ideal of $A_K$
. Let $\varPhi (x)$
and $V$
be same as in Case A8. Keeping in mind (3.5), one can easily check that the $V$
-Newton polygon of $f(x)$
of second order being the lower convex hull of the points $(0,\,5)$
, $(1,\,8)$
, $(2,\,11)$
has a single edge of the positive slope and the residual polynomial attached to this edge is $Y^2+Y+\bar 1\in \mathbb {F}_2[Y]$
. So $2A_K=\mathfrak {p_1}\mathfrak {p_2^2}$
, where the residual degree of $\mathfrak {p_2}$
is $2$
. Hence, $2\nmid i(K)$
.
Case A11: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)\ge 6$
, $a_2\equiv 5(8)$
. Proceeding as in Case A5, one can verify that $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak p_1$
is a prime ideal with residual degree $1$
and $\mathfrak a$
is an ideal of $A_K$
. Take $\varPhi (x)$
and $V$
same as in Case A8. Using the $\varPhi (x)$
expansion of $f(x)$
given in (3.5), we see that the $V$
-Newton polygon of $f(x)$
of second order being the lower convex hull of the points $(0,\,5)$
, $(1,\,8)$
, $(2,\,10)$
has a single edge of positive slope. The residual polynomial attached to this edge is linear. So $2A_K=\mathfrak {p_1}\mathfrak {p_2^4}$
, where the residual degree of $\mathfrak { p_2}$
is $1$
. Thus, $2\nmid i(K)$
.
Keeping in mind (3.5), it is easy to observe that when $a_2\equiv 1(8)$
and $m\ge 2$
, then the $V$
-Newton polygon of $f(x)$
of second order being the lower convex hull of the points $(0,\,5),\,(1,\,8),\,(2,\,9)$
has a single edge of positive slope. The residual polynomial attached with this edge is not square-free. So we shall use a different key polynomial.
Case A12: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)=5$
, $a_2\equiv 1(8)$
. Here $f(x)\equiv x^5(2)$
. Set $\phi (x)=x$
. Arguing as in Case A5, we have $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak p_1$
is a prime ideal of residual degree $1$
and $\mathfrak a$
is an ideal of $A_K$
. Let $\varPhi (x)=x^2-2x+2$
. As in (2.1), we can define valuation $V$
of second order such that $V(x)=1$
, $V(2)=2$
and $V(\varPhi (x))=2$
. The $\varPhi$
-expansion of $f(x)$
is given by
Keeping in mind the above expansion of $f(x)$
, we see that the $V$
-Newton polygon of $f(x)$
of second order has a single edge of positive slope. The residual polynomial attached with this edge is linear. So $2A_K=\mathfrak {p_1}\mathfrak {p_2^4}$
, where the residual degree of $\mathfrak { p_2}$
is $1$
. Thus, $2\nmid i(K)$
.
Case A13: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)\ge 6$
, $a_2\equiv 1(16)$
. Proceeding as in Case A5, we observe that $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak p_1$
is a prime ideal of residual degree $1$
and $\mathfrak a$
is an ideal of $A_K$
. Take $\varPhi (x)$
and $V$
same as in Case A12. Keeping in mind (3.6), we see that the $V$
-Newton polygon of $f(x)$
being the lower convex hull of the points of the set $T=\{(0,\,5),\,(1,\,8),\,(2,\,\min \{5+2v_2(a_2-1),\,2v_2(b)\})\}$
has two edges of positive slope. The residual polynomial attached to each edge is linear. So $2A_K=\mathfrak {p_1}\mathfrak {p_2^2}\mathfrak {p_3^2}$
, where the residual degree of each prime ideal $\mathfrak {p_i}$
for $i=2,\,3$
is $1$
. Thus, $2\mid i(K)$
.
