NONMONOGENITY OF NUMBER FIELDS DEFINED BY TRUNCATED EXPONENTIAL POLYNOMIALS

Abstract

Let p be a prime number. Let $n\geq 2$ be an integer given by $n = p^{m_1} + p^{m_2} + \cdots + p^{m_r}$, where $0\leq m_1 < m_2 < \cdots < m_r$ are integers. Let $a_0, a_1, \ldots , a_{n-1}$ be integers not divisible by p. Let $K = \mathbb Q(\theta )$ be an algebraic number field with $\theta \in {\mathbb C}$ a root of an irreducible polynomial $f(x) = \sum _{i=0}^{n-1}a_i{x^i}/{i!} + {x^n}/{n!}$ over the field $\mathbb Q$ of rationals. We prove that p divides the common index divisor of K if and only if $r>p$. In particular, if $r>p$, then K is always nonmonogenic. As an application, we show that if $n \geq 3$ is an odd integer such that $n-1\neq 2^s$ for $s\in {\mathbb Z}$ and K is a number field generated by a root of a truncated exponential Taylor polynomial of degree n, then K is always nonmonogenic.

1 Introduction

Let $K = \mathbb Q(\theta )$ be an algebraic number field with $\theta $ in the ring ${\mathbb Z}_K$ of algebraic integers of K. Let $f(x)$ be the minimal polynomial of $\theta $ having degree n over the field $\mathbb Q$ of rational numbers. It is well known that ${\mathbb Z}_K$ is a free abelian group of rank n. A number field K is said to be monogenic if there exists some $\beta \in {\mathbb Z}_K$ such that ${\mathbb Z}_K = {\mathbb Z}[\beta ].$ In this case, $\{1, \beta ,\ldots ,\beta ^{n-1}\}$ 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 nonmonogenic. Quadratic and cyclotomic fields are monogenic. In algebraic number theory, it is important to know whether a number field is monogenic or not. The first example of a nonmonogenic number field was given by Dedekind in 1878; he proved that the cubic field $\mathbb Q(\eta )$ is not monogenic when $\eta $ is a root of the polynomial $x^3-x^2-2x-8$ (see [15, page 64]). The problems of testing the monogenity of number fields and constructing power integral bases have been intensively studied (see [7] for an overview of the latest developments).

Throughout this paper, $\mathop {\mathrm {ind}} \theta $ denotes the index of the subgroup ${\mathbb Z}[\theta ]$ in ${\mathbb Z}_K $ and $i(K)$ stands for the index of the field K defined by $ i(K) = \gcd \{\mathop {\mathrm {ind}} \alpha \mid {\text {} K=\mathbb Q(\alpha ) {\text { and }} \alpha \in {\mathbb Z}_K} \}.$ 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 nonmonogenic. However, there exist nonmonogenic number fields having $i(K)=1$ , for example, $K=\mathbb Q(\sqrt [3]{175})$ is not monogenic and has $i(K)=1$ . Nakahara [14] studied the index of noncyclic but abelian biquadratic number fields. Gaál et al. [8] characterised the field indices of biquadratic number fields having Galois group $V_4$ . Ahmad et al. [1, 2] proved that for a square free integer m not congruent to $\pm 1\bmod 9$ , a pure field $\mathbb Q(m^{1/6})$ having degree $6$ over $\mathbb Q$ is monogenic when $m\equiv 2$ or $3\bmod 4$ and it is nonmonogenic when $m\equiv 1\bmod 4$ . Gaál and Remete [9] studied monogenity of number fields of the type $\mathbb Q(m^{1/n})$ where $3\leq n\leq 9$ and m is square free. Gaál [6] and Jakhar and Kaur [10] studied monogenity of number fields defined by some sextic irreducible trinomials.

Let $a_0, \ldots , a_{n-1}$ be integers. It is known that the polynomial

(1.1) $$ \begin{align} f(x) = a_0 + a_1x+ a_2\frac{x^2}{2!} + \cdots +a_{n-1}\frac{x^{n-1}}{(n-1)!} + \frac{x^n}{n!} \end{align} $$

of degree n is irreducible over $\mathbb Q$ if one of the following conditions is satisfied:

1. (1) $\gcd (a_0, n!) = 1$ (see [5, 16]);

2. (2) $\gcd (a_0a_1\cdots a_{n-1}, n) = 1$ (see [11, Theorem 1.2]).

