NON-PÓLYA FIELDS WITH LARGE PÓLYA GROUPS ARISING FROM LEHMER QUINTICS

Abstract

We construct a new family of quintic non-Pólya fields with large Pólya groups. We show that the Pólya number of such a field never exceeds five times the size of its Pólya group. Finally, we show that these non-Pólya fields are nonmonogenic of field index one.

1 Introduction

Let K be an algebraic number field and $\mathcal {O}_K$ be its ring of integers. Let ${Int}(\mathcal {O}_K)=\{f\in K[X] \mid f(\mathcal {O}_K)\subseteq \mathcal {O}_K\}$ be the ring of integer-valued polynomials on $\mathcal {O}_K$ . Then the number field K is said to be a Pólya field if the $\mathcal {O}_K$ -module ${Int}(\mathcal {O}_K)$ has a regular basis, that is, a basis $(f_n)$ such that for each $n\in \mathbb {N}\cup \{0\}$ , degree $(f_n)=n$ (see [26]). For each $n\in \mathbb {N}$ , the leading coefficients of the polynomials in ${Int}(\mathcal {O}_K)$ of degree n together with zero form a fractional ideal of $\mathcal {O}_K$ , denoted by $\mathfrak {J}_n(K)$ . The following result establishes a connection between $\mathfrak {J}_n(K)$ and the Pólya-ness of the number field K.

Proposition 1.1 [1, Proposition II.1.4].

A number field K is a Pólya field if and only if $\mathfrak {J}_n(K)$ is principal for every integer $n\geq 1$ .

It follows immediately from Proposition 1.1 that if the class number of K is one, then K is a Pólya field. However, the converse is not valid in general. That is, if the class number $h_K$ of K is not one, then we cannot decide whether K is a Pólya field or not: for instance, every cyclotomic field is a Pólya field (see [26, Proposition 2.6]).

Let ${Cl}(K)$ denote the ideal class group of K. For each integer $n\geq 1$ , let $[\mathfrak {J}_n(K)]$ be the ideal class in ${Cl}(K)$ corresponding to the fractional ideal $\mathfrak {J}_n(K)$ . The Pólya group ${Po}(K)$ of K is defined to be the subgroup of ${Cl}(K)$ generated by the elements $[\mathfrak {J}_n(K)]$ in ${Cl}(K)$ . Therefore, K is a Pólya field if and only if ${Po}(K)=\{1\}$ .

It is an interesting problem to look for explicit families of number fields that are Pólya or non-Pólya (for example, see [9, 10, 19, 23]). The classification of Pólya fields of low degree is of significant interest. Towards this, Zantema [26] completely characterised quadratic Pólya fields.

Proposition 1.2 [26, Example 3.3].

Let p and q be two distinct odd primes. A quadratic field $\mathbb {Q}(\sqrt {d})$ is a Pólya field if and only if d is of one of the following forms:

1. (1) $d=2$ , or $d=-1$ , or $d=-2$ , or $d=-p$ where $p\equiv 3\pmod 4$ , or $d=p$ ;

2. (2) $d=2p$ , or $d=pq$ where $pq\equiv 1\pmod 4$ and, in both cases, the fundamental unit has norm $1$ if $p\equiv 1\pmod 4$ .

Remark 1.3. The classification of quadratic Pólya fields can also be obtained from Hilbert’s theorems (see [13, Satz 105 and Satz 106]).

Leriche [17] completely classified Galois cubic Pólya fields. In the same article, Leriche also classified cyclic quartic and cyclic sextic Pólya fields. Moreover, she obtained similar classifications for some more families of bi-quadratic and sextic fields (see [17, Theorems 5.1 and 6.2]). Recently, there have been several attempts to provide families of Pólya and non-Pólya fields in the remaining cases of bi-quadratic extensions (see [9, 10, 19, 23]). In [3], the authors constructed a new family of totally real bi-quadratic fields with large Pólya groups.

In this article, we characterise the Pólya-ness of a special family of quintic fields arising from Lehmer qunitics. For each integer $n \in \mathbb {Z}$ , the Lehmer quintic $f_n(x)\in \mathbb {Z}[x]$ is defined by

$$ \begin{align*} f_n(x) & = x^5+n^2x^4-(2n^3+6n^2+10n+10)x^3+(n^4+5n^3+11n^2+15n+5)x^2\\ &\quad +(n^3+4n^2+10n+10)x+1. \end{align*} $$

Let $\theta _n\in \mathbb {C}$ be a root of $f_n(x)=0$ . If we set $K_n=\mathbb {Q}(\theta _n)$ , then $[K_n : \mathbb {Q}]=5$ and the fields $K_n$ are called Lehmer quintic fields [16]. Our main theorem is the following result.

