Some Properties of Euler’s Function and of the Function τ and Their Generalizations in Algebraic Number Fields

Abstract

In this paper, we find some inequalities which involve Euler’s function, extended Euler’s function, the function $\tau$, and the generalized function $\tau$ in algebraic number fields.

1. Introduction and Preliminaries

Let the function $\phi :{\mathbb{N}}^{*}\to {\mathbb{N}}^{*},$ $\phi \left(n\right)=\left|\left\{k\in {\mathbb{N}}^{*}|k\le n,\left(k,n\right)=1\right\}\right|,$ $\left(\forall \right)$ $n\in {\mathbb{N}}^{*}.$ $\phi$ is called Euler’s function or Euler’s totient function. We remark that $\phi \left(n\right)$ is the number of invertible elements in the unitary ring $\mathbb{Z}/n\mathbb{Z}.$ If $n\in \mathbb{N},n\ge 2,$ the following formula to calculate $\phi \left(n\right)$ is known: if $l\in {\mathbb{N}}^{*},$ $n=p_1^{{\alpha }_1}·p_2^{{\alpha }_2}·\dots p_l^{{\alpha }_l}$, where $p_1,p_2\dots p_l\in \mathbb{N}$ are unique distinct prime numbers and ${\alpha }_i\in {\mathbb{N}}^{*}$, $i=\overline{1,l}$ then $\phi \left(n\right)=n·\left(1-\frac{1}{p_1}\right)·\left(1-\frac{1}{p_2}\right)·\dots ·\left(1-\frac{1}{p_l}\right).$ A known property is that $\phi$ is a multiplicative function, but it is immediately noticed that $\phi$ is not a completely multiplicative function. An important number of monographs in number theory studied these types of functions [1,2,3,4,5].

Let the function $\tau :{\mathbb{N}}^{*}\to {\mathbb{N}}^{*},$ $\tau \left(n\right)=\left|\left\{k\in {\mathbb{N}}^{*}\left|k\right|n\right\}\right|,$ $\left(\forall \right)$ $n\in {\mathbb{N}}^{*}.$ $\tau \left(1\right)=1.$ If $n\in \mathbb{N},n\ge 2,,$ $n=p_1^{{\alpha }_1}·p_2^{{\alpha }_2}·\dots p_l^{{\alpha }_l}$, where $p_1,p_2\dots p_l\in \mathbb{N}$ are unique distinct prime numbers and ${\alpha }_i\in {\mathbb{N}}^{*}$, then $\tau \left(n\right)=\left({\alpha }_1+1\right)·\left({\alpha }_2+1\right)·\dots ·\left({\alpha }_l+1\right).$ $\tau$ is a multiplicative function, but $\tau$ is not a completely multiplicative function.

Many analytics properties of these functions can be found in [6,7,8]. In [9] Rassias introduced the function

$$\varphi \left(n,A,B\right)=\sum _{A\le k\le B,\left(n,k\right)=1}1,$$

as a generalized totient function. He proved that:

$$\varphi \left(n,A,B\right)=\sum _{d|n}\mu \left(d\right)·\left(\left[\frac{B}{d}\right]-\left[\frac{A}{d}\right]\right),$$

where $\mu$ is Möbius’ function (see Lemma 5.22 from [9]) and

$$\varphi \left(n,A,B\right)=\frac{B-A}{n}·\phi \left(n\right)+{\delta }_{n,A}+\mathit{O}\left(\sum _{d|n}\mu {\left(d\right)}^2\right),$$

for each $n,A,B$∈$\mathbb{N},$ $n>1,$ where ${\delta }_{n,A}=1$ if $\left(n,A\right)=1$, and 0 otherwise (see Proposition 5.23 from [9]).

Let n be a positive integer, $n\ge 2$, and let K be an algebraic number field, with degree $\left[K:\mathbb{Q}\right]=n.$ Let ${\mathcal{O}}_K$ be the ring of integers of the field K, and let Spec $\left({\mathcal{O}}_K\right)$ be the set of the prime ideals of the ring ${\mathcal{O}}_K.$ It is known that the ring of integers of an arbitrary algebraic number field K is a Dedekind domain. Let $\mathbb{J}$ be the set of ideals of the ring ${\mathcal{O}}_K.$

