Some generalizations of the functions $\tau$ and $\tau^{\left(e\right)}$ in algebraic number fields

TL;DR

This paper extends the definitions of the arithmetic functions  and ^{(e)} in algebraic number fields and explores their properties, contributing to the theoretical understanding of these functions in algebraic number theory.

Contribution

The paper introduces generalized versions of  and ^{(e)} in algebraic number fields and analyzes their properties, expanding the theoretical framework of these functions.

Findings

Generalized  and ^{(e)} functions in algebraic number fields.

Derived properties of the generalized functions.

Enhanced understanding of arithmetic functions in algebraic number theory.

Abstract

In this paper, we generalize the arithmetic functions $\tau$ and $\tau^{\left(e\right)}$ in algebraic number fields and we find some properties of these functions.

Equations

$$p{\mathcal{O}}_K=P_1^{e_1}\cdot P_2^{e_2}\cdot ...\cdot P_g^{e_g}.$$

$$\sum _{i=1}^g{e}_i{f}_i=[K:\mathbb{Q}]=n$$

$$p\mathbb{Z}\left[\xi \right]=P_1\cdot P_2\cdot ...\cdot P_g,$$

$${\tau }^{\left(e\right)}\left(I\right)=\sum _{d_I^{\left(e\right)}{|}_{\left(e\right)}I}1.$$

$$\tau \left(I\right)=\left(e_1+1\right)\cdot \left(e_2+1\right)\cdot ...\cdot \left(e_g+1\right)$$

$${\tau }^{\left(e\right)}\left(I\right)=\tau \left(e_1\right)\cdot \tau \left(e_2\right)\cdot ...\cdot \tau \left(e_g\right).$$

$${\tau }^{\left(e\right)}\left(p{\mathcal{O}}_K\right)={\left(\tau \left(e_1\right)\right)}^g.$$

$${\tau }^{\left(e\right)}\left(p{\mathcal{O}}_K\right)\in \{1,2\},$$

$$2^{\omega \left(n\right)}\le {\tau }^{\left(e\right)}\left(n\right)\le 2^{\Omega \left(n\right)},$$

$$2^{\omega \left(I^2\right)}\le {\tau }^{\left(e\right)}\left(I^2\right)<2^{\Omega \left(I^2\right)},$$

$$\tau \left(n\right)\le 2^{n-1}.$$

$$2^{\omega \left(I^2\right)}\le {\tau }^{\left(e\right)}\left(I^2\right)\le 2^{\Omega \left(I^2\right)-\omega \left(I^2\right)}.$$

$${\left(m+1\right)}^n+1\ge {\left[\tau \left(m\right)\right]}^n+2^n.$$

$$\tau \left(n\right)+1\ge {\tau }^{\left(e\right)}\left(n\right)+{\tau }^{\ast }\left(n\right)$$

$$\tau \left(p{\mathcal{O}}_K\right)+1={\tau }^{\left(e\right)}\left(p{\mathcal{O}}_K\right)+2^{\omega \left(p{\mathcal{O}}_K\right)}.$$

$$\tau \left(p{\mathcal{O}}_K\right)+1\ge {\tau }^{\left(e\right)}\left(p{\mathcal{O}}_K\right)+2^{\omega \left(p{\mathcal{O}}_K\right)}.$$

$${\left(e+1\right)}^g+1\ge {\left(\tau \left(e\right)\right)}^g+2^g.$$

$$\tau \left(p{\mathcal{O}}_K\right)+1\ge {\tau }^{\left(e\right)}\left(p{\mathcal{O}}_K\right)+2^{\omega \left(p{\mathcal{O}}_K\right)}.$$

$$\tau \left(p{\mathcal{O}}_K\right)+1>{\tau }^{\left(e\right)}\left(p{\mathcal{O}}_K\right)+2^{\omega \left(p{\mathcal{O}}_K\right)}.$$

$$\tau \left(n\right)\ge {\tau }^{\left(e\right)}\left(n\right)+\frac{\tau \left(n\right)}{\omega \left(n\right)}\cdot \left(\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_r+1}\right).$$

$$\tau \left(p{\mathcal{O}}_K\right)\ge {\tau }^{\left(e\right)}\left(p{\mathcal{O}}_K\right)+\frac{\tau \left(p{\mathcal{O}}_K\right)}{\omega \left(p{\mathcal{O}}_K\right)}\cdot \frac{g}{e+1}.$$

$${\left(e+1\right)}^g\stackrel{?}{\ge }{\left(\tau \left(e\right)\right)}^g+{\left(e+1\right)}^{g-1}\iff$$

$${\left(e+1\right)}^{g-1}\cdot e\stackrel{?}{\ge }{\left(\tau \left(e\right)\right)}^g.$$

$$\tau \left(x\right)<x+1,\left(\forall \right)x\in {\mathbb{N}}^{\ast },$$

$${\left(\tau \left(e\right)\right)}^g<{\left(e+1\right)}^{g-1}\cdot e.$$

$$\tau \left(p{\mathcal{O}}_K\right)>{\tau }^{\left(e\right)}\left(p{\mathcal{O}}_K\right)+\frac{\tau \left(p{\mathcal{O}}_K\right)}{\omega \left(p{\mathcal{O}}_K\right)}\cdot \frac{g}{e+1},\left(\forall \right)e,g\in {\mathbb{N}}^{\ast },g\ge 2.$$

$$\frac{e}{e+1}\ge \frac{{\tau }^{\left(e\right)}\left(p{\mathcal{O}}_K\right)}{\tau \left(p{\mathcal{O}}_K\right)}+\frac{1}{\omega \left(p{\mathcal{O}}_K\right)}\cdot \frac{g}{e+1}.$$

$$\frac{e}{e+1}\cdot {\left(e+1\right)}^g\ge {\left(\tau \left(e\right)\right)}^g+\frac{{\left(e+1\right)}^g}{g}\cdot \frac{g}{e+1}\iff$$

$${\left(e+1\right)}^{g-1}\cdot \left(e-1\right)\ge {\left(\tau \left(e\right)\right)}^g.$$