Theorem 1.4. Let $\{K_n\}$ be the family of Lehmer quintic fields. If $m_n=n^4+5n^3+15n^2+25n+25$ is cube free, then:

1. (1) $K_n$ is a Pólya field if and only if $m_n$ is a prime or $m_n=25$ ;

2. (2) ${Po}(K_n)\simeq (\mathbb {Z}/5\mathbb {Z})^{\omega ({m_n})-1}$ , where $\omega (t)$ is the number of distinct prime divisors of t.

Moreover, there are infinitely many non-Pólya fields in the family $\{K_n\}$ .

Let G be a finite group. If $m>1$ is an integer, then the m-rank of G is the maximal integer r such that $(\mathbb {Z}/m\mathbb {Z})^r$ is a subgroup of G. The following folklore conjecture is widely believed to be true but it is still open.

Conjecture 1.5 [5, Conjecture 1.1].

Let $d>1$ and $m>1$ be two integers. Then the m-rank of the class group of K is unbounded when K runs through the number fields of degree $[K:\mathbb {Q}]=d$ .

It is known that when $m=d$ , or more generally when $m\mid d$ , this conjecture follows from class field theory (see [5, Conjecture 1.1]). The following corollary to Theorem 1.4 gives an alternative and elementary proof of the conjecture for the case $m=d=5$ .

Corollary 1.6. The $5$ -ranks of the class groups of the non-Pólya Lehmer quintic fields $K_n$ are unbounded.

Recall the classical embedding problem: is every number field K contained in a field L with class number one? In 1964, Golod and Shafarevich [6] gave a negative answer to this question. The corresponding embedding problem for Pólya fields was confirmed affirmatively by Leriche [18]. Leriche proved that every number field is contained in a Pólya field, namely its Hilbert class field (see, [18, Theorem 3.3]).

A minimal Pólya field over K is a field extension L of K which is a Pólya field and such that no intermediate field $K\subseteq M\subsetneq L$ is a Pólya field.

Definition 1.7. [18, Definition 6.1] The Pólya number of K is

$$ \begin{align*}{po}_K=\min\{[L : K]\mid K\subseteq L,\: L\text{ is a P\'olya field}\}.\end{align*} $$

We study the Pólya number ${po}_{K_n}$ of the non-Pólya field $K_n$ and obtain an upper bound for ${po}_{K_n}$ in terms of the size of the corresponding Pólya group ${ Po}(K_n)$ .

Theorem 1.8. Let $K_n$ be the family of Lehmer quintic fields such that $m_n$ is cube-free. Then ${po}_{K_n}\leq 5|{Po}(K_n)|$ .

Let K be a number field and $\theta \in \mathcal {O}_K$ be a primitive element. The index $[\mathcal {O}_K:\mathbb {Z}[\theta ]]$ is called the index of $\theta $ in K and is denoted by $I(\theta )$ . The index of the number field K is defined by $I(K)=\gcd \{I(\theta )\mid \theta \in \mathcal {O}_K \mbox { and }K=\mathbb {Q}(\theta )\}$ . If $I(K)>1$ , then the number field K is not monogenic, that is, $\mathcal {O}_K\neq \mathbb {Z}[\theta ]$ for any $\theta \in K$ . However, the converse is not true in general. That is, there exist nonmonogenic number fields K with $I(K)=1$ . These are basically fields which are not monogenic, but not for a local reason (see [25] for more details). In this direction, we prove the following result.

Theorem 1.9. Let $\{K_n\}$ be the family of non-Pólya Lehmer quintic fields such that $m_n$ is cube-free. Then $K_n$ is not monogenic and $I(K_n)=1$ .

In Section 2, we develop some preliminaries required to prove Theorem 1.4. Section 3 contains the proofs of Theorem 1.4 and Corollary 1.6. In Section 4, we study the Pólya numbers of Lehmer quintic fields and prove Theorem 1.8. In the same section, we also study the monogenicity of the non-Pólya number fields $K_n$ and give the proof of Theorem 1.9. Finally, in Section 5, we present some computations performed with SageMath.

2 Preliminaries

In this section, we assume that the number field K is a finite Galois extension of $\mathbb {Q}$ and for any prime number p, we denote the ramification index of p in $K/\mathbb {Q}$ by $e_p$ .