It is known that Euler’s function was extended to the set $\mathbb{J}$ like this: let I be an ideal from the set $\mathbb{J}.$ Taking into account that ${\mathcal{O}}_K$ is a Dedekind domain and the fact that in any Dedekind domain, there is a factorization theorem for ideals similar to the fundamental theorem of arithmetics in the set of integer numbers, $I=P_1^{{\alpha }_1}·P_2^{{\alpha }_2}·\dots ·P_l^{{\alpha }_l},$ where $P_1,$ $P_2,$ …, $P_l$ are unique different prime ideals in the ring ${\mathcal{O}}_K$ and ${\alpha }_i\in {\mathbb{N}}^{*}$ $i=\overline{1,l}$, and then

$${\phi }_{ext}:\mathbb{J}\to {\mathbb{N}}^{*},$$

$${\phi }_{ext}\left(I\right)=N\left(I\right)·\left(1-\frac{1}{N\left(P_1\right)}\right)·\left(1-\frac{1}{N\left(P_2\right)}\right)·\dots ·\left(1-\frac{1}{N\left(P_l\right)}\right),$$

where $N\left(I\right)$ is the norm of the ideal $I.$ We recall the norm of an ideal I is defined as follows $N\left(I\right)=\left[{\mathcal{O}}_K:I\right].$ The following properties of the norm function are known:

Proposition 1.

$$N\left(I_1·I_2\right)=N\left(I_1\right)·N\left(I_2\right),$$

for $\left(\forall \right)$ nonzero ideals $I_1,$ $I_2$ from the set $\mathbb{J}.$

Proposition 2.

If I is an ideal from $\mathbb{J}$ with the property $N\left(I\right)$ as a prime number, then I ∈ Spec$\left({\mathcal{O}}_K\right).$

Proposition 3.

If P ∈ Spec$\left({\mathcal{O}}_K\right)$ and p is a prime positive integer such that the ideal P divides the ideal $p{\mathcal{O}}_K$, then $N\left(P\right)=p^f$, where f ∈ ${\mathbb{N}}^{*}$ is the residual degree of the ideal $P.$

Proposition 4.

The norm function $N:\mathbb{J}\to {\mathbb{N}}^{*}$ is not injective.

We recall that:

Proposition 5.

If $I_1$ and $I_2$ are nonzero ideals from $\mathbb{J}$ such that $I_1+I_2={\mathcal{O}}_K,$ then

$${\phi }_{ext}\left(I_1·I_2\right)={\phi }_{ext}\left(I_1\right)·{\phi }_{ext}\left(I_2\right).$$

These results can be found in [6,10,11,12,13,14,15,16,17].

In the paper [18], the authors extended the function $\tau$ to the set $\mathbb{J}$ of the ideals of the ring ${\mathcal{O}}_K.$ We denote this function with ${\tau }_{ext}$ to distinguish it from the function $\tau :{\mathbb{N}}^{*}\to {\mathbb{N}}^{*}$. Thus, ${\tau }_{ext}:\mathbb{J}\to {\mathbb{N}}^{*},$ ${\tau }_{ext}\left(I\right)=$ the number of ideals from $\mathbb{J}$, which divide the ideal $I.$ Using the above notations, we have:

$${\tau }_{ext}\left(I\right)=\left({\alpha }_1+1\right)·\left({\alpha }_2+1\right)·\dots ·\left({\alpha }_l+1\right).$$

Quickly, we obtain that:

Proposition 6.

$${\tau }_{ext}\left(I_1·I_2\right)={\tau }_{ext}\left(I_1\right)·{\tau }_{ext}\left(I_2\right),$$

for any $I_1$ and $I_2$ which are nonzero ideals from $\mathbb{J}$, such that $I_1+I_2={\mathcal{O}}_K.$

In this article, we obtain certain inequalities involving the functions $\tau ,$ ${\tau }_{ext},$ $\phi ,$ ${\phi }_{ext}.$

2. Results

Popovici (in [19]) obtained the following inequality:

$${\phi }^2\left(a·b\right)\le \phi \left(a^2\right)·\phi \left(b^2\right)\left(\forall \right)a,b\in {\mathbb{N}}^{*},$$

where $\phi$ is Euler’s function. In [20] (Proposition 3.4), Minculete and Savin proved a similar inequality, for extended Euler’s function:

Proposition 7.

Let n be a positive integer, $n\ge 2$, and let K be an algebraic number field of degree $[K:\mathbb{Q}]=n.$ Then:

$${\phi }_{ext}^2\left(I·J\right)\le {\phi }_{ext}\left(I^2\right)·{\phi }_{ext}\left(J^2\right),\left(\forall \right)idealsIandJof{\mathcal{O}}_K.$$

We ask ourselves if the functions $\tau$ and ${\tau }_{ext}$ satisfy a similar inequality. We obtain that these functions satisfy the opposite inequality.

Proposition 8.

Let K be an algebraic number field. Then:

$${\tau }_{ext}^2\left(I·J\right)\ge {\tau }_{ext}\left(I^2\right)·{\tau }_{ext}\left(J^2\right),\left(\forall \right)idealsIandJof{\mathcal{O}}_K.$$

Proof. 