Full text

1. Preliminaries

Let $n\geq 1$ be a natural number and we define by $\tau\left(n\right)$ the number of divisors of $n$. We note that $\tau\left(1\right)=1$; if $p$ is a prime number, then $\tau\left(p\right)=2$ and $\tau\left(p^{a}\right)=a+1$. It is known that the function $\tau$ is multiplicative, but not completely multiplicative and for $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdot\cdot\cdot p_{r}^{a_{r}}$, $n>1$, we have the relation $\tau\left(n\right)=\left(a_{1}+1\right)\left(a_{2}+1\right)\cdot\cdot\cdot\left(a_{r}+1\right).$

A natural number $d$ is a unitary divisor of a number $n$ if $d$ is a divisor of $n$ and if $d$ and $\frac{n}{d}$ are coprime, so they have no common factor other than $1.$ If $\tau^{*}\left(n\right)$ is the number of the unitary divisors of $n$, then for $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdot\cdot\cdot p_{r}^{a_{r}}$, $n>1$, we obtain the relation $\tau^{*}\left(n\right)=2^{r}=2^{\omega\left(n\right)},$ where $\omega\left(n\right)$ denote the number of distinct prime factors of $n$.

The notion of exponential divisor was introduced by Subbarao [9] in the following way: if we consider a natural number $n>1$ which can be written in canonical form as $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdot\cdot\cdot p_{r}^{a_{r}}$, the natural number $d=\prod_{i=1}^{r}p_{i}^{b_{i}}$ is called exponential divisor or e-divisor of $n=\prod_{i=1}^{r}p_{i}^{a_{i}}>1$ if $b_{i}|a_{i}$, for every $i\in\{1,...,r\}.$ We write $d|_{\left(e\right)}n$.

Let $\tau^{\left(e\right)}\left(n\right)$ be the number of exponential divisors of $n$. By convention $\tau^{\left(e\right)}\left(1\right)=1.$ It is easy to see that the function $\tau^{\left(e\right)}$ is multiplicative and for $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdot\cdot\cdot p_{r}^{a_{r}}>1$ we have $\tau^{\left(e\right)}\left(n\right)=\tau\left(a_{1}\right)\tau\left(a_{2}\right)\cdot\cdot\cdot\tau\left(a_{r}\right).$ It is obvious that $\tau^{\left(e\right)}\left(n\right)\leq\tau\left(n\right)$, for all $n\in\mathbb{N}^{*}.$

Other properties of the number of the exponential divisors of $n$ can be found in the papers [4], [6] and [10].

Let n\in$$\mathbb{N}, $n\geq 2$ and let $K$ be an algebraic number field of degree $[K:\mathbb{Q}]=n.$ Let $p$ be a prime positive integer. Let $\mathcal{O}_{K}$ be the ring of integers of the field $K$. It is known that $\mathcal{O}_{K}$ is a Dedeking ring. According to the fundamental theorem about Dedekind rings, there exist positive integers $g$ and $e_{i},$ $i=\overline{1,g}$ and the different ideals $P_{1},$ $P_{2},$…, P_{g}$$\inSpec$\left(\mathcal{O}_{K}\right)$ such that

[TABLE]

This decomposition is unique, except the order of factors. The integer $e_{i}$ ($i=\overline{1,g}$) is called the ramification index of $p$ at the ideal $P_{i}$ and the degree $f_{i}$ of the following extension of fields $\left(\mathcal{O}_{K}/P_{i}\right)/\left(\mathbb{Z}/p\mathbb{Z}\right)$ i. e. $f_{i}=\left[\mathcal{O}_{K}/P_{i}:\mathbb{Z}/p\mathbb{Z}\right]$ is called the residual degree of $p$ (see [2]). There are the following formulas (see [2]):

Proposition 1.1**.**

a) With the above notations, we have

[TABLE]

b) If $\mathbb{Q}\subset K$ is a Galois extension, then $e_{1}=e_{2}=...=e_{g}$ (say to $e$), $f_{1}=f_{2}=...=f_{g}$ (say to $f$). So, $efg=n.$

We recall now how a prime ideal of $\mathbb{Z}$ is decomposing into the ring of integers of a quadratic field, into the ring of integers of a cyclotomic field, respectively, into the ring of integers of a Kummer field, the results we will use in the next section.