Case A14: $a\equiv 0(2)$
, $b\equiv 0(2)$
, $5v_2(a)<4v_2(b)$
, $v_2(a)=2$
, $v_2(b)\ge 6$
, $a_2\equiv 9(16)$
. Arguing as in Case A5, we see that $2A_K=\mathfrak p_1\mathfrak a$
, where $\mathfrak p_1$
is a prime ideal of residual degree $1$
and $\mathfrak a$
is an ideal of $A_K$
. Let $\varPhi (x)$
and $V$
be same as in Case A12. Using the $\varPhi (x)$
expansion of $f(x)$
given in (3.6), we see that the $V$
-Newton polygon of $f(x)$
being the lower convex hull of the points $(0,\,5)$
, $(1,\,8)$
, $(2,\,11)$
has one edge of positive slope. In this case, $2A_K=\mathfrak {p_1}\mathfrak {p_2^2}$
, where the residual degree of $\mathfrak {p_2}$
is $2$
. So $2\nmid i(K)$
.
Case A15: $a\equiv 1(4)$
, $b\equiv 2(4)$
. In this case, $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x+1$
. The $\phi _1$
-Newton polygon of $f$
has a single edge, say $S$
, of positive slope joining the points $(4,\, 0)$
and $(5,\,1)$
. The residual polynomial of $f(x)$
with respect to $(x,\,S)$
is linear. Thus, $\phi _1$
provides one prime ideal of residual degree $1$
. The $\phi _2$
-expansion of $f(x)$
can be written as
The $\phi _2$
-Newton polygon of $f(x)$
is the lower convex hull of the points of the set $T=\{(0,\,0),\, (1,\,0),\, (2,\,1) ,\,(3,\,1),\, (4,\,1),\, (5,\,v_2(b-a-1))\}$
. By hypothesis, $v_2(b-a-1)\ge 2$
. Therefore, $\phi _2$
-Newton polygon of $f(x)$
has two edges of positive slope. The first edge, say $S_1,$
is the line segment joining the point $(1,\, 0)$
to $(4,\, 1)$
and the second edge, say $S_2,$
join the point $(4,\, 1)$
to $(5,\, v_2(b-a-1))$
. For each $i=1,\,2$
, the residual polynomial of $f(x)$
with respect to $(\phi _2,\,S_i)$
is linear. So $\phi _2$
provides two prime ideals, say $\mathfrak { p_2}$
and $\mathfrak { p_3}$
, of residual degree $1$
each. Using Theorem 2.6, we have $2A_K=\mathfrak { p_1}\mathfrak { p_2^3}\mathfrak { p_3}$
. Hence, $2\mid i(K)$
.
Case A16: $a\equiv 1(4)$
, $b\equiv 0(4)$
. Here $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x+1$
. Proceeding as in Case A15, it is easy to verify that $\phi _1$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
and $\phi _2$
provides one prime ideal, say $\mathfrak { p_2}$
, of residual degree $1$
. In this case, we have $2A_K=\mathfrak { p_1}\mathfrak { p_2^4}$
. Hence, $2\nmid i(K)$
.
Case A17: $a\equiv 3(8)$
, $b\equiv 2(4)$
. We have $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x+1$
. Arguing as in Case A15, we see that $\phi _1$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
and $\phi _2$
gives one prime ideal, say $\mathfrak { p_2}$
, of residual degree $1$
. Here $2A_K=\mathfrak { p_1}\mathfrak { p_2^4}$
. So $2\nmid i(K)$
in this case.
Case A18: $a\equiv 3(8)$
, $b\equiv 4(8)$
, $v_2(D)$
is even. In this case, $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x+1$
. Arguing as in Case A15, we see that $\phi _1$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
. Keeping in mind (3.7) and the hypothesis, we note that the number of edges of the $(x+1)$
-Newton polygon of $f(x)$
is not specific. So we choose a rational number $\beta$
of zero valuation such that $f(x)$
is regular with respect to $x-\beta$
. In this case, one can easily check that $v_2(D)\geq 12$
. Define $\beta =\frac {-5b}{4a}$
. Set $\phi _2(x)=x-\beta$
. The $\phi _2$
-expansion of $f(x)$
is
Keeping in mind (1.3), it can be verified that
Since $v_2(a)=0$
and $v_2(b)=2$
, so $v_2(f(\beta ))=v_2(f'(\beta ))=v_2(D)-8$
. The $\phi _2$
-Newton polygon of $f(x)$
is the lower convex hull of the points $(0,\,0),\,(1,\,0),\,(2,\,1),\,(3,\,1), (4,\,v_2(D)-8)$
and $(5,\,v_2(D)-8)$
. As $v_2(D)\ge 12$
and even, therefore, $v_2(D)-8\ge 4$
. Thus, the $\phi _2$
- Newton polygon of $f(x)$
has two edges of positive slope. The first edge, say $S_1$
, is the line segment joining the point $(1,\,0)$
with $(3,\,1)$
and has slope $\frac {1}{2}$
. The second edge, say $S_2$
, is the line segment joining the point $(3,\,1)$
with $(5,\,v_2(D)-8)$
and has slope $\frac {v_2(D)-9}{2}$
. For each $i=1,\,2$
, the residual polynomial of $f(x)$
with respect to $(\phi _2,\,S_i)$
is linear. Thus $2A_K= \mathfrak { p_1}\mathfrak { p_2^2} \mathfrak { p_3^2}$
, where for each $i=1,\,2,\,3$
, the residual degree of $\mathfrak { p_i}$
is $1$
. Hence, $2\mid i(K)$
.