Let I and J be two nonzero ideals in the ring ${\mathcal{O}}_K.$ Applying the fundamental theorem of Dedekind rings, $\left(\exists !\right)l,r\in {\mathbb{N}}^{*},$ the different prime ideals $P_1,$ $P_2,$ …, $P_l,$ $P_1^{″},$ $P_2^{″},$ …, $P_r^{″}$ of the ring ${\mathcal{O}}_K$ and ${\alpha }_1,{\alpha }_2,\dots {\alpha }_l,{\gamma }_1,{\gamma }_2,\dots ,{\gamma }_r\in {\mathbb{N}}^{*}$ such that $I=P_1^{{\alpha }_1}·P_2^{{\alpha }_2}·\dots ·P_l^{{\alpha }_l}·{\left(P_1^{″}\right)}^{{\gamma }_1}·\dots ·{\left(P_r^{″}\right)}^{{\gamma }_r}$ and $\left(\exists !\right)m\in {\mathbb{N}}^{*},$ the different prime ideals $P_1^{\prime },$ $P_2^{\prime },$ …, $P_m^{\prime }$ of the ring ${\mathcal{O}}_K$ and ${\beta }_1,{\beta }_2,\dots ,{\beta }_m,{\gamma }_{r+1},{\gamma }_{r+2},\dots ,{\gamma }_{2r}\in {\mathbb{N}}^{*}$ such that $J={\left(P_1^{\prime }\right)}^{{\beta }_1}·\dots ·{\left(P_m^{\prime }\right)}^{{\beta }_m}·{\left(P_1^{″}\right)}^{{\gamma }_{r+1}}·\dots ·{\left(P_r^{″}\right)}^{{\gamma }_{2r}}.$ It results that

$${\tau }_{ext}^2\left(I·J\right)={\left({\alpha }_1+1\right)}^2·{\left({\alpha }_2+1\right)}^2·\dots ·{\left({\alpha }_l+1\right)}^2·$$

$$·{\left({\beta }_1+1\right)}^2·{\left({\beta }_2+1\right)}^2·...·{\left({\beta }_m+1\right)}^2·{\left({\gamma }_1+{\gamma }_{r+1}+1\right)}^2·{\left({\gamma }_2+{\gamma }_{r+2}+1\right)}^2·\dots ·{\left({\gamma }_r+{\gamma }_{2r}+1\right)}^2$$

and

$${\tau }_{ext}\left(I^2\right)·{\tau }_{ext}\left(J^2\right)=\left(2{\alpha }_1+1\right)·\left(2{\alpha }_2+1\right)·...·\left(2{\alpha }_l+1\right)·\left(2{\beta }_1+1\right)·\left(2{\beta }_2+1\right)·\dots ·\left(2{\beta }_m+1\right)·$$

$$·\left(2{\gamma }_1+1\right)·\left(2{\gamma }_2+1\right)·\dots ·\left(2{\gamma }_r+1\right)·\left(2{\gamma }_{r+1}+1\right)·\left(2{\gamma }_{r+2}+1\right)·\dots ·\left(2{\gamma }_{2r}+1\right).$$

It immediately follows that

$${\left({\alpha }_i+1\right)}^2\ge 2{\alpha }_i+1,\left(\forall \right)i=\overline{1,l},$$

$${\left({\beta }_i+1\right)}^2\ge 2{\beta }_i+1,\left(\forall \right)i=\overline{1,m}$$

and

$${\left({\gamma }_i+{\gamma }_{r+i}+1\right)}^2\ge \left(2{\gamma }_i+1\right)·\left(2{\gamma }_{r+i}+1\right)\iff {\left({\gamma }_i-{\gamma }_{r+i}\right)}^2\ge 0\left(\forall \right)i=\overline{1,r}.$$

Thus, we obtain that

$${\tau }_{ext}^2\left(I·J\right)\ge {\tau }_{ext}\left(I^2\right)·{\tau }_{ext}\left(J^2\right),\left(\forall \right)idealsIandJof{\mathcal{O}}_K.$$

□

Sivaramakrishnan (in [21]) obtained the following inequality involving Euler’s function and the function $\tau$:

Proposition 9.

For any positive integer $n,$ the following inequality is true

$$\phi \left(n\right)·\tau \left(n\right)\ge n.$$

Now, we generalize Proposition 9, for an extended Euler’s function and the function ${\tau }_{ext}.$

Proposition 10.

Let K be an algebraic number field and let ${\mathcal{O}}_K$ be the ring integers of the field $K.$ Then, the following inequality is true:

$${\phi }_{ext}\left(I\right)·{\tau }_{ext}\left(I\right)\ge N\left(I\right),\left(\forall \right)anonzeroidealIof{\mathcal{O}}_K.$$

Proof. 