Theorem 1.2** ([2]).**

Let $d\neq 0,1$ be a square free integer. Let $\mathcal{O}_{K}$ be the ring of integers of the quadratic field $K=\mathbb{Q}(\sqrt{d})$ and $\Delta_{K}$ be the discriminant of $K.$ Let $p$ be an odd prime integer. Then, we have:

1. (i)

$p$* is ramified in $\mathcal{O}_{K}$ if and only if $p|\Delta_{K}$. In this case $p\mathcal{O}_{K}=(p,\sqrt{d})^{2}$;* 2. (ii)

$p$* splits totally in $\mathcal{O}_{K}$ if and only if the Legendre symbol $(\frac{\Delta_{K}}{p})=1$. In this case $p\mathcal{O}_{K}=P_{1}\cdot{P_{2}},$ where $P_{1}$ and $P_{2}$ are distinct prime ideals in $\mathcal{O}_{K};$* 3. (iii)

$p$* is inert in $\mathcal{O}_{K}$ if and only if the Legendre symbol $(\frac{\Delta_{K}}{p})=-1$;* 4. (iv)

the prime 2 is ramified in $\mathcal{O}_{K}$ if and only if $d\equiv 2$ (mod $4$) or $d\equiv 3$ (mod $4$) In the first case $2\mathcal{O}_{K}=(2,\sqrt{d})^{2}$, while in the second case $2\mathcal{O}_{K}=(2,1+\sqrt{d})^{2};$ 5. (v)

the prime $2$ splits totally in $\mathcal{O}_{K}$ if and only if $d\equiv 1$ (mod 8). In this case $2\mathcal{O}_{K}=P_{1}\cdot P_{2}$, where $P_{1},P_{2}$ are distinct prime ideals in $\mathcal{O}_{K}$, with $P_{1}=(2,\frac{1+\sqrt{d}}{2})$; 6. (vi)

the prime $2$ is inert in $\mathcal{O}_{K}$ if and only if $d\equiv 5$ (mod $8$).

Let $n$ be a positive integer, $n\geq 2$ and let $\xi$ be a primitive root of unity of degree $n.$ It is known that the ring of integers of the cyclotomic field $\mathbb{Q}\left(\xi\right)$ is $\mathbb{Z}\left[\xi\right].$

We recall now a result of Dedekind.

Theorem 1.3** ([2], [7]).**

Let K be an algebraic number field and p a natural prime number. Then p is ramified in $\mathcal{O}_{K}$ if and only if $p|\Delta_{K}.$

Theorem 1.4** ([1], [2]).**

Let $n$ be an integer, $n\geq 3,$ $\xi$ be a primitive root of unity of order $n$ and let the cyclotomic field $\mathbb{Q}\left(\xi\right)$. If $p$ is a prime integer which does not divide $n$ and $f$ is the smallest positive integer such that p^{f}$$\equiv$$1 (mod $n$), then

[TABLE]

where $g=\frac{\varphi\left(n\right)}{f}$ and $P_{j}$, $j=\overline{1,g}$ are different prime ideals in the ring $\mathbb{Z}\left[\xi\right]$ and $\varphi$ is the Euler’s function.

Theorem 1.5** ([1], [2]).**

Let $p$ be an odd prime positive integer and let $\xi$ be a primitive root of unity of order $p$. Then the ideal $p\mathbb{Z}\left[\xi\right]$ is ramified totally in $\mathbb{Z}\left[\xi\right]$.

Theorem 1.6** ([1]).**

*Let $\xi$ be a primitive root of the unity of $l-$order, where $l$ is a prime natural number and let $\mathcal{O}_{L}.$ be the ring of integers of the Kummer field $\mathbb{Q}(\xi,\sqrt[l]{\mu})$ . A prime ideal $P$ in the ring $\mathbb{Z}[\xi]$ is in $\mathcal{O}_{L}$ in one of the situations:

i) It is equal with the $l-$power of a prime ideal from $\mathcal{O}_{L},$ if the $l-$power character $\left(\frac{\mu}{P}\right)_{l}=0;$

ii) It is a prime ideal in $\mathcal{O}_{L}$, if $\left(\frac{\mu}{P}\right)_{l}=$ a root of order $l$ of unity, different from $1$.

iii) It decomposes in $l$ different prime ideals from $\mathcal{O}_{L}$if $\left(\frac{\mu}{P}\right)_{l}=1.$*

2. The generalization of the arithmetic functions $\tau$ and $\tau^{\left(e\right)}$ in algebraic number fields

Let $K$ be an algebraic number field and let $\mathcal{O}_{K}$ be the ring of integers of $K$. We denote with $\mathbb{J}$ the set of ideals of the ring $\mathcal{O}_{K}.$ Let $I$ be an ideal of the ring $\mathcal{O}_{K}.$ Since $\mathcal{O}_{K}$ is a Dedekind ring, $\left(\exists!\right)$ g$$\in$$\mathbb{N}^{*}, the different ideals $P_{1},$ $P_{2},$…, P_{g}$$\inSpec$\left(\mathcal{O}_{K}\right)$ and the positive integer numbers $e_{1},$ $e_{2},$…,e_{g}$$\in$$\mathbb{N}^{*} such that $I=P^{e_{1}}_{1}\cdot P^{e_{2}}_{2}\cdot...\cdot P^{e_{g}}_{g}.$ A divisor of the ideal $I$ is of the following form $d_{I}=P^{a_{1}}_{1}\cdot P^{a_{2}}_{2}\cdot...\cdot P^{a_{g}}_{g},$ where $a_{1},$ $a_{2},$…,a_{g}$$\in$$\mathbb{N}, with a_{i}$$\leq$$e_{i}, for $\left(\forall\right)$ $i=\overline{1,g}.$ An exponential divisor of the ideal $I$ is of the following form $d^{\left(e\right)}_{I}=P^{b_{1}}_{1}\cdot P^{b_{2}}_{2}\cdot...\cdot P^{b_{g}}_{g},$ where $b_{1},$ $b_{2},$…,b_{g}$$\in$$\mathbb{N}^{*}, with $b_{i}$ $|$ $e_{i},$ for $\left(\forall\right)$ $i=\overline{1,g}.$