In [2], Chabert obtained a nice description for the cardinality of ${Po}(K)$ for cyclic extensions $K/\mathbb {Q}$ .

Proposition 2.1 [2, Corollary 3.11].

Assume that the extension $K/\mathbb {Q}$ is cyclic of degree n.

1. (1) If K is real and $N(\mathcal {O}_K^{\times })=\{1\}$ , then $|{Po}(K)|=(\prod _p e_p)/(2n)$ .

2. (2) In all other cases, $|{Po}(K)|=(\prod _p e_p)/n$ .

When K is a cyclic number field of odd degree, all ramification indices $e_p$ are odd and case (1) of Proposition 2.1 does not occur. We record this in the following corollary.

Corollary 2.2. If $K/\mathbb {Q}$ is a cyclic extension of degree n and n is odd, then $|{Po}(K)|=(\prod _p e_p)/n$ .

Zantema [26, Section 3] showed that ${Po}(K)$ is the subgroup of ${Cl}(K)$ generated by the classes of the ambiguous ideals of K. In other words,

$$ \begin{align*} {Po}(K)=\{[\mathfrak{a}]\in {Cl}(K) : \mathfrak{a}^\tau=\mathfrak{a}\text{ for all }\tau\in \mathrm{Gal}(K/\mathbb{Q})\}. \end{align*} $$

Next, we state some results on Lehmer quintics and their discriminants. In [22], Schoof and Washington showed that $f_n(x)$ is irreducible for all $n\in \mathbb {Z}$ and its discriminant is $(n^3+5n^2+10n+7)^2(n^4+5n^3+15n^2+25n+25)^4$ . Let $\theta _n\in \mathbb {C}$ be a root of $f_n(x)=0$ . If we set $K_n=\mathbb {Q}(\theta _n)$ , then $K_n$ is a cyclic field for all $n\in \mathbb {Z}$ [22, Theorem 3.5]. We denote the ring of integers of $K_n$ by $\mathcal {O}_{K_n}$ . Now we recall some results of Jeannin [15] on the discriminant $d(K_n)$ of $K_n$ .

Lemma 2.3 [15, Lemme 2.1.1].

All the prime divisors $p \neq 5$ of $n^4+5n^3+15n^2+25n+25$ satisfy $p \equiv 1 \pmod 5$ .

Lemma 2.4 [15, Théorème 2.2.1].

The discriminant $d(K_n)=f(K_n)^4$ , where the conductor $f(K_n)$ of $K_n$ is given by

(2.1) $$ \begin{align} f(K_n)=5^b\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\prod_{\substack{p\equiv1\pmod5\\ v_p(n^4+5n^3+15n^2+25n+25)\not\equiv0\pmod5}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!p.\end{align} $$

Here $v_p(k)$ denotes the exponent of the highest power of the prime p dividing a nonzero integer k and

(2.2) $$ \begin{align} b= \begin{cases} 0 & \text{ if } 5\nmid n,\\ 2 & \text{ if } 5\mid n. \end{cases} \end{align} $$

We quote the following result due to Erdős [4] which plays a crucial role in the proof of our main theorem.

Theorem 2.5 [4, Section 1].

Let $f(x)$ be a polynomial of degree $d\geq 3$ whose coefficients are integers with highest common factor $1$ and whose leading coefficient is positive. Assume that $f(x)$ is not divisible by the $(d-1)$ th power of a linear polynomial with integer coefficients. Then there are infinitely many positive integers n for which $f(n)$ is $(d-1)$ th power free.

Next we state a deep result on power-free values of polynomials (see [11, 12, 21]).

Theorem 2.6 [21, Theorem 1].

Let $f(x)\in \mathbb {Z}[x]$ be an irreducible polynomial of degree $d \geq 3$ and assume that f has no fixed $(d-1)$ th power prime divisor. Define

$$ \begin{align*}N_f'(X)=\#\{p\leq X : p \text{ prime}, f(p) \text{ is } (d-1)\text{-free}\}.\end{align*} $$

Then, for any $C>1$ ,

$$ \begin{align*}N_f'(X)=c_f'\pi(X)+O_{C,f}\bigg(\frac{X}{(\log X)^C}\bigg),\end{align*} $$

as $x\rightarrow \infty $ , where

$$ \begin{align*}c_f'=\prod_{p}\bigg(1-\frac{\rho'(p^{d-1})}{\phi(p^{d-1})}\bigg)\end{align*} $$

and $\rho '(d)=\#\{n \pmod d : (d,n)=1, d \mid f(n) \}$ .