Let I be a nonzero ideal of the ring ${\mathcal{O}}_K.$ According to the fundamental theorem of Dedekind rings, $\left(\exists !\right)l\in {\mathbb{N}}^{*},$ the different ideals $P_1,$ $P_2,$..., $P_l$ ∈ Spec$\left({\mathcal{O}}_K\right)$ and ${\alpha }_1,{\alpha }_2,\dots {\alpha }_l\in {\mathbb{N}}^{*}$ such that $I=P_1^{{\alpha }_1}·P_2^{{\alpha }_2}·...·P_l^{{\alpha }_l}.$ Using the properties of the functions ${\phi }_{ext},$ N and ${\tau }_{ext}$ which we specified in the introduction and preliminaries section, we have:

$${\phi }_{ext}\left(I\right)·{\tau }_{ext}\left(I\right)=\prod _{i=1}^l{\phi }_{ext}\left(P_i^{{\alpha }_i}\right)·{\tau }_{ext}\left(P_i^{{\alpha }_i}\right)=$$

$$\prod _{i=1}^l{\left(N\left(P_i\right)\right)}^{{\alpha }_i}·\left(1-\frac{1}{N\left(P_i\right)}\right)·\left({\alpha }_i+1\right).$$

It results that

$${\phi }_{ext}\left(I\right)·{\tau }_{ext}\left(I\right)=N\left(I\right)·\prod _{i=1}^l\frac{N\left(P_i\right)-1}{N\left(P_i\right)}·\left({\alpha }_i+1\right).$$

(1)

It is easy to see that

$$\frac{N\left(P_i\right)-1}{N\left(P_i\right)}·\left({\alpha }_i+1\right)\ge \frac{N\left(P_i\right)-1}{N\left(P_i\right)}·2\ge 1,\left(\forall \right){\alpha }_i\in {\mathbb{N}}^{*},\left(\forall \right)P_i\in Spec\left({\mathcal{O}}_K\right).$$

(2)

From (1) and (2), it results that

$${\phi }_{ext}\left(I\right)·{\tau }_{ext}\left(I\right)\ge N\left(I\right),\left(\forall \right)anonzeroidealIof{\mathcal{O}}_K.$$

□

We are giving another result involving Euler’s function and the function $\tau .$

Proposition 11.

For any positive integer $n,$ the following inequality

$$\frac{3\sqrt{15}}{2}·\phi \left(n\right)\ge \tau \left(n\right)·\sqrt{n}$$

(3)

holds. The equality is obtained only for $n=60.$

Proof. 

For $n=1$, we have $\frac{3\sqrt{15}}{2}\phi \left(1\right)=\frac{3\sqrt{15}}{2}>1=\tau \left(1\right)\sqrt{1}.$ We take $n\ge 2$. By mathematical induction, we proved the inequality

$$\sqrt{p^d}\left(1-\frac{1}{p}\right)\ge d+1,$$

for every $d\ge 1$, where $p\ge 7$ is a prime number. This inequality is in fact the following:

$$\phi \left(p^d\right)\ge \tau \left(p^d\right)\sqrt{p^d}.$$

(4)

We consider the decomposition in prime factors of n given by $n=2^a{3}^b{5}^c{\prod }_{i=1}^s{p}_i^{a_i}$, $p_i\ne 2,3,5$. We know that if the functions $\phi$ and $\tau$ are multiplicative arithmetic functions, then the inequality of the statement becomes

$$\frac{3\sqrt{15}}{2}\phi \left(2^a\right)\phi \left(3^b\right)\phi \left(5^c\right)\prod _{i=1}^s\phi \left(p_i^{a_i}\right)\ge \tau \left(2^a\right)\sqrt{2^a}\tau \left(3^b\right)\sqrt{3^b}\tau \left(5^c\right)\sqrt{5^c}\prod _{i=1}^s\tau \left(p_i^{a_i}\right)\sqrt{p_i^{a_i}}$$

$$=(a+1)(b+1)(c+1)\sqrt{2^a{3}^b{5}^c}\prod _{i=1}^s(a_i+1)\sqrt{p_i^{a_i}}.$$

It is easy to see, by mathematical induction, that for every $a,b,c\ge 1$, we have the following inequalities:

$$2^a\ge \frac{4}{9}{(a+1)}^2,3^b\ge \frac{3}{4}{(b+1)}^2,5^c\ge \frac{5}{4}{(c+1)}^2,$$

which are equivalent to

$$3\sqrt{2^a}\frac{1}{2}\ge a+1,\sqrt{3}\sqrt{3^b}\frac{2}{3}\ge b+1,\frac{\sqrt{5}}{2}\sqrt{5^c}\frac{4}{5}\ge c+1,$$

which means that

$$3\phi \left(2^a\right)\ge \tau \left(2^a\right)\sqrt{2^a},\sqrt{3}\phi \left(3^b\right)\ge \tau \left(3^b\right)\sqrt{3^b},\frac{\sqrt{5}}{2}\phi \left(5^c\right)\ge \tau \left(5^c\right)\sqrt{5^c}.$$