Let p be a prime number. Let $n\geq 2$ be an integer given by $n = p^{m_1} + p^{m_2} + \cdots + p^{m_r}$ , where $0\leq m_1 < m_2 < \cdots < m_r$ are integers. Let $K = \mathbb Q(\theta )$ with $\theta $ a root of an irreducible polynomial $f(x)$ over $\mathbb Q$ , where $f(x)$ is given by (1.1) and $a_0, \ldots , a_{n-1}$ are integers not divisible by p. We provide necessary and sufficient conditions so that $p \mid i(K)$ for $n\geq 2.$ As an application, we give a family of number fields which are nonmonogenic. Precisely stated, we prove the following result.

Theorem 1.1. Let p be a prime number. Let $n\geq 2$ be an integer given by ${n = p^{m_1} + p^{m_2} + \cdots + p^{m_r}}$ , where $0\leq m_1 < m_2 < \cdots < m_r$ are integers. Let $a_0, a_1, \ldots , a_{n-1}$ be integers not divisible by p. Let $K = \mathbb Q(\theta )$ be an algebraic number field with $\theta $ a root of an irreducible polynomial $f(x) = x^n+ n! \sum _{i=0}^{n-1}a_i x^i/i!$ over $\mathbb Q$ . Then:

1. (1) $p{\mathbb Z}_K = \wp _1^{e_1}\cdots \wp _r^{e_r},$ where the $\wp _i$ are distinct prime ideals lying above the prime p with index of ramification $e_i = p^{m_i}$ and residual degree one for each i;

2. (2) p divides $i(K)$ if and only if $r>p$ .

In particular, if $r>p$ , then K is always nonmonogenic.

The following corollary is an immediate consequence of the theorem.

Corollary 1.2. Let $n\geq 2$ be an integer with $2$ -adic expansion $n = 2^{m_1} + 2^{m_2} + \cdots + 2^{m_r}$ , where $0\leq m_1 < m_2 < \cdots < m_r$ . Let $a_0, a_1, \ldots , a_{n-1}$ be odd integers. Let $K = \mathbb Q(\theta )$ be an algebraic number field with $\theta $ a root of an irreducible polynomial $f(x) = x^n+ n! \sum _{i=0}^{n-1}a_i x^i/i!$ over $\mathbb Q$ . If $r> 2$ , then K is nonmonogenic.

As an application of this corollary, we obtain the following result.

Corollary 1.3. Let $n\geq 2$ be an integer with $2$ -adic expansion $n = 2^{m_1} + 2^{m_2} + \cdots + 2^{m_r}$ , where $0\leq m_1 < m_2 < \cdots < m_r$ . Let $K = \mathbb Q(\theta )$ be an algebraic number field with $\theta $ a root of a truncated exponential Taylor polynomial $f(x) = 1 + x + x^2/2! + \cdots + x^n/n!$ . Assume that $r\geq 3$ . Then K is always nonmonogenic.

Example 1.4. This example provides a family of nonmonogenic algebraic number fields. Let $n\geq 3$ be an odd integer such that $n-1 \neq 2^s$ for any $s\in {\boldsymbol N}$ . If $K = \mathbb Q(\theta )$ is an algebraic number field with $\theta \in {\mathbb C}$ a root of $f(x) = \sum _{i=0}^{n}{x^i}/{i!}$ , then K is nonmonogenic by Corollary 1.3.