Let $f(x)$ be an irreducible polynomial with integral coefficients and $f(m)>0$ for $m=1,2,\ldots .$ Let $\omega (m)$ denote the number of distinct primes dividing m. For primes p, the following result due to Halberstam [8] determines the distribution of values of $\omega (f(p))$ .

Theorem 2.7 [8, Theorem 2].

Let $f(X) \in \mathbb {Z}[X]$ be any nonconstant polynomial. For all but $o(X/\! \log X)$ primes $p \leq X$ ,

$$ \begin{align*} \omega(f(p))=(1+o(1)){\log \log X}. \end{align*} $$

Now we state some results on the number of integral solutions of a Diophantine equation of the type

(2.3) $$ \begin{align} Y^m=f(X). \end{align} $$

When $m=2$ and $f(x)$ is a monic quartic polynomial, the following result due to Masser [20] gives a specific bound for integral points on the curve.

Theorem 2.8 [20].

Consider the Diophantine equation $Y^2 = f(X)$ , where $f(X)$ is a polynomial of degree four with integer coefficients. Assume that $f(X)$ is monic and its discriminant is not a perfect square. Then any integer solution $(x, y)$ of the equation satisfies $|x| \leq 26 H(f)^3$ , where $H(f)$ denotes the maximum of the absolute values of the coefficients of $f(X)$ .

3 Proof of Theorem 1.4

Proof. We consider the set

$$ \begin{align*}\mathcal{P}=\{n \in \mathbb{Z}: m_n=n^4+5n^3+15n^2+25n+25 \mbox{ is a cube-free integer} \}.\end{align*} $$

For $n \in \mathcal {P}$ ,

(3.1) $$ \begin{align} m_n=5^bAB^2. \end{align} $$

Here $b=0$ if n is not divisible by $5$ and $b=2$ otherwise, and $A,B$ are square-free natural numbers which are relatively prime and $5 \nmid AB$ . From Lemma 2.3 and (2.1),

(3.2) $$ \begin{align} f(K_n)=5^bAB \quad\mbox{and}\quad d(K_n)=(5^bAB)^4. \end{align} $$

Since $K_n / \mathbb {Q}$ is Galois and of degree $5$ , we see that for any prime p, the ramification index $e_p$ of p in $K_n$ is given by

$$ \begin{align*}e_p= \begin{cases} 5 & \mbox{ if } p\mid5AB,\\ 1 & \mbox{ otherwise}. \end{cases} \end{align*} $$

Thus,

(3.3) $$ \begin{align} \prod_pe_p=5^{\omega(d(K_n))}=5^{\omega(m_n)}. \end{align} $$

Now from Corollary 2.2,

(3.4) $$ \begin{align} |{Po}(K_n)|=5^{\omega(m_n)-1}. \end{align} $$

Thus, for $n \in \mathcal {P}$ , the Lehmer quintic field $K_n$ is a Pólya field if and only if $m_n$ is a prime or a square of a prime. We claim that $m_n$ is a square of a prime if and only if $m_n=25$ . This claim will prove Theorem 1.4(1). To prove the claim, consider the curve

(3.5) $$ \begin{align} Y^2=f(X)=X^4+5X^3+15X^2+25X+25. \end{align} $$

From Theorem 2.8, any integral solution $(x,y)$ of (3.5) satisfies

$$ \begin{align*}|x|\leq 26\times25^3=406250.\end{align*} $$

Using a SageMath program, we find that for $x\in [-406250,406250]$ , the only integral point on the curve $Y^2=f(X)$ is $(X,Y)=(0,5)$ . In other words, $m_n$ is not a square for any nonzero integer n unless $m_n=25$ . This establishes the claim.

We have $\mathrm {Gal}(K_n/\mathbb {Q})\simeq \mathbb {Z}/5\mathbb {Z}$ . Let $\sigma $ be a generator of $\mathrm {Gal}(K_n/\mathbb {Q})$ and $[\mathfrak {I}]\neq [1]$ be an ambiguous ideal class in $K_n$ . Then

$$ \begin{align*} [\mathfrak{I}]^5=[\mathfrak{I}][\mathfrak{I}][\mathfrak{I}][\mathfrak{I}][\mathfrak{I}] =[\mathfrak{I}][\mathfrak{I}]^\sigma[\mathfrak{I}]^{\sigma^2}[\mathfrak{I}]^{\sigma^3}[\mathfrak{I}]^{\sigma^4} =[N(\mathfrak{I})] =[1], \end{align*} $$

where $N(\mathfrak {I})\in \mathbb {Q}$ denotes the norm of the ideal $\mathfrak {I}$ . We conclude that the order of any nontrivial ambiguous ideal class in the class group of $K_n$ is $5$ . From the structure theorem for abelian groups,

$$ \begin{align*}{Po}(K_n)\simeq\bigg(\frac{\mathbb{Z}}{5\mathbb{Z}}\bigg)^{\omega({m_n})-1}.\end{align*} $$

This completes the proof of Theorem 1.4(2).