Using the above inequalities and (4), we deduce the inequality of the statement. In the case when $c=0$, we have $n=2^a{3}^b{\prod }_{i=1}^s{p}_i^{a_i}$, so the inequality of the statement becomes

$$\frac{3\sqrt{15}}{2}\phi \left(2^a\right)\phi \left(3^b\right)\prod _{i=1}^s\phi \left(p_i^{a_i}\right)=\frac{\sqrt{5}}{2}3\phi \left(2^a\right)\sqrt{3}\phi \left(3^b\right)\prod _{i=1}^s\phi \left(p_i^{a_i}\right)\ge \frac{\sqrt{5}}{2}\tau \left(n\right)\sqrt{n}>\tau \left(n\right)\sqrt{n}.$$

Analogously, the cases are treated when at least one of the numbers $a,b,c$ is equal to 0. Therefore, the inequality of the statement is true.

Now, we prove that the equality in (3) is obtained only for $n=60.$ For this, we study the equality

$$\frac{3\sqrt{15}}{2}·\phi \left(n\right)=\tau \left(n\right)·\sqrt{n}.$$

(5)

If $n\not\equiv 0$ (mod 15), then $n=15k+r,$ where $k\in$ $\mathbb{N},$ $r\in$ $\left\{1,\dots ,14\right\},$ which means that

$$\sqrt{\frac{15}{15k+r}}=\frac{2\tau \left(15k+r\right)}{3\phi \left(15k+r\right)}\in \mathbb{Q},$$

which is false, because $\sqrt{\frac{15}{15k+r}}$∉$\mathbb{Q}.$ To prove this, we assume by absurdity that $\sqrt{\frac{15}{15k+r}}$∈$\mathbb{Q},$ so there are $a,b$∈${\mathbb{N}}^{*},$ $\left(a,b\right)=1$ such that $\sqrt{\frac{15}{15k+r}}=\frac{b}{a}.$ This implies the equality

$$b^2·\left(15k+r\right)=15a^2.$$

Since $\left(15k+r,15\right)\in \left\{1,3,5\right\}$, we obtain that $b\equiv 0$ (mod 15) when $\left(15k+r,15\right)=1,$ $b\equiv 0$ (mod 5) when $\left(15k+r,15\right)=3,$ respectively, $b\equiv 0$ (mod 3) when $\left(15k+r,15\right)=5.$ In the first case, when $\left(15k+r,15\right)=1,$ we find $b=15b^{\prime },$ where $b^{\prime }$∈$\mathbb{N}.$ Therefore, the above equality becomes

$$15{\left(b^{\prime }\right)}^2·\left(15k+r\right)=a^2.$$

It results that $a\equiv 0$ (mod 15), which is false because $\left(a,b\right)=1.$

In the second case, when $\left(15k+r,15\right)=3,$ we find $r=3r^{\prime }$ and $b=5b^{\prime },$ where $r\in \left\{3,6,9,12\right\},$ $b^{\prime }$∈$\mathbb{N}.$ We obtain that

$$5{\left(b^{\prime }\right)}^2·\left(5k+r^{\prime }\right)=a^2.$$

It results that $a\equiv 0$ (mod 5), which is false because $\left(a,b\right)=1.$

Analogously, we obtain a contradiction in the third case, when $\left(15k+r,15\right)=5.$

If $n\equiv 0$ (mod 15), then we have $n=15k,$ with $k\in$ ${\mathbb{N}}^{*}.$ Replacing it in equality (5), we obtain

$$2\sqrt{k}·\tau \left(15k\right)=3\phi \left(15k\right),$$

which can be written as

$$\sqrt{k}=\frac{3\phi \left(15k\right)}{2\tau \left(15k\right)}\in \mathbb{Q},$$

thus, there exists $q\in {\mathbb{N}}^{*}$ such that $k=q^2.$ Replacing it in the above equality, we deduce the following relation

$$3\phi \left(15q^2\right)=2q\tau \left(15q^2\right).$$

(6)

If $\left(15,q\right)=1,$ then relation (6) becomes

$$3\phi \left(q^2\right)=q\tau \left(q^2\right).$$

(7)

We study equality (7) in two cases:

Case I: when q is a prime number, we obtain

$$3q\left(q-1\right)=3q,$$

it follows that $q=2,$ so $k=4.$ Therefore, we have $n=60.$

Case II: when q is a compose number,

$\phi \left(q^2\right)$ is an even number and $\tau \left(q^2\right)$ is an odd number. It follows from relation (7) that q is an even number, so $q=2^s·v,$ where $s,v\in {\mathbb{N}}^{*}$, v is an odd number. Relation (7) becomes

$$3\phi \left(2^{2s}·v^2\right)=2^s·v·\tau \left(2^{2s}·v^2\right),$$

which implies, taking into account that $(2,v)=1$, the following inequality holds:

$$3·2^{s-1}\phi \left(v^2\right)=v(2s+1)·\tau \left(v^2\right).$$

(8)