We extend the functions $\tau$ and $\tau^{\left(e\right)}$ to ideals in the following way: $\tau:\mathbb{J}\rightarrow\mathbb{C},\tau\left(I\right)$ is the number of divisors of the ideal $I$ and $\tau^{\left(e\right)}:\mathbb{J}\rightarrow\mathbb{C}$, $\tau^{\left(e\right)}\left(I\right)$ is the number of exponential divisors of the ideal $I$. So, we have

[TABLE]

We quickly obtain the following formulas:

Proposition 2.1**.**

With the above notations, we have

[TABLE]

and

[TABLE]

If $p$ is a prime positive integer, then we study the number of the exponential divisors of the ideal $p\mathcal{O}_{K}$ and and their shape.

Let $\mathbb{Q}\subset K$ be a Galois extension of degree $\left[K:\mathbb{Q}\right]=n<\infty$, let $p$ be a prime positive integer and let $\mathcal{O}_{K}$ be the ring of integers of the field $K$. If $p\mathcal{O}_{K}$ has decomposition into product of prime ideals, given by $p\mathcal{O}_{K}=P^{e_{1}}_{1}\cdot P^{e_{2}}_{2}\cdot...\cdot P^{e_{g}}_{g}$, where $e_{i}\in\mathbb{N}^{*},$ $i=\overline{1,g}$ and the different ideals $P_{1},$ $P_{2},$…, P_{g}$$\inSpec$\left(\mathcal{O}_{K}\right),$ then from the fact that $\mathbb{Q}\subset K$ is a Galois extension we obtain $e_{1}=e_{2}=...=e_{g}$. Therefore, we deduce $p\mathcal{O}_{K}=P^{e_{1}}_{1}\cdot P^{e_{1}}_{2}\cdot...\cdot P^{e_{1}}_{g}$, so we find the relation

[TABLE]

We consider $K$ an algebraic number field of degree $[K:\mathbb{Q}]=q$, where $q$ is a prime positive integer. If $\mathbb{Q}\subset K$ is a Galois extension and $\mathcal{O}_{K}$ is the ring of integers of the field $K$, then

[TABLE]

because applying Propositon 1.1 (b) we distinguish the following cases: i) $p\mathcal{O}_{K}\in\text{Spec}\left(\mathcal{O}_{K}\right)$, implies that $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=1;$ ii) $p\mathcal{O}_{K}=P_{1}\cdot P_{2}\cdot\cdot\cdot P_{q}$, where $P_{i}\in\text{Spec}\left(\mathcal{O}_{K}\right)$, for all $i=\overline{1,q}$, prove that $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=1$ and iii) $p\mathcal{O}_{K}=P^{q}$, with $P\in\text{Spec}\left(\mathcal{O}_{K}\right)$, implies that $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=2,$ because the ideal $p\mathcal{O}_{K}$ has only exponential divisors $P$ and $P^{q}$.

Sándor ([5]) established the following inequality:

[TABLE]

for all natural numbers $n\geq 2$, such that $n$ is a perfect square, where $\omega\left(n\right)$ represents the number of distinct prime factors of $n$ and $\Omega\left(n\right)$ represents the number of distinct prime factors of $n$ together with their multiplicities. We study this inequality for ideals in the rings of integers of an algebraic number field.

Proposition 2.2**.**

Let $K$ be an algebraic number field of degree $[K:\mathbb{Q}]=n<\infty$ and its ring of integers $\mathcal{O}_{K}$. Then, for every ideal I of the ring $\mathcal{O}_{K}$, we have

[TABLE]

where $\omega\left(I^{2}\right)$ represents the number of distinct prime factors of $I^{2}$ ($\omega\left(I^{2}\right)=\omega\left(I\right)$) and $\Omega\left(I^{2}\right)$ represents the number of distinct prime factors of $I^{2}$ together with their multiplicities.

Proof.