From (3.1), we see that $m_n$ cannot be a prime whenever $5\mid n$ . Thus, $K_n$ is a non-Pólya field whenever $n \neq 0, 5\mid n$ and $n \in \mathcal {P}$ . Next we show that there are infinitely many such n. To do this, we show that there are infinitely many k such that $m_{5k}$ is cube-free. Note that

(3.6) $$ \begin{align} m_{5k}&=(5k)^4+5(5k)^3+15(5k)^2+25(5k)+25\nonumber\\ &=25(25k^4+25k^3+15k^2+5k+1)=25g(k). \end{align} $$

If $h(k)=ak+b$ is a linear polynomial such that $h(k)^3\mid g(k)$ , then for $t=-b/a\in \mathbb {Q}$ ,

(3.7) $$ \begin{align} \kern10.5pt g(t)&=0\!\implies\! 25t^4+25t^3+15t^2+5t+1=0, \end{align} $$

(3.8) $$ \begin{align} \kern2pt g'(t)&=0\!\implies\! 100t^3+75t^2+30t+5=0,\ \quad \end{align} $$

(3.9) $$ \begin{align} g"(t)&=0\!\implies\! 300t^2+150t+30=0.\qquad\quad \end{align} $$

This contradicts the fact that $t\in \mathbb {Q}$ . Thus, from Theorem 2.5, it follows that $g(k)$ is cube-free for infinitely many k. Since $5\nmid g(k)$ for all k, it follows that $m_{5k}$ is cube-free for infinitely many integers k. This proves that $\mathcal {P}$ is an infinite set and completes the proof of the theorem.

Remark 3.1. From the proof of Theorem 1.4, it follows that for any $n \neq 0$ , the Lehmer quintic field $K_{5n}$ is non-Pólya whenever $m_{5n}$ is cube-free. However, there are non-Pólya fields $K_{5n}$ with $m_{5n}$ not being cube-free (see the entry for $n=-53$ in Table 1).

Table 1 Family of non-Pólya fields.

Conjecturally, there are infinitely many $n \in \mathbb {Z}$ such that $m_n$ is prime and thus the family $\{K_n\}$ should have infinitely many Pólya fields. Under the assumption that $m_n$ is cube-free, Theorem 1.4 asserts that there are infinitely many Pólya fields in the family $K_n$ only if there are infinitely many primes of the form $m_n$ .

Proof of Corollary 1.6.

From the above remark, it is enough to find integers n such that $m_{5n}$ is cube-free and $\omega (m_{5n})$ goes to infinity as n goes to infinity. Let $g(k)=(25k^4+25k^3+15k^2+5k+1)$ . From Theorem 2.6, for a positive proportion of primes p, we see that $g(p)$ is cube-free. Consequently, $m_{5p}$ is cube-free for a positive proportion of prime numbers p. Now, we only consider those primes p such that $m_{5p}$ is cube-free. There is a positive constant c such that for any large real number X, there are at least $c {X}/{\log X}$ many primes. From Theorem 2.7, it follows that $\omega (m_{5p})$ goes to infinity as p goes to infinity.

4 Pólya numbers and monogenicity of Lehmer quintic fields

The genus field (respectively, genus field in the narrow sense) of K is the maximal abelian extension $\Gamma _K$ (respectively, $\Gamma _K'$ ) of K which is a compositum of K with an absolute abelian number field and is unramified over K at all places (respectively, all finite places) of K. The genus number of K is defined to be the degree ${g_K=[\Gamma _K : K]}$ . If K is abelian, then Leriche showed that the genus field $\Gamma _K$ is Pólya and hence

(4.1) $$ \begin{align} {po}_K\leq g_K, \end{align} $$

where ${po}_K$ denotes the Pólya number of K. However, Zantema proved that both the cyclotomic and real cyclotomic fields are Pólya fields [26]. Thus, for abelian number fields K, if f is the conductor of K and $\phi (f)$ is the value of the Euler totient function, then