Note that if $a\equiv 3(8)$
, $b\equiv 4(8)$
and $v_2(D)$
is odd, then the residual polynomial of $f(x)$
with respect to $(x-\beta,\,S_2)$
is not square-free and $f(x)$
is not $(x-\beta )$
-regular. So we choose another rational number $\delta$
such that $f(x)$
is $(x-\delta )$
-regular.
In what follows, when $a\equiv 3(8)$
, $b\equiv 4(8)$
and $v_2(D)$
is odd, then $\delta$
will stand for $2^u-\frac {5b_2}{a}$
, where $u=\frac {v_2(D)-9}{2}$
. It can be verified that $v_2(f(\delta ))\ge 2u+2$
and $v_2(f'(\delta ))=u+2$
.
Case A19: $a\equiv 3(8)$
, $b\equiv 4(8)$
, $v_2(D)\neq 11$
is odd, $v_2(f(\delta ))=2u+2$
. In this case, $f(x)\equiv x(x+1)^4(2)$
and $v_2(D)\ge 13$
. Set $\phi _1(x)=x$
. Then, $\phi _1$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
. Arguing as in Case A18, we see that $f(x)$
is not $2$
-regular always. Consider $\phi _2(x)=x-\delta$
. The $\phi _2(x)$
-expansion of $f(x)$
is
The $\phi _2$
-Newton polygon of $f(x)$
is the lower convex hull of the points $(0,\,0)$
, $(1,\,0)$
, $(2,\,1)$
, $(3,\,1)$
, $(4,\,v_2(f'(\delta ))$
and $(5,\,v_2(f(\delta ))$
. Since $v_2(f(\delta ))=2u+2$
and $v_2(f'(\delta ))=u+2$
, the $\phi _2$
-Newton polygon of $f(x)$
has two edges, say $S_1$
and $S_2$
, of positive slope. For each $i=1,\,2$
, the residual polynomial of $f(x)$
with respect to $(\phi _2,\,S_i)$
is linear. Thus, $\phi _2(x)$
provides two prime ideals, say $\mathfrak { p_2}$
and $\mathfrak { p_3}$
, of residual degree $1$
each. Therefore, $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}\mathfrak { p_3^2}$
. Hence, $2\mid i(K)$
.
Case A20: $a\equiv 3(8)$
, $b\equiv 4(8)$
, $v_2(D)\neq 11$
is odd, $v_2(f(\delta ))=2u+3$
. Here $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
. Then $\phi _1$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
. Take $\phi _2(x)=x-\delta$
. Arguing as in Case A19, it can be checked that $\phi _2$
-Newton polygon of $f(x)$
being the lower convex hull of the points $(0,\,0)$
, $(1,\,0)$
, $(2,\,1)$
, $(3,\,1)$
, $(4,\,u+2)$
and $(5,\,2u+3)$
has two edges of positive slope. The first edge, say $S_1$
, is the line segment joining the points $(1,\,0)$
and $(3,\,1)$
. The second edge, say $S_2$
, is the line segment joining the points $(3,\,1)$
and $(5,\,2u+3)$
with a lattice point $(4,\,u+2)$
lying on it. The residual polynomial of $f(x)$
with respect to $(\phi _2,\,S_1$
) and $(\phi _2,\,S_2)$
is $Y+\bar {1}$
and $Y^2+Y+\bar {1}$
respectively. Thus, $\phi _2$
provides two prime ideals, say $\mathfrak { p_2}$
and $\mathfrak { p_3}$
, of residual degree $1$
and $2$
respectively. So $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}\mathfrak { p_3}$
and $2\nmid i(K)$
.