Since the ring of integers $\mathcal{O}_{K}$ is a Dedekind ring and $I$ is a ideal of $\mathcal{O}_{K}$ there are unique $P_{1},P_{2},...,P_{g}\in\text{Spec}\left(\mathcal{O}_{K}\right)$ and $a_{1},a_{2},...,a_{g}\in\mathbb{N}^{*}$ such that $I=P^{a_{1}}_{1}\cdot P^{a_{1}}_{2}\cdot...\cdot P^{a_{g}}_{g}$. It follows that $I^{2}=P^{2a_{1}}_{1}\cdot P^{2a_{1}}_{2}\cdot...\cdot P^{2a_{g}}_{g}$, so $\tau^{\left(e\right)}\left(I^{2}\right)=\tau\left(2a_{1}\right)\cdot\tau\left(2a_{2}\right)\cdot...\cdot\tau\left(2a_{g}\right).$ But, we have $\tau\left(2a_{i}\right)\geq 2,$ for all $i=\overline{1,g}$. Therefore we find $\tau^{\left(e\right)}\left(I^{2}\right)\geq 2^{g}$ and because $\omega\left(I^{2}\right)=g$, we deduce the inequality $\tau^{\left(e\right)}\left(I^{2}\right)\geq 2^{\omega\left(I^{2}\right)}.$ By mathematical induction, it easy to see that $\tau\left(n\right)<2^{2n-1}$, for every $n\geq 1$. Since $\tau\left(mn\right)\leq\tau\left(m\right)\tau\left(n\right)$, for every $m,n\in\mathbb{N}^{*}$, we deduce that $\tau\left(2a_{i}\right)\leq\tau\left(2\right)\tau\left(a_{i}\right)=2\tau\left(a_{i}\right)<2^{2a_{i}}$, for $a_{i}\geq 1$ and $i=\overline{1,g}$. This means that $\tau^{\left(e\right)}\left(I^{2}\right)=\tau\left(2a_{1}\right)\cdot\tau\left(2a_{2}\right)\cdot...\cdot\tau\left(2a_{g}\right)<2^{2a_{1}+2a_{2}+...+2a_{g}}=2^{\Omega\left(I^{2}\right)}.$ Therefore, the statement is true. ∎

Proposition 2.3**.**

For every number $n\in\mathbb{N}^{*}$, we have

[TABLE]

Proof.

We consider the proposition $P\left(n\right):\tau\left(n\right)\leq 2^{n-1},$ for all $n\geq 1$. Verify that $P\left(1\right)$ is true: $\tau\left(1\right)=1\leq 1=2^{1-1}.$ We assume that $P\left(k\right)$ is true, for all $k\leq n-1$. We prove that the proposition $P\left(n\right)$ is true. We identify two cases: i) If $n=q$, where $q$ is a prime number, then we have $\tau\left(n\right)=2\leq 2^{n-1}$; ii) If $n>2$ is a composite number, then we deduce that there are the numbers $d_{1},d_{2}\in\mathbb{N}$, $2\leq d_{1},d_{2}\leq n-1$, $\left(d_{1},d_{2}\right)=1$, such that $n=d_{1}d_{2}$. Since $d_{1},d_{2}\leq n-1$, we get $\tau\left(d_{1}\right)\leq 2^{d_{1}-1}$, $\tau\left(d_{2}\right)\leq 2^{d_{2}-1}$. Because the function $\tau$ is multiplicative, then we obtain $\tau\left(n\right)=\tau\left(d_{1}\right)\cdot\tau\left(d_{2}\right)\leq 2^{d_{1}-1}\cdot 2^{d_{2}-1}=2^{d_{1}+d_{2}-2}\leq 2^{d_{1}\cdot d_{2}-1}=2^{n-1}$, so $P\left(n\right)$ is true. According to the principle of mathematical induction, we have that $P\left(n\right)$ is true, for every $n\geq 1$. ∎

Next, we obtain a refinement of Proposition 2.2, given by the following:

Proposition 2.4**.**

Let $K$ be an algebraic number field of degree $[K:\mathbb{Q}]=n<\infty$ and its ring of integers $\mathcal{O}_{K}$. Then, for every ideal I of the ring $\mathcal{O}_{K}$, we have

[TABLE]

Proof.

Similarly, as in the proof from Proposition 2.2 and using the inequality from Proposition 2.3, we find the statement. ∎

We also obtain the following inequality involving the function $\tau$, the inequality that we will apply later.

Proposition 2.5**.**

For every number $m,n\in\mathbb{N},m\geq 2$, we have

[TABLE]

Proof.

We consider the proposition $P\left(n\right):\left(m+1\right)^{n}+1\geq[\tau\left[\left(m\right)\right]^{n}+2^{n},$ for all $n\in\mathbb{N}$. It is obviously that $P\left(0\right)$ is true. We check that $P\left(1\right)$ is true: the inequality of the statement becomes $m\geq\tau\left(m\right)$, which is true, for all $m\geq 2$. We assume that $P\left(n\right)$ is true, i. e. $\left(m+1\right)^{n}+1\geq[\tau\left[\left(m\right)\right]^{n}+2^{n}.$ Since, we have $\left(m+1\right)^{n+1}+1=\left(m+1\right)\left(m+1\right)^{n}+1\geq\left(m+1\right)\cdot\{\left[\tau\left(m\right)\right]^{n}+2^{n}-1\}+1=m\cdot\left[\tau\left(m\right)\right]^{n}+\left[\tau\left(m\right)\right]^{n}+2^{n}+m\left(2^{n}-1\right)\geq\tau\left(m\right)\cdot\left[\tau\left(m\right)\right]^{n}+2^{n}+2^{n}=\left[\tau\left(m\right)\right]^{n+1}+2^{n+1}$, because $m\left(2^{n}-1\right)\geq 0$, for every $n\geq 1$ and $m\geq\tau\left(m\right)\geq 2$, for all $m\geq 2$. According to the principle of mathematical induction, we have that $P\left(n\right)$ is true, for every $n\in\mathbb{N}$. ∎

In [3], the first author gave the following result:

Proposition 2.6**.**

For any $n\geq 1$, the following inequality

[TABLE]

occurs.