(4.2) $$ \begin{align} {po}_K\leq\frac{\phi(f)}{[K : \mathbb{Q}]}\quad\text{and}\quad {po}_K\leq\frac{\phi(f)}{2[K : \mathbb{Q}]}\quad\text{if}\ K\ \text{is real}. \end{align} $$

For the general case, when K is a Galois number field (not necessarily abelian) with class number $h_K$ ,

(4.3) $$ \begin{align} {po}_K\leq h_K. \end{align} $$

To prove Theorem 1.8, we need the following result due to Ishida [14] on the genus number of a cyclic number field of prime degree.

Theorem 4.1 [14, Theorem 5].

Let K be a cyclic number field of degree q, where q is an odd prime. If t is the number of primes p such that p is totally ramified in K, then the genus number $g_K$ of K is

$$ \begin{align*}g_K=\begin{cases} q^t & \text{if } q \text{ is totally ramified in }K, \\ q^{t-1} & \text{otherwise. } \end{cases}\end{align*} $$

Proof of Theorem 1.8.

We have already seen in the proof of Theorem 1.4 that the number of primes p such that p is totally ramified in $K_n$ is $\omega (m_n)$ . Now applying Theorem 4.1 to the family of number fields $K_n$ ,

(4.4) $$ \begin{align} g_{K_n}=5^{\omega(m_n)} \implies {po}_{K_n}\leq 5^{\omega(m_n)}. \end{align} $$

Substituting (3.4) in (4.4),

$$ \begin{align*} {po}_{K_n}\leq 5|Po(K_n)|.\\[-38pt]\end{align*} $$

Remark 4.2. Generally, ${po}_{K}$ and ${Po}(K)$ are mutually independent, but here in the case of non-Pólya Lehmer quintic fields, we have an unexpected relation.

To prove Theorem 1.9, we need the following result of Gras [7] on the monogenicity of cyclic number fields of prime degree.

Proposition 4.3 [7, Section 5].

If K is a cyclic number field of prime degree $p\geq 5$ , then K is monogenic only if $2p+1$ is prime and it is the maximal real subfield of the cyclotomic field $\mathbb {Q}(\zeta _{2p+1})$ .

Lastly, we recall a result of von Zylinski [24].

Proposition 4.4 [24].

If K is a number field of degree n, then $I(K)$ has only prime divisors p satisfying $p<n$ .

Proof of Theorem 1.9.

Let $K_n$ be the family of non-Pólya Lehmer quintic fields. We know that $\mathrm {Gal}(K_n/\mathbb {Q})\simeq \mathbb {Z}/5\mathbb {Z}$ . From Theorem 1.4,

(4.5) $$ \begin{align} |{Po}(K_n)|\geq5. \end{align} $$

Since real cyclotomic fields are Pólya fields, $K_n$ never occurs as the maximal real subfield of a cyclotomic field. From Proposition 4.3, it follows that $K_n$ is not monogenic.

Next, we aim to show $I(K_n)=1$ for all nonzero n for which $K_n$ is non-Pólya. We recall the relation

(4.6) $$ \begin{align} d(\theta_n)=[I(\theta_n)]^2d(K_n). \end{align} $$

As mentioned earlier,

$$ \begin{align*}d(\theta_n)=(n^3+5n^2+10n+7)^2(n^4+5n^3+15n^2+25n+25)^4.\end{align*} $$

From Lemma 2.3, $(n^4+5n^3+15n^2+25n+25)$ is not divisible by $2$ or $3$ . It is easily seen that $(n^3+5n^2+10n+7)$ is also not divisible by $2$ or $3$ . Consequently, we conclude that $I(\theta _n)$ is not divisible by $2$ or $3$ . Now, from the result of Zylinski, it follows that $I(\theta _n)=1$ .

5 Computation

The computations summarised in Table 1 show that there are many non-Pólya fields in the family $K_{5n}$ . For $n \in \{-60,-59, \ldots , 59,60\}$ , we see that $m_{5n}$ is cube-free with the only exception occurring at $n=-53$ . For all n in this range, the Pólya group is nontrivial. In fact, for $n=-53$ , $m_{5n}$ is not cube-free but the field $K_{5n}$ is non-Pólya. In Table 1, $C_{m_{5n}}$ denotes the cube part of $m_{5n}$ . We performed the computations using the SageMath software. The program can be obtained by writing to the authors.