Remark 1.5. If we take $r < 3$ , then K can be monogenic. For example, consider $n = 3$ , $r = 2$ and $f(x) = x^3 + 3x^2+ 6x + 6$ in Corollary 1.3. It can be easily checked that the discriminant of $f(x)$ is $-2^3\cdot 3^3.$ Let $K = \mathbb Q(\theta )$ with $\theta $ a root of $f(x)$ . Since $f(x)$ is an Eisenstein polynomial with respect to $3$ , in view of a basic result [12, Theorem 2.18], we see that $3\nmid [{\mathbb Z}_K : {\mathbb Z}[\theta ]].$ Further note that $f(x) \equiv x^2(x+1) \pmod 2$ . Hence, using Dedekind’s criterion [12, page 78], it is easy to see that $2\nmid [{\mathbb Z}_K : {\mathbb Z}[\theta ]]$ . Therefore, in view of the formula $D_f = [{\mathbb Z}_K : {\mathbb Z}[\theta ]]^2d_K$ , where $D_f$ denotes the discriminant of the polynomial $f(x)$ and $d_K$ denotes the discriminant of K, it follows that ${\mathbb Z}_K = {\mathbb Z}[\theta ].$ Hence, K is monogenic.

2 Preliminary results

Let $K=\mathbb Q(\theta )$ be an algebraic number field with $\theta $ a root of a monic irreducible polynomial $f(x)$ belonging to ${\mathbb Z}[x]$ . In what follows, ${\mathbb Z}_K$ stands for the ring of algebraic integers of K. For a rational prime p, let ${\mathbb F}_p$ be the finite field with p elements and ${\mathbb Z}_p$ denote the ring of p-adic integers. Throughout the paper, $f(x)\rightarrow \overline {f(x)}$ stands for the canonical homomorphism from ${\mathbb Z}_p[x]$ onto ${\mathbb F}_p[x]$ . For a prime p and a nonzero m belonging to the ring ${\mathbb Z}_p$ of p-adic integers, $v_p(m)$ denotes the highest power of p dividing m. The following lemma will play an important role in the proof of the theorem.

Lemma 2.1 [15, Theorem 4.34].

Let K be an algebraic number field and p be a rational prime. Then p is a prime common index divisor of K if and only if for some positive integer h, the number of distinct prime ideals of ${\mathbb Z}_K$ lying above p having residual degree h is greater than the number of monic irreducible polynomials of degree h in ${\mathbb F}_p[x]$ .

The following simple result will also be used. Its proof is omitted.

Lemma 2.2. Let p be a prime number. If $n = c_0 + c_1p + \cdots + c_rp^r$ is the representation of the positive integer n in base p with $0\leq c_i < p$ for each i, then

$$ \begin{align*}v_p(n!) = \frac{n-(c_0+c_1+\cdots + c_r)}{p-1}.\end{align*} $$

3 A short introduction to prime ideal factorisation based on Newton polygons

In 1894, Hensel developed a powerful approach for finding prime ideals of ${\mathbb Z}_K$ over a rational prime p. He showed that for every prime p, the prime ideals of ${\mathbb Z}_K$ lying above p are in one-to-one correspondence with monic irreducible factors of $f(x)$ in $\mathbb Q_p[x]$ . Newton polygons are very helpful for finding the factors of $f(x)$ in $\mathbb Q_p[x]$ . This is a standard method which is rather technical but efficient to apply. Therefore, we first introduce the notion of Gauss valuation and $\phi $ -Newton polygon, where $\phi (x)$ belonging to ${\mathbb Z}_p[x]$ is a monic polynomial with $\overline {\phi }(x)$ irreducible over ${\mathbb F}_p$ .

Definition 3.1. The Gauss valuation of the field $\mathbb Q_p(x)$ of rational functions in an indeterminate x extends the valuation $v_p$ of $\mathbb Q_p$ and is defined on $\mathbb Q_p[x]$ by

$$ \begin{align*} v_{p,x}(a_0+a_1x+a_2x^2+\cdots+a_sx^s)= \displaystyle\min_{1\leq i\leq s}\{v_p(a_i)\}, \quad a_i\in \mathbb Q_p. \end{align*} $$

Definition 3.2. Let p be a rational prime. Let $\phi (x)\in {\mathbb Z}_p[x]$ be a monic polynomial which is irreducible modulo p and $f(x)\in {\mathbb Z}_p[x]$ be a monic polynomial not divisible by $\phi (x)$ . Let $\sum _{i=0}^{n}a_i(x)\phi (x)^i$ , with $\deg a_i(x)<\deg \phi (x)$ , $a_n(x)\neq 0$ , be the $\phi (x)$ -expansion of $f(x)$ obtained by dividing $f(x)$ 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 $P_i$ to $P_j$ if $a_{n-i}(x)a_{n-j}(x)\neq 0$ . Let $i_1$ be the largest positive index not exceeding n such that

$$ \begin{align*}\mu_{0i_1}=\min\{\mu_{0j}~|~0<j\leq n,~a_{n-j}(x)\neq0\}.\end{align*} $$