Case A21: $a\equiv 3(8)$
, $b\equiv 4(8)$
, $v_2(D)\neq 11$
is odd, $v_2(f(\delta ))\ge 2u+4$
. Here $f(x)\equiv x(x+1)^4(2)$
. Let $\phi _1(x)=x$
. Clearly, $\phi _1$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
. Let $\phi _2=x-\delta$
. Keeping in mind (3.10), it is easy to verify that the $\phi _2$
-Newton polygon of $f(x)$
has three edges of positive slope. The first edge, say $S_1$
, is the line segment joining the points $(1,\,0)$
and $(3,\,1)$
. The second edge, say $S_2$
, is the line segment joining the points $(3,\,1)$
and $(4,\,u+2)$
. The third edge, say $S_3$
, is the line segment joining the points $(4,\,u+2)$
and $(5,\,v_2(f(\delta )))$
. For each $i=1,\,2,\,3$
, the residual polynomial of $f(x)$
with respect to $(\phi _2,\,S_i)$
is linear. So $\phi _2$
provides three prime ideals, say $\mathfrak { p_2}$
, $\mathfrak { p_3}$
and $\mathfrak { p_4}$
, of residual degree $1$
each. Thus, $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}\mathfrak { p_3}\mathfrak { p_4}$
. Hence, $2\mid i(K)$
.
Note that when $a\equiv 3(8)$
, $b\equiv 4(8)$
and $v_2(D)=11$
, then $u=\frac {v_2(D)-9}{2}=1$
. So $v_2(f(\delta ))\ge 4$
and $v_2(f'(\delta ))=3$
.
Case A22: $a\equiv 3(8)$
, $b\equiv 4(8)$
, $v_2(D)=11$
, $v_2(f(\delta ))= 4$
. We have $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x-\delta$
. Clearly $\phi _1$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
. Keeping in mind (3.10), we see that the $\phi _2$
-Newton polygon of $f(x)$
has two edges, say $S_1$
and $S_2$
, of positive slope. The residual polynomial of $f(x)$
with respect to $(\phi _2,\,S_i)$
, for $i=1,\,2$
, is linear. Therefore $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}\mathfrak { p_3^2}$
, where the residual degree of each prime ideal $\mathfrak { p_i}$
is $1$
, for each $i=2,\,3$
. Hence, $2\mid i(K)$
.
Case A23: $a\equiv 3(8)$
, $b\equiv 4(8)$
, $v_2(D)=11$
, $v_2(f(\delta ))= 5$
. In this case, $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x-\delta$
. Note that $\phi _1$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
. In view of (3.10), we see that the $\phi _2$
-Newton polygon of $f(x)$
has two edges, say $S_1$
and $S_2$
, of positive slope. The residual polynomial of $f(x)$
with respect to $(\phi _2,\,S_1)$
is linear. The residual polynomial of $f(x)$
with respect to $(\phi _2,\,S_2)$
is $Y^2+Y+\bar {1}$
. Therefore, $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}\mathfrak { p_3}$
, where the residual degree of $\mathfrak { p_2}$
and $\mathfrak { p_3}$
is $1$
and $2$
respectively. Hence, $2\nmid i(K)$
.
Case A24: $a\equiv 3(8)$
, $b\equiv 4(8)$
, $v_2(D)=11$
, $v_2(f(\delta ))\ge 6$
. In this case, $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x-\delta$
. Here $\phi _1$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
. In view of (3.10), we observe that the $\phi _2$
-Newton polygon of $f(x)$
being the lower convex hull of the points $(0,\,0)$
, $(1,\,0)$
, $(2,\,1)$
, $(3,\,1)$
, $(4,\,3)$
, $(5,\,v_2(f(\delta )))$
has three edges, say $S_1$
, $S_2$
and $S_3$
, of positive slope. The residual polynomial of $f(x)$
with respect to $(\phi _2,\,S_i)$
, for $i=1,\,2,\,3$
, is linear. So $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}\mathfrak { p_3}\mathfrak { p_4}$
, where the residual degree of prime ideal $\mathfrak { p_i}$
is $1$
, for each $i=2,\,3,\,4$
. Thus, $2\mid i(K)$
.
