2 Preliminaries

Let K be a totally real number field with degree d and let $\sigma _1,\ldots ,\sigma _d$ be the embeddings of K in an algebraic closure. By the symbols $\text {Tr}(\alpha )$ and $N(\alpha )$ , we mean the algebraic trace and norm of $\alpha \in K$ over K. An element $\alpha \in K$ is totally positive if $\sigma _i(\alpha )>0$ for $1\leq i\leq d$ . Let $\mathcal {O}_K$ be the ring of integers of K. The subset of totally positive elements of $\mathcal {O}_K$ will be denoted by $\mathcal {O}_K^+$ . An element $\alpha \in \mathcal {O}_K^+$ is indecomposable if $\alpha \neq \beta +\gamma $ for any $\beta ,\gamma \in \mathcal {O}_K^+$ .

A quadratic form over a number field K is a homogeneous polynomial of degree two with coefficients in its ring of integers $\mathcal {O}_K$ , that is, a polynomial of the form

$$ \begin{align*}Q(x_1,\ldots,x_n)=\sum_{r,s=1}^na_{rs}x_rx_s \quad \mbox{for every }a_{rs}\in\mathcal{O}_K.\end{align*} $$

If $a_{rs}\in \mathbb {Z}$ for $1\leq r,s\leq n$ , we say that Q is a $\mathbb {Z}$ -form. From now on, it is assumed that all quadratic forms Q are totally positive definite, that is, $Q(x_1,\ldots ,x_n)$ is totally positive as an element of $\mathcal {O}_K$ for every $(x_1,\ldots ,x_n)\in \mathcal {O}_K^n$ with $(x_1,\ldots ,x_n)\neq (0,\ldots ,0)$ . Such a form is universal if it represents all elements in $\mathcal {O}_K^+$ .

A simplest cubic field is a number field of the form $K=\mathbb {Q}(\rho )$ , where $\rho $ is a root of the polynomial

$$ \begin{align*}f(x)=x^3-ax^2-(a+3)x-1 \quad \mbox{for every }a\in\mathbb{Z}_{\geq -1}.\end{align*} $$

The other roots of f are

$$ \begin{align*}\rho'=-1-\frac{1}{\rho}=a+2+a\rho-\rho^2,\quad\rho"=-\frac{1}{1+\rho}=-2-(a+1)\rho+\rho^2.\end{align*} $$

We see immediately that K is a cyclic cubic extension of $\mathbb {Q}$ (and, therefore, a totally real number field). Using the above relations and Vieta’s formulas, it is easy to compute the coefficients of the minimal polynomial for every $\alpha \in {\mathbb Q}(\rho )$ . If this minimal polynomial is of the form $x^3-Ax^2+Bx-C$ for some $A,B,C>0$ , Descartes’ rule of signs implies that $\alpha $ is totally positive. If $a\geq 7$ , we can assume without loss of generality that

$$ \begin{align*}a+1<\rho<a+1+\frac{2}{a},\quad-1-\frac{1}{a+1}<\rho'<-1-\frac{1}{a+2},\quad-\frac{1}{a+2}<\rho"<-\frac{1}{a+3}.\end{align*} $$

The first and third estimates come from [Reference Lemmermeyer and Pethö9]; the second can be easily verified.

Let us denote and write for the index of $\rho $ in K. With this notation, $\mathrm {disc}(f)=\Delta ^2$ . Recall that $\mathrm {disc}(f)=\delta ^2\mathrm {disc}(K)$ , where we write $\mathrm {disc}(K)$ for the discriminant of K. The conductor-discriminant formula is $\mathrm {disc}(K)=\mathfrak {c}^2$ , where $\mathfrak {c}$ is the conductor of K. Hence, the index of $\rho $ may be expressed as

$$ \begin{align*}\delta=\frac{\Delta}{\mathfrak{c}}.\end{align*} $$

In statement (3), $v_p(x)$ stands for the p-adic valuation of the integer number x: the greatest power of p that is a divisor of x.

It is possible to determine explicitly the value of the conductor $\mathfrak {c}$ from the factorisation of $\Delta $ . We write $\Delta =bc^3$ with $b,c>0$ integers and b cube-free. From [Reference Kashio and Sekigawa7, Remark 1.5],

$$ \begin{align*}\mathfrak{c}=\begin{cases}\prod_{p\mid b}p & \mbox{if }3\nmid a\mbox{ or }a\equiv12\ (\mathrm{mod}\,27),\\ 3^2\prod_{p\mid b,p\neq3}p & \mbox{otherwise}.\end{cases}\end{align*} $$

3 Bases of nonmonogenic simplest cubic fields

Let $K=\mathbb {Q}(\rho )$ be a nonmonogenic simplest cubic field. An integral basis of K can be taken of the form

(3.1) $$ \begin{align} B_{n_1,n_2}(j,k,l)=\bigg\{1,\frac{j+\rho}{n_1},\frac{k+l\rho+\rho^2}{n_2}\bigg\}=\{g_1,g_2,g_3\}, \end{align} $$

for $n_1,n_2\in \mathbb {Z}_{>0}$ , $n_1n_2=\delta $ , $0\leq j\leq n_1-1$ and $1\leq k,l\leq n_2-1$ . Hashimoto and Aoki [Reference Hashimoto and Aoki3, Proposition 1] derived another form for an integral basis of simplest cubic fields.

In this section, we determine the possible values for $\delta $ and, for each of these, how to determine the explicit form of an integral basis as in (3.1). We already know the answer when $\delta $ is a prime number p. If $p\neq 3$ , then, from the form of $\mathfrak {c}$ and $\Delta =p\mathfrak {c}$ , it follows that $p^2\mid \Delta $ . This is true only for $p\equiv 1\ (\mathrm {mod}\,6)$ (see [Reference Gil-Muñoz and Tinková2, Proposition 3.15]).