If $i_1<n,$ let $i_2$ be the largest index such that $i_1<i_2\leq n$ with

$$ \begin{align*}\mu_{i_1i_2}=\min\{\mu_{i_1j}~|~i_1<j\leq n,~a_{n-j}(x)\neq0\},\end{align*} $$

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 nonnegative as $f(x)$ is a monic polynomial with coefficients in ${\mathbb Z}_p$ .

Definition 3.3. Let $\phi (x) \in {\mathbb Z}_p[x]$ be a monic polynomial which is irreducible modulo a rational prime p having a root $\alpha $ in the algebraic closure $\widetilde {\mathbb Q}_{p}$ of $\mathbb Q_p$ . Let $f(x) \in {\mathbb Z}_p[x]$ be a monic polynomial not divisible by $\phi (x)$ whose $\phi (x)$ -expansion is given by $\phi (x)^n + a_{n-1}(x)\phi (x)^{n-1} + \cdots + a_0(x)$ and such that $\overline {f}(x)$ is a power of $\overline {\phi }(x)$ . Suppose that the $\phi $ -Newton polygon of $f(x)$ with respect to p consists of a single edge, say S, having positive slope ${l}/{e}$ with $l, e$ coprime, that is,

$$ \begin{align*}\min\bigg\{\frac{v_{p,x}(a_{n-i}(x))}{i} \ \bigg|\ 1\leq i\leq n\bigg\} = \frac{v_{p,x}(a_0(x))}{n} = \frac{l}{e},\end{align*} $$

so that n is divisible by e, say $n=et$ , and $v_{p,x}(a_{n-ej}(x)) \geq lj$ with $1\leq j\leq t$ . Thus, the polynomial $b_j(x):={a_{n-ej}(x)}/{p^{lj}}$ has coefficients in ${\mathbb Z}_p$ and $b_j(\alpha )\in {\mathbb Z}_p[\alpha ]$ for $1\leq j \leq t$ . The polynomial $T(Y)$ in the indeterminate Y defined by $T(Y) = Y^t + \sum _{j=1}^{t} \overline {b_j}(\overline {\alpha })Y^{t-j}$ with coefficients in ${\mathbb F}_p[\overline {\alpha }]\cong {{\mathbb F}_p[x]}/{\langle \phi (x)\rangle }$ is called the residual polynomial of $f (x)$ with respect to $(\phi ,S)$ .

The following weaker version of the theorem of the product, originally due to Ore, will be used in the proof of main result (see [4, Theorem 1.5], [13, Theorem 1.1]).

Theorem 3.4. Let $\phi (x) \in {\mathbb Z}_p[x]$ be a monic polynomial which is irreducible modulo a rational prime p having a root $\alpha $ in the algebraic closure $\widetilde {\mathbb Q}_{p}$ of $\mathbb Q_p$ . Let $g(x) \in {\mathbb Z}_p[x]$ be a monic polynomial not divisible by $\phi (x)$ whose $\phi (x)$ -expansion is given by $\phi (x)^n + a_{n-1}(x)\phi (x)^{n-1} + \cdots + a_0(x)$ and such that $\overline {f}(x)$ is a power of $\overline {\phi }(x)$ . Suppose that the $\phi $ -Newton polygon of $g(x)$ with respect to the prime p has k edges $S_1, \ldots , S_k$ having slopes $\lambda _1 < \cdots < \lambda _k$ . Then:

1. (1) $g(x) = g_1(x)\cdots g_k(x)$ , where each $g_i(x) \in {\mathbb Z}_{{p}}[x]$ is a monic polynomial of degree $\ell _i\deg (\phi (x))$ and whose $\phi $ -Newton polygon has a single edge, say $S_i'$ , which is a translate of $S_i$ such that $\ell _i$ is the length of the horizontal projection of $S_i$ ;

2. (2) the residual polynomial $T_i(Y) \in {{\mathbb F}}_{p}[\overline {\alpha }][Y]$ of $g_i(x)$ with respect to ( $\phi ,~S_i'$ ) has degree $\ell _i/e_i$ , where $e_i$ is the smallest positive integer such that $e_i\lambda _i \in {\mathbb Z}.$

The next definition extends the notion of residual polynomial to more general polynomials $f(x)$ .