Next, we extend the above result to the followings:

Proposition 2.7**.**

Let $K$ be an algebraic number field of degree $[K:\mathbb{Q}]=n<\infty$ such that $\mathbb{Q}\subset K$ is a Galois extension. Let $\Delta_{K}$ be the discriminant of the field $K.$ If $p$ is a prime positive integer such that $p\not{|}\Delta_{K}$, then

[TABLE]

Proof.

Since $p\not{|}\Delta_{K}$ it follows that $p$ is unramified in $\mathcal{O}_{K}$ (according to Theorem 1.3). Therefore there are unique $P_{1},P_{2},...,P_{g}\in\text{Spec}\left(\mathcal{O}_{K}\right)$, $P_{i}\neq P_{j}$, for all $i,j=\overline{1,g},i\neq j$, such that $p\mathcal{O}_{K}=P_{1}\cdot P_{2}\cdot...\cdot P_{g}$. This means that $\tau\left(p\mathcal{O}_{K}\right)=\left(1+1\right)\cdot\left(1+1\right)\cdot...\cdot\left(1+1\right)=2^{g},2^{\omega\left(p\mathcal{O}_{K}\right)}=2^{g}$ and $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=1$. Therefore, we deduce the statement. ∎

Proposition 2.8**.**

Let $K$ be quadratic field. If $p$ is a prime positive integer, then

[TABLE]

Proof.

According Theorem 1.2, we have three cases: i) $p\mathcal{O}_{K}=P_{1}\cdot P_{2},$ where P_{1},P_{2}$$\in\text{Spec}\left(\mathcal{O}_{K}\right). This means that $\tau\left(p\mathcal{O}_{K}\right)=4,2^{\omega\left(p\mathcal{O}_{K}\right)}=4$ and $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=1$. Therefore, the inequality of the statement is true;

ii) $p\mathcal{O}_{K}=P_{1}^{2}$, where P_{1}$$\in\text{Spec}\left(\mathcal{O}_{K}\right). It follows that $p$ is ramified totally in $\mathcal{O}_{K}.$ This implies $\tau\left(p\mathcal{O}_{K}\right)=3,2^{\omega\left(p\mathcal{O}_{K}\right)}=2$ and $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=2$. Therefore, the equality of the statement is true;

iii) $p\mathcal{O}_{K}=P\in\text{Spec}\left(\mathcal{O}_{K}\right)$, which proves that $p$ is inert in $\mathcal{O}_{K}$. Consequently, we obtain $\tau\left(p\mathcal{O}_{K}\right)=2,2^{\omega\left(p\mathcal{O}_{K}\right)}=2$ and $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=\tau\left(1\right)=1$. Therefore, the equality of the statement is true. ∎

Now we study what happens when $p|\Delta_{K}.$ We obtain the following result.

Proposition 2.9**.**

Let $K$ be an algebraic number field of degree $[K:\mathbb{Q}]=n<\infty,$ $n\geq 3$ such that $\mathbb{Q}\subset K$ is a Galois extension. Let $\Delta_{K}$ be the discriminant of the field $K.$ If $p$ is a prime positive integer such that $p|\Delta_{K},$ then

[TABLE]

Proof.

Since $p|\Delta_{K}$ it follows that $p$ is ramified in $\mathcal{O}_{K}$ (according to Theorem 1.3). Therefore there are unique $P_{1},P_{2},...,P_{g}\in\text{Spec}\left(\mathcal{O}_{K}\right)$, $P_{i}\neq P_{j}$, for all $i,j=\overline{1,g},i\neq j$ and there is e$$\in$$\mathbb{N}, $e\geq 2$ such that $p\mathcal{O}_{K}=P^{e}_{1}\cdot P^{e}_{2}\cdot...\cdot P^{e}_{q}$ and $efg=n$. This implies that $\tau\left(p\mathcal{O}_{K}\right)=\left(e+1\right)^{g},2^{\omega\left(p\mathcal{O}_{K}\right)}=2^{g}$ and $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=\left(\tau\left(e\right)\right)^{g}$. Applying Proposition 2.5 we obtain:

[TABLE]

This inequality is equivalent with

[TABLE]

∎

If $p$ is totally ramified in $\mathcal{O}_{K}$, then we obtain the strict inequality, which is true in any extension of fields, not only in the Galois extension with the degree $\geq 3$.

Proposition 2.10**.**

Let $K$ be an algebraic number field of degree $[K:\mathbb{Q}]=n\geq 3$. If p is a prime positive integer such that $p$ is totally ramified in $\mathcal{O}_{K}$, then we have the following:

[TABLE]

Proof.