For $s\ge 2$, the term from the left part of the equality (8) is an even number and the term $(2s+1)·\tau \left(v^2\right)$ is an odd number, so v is an even number, which is false, because $(2,v)=1$. The case $q=2v$ then remains, where $v\ge 3$ is an odd number. Relation (7) becomes

$$\phi \left(v^2\right)=v\tau \left(v^2\right),$$

but $\phi \left(v^2\right)$ is an even number and $\tau \left(v^2\right)$ is an odd number; thus, we deduce that the number v is an even number, which is false.

If $\left(15,q\right)\ne 1,$ then $q=3^a{5}^b$ or $q=3^a{5}^b{\prod }_{pprimep\ge 7}{p}^c$, where $a,b\in \mathbb{N}$, with $a+b\ge 1$ and $c\in {\mathbb{N}}^{*}$. We note $P={\prod }_{pprime,p\ge 7}{p}^c$. For $q=3^a{5}^b$, relation (6) becomes

$$3\phi \left(3^{2a+1}{5}^{2b+1}\right)=2·3^a{5}^b\tau \left(3^{2a+1}{5}^{2b+1}\right),$$

which is equivalent to $3^{a+1}{5}^b=(a+1)(b+1)$, which is false, because $3^{a+1}{5}^b>(a+1)(b+1)$, and it is easy to see by mathematical induction for $a,b\in \mathbb{N}$, with $a+b\ge 1$. For $q=3^a{5}^{b}P$, relation (6) becomes

$$3\phi \left(3^{2a+1}{5}^{2b+1}{P}^2\right)=2·3^a{5}^{b}P\tau \left(3^{2a+1}{5}^{2b+1}{P}^2\right),$$

which is equivalent to $3^{a+1}{5}^b\phi \left(P^2\right)=(a+1)(b+1)P\tau \left(P^2\right)$, so we obtain

$$3^{a+1}{5}^b\prod _{pprime,p\ge 7}\left(p^c-1\right)=(a+1)(b+1)\prod (2c+1).$$

However, by mathematical induction, we have $p^c-1>2c+1$, where $p\ge 7$ and $c\ge 1$. Combining the above inequalities, we prove that $3^{a+1}{5}^b{\prod }_{pprime,p\ge 7}\left(p^c-1\right)>(a+1)(b+1)\prod (2c+1)$. Consequently, the statement is true. □

Now, we generalize Proposition 11, for extended Euler’s function and the function ${\tau }_{ext}.$

Proposition 12.

Let K be an algebraic number field of degree $[K:\mathbb{Q}]=n,$ where n is a positive integer, $n\ge 2.$ Then:

$$3^{r_1}·{\sqrt{3}}^{r_2}·{\left(\frac{4}{3}\right)}^{r_3}·{\left(\frac{\sqrt{5}}{2}\right)}^{r_4}·{\phi }_{ext}\left(I\right)\ge {\tau }_{ext}\left(I\right)·\sqrt{N\left(I\right)},\left(\forall \right)anonzeroidealIof{\mathcal{O}}_K,$$

where $r_1$ is the number of prime ideals of norm $2,$ which divides $I;$ $r_2$ is the number of prime ideals of norm $3,$ which divides $I;$ $r_3$ is the number of prime ideals of norm $4,$ which divides $I;$ and $r_4$ is the number of prime ideals of norm $5,$ which divides $I.$

Proof. 

Let I be a nonzero ideal of the ring ${\mathcal{O}}_K.$ Applying the fundamental theorem of Dedekind rings, Propositions 3 and 4, it results that $\left(\exists !\right)r_1,r_2,r_3,r_4,l\in \mathbb{N},$ $l\ge 5,$ the different prime ideals $P_{11},$...,$P_{1r_1},$ $P_{21},$...,$P_{2r_2},$ $P_{31},$...,$P_{3r_3},$ $P_{41},$...,$P_{4r_4},$ $P_5,$ $P_6,$..., $P_l$ of the ring ${\mathcal{O}}_K$ and ${\alpha }_{11},...,{\alpha }_{1r_1},{\alpha }_{21},...,{\alpha }_{2r_2},{\alpha }_{31},...,{\alpha }_{3r_3},{\alpha }_{41},...,{\alpha }_{4r_4}\in \mathbb{N},$ ${\alpha }_5,\dots {\alpha }_l\in {\mathbb{N}}^{*}$ such that