Definition 3.5. Let $p, \phi (x), \alpha $ be as in Definition 3.3. Let $g(x)\in {\mathbb Z}_p[x]$ 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 Theorem 3.4, we can write $g(x) = g_1(x)\cdots g_k(x)$ , where the $\phi $ -Newton polygon of $g_i(x) \in {\mathbb Z}_{{p}}[x]$ 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 ( $\phi ,~S_i'$ ) as in Definition 3.3. 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 ${\mathbb Z}_p[x]$ is a monic polynomial and $\overline {f}(x) = \overline {\phi }_{1}(x)^{e_1}\cdots \overline {\phi }_r{(x)}^{e_r}$ is its factorisation modulo p into irreducible polynomials with each $\phi _i(x)$ belonging to ${\mathbb Z}_p[x]$ monic and $e_i> 0$ , then by Hensel’s lemma [3, Ch. 4, Section 3], there exist monic polynomials $f_1(x), \ldots , f_r(x)$ belonging to ${\mathbb Z}_{{p}}[x]$ 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 $\phi _1, \ldots , \phi _r$ ) if each $f_i(x)$ is ${p}$ -regular with respect to $\phi _i$ .

We provide a simple example of a p-regular polynomial with respect to any monic polynomial $\phi (x) \in {\mathbb Z}[x]$ which is irreducible modulo a prime p.

Example 3.6. If $p, \phi (x)$ are as above and $g(x) \neq \phi (x)$ belonging to ${\mathbb Z}_p[x]$ is a monic polynomial with $\overline {g}(x) = \overline {\phi }(x)$ , then the $\phi $ -Newton polygon of $g(x)$ with respect to p is a line segment S joining the point $(0,0)$ with $(1,b)$ for some $b>0$ . Consequently, the polynomial associated to $g(x)$ with respect to $(\phi , S)$ is linear and $g(x)$ is p-regular with respect to $\phi $ .

To determine the number of distinct prime ideals of ${\mathbb Z}_K$ lying above a rational prime p, we will use the following theorem which is a weaker version of [13, Theorem 1.2].

Theorem 3.7. Let $L=\mathbb Q(\xi )$ be an algebraic number field with $\xi $ satisfying an irreducible polynomial $g(x)\in {\mathbb Z}[x]$ and p be a rational prime. Let $ \overline {\phi }_{1}(x)^{e_1}\cdots \overline {\phi }_r{(x)}^{e_r}$ be the factorisation 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 ${\mathbb Z}[x]$ monic. Suppose that the $\phi _i$ -Newton polygon of $g(x)$ has $k_i$ edges, say $S_{ij}$ , having slopes $\lambda _{ij}={l_{ij}}/{e_{ij}} $ with $\gcd (l_{ij},~e_{ij})=1$ for $1\leq j\leq k_i$ . If $T_{ij}(Y) = \prod _{s=1}^{s_{ij}}U_{ijs}(Y)$ is the factorisation 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

$$ \begin{align*}p{\mathbb Z}_L=\displaystyle\prod_{i=1}^{r}\displaystyle\prod_{j=1}^{k_i}\displaystyle\prod_{s=1}^{s_{ij}}\mathfrak p_{ijs}^{e_{ij}},\end{align*} $$

where $\mathfrak p_{ijs}$ are distinct prime ideals of ${\mathbb Z}_L$ having residual degree $\deg \phi _i(x)\cdot \deg U_{ijs}(Y).$