Since $p$ is ramified totally in $\mathcal{O}_{K}$, we get $p\mathcal{O}_{K}=P^{g}$, where $P\in\text{Spec}\left(\mathcal{O}_{K}\right)$. If $f$ is the residual degree, then using Proposition 1.1, we find $e=n,g=1,f=1$, which implies $p\mathcal{O}_{K}=P$, with $P\in\text{Spec}\left(\mathcal{O}_{K}\right)$. Consequently, we obtain $\tau\left(p\mathcal{O}_{K}\right)=n+1$, $2^{\omega\left(p\mathcal{O}_{K}\right)}=2$ and $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=\tau\left(n\right)$. It is easy to see that $\tau\left(n\right)<n$, for $n\geq 3$, this shows that the inequality of the statement is true. ∎

In [3], N. Minculete (the first author) obtained the following result:

Theorem 2.11**.**

For each integer number n=p^{a_{1}}_{1}p^{a_{2}}_{2}...p^{a_{r}}_{r}$$>$$1 (where $p_{1},p_{2},...,p_{r}$ are different prime positive integers and a_{1},a_{2},...,a_{r}$$\in$$\mathbb{N}^{*}), we have the following inequality:

[TABLE]

Now, we generalize this result for ideals in the ring of integers of an algebraic number field $K,$ when $\mathbb{Q}\subset K$ is a Galois extension. We obtain:

Theorem 2.12**.**

Let n$$\in$$\mathbb{N}, $n\geq 2$ and let $K$ be an algebraic number field of degree $[K:\mathbb{Q}]=n$ such that $\mathbb{Q}\subset K$ is a Galois extension. Let $p$ be a prime positive integer and let the ideal $p\mathcal{O}_{K}=P^{e}_{1}\cdot P^{e}_{2}\cdot...\cdot P^{e}_{g},$ where g$$\in$$\mathbb{N}^{*}, e$$\in$$\mathbb{N}^{*}, and $P_{1},$ $P_{2},$…, $P_{g}$ are prime different ideals in the ring of integers $\mathcal{O}_{K}.$ Then, we have the following inequality:

[TABLE]

Moreover, the equality occurs if and only if $p\mathcal{O}_{K}$ is a prime ideal or $p\mathcal{O}_{K}=P^{2},$ with P$$\inSpec$\left(\mathcal{O}_{K}\right).$

Proof.

We have: $\tau\left(p\mathcal{O}_{K}\right)=\left(e+1\right)^{g},$ $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=\left(\tau\left(e\right)\right)^{g},$ $\omega\left(p\mathcal{O}_{K}\right)=g.$ The inequality (2.1) is equivalent with

[TABLE]

[TABLE]

In order to prove the inequality (2.2), we consider three cases:

Case 1. If p\mathcal{O}_{K}$$\inSpec$\left(\mathcal{O}_{K}\right),$ then $e=g=1$ and we obtain the equality in (2.2) and also in (2.1).

Case 2. If $p\mathcal{O}_{K}=P^{2},$ with P$$\inSpec$\left(\mathcal{O}_{K}\right),$ it means that $e=2$ and $g=1.$ Since $\tau\left(2\right)=2,$ we obtain the equality in (2.2) and also in (2.1).

Case 3. If p\mathcal{O}_{K}$$\notinSpec$\left(\mathcal{O}_{K}\right)$ and $p\mathcal{O}_{K}\neq P^{2},$ with P$$\inSpec$\left(\mathcal{O}_{K}\right),$ it results that we have neither the possibility $e=g=1$ nor the possibility $e=2,$ $g=1.$ Applying these and the obvious inequality

[TABLE]

which implies \tau\left(x\right)\leq x\left(\forall\right)x$$\in$$\mathbb{N}^{*}, we are proving inequality (2.2) considering two subcases:

Subcase 3.a) If e$$\in$$\mathbb{N}^{*}, $e\geq 3$ and $g=1,$ inequality (2.2) is equivalent to $e\geq\tau\left(e\right),$ which is true. In fact, we have $e>\tau\left(e\right),$ for each e$$\in$$\mathbb{N}, $e\geq 3.$ Therefore, we obtain:

$\tau\left(p\mathcal{O}_{K}\right)>\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)+\frac{\tau\left(p\mathcal{O}_{K}\right)}{\omega\left(p\mathcal{O}_{K}\right)}\cdot\frac{g}{e+1},$ \left(\forall\right)e$$\in$$\mathbb{N}^{*}, $e\geq 3$ and $g=1.$

Subcase 3.b) If e;g$$\in$$\mathbb{N}^{*}, $g\geq 2,$ we have: $0<\tau\left(e\right)<e+1.$ Since $g\geq 2,$ it results $0<\left(\tau\left(e\right)\right)^{g-1}<\left(e+1\right)^{g-1}.$ Multiplying this inequality with inequality $0<\tau\left(e\right)\leq e,$ \left(\forall\right)e$$\in$$\mathbb{N}^{*}, we obtain

[TABLE]

The above inequality is equivalent with

[TABLE]

∎

When the extension of fields $\mathbb{Q}\subset K$ is a nontrivial Galois extension of degree odd, we are improving the Theorem 2.12 like this:

Theorem 2.13**.**

Let n$$\in$$\mathbb{N}^{*} and let $K$ be an algebraic number field of degree odd $[K:\mathbb{Q}]=n\geq 3$ such that $\mathbb{Q}\subset K$ is a Galois extension. Let $p$ be a prime positive integer and let the ideal $p\mathcal{O}_{K}=P^{e}_{1}\cdot P^{e}_{2}\cdot...\cdot P^{e}_{g},$ where g$$\in$$\mathbb{N}^{*}, e$$\in$$\mathbb{N}^{*}, and $P_{1},$ $P_{2},$…, $P_{g}$ are prime different ideals in the ring of integers $\mathcal{O}_{K}.$ If the ideal $p\mathcal{O}_{K}$ is ramified, then, we have the following inequality:

[TABLE]

Proof.