$$I=\prod _{i=1}^{r_1}{P}_{1i}^{{\alpha }_{1i}}·\prod _{i=1}^{r_2}{P}_{2i}^{{\alpha }_{2i}}·\prod _{i=1}^{r_3}{P}_{3i}^{{\alpha }_{3i}}·\prod _{i=1}^{r_4}{P}_{4i}^{{\alpha }_{4i}}·P_5^{{\alpha }_5}·...·P_l^{{\alpha }_l},$$

with $N\left(P_{1i}\right)=2,$ $i=\overline{1,r_1},$ $N\left(P_{2i}\right)=3,$ $i=\overline{1,r_2},$ $N\left(P_{3i}\right)=4,$ $i=\overline{1,r_3},$ $N\left(P_{4i}\right)=5$, $i=\overline{1,r_4}$ and $N\left(P_i\right)\ge 7,$ $\left(\forall \right)$ $i=\overline{5,l}.$

Applying the inequality $\sqrt{a^d}·\left(1-\frac{1}{a}\right)\ge d+1,$ $\left(\forall \right)$ $d,a$∈${\mathbb{N}}^{*},$ $a\ge 7$ for $a=N\left(P_i\right)$ we obtain:

$$\sqrt{{\left(N\left(P_i\right)\right)}^d}·\left(1-\frac{1}{N\left(P_i\right)}\right)\ge d+1,\left(\forall \right)d\in {\mathbb{N}}^{*},\left(\forall \right)P_i\in \mathbb{J},N\left(P_i\right)\ge 7.$$

The last inequality is equivalent with

$$\sqrt{{\left(N\left(P_i\right)\right)}^{{\alpha }_i}}·\left(1-\frac{1}{N\left(P_i\right)}\right)\ge {\alpha }_i+1,\left(\forall \right)P_i|I,\left(\forall \right)i=\overline{5,l}.$$

It results that

$$\prod _{i=5}^l\sqrt{{\left(N\left(P_i\right)\right)}^{{\alpha }_i}}·\left(1-\frac{1}{N\left(P_i\right)}\right)\ge \prod _{i=5}^l{\tau }_{ext}\left(P_i^{{\alpha }_i}\right).$$

(9)

Applying the inequality $3\sqrt{2^d}·\left(1-\frac{1}{2}\right)\ge d+1,$ $\left(\forall \right)$d∈${\mathbb{N}}^{*},$ for $N\left(P_{1i}\right)=2$ and for $d={\alpha }_{1i},$ $\left(\forall \right)i=\overline{1,r_1},$ we obtain:

$$3\sqrt{{\left(N\left(P_{1i}\right)\right)}^{{\alpha }_{1i}}}·\left(1-\frac{1}{N\left(P_{1i}\right)}\right)\ge {\alpha }_{1i}+1,\left(\forall \right)i=\overline{1,r_1}.$$

From this last inequality, it results that

$$3^{r_1}·\prod _{i=1}^{r_1}\sqrt{{\left(N\left(P_{1i}\right)\right)}^{{\alpha }_{1i}}}·\left(1-\frac{1}{N\left(P_{1i}\right)}\right)\ge \prod _{i=1}^{r_1}{\tau }_{ext}\left(P_{1i}^{{\alpha }_{1i}}\right).$$

(10)

Applying the inequality $\sqrt{3}·\sqrt{3^d}·\left(1-\frac{1}{3}\right)\ge d+1,$ $\left(\forall \right)$d∈${\mathbb{N}}^{*},$ for $N\left(P_{2i}\right)=3$ and for $d={\alpha }_{2i},$ $\left(\forall \right)i=\overline{1,r_2},$ we obtain:

$$\sqrt{3}\sqrt{{\left(N\left(P_{2i}\right)\right)}^{{\alpha }_{2i}}}·\left(1-\frac{1}{N\left(P_{2i}\right)}\right)\ge {\alpha }_{2i}+1,\left(\forall \right)i=\overline{1,r_2}.$$

From this last inequality, it results that

$${\sqrt{3}}^{r_2}·\prod _{i=1}^{r_2}\sqrt{{\left(N\left(P_{2i}\right)\right)}^{{\alpha }_{2i}}}·\left(1-\frac{1}{N\left(P_{2i}\right)}\right)\ge \prod _{i=1}^{r_2}{\tau }_{ext}\left(P_{2i}^{{\alpha }_{2i}}\right).$$

(11)

Applying the inequality $\frac{4}{3}·\sqrt{4^d}·\left(1-\frac{1}{4}\right)\ge d+1,$ $\left(\forall \right)$d∈${\mathbb{N}}^{*},$ for $N\left(P_{3i}\right)=4$ and for $d={\alpha }_{3i},$ $\left(\forall \right)i=\overline{1,r_3},$ we obtain:

$$\frac{4}{3}\sqrt{{\left(N\left(P_{3i}\right)\right)}^{{\alpha }_{3i}}}·\left(1-\frac{1}{N\left(P_{3i}\right)}\right)\ge {\alpha }_{3i}+1,\left(\forall \right)i=\overline{1,r_3}.$$