4 Proof of Theorem 1.1

Proof. Observe that $p\leq n.$ We first show that x is the only repeated factor of $f(x)$ modulo p. If $p \mid n$ , then clearly $f(x) \equiv x^n \pmod p$ . If $p\nmid n$ , then assume that $j, 0\leq j\leq n-2$ , is the smallest index such that p divides $n-j$ . Keeping in mind that $p\nmid a_i$ , we see that $f(x)$ is congruent to

$$ \begin{align*} & x^n + na_{n-1}x^{n-1} + \cdots + a_{n-j}\frac{n!}{(n-j)!}x^{n-j} \\ &\quad \equiv x^{n-j}\bigg(x^j +na_{n-1}x^{j-1} + \cdots + a_{n-j}\frac{n!}{(n-j)!}\bigg) \bmod p. \end{align*} $$

Note that $p\nmid j$ . Otherwise, if $p \mid j$ , then since $p \mid (n-j)$ , we have $p \mid n$ , which is a contradiction. Hence, the polynomial $x^j +\overline {n}\overline {a}_{n-1}x^{j-1} + \cdots + \overline {a}_{n-j}\overline {{n!}/{(n-j)!}}$ belonging to ${\mathbb Z}/p{\mathbb Z}[x]$ is a separable polynomial. It follows that x is the only repeated factor of $f(x)$ modulo p.

Now we show that $f(x)$ is p-regular with respect to $\phi (x) = x$ . Recall that $p\nmid a_i$ . By the definition of the p-Newton polygon, we see that it will be the polygonal path formed by the lower edges along the convex hull of the points of the set S defined by

$$ \begin{align*}S = \bigg\{\bigg(i, v_p\bigg(\frac{n!}{(n-i)!}\bigg)\bigg) \ \bigg|\ 0\leq i\leq n\bigg\}.\end{align*} $$

By hypothesis, $n = p^{m_1} + p^{m_2} + \cdots + p^{m_r}$ , where $0\leq m_1 < m_2 < \cdots <m_r$ . Let $\ell _i$ denote the integer

$$ \begin{align*} \ell_i = p^{m_1} + \cdots + p^{m_i}, ~~~1\leq i\leq r. \end{align*} $$

Set $\ell _0 = 0.$ As in [5], using Lemma 2.2 and keeping in mind that $v_p(a_i) = 0$ for each i, it can be easily checked that the p-Newton polygon of $f(x)$ consists of r edges, and the ith edge is the line segment having vertices $(\ell _{i-1}, v_p(n!/(n-\ell _{i-1})!))$ and ${(\ell _{i}, v_p(n!/(n-\ell _{i})!))}$ . So by Lemma 2.2, the slope $\lambda _i$ of the ith edge of the p-Newton polygon of $f(x)$ is

$$ \begin{align*} \lambda_i = \frac{-v_p((n-\ell_i)!) + v_p((n-\ell_{i-1})!)}{\ell_i - \ell_{i-1}} = \frac{\ell_i - \ell_{i-1} -1}{(\ell_i - \ell_{i-1})(p-1)} = \frac{p^{m_i} - 1}{p^{m_i}(p-1)}. \end{align*} $$

Observe that $f(x)$ can have an edge with slope zero if and only if $m_1 = 0$ . Also, $m_1$ can be zero only when $p\nmid n$ . Therefore, in view of Hensel’s lemma and Theorem 3.4, we can write $f(x) = g_1(x)\cdots g_r(x)$ , where $g_i(x) \in {\mathbb Z}_p[x]$ has degree $\ell _i - \ell _{i-1} = p^{m_i}$ and the p-Newton polygon of $g_i(x)$ has a single edge, say $S_i$ , with slope $\lambda _i$ . When $\lambda _i> 0$ , the polynomial, say $T_i(y) \in {\mathbb F}_p[y]$ , associated to $g_i(x)$ with respect to $(x, S_i)$ is linear. Hence, $f(x)$ is p-regular with respect to $\phi (x) = x.$ So, by Theorem 3.7,

$$ \begin{align*}p{\mathbb Z}_K = \wp_1^{e_1}\cdots \wp_r^{e_r},\end{align*} $$

where the $\wp _i$ are distinct prime ideals lying above prime p with index of ramification $e_i = p^{m_i}$ and residual degree one for each i. Hence, by Lemma 2.1, $p \mid i(K)$ if and only if $r> p.$ This completes the proof of the theorem.