Case A25: $a\equiv 3(8)$
, $b\equiv 0(8)$
. Here $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x+1$
. The $\phi _1$
-Newton polygon of $f(x)$
has a single edge of positive slope, say $S$
, joining the points $(4,\,~0)$
and $(5,\,~v_2(b))$
. So the residual polynomial attached to $(\phi _1,\,~S)$
is linear. Keeping in mind the $\phi _2$
expansion of $f(x)$
given in (3.7), one can see that the $\phi _2$
-Newton polygon of $f(x)$
has a single edge, say $S'$
, joining the points $(1,\,~ 0)$
and $(5,\, ~2)$
with point $(3,\,~ 1)$
lying on it. The polynomial associated to $f(x)$
with respect to ($\phi _2,\,~S')$
is $Y^2+Y+\overline {1}\in \mathbb {F}_2[Y]$
having no repeated roots. Thus, $f(x)$
is $2$
-regular. Using Theorem 2.6, one can check that $\phi _1(x)$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
and $\phi _2(x)$
provides one prime ideal, say $\mathfrak { p_2}$
, of residual degree $2$
. Therefore, $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}$
and $2\nmid i(K)$
.
Case A26: $a\equiv 7(8)$
, $b\equiv 2(4)$
. In this case, $f(x)\equiv x(x+1)^4(2)$
. Let $\phi _1(x)=x$
and $\phi _2(x)=x+1$
. Arguing as in Case A25, it can be easily seen that $2A_K=\mathfrak { p_1}\mathfrak { p_2^4}$
, where the residual degree of $\mathfrak { p_1}$
and $\mathfrak { p_2}$
is $1$
. Hence, $2\nmid i(K)$
.
Case A27: $a\equiv 7(8)$
, $b\equiv 4(8)$
. Here $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x+1$
. Proceeding as in Case A25, we can check that $\phi _1(x)$
gives one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
and $\phi _2(x)$
provides one prime ideal, say $\mathfrak { p_2}$
, of residual degree $2$
. So $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}$
. Hence, $2\nmid i(K)$
.
Case A28: $a\equiv 7(8)$
, $b\equiv 0(8)$
, $v_2(b-a-1)=3$
. In this case, $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x+1$
. Arguing as in Case A25, it can be verified that $\phi _1(x)$
provides one prime ideal, say $\mathfrak { p_1}$
, of residual degree $1$
and $\phi _2$
provides two prime ideals say, $\mathfrak { p_2}$
and $\mathfrak { p_3}$
, of residual degree $1$
and $2$
, respectively. Thus, $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}\mathfrak { p_3}$
. Hence, $2\nmid i(K)$
.
Case A29: $a\equiv 7(8)$
, $b\equiv 0(8)$
, $v_2(b-a-1)\ge 4$
. In this case, $f(x)\equiv x(x+1)^4(2)$
. Set $\phi _1(x)=x$
and $\phi _2(x)=x+1$
. Arguing as in Case A25, we have $2A_K=\mathfrak { p_1}\mathfrak { p_2^2}\mathfrak { p_3}\mathfrak { p_4}$
, where the residual degree of each $\mathfrak {p_i},\,~1\leq i\leq 4$
is $1$
. Hence, $2\mid i(K)$
. This completes the proof of the theorem.
Acknowledgements
The first author is thankful to SERB grant SRG/2021/000393. The second author is grateful to the Council of Scientific and Industrial Research, New Delhi for providing financial support in the form of Senior Research Fellowship through Grant No. $09/135(0878)/2019$
-EMR-$1$
. The third author is grateful to the University Grants Commission, New Delhi for providing financial support in the form of Junior Research Fellowship through Ref No.1129/(CSIR-NET JUNE 2019).