From this last inequality, it results that

$${\left(\frac{4}{3}\right)}^{r_3}·\prod _{i=1}^{r_3}\sqrt{{\left(N\left(P_{3i}\right)\right)}^{{\alpha }_{3i}}}·\left(1-\frac{1}{N\left(P_{3i}\right)}\right)\ge \prod _{i=1}^{r_3}{\tau }_{ext}\left(P_{3i}^{{\alpha }_{3i}}\right).$$

(12)

Applying the inequality $\frac{\sqrt{5}}{2}·\sqrt{5^d}·\left(1-\frac{1}{5}\right)\ge d+1,$ $\left(\forall \right)$d∈${\mathbb{N}}^{*},$ for $N\left(P_{4i}\right)=5$ and for $d={\alpha }_{4i},$ $\left(\forall \right)i=\overline{1,r_4},$ we obtain:

$$\frac{\sqrt{5}}{2}\sqrt{{\left(N\left(P_{4i}\right)\right)}^{{\alpha }_{4i}}}·\left(1-\frac{1}{N\left(P_{4i}\right)}\right)\ge {\alpha }_{4i}+1,\left(\forall \right)i=\overline{1,r_4}.$$

From this last inequality, it results that

$${\left(\frac{\sqrt{5}}{2}\right)}^{r_4}·\prod _{i=1}^{r_4}\sqrt{{\left(N\left(P_{4i}\right)\right)}^{{\alpha }_{4i}}}·\left(1-\frac{1}{N\left(P_{4i}\right)}\right)\ge \prod _{i=1}^{r_4}{\tau }_{ext}\left(P_{4i}^{{\alpha }_{4i}}\right).$$

(13)

Multiplying member-by-member inequalities (9)–(13) and applying Propositions 5 and 6, we obtain that

$$3^{r_1}·{\sqrt{3}}^{r_2}·{\left(\frac{4}{3}\right)}^{r_3}·{\left(\frac{\sqrt{5}}{2}\right)}^{r_4}·{\phi }_{ext}\left(I\right)\ge {\tau }_{ext}\left(I\right)·\sqrt{N\left(I\right)},$$

$\left(\forall \right)$ a nonzero ideal I of the ring ${\mathcal{O}}_K.$

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3. Conclusions

Regarding the Number Theory, many papers studied the properties of the Euler totient function and the function that characterizes the number of divisors of a natural number. In this paper, we have presented some arithmetic inequalities that can be extended to inequalities in the algebraic fields theory. If K is an algebraic number field, then we deduce:

$${\tau }_{ext}^2\left(I·J\right)\ge {\tau }_{ext}\left(I^2\right)·{\tau }_{ext}\left(J^2\right),\left(\forall \right)idealsIandJof{\mathcal{O}}_K.$$

For any positive integer $n,$ the following inequality $\phi \left(n\right)·\tau \left(n\right)\ge n$ holds. This inequality has been extended to an algebraic number field K:

$${\phi }_{ext}\left(I\right)·{\tau }_{ext}\left(I\right)\ge N\left(I\right),\left(\forall \right)anonzeroidealIof{\mathcal{O}}_K,$$

where ${\mathcal{O}}_K$ is the ring integers of the field $K.$ Another interesting arithmetic inequality is proven, namely:

$$\frac{3\sqrt{15}}{2}·\phi \left(n\right)\ge \tau \left(n\right)·\sqrt{n},$$

for any positive integer n. This generates the following inequality in an algebraic number field K with the degree $[K:\mathbb{Q}]=n,n\ge 2$:

$$3^{r_1}·{\sqrt{3}}^{r_2}·{\left(\frac{4}{3}\right)}^{r_3}·{\left(\frac{\sqrt{5}}{2}\right)}^{r_4}·{\phi }_{ext}\left(I\right)\ge {\tau }_{ext}\left(I\right)·\sqrt{N\left(I\right)},$$

for all nonzero ideal I of ${\mathcal{O}}_K,$ where $r_1$ is the number of prime ideals of norm $2,$ which divides $I,$ $r_2$ is the number of prime ideals of norm $3,$ which divides $I,$ $r_3$ is the number of prime ideals of norm $4,$ which divides $I,$ and $r_4$ is the number of prime ideals of norm $5,$ which divides $I.$

In future research, we will search for other arithmetic inequalities that can extend to an algebraic field. We can see how some calculations are transferred from the elementary number theory to algebraic fields theory. It should be mentioned that these calculations cannot always be done by analogy.