Let $f$ be the the residual degree of $p.$ Since $\mathbb{Q}\subset K$ is a Galois extension, we have $f=f_{i}=\left[\mathcal{O}_{K}/P_{i}:\mathbb{Z}/p\mathbb{Z}\right],$ $\left(\forall\right)$ $i=\overline{1,g}.$ Since the ideal $p\mathcal{O}_{K}$ is ramified, it results $e>1.$ But $efg=n\geq 3$ and $n$ is an odd number, it results that $e\geq 3.$

We have: $\tau\left(p\mathcal{O}_{K}\right)=\left(e+1\right)^{g},$ $\tau^{\left(e\right)}\left(p\mathcal{O}_{K}\right)=\left(\tau\left(e\right)\right)^{g},$ $\omega\left(p\mathcal{O}_{K}\right)=g.$ Therefore, the inequality (2.3) is equivalent with

[TABLE]

[TABLE]

In order to prove this last inequality, we are using the fact that $\tau\left(e\right)\leq e-1,$ \left(\forall\right)e$$\in$$\mathbb{N}^{*}, $e\geq 3.$ Since g$$\in$$\mathbb{N}^{*}, it results that $\left(\tau\left(e\right)\right)^{g}\leq\left(e-1\right)^{g}.$ But

[TABLE]

So, by transitivity, we obtain:

[TABLE]

This last inequality is equivalent with inequality (2.3).

We remark that when $e=3,g=1$ we obtain the equality in (2.3).

∎

Now, we give some examples of fields extensions and of ideals which verify Theorem 2.13:

1. Let $\epsilon$ be a primitive root of order $3$ of the unity, let $p$ be a prime positive integer, p$$\equiv$$2 (mod $3$) and let $l$ be an integer such that $p|l.$ Let the cyclotomic field $K=\mathbb{Q}\left(\epsilon\right)$ and let the Kummer field $L=\mathbb{Q}\left(\epsilon;\sqrt[3]{l}\right).$ The fields extension $K\subset L$ is a Galois extension of degree $3.$ Since p$$\equiv$$2 (mod $3$), applying Theorem 1.4, it results that $p\mathbb{Z}\left[\epsilon\right]$ is a prime ideal. Since $p|l,$ it results that the cubic character $\left(\frac{l}{p\mathbb{Z}\left[\epsilon\right]}\right)_{3}=0$ and according to Theorem 1.6, we obtain that the ideal $p\mathcal{O}_{L}=P^{3}_{1},$ where P_{1}$$\inSpec($\mathcal{O}_{L}$). So, the ideal $p\mathcal{O}_{L}$ is totally ramified in $\mathcal{O}_{L}$. Using the notations from Theorem 2.13, we have: $g=1,$ $e=3,$ $\tau\left(p\mathcal{O}_{L}\right)=\left(e+1\right)^{g}=4,$ \tau^{\left(e\right)}\left(p\mathcal{O}_{L}\right)=\left(\tau\left(e\right)\right)^{g}$$=2, $\omega\left(p\mathcal{O}_{L}\right)=g=1.$ It results $\frac{e}{e+1}=\frac{3}{4}$ and $\frac{\tau^{\left(e\right)}\left(p\mathcal{O}_{L}\right)}{\tau\left(p\mathcal{O}_{L}\right)}+\frac{1}{\omega\left(p\mathcal{O}_{L}\right)}\cdot\frac{g}{e+1}=\frac{3}{4}.$ So, we have equality in (2.3).

2. Let $\xi$ be a primitive root of order $5$ of the unity, let $p$ be a prime positive integer, p$$\equiv$$3 (mod $5$) and let $l$ be an integer such that $p|l.$ Let the cyclotomic field $K=\mathbb{Q}\left(\xi\right)$ and let the Kummer field $L=\mathbb{Q}\left(\xi;\sqrt[5]{l}\right).$ The fields extension $K\subset L$ is a Galois extension of degree $5.$ Since p$$\equiv$$3 (mod $5$), and applying Theorem 1.4, it results that $p\mathbb{Z}\left[\xi\right]$ is a prime ideal. Since $p|l,$ it results that the the $5$-power character $\left(\frac{l}{p\mathbb{Z}\left[\xi\right]}\right)_{5}=0$ and according to Theorem 1.6, we obtain that the ideal $p\mathcal{O}_{L}=P^{5}_{1},$ where P_{1}$$\inSpec($\mathcal{O}_{L}$). So, the ideal $p\mathcal{O}_{L}$ is ramified totally in $\mathcal{O}_{L}$. Using the notations from Theorem 2.13, we have: $g=1,$ $e=5,$ $\tau\left(p\mathcal{O}_{L}\right)=\left(e+1\right)^{g}=6,$ \tau^{\left(e\right)}\left(p\mathcal{O}_{L}\right)=\left(\tau\left(e\right)\right)^{g}$$=2, $\omega\left(p\mathcal{O}_{L}\right)=g=1.$ It results $\frac{e}{e+1}=\frac{5}{6}$ and $\frac{\tau^{\left(e\right)}\left(p\mathcal{O}_{L}\right)}{\tau\left(p\mathcal{O}_{L}\right)}+\frac{1}{\omega\left(p\mathcal{O}_{L}\right)}\cdot\frac{g}{e+1}=\frac{1}{2}.$ So, we have

[TABLE]

The same two examples verify Proposition 2.9.