Next, we move on to the general case. We need the following lemmas.

Proof. If $p=3$ and $s=3$ , then, necessarily, $v_3(\Delta )=3$ and $a\equiv 12\ (\mathrm {mod}\,27)$ (see [Reference Kashio and Sekigawa7, Proposition 3.2]). It can be easily checked that ${(2+\rho )}/{3}$ is a root of

$$ \begin{align*} x^3-\bigg(\frac{a}{3}+2\bigg)x^2+\bigg(\frac{a}{3}+1\bigg)x-\frac{2a+3}{27}. \end{align*} $$

Thus, it is clearly an algebraic integer for these cases of a.

We will proceed with $p\neq 3$ . Assume that $\lfloor {s}/{3}\rfloor \geq 1$ since, otherwise, our claim is trivial. The trace of ${(j+\rho )}/{p^{s_1}}$ is equal to ${(a+3j)}/{p^{s_1}}$ , which is a rational integer if $j\equiv -3^{-1}a\;(\text {mod }p^{s_1})$ . For these cases of p, the inverse of $3$ modulo $p^{s_1}$ exists and we denote it by $3^{-1}$ . The next coefficient of the minimal polynomial of ${(j+\rho )}/{p^{s_1}}$ is

$$ \begin{align*} \frac{-a-3+2aj+3j^2}{p^{2s_1}}. \end{align*} $$

Let $j=-3^{-1}a+Up^{s_1}$ for some $U\in {\mathbb Z}$ . Then $-a-3+2aj+3j^2=-3^{-1}\Delta +3Up^{2s_1}$ . This is clearly $0$ modulo $p^{2s_1}$ since $v_p(\Delta )\in \{s,s+1\}$ and $s_1\leq \lfloor {s}/{3}\rfloor $ . Thus, this coefficient is a rational integer. Finally, the norm of ${(j+\rho )}/{p^{s_1}}$ is

$$ \begin{align*} \frac{1-(a+3)j+aj^2+j^3}{p^{3s_1}}. \end{align*} $$

For $j=-3^{-1}a+Up^{s_1}$ , we obtain

$$ \begin{align*} 1-(a+3)j+aj^2+j^3=3^{-3}(2a+3)\Delta-3^{-1}\Delta Up^{s_1}+U^3p^{3s_1} \equiv 0 \pmod{p^{3s_1}}, \end{align*} $$

since $p^{3s_1} \mid \Delta $ . Therefore, for $j\equiv -3^{-1}a\;(\text {mod }p^{s_1})$ , we get an algebraic integer.

Proof. If $p=3$ and $s=3$ , we already know that $a\equiv 12\;(\text {mod }27)$ . The element ${(1+7\rho +\rho ^2)}/{9}$ is a root of

$$ \begin{align*} x^3-\frac{a^2+9a+9}{9}x^2-\frac{4(a+6)^2}{3^4}x+\frac{5a^2+15a-171}{3^6}. \end{align*} $$

The coefficient ${(a^2+9a+9)}/{9}$ is obviously a rational integer. The same is true for ${(4(a+6)^2)}/{3^4}$ since $a+6\equiv 18\;(\text {mod }27)$ . Let $a=12+27\overline {a}$ for some $\overline {a}\in {\mathbb Z}$ . Then

$$ \begin{align*} 5a^2+15a-171=3^6(5\overline{a}^2+5\overline{a}+1). \end{align*} $$

Therefore, the norm is also a rational integer. The proof for the remaining two elements is similar.

Now, let $p\neq 3$ . Here, we can closely follow [Reference Gil-Muñoz and Tinková2, Proposition 3.8]. The expression for the trace of ${(k+l\rho +\rho ^2)}/{p^{s_2}}$ leads to the condition

$$ \begin{align*} k=-3^{-1}(al+a^2+2a+6)+Up^{s_2} \end{align*} $$

for some $U\in {\mathbb Z}$ . For this k, the numerator of the second coefficient of the minimal polynomial is

$$ \begin{align*} -3^{-1}(a^2+a+1+(2a+1)l+l^2)\Delta+3U^2p^{2s_2}. \end{align*} $$

Therefore, we must have $p^{2s_2-v_p(\Delta )} \mid a^2+a+1+(2a+1)l+l^2$ . As in [Reference Gil-Muñoz and Tinková2, Proposition 3.8], the numerator of the norm of ${(k+l\rho +\rho ^2)}/{p^{s_2}}$ can be expressed as

$$ \begin{align*} 27^{-1}g(l)-3^{-1}\Delta U(a^2+a+1+(2a+1)l+l^2)p^{s_2}+U^3p^{3s_2}, \end{align*} $$

where

$$ \begin{align*} g(l) &=\Delta((2a^2-3)\Delta+8(2a+3)\\ &\quad +((6a-3)\Delta-12a+36)l +(6\Delta-6a-36)l^2+(2a+3)l^3). \end{align*} $$

Clearly, we still need $p^{2s_2-v_p(\Delta )} \mid a^2+a+1+(2a+1)l+l^2$ and, moreover, $p^{3s_2}|g(l)$ . The latter is true if

$$ \begin{align*} l\equiv -(2a+3)^{-1}(2a^2+4a+6)\;(\text{mod }p^{3s_2-v_p(\Delta)}). \end{align*} $$

It remains to check the first part. If $l=-(2a+3)^{-1}(2a^2+4a+6)+Vp^{3s_2-v_p(\Delta )}$ for some $V\in {\mathbb Z}$ , then

$$ \begin{align*} a^2+a+1+(2a+1)l+l^2=3(2a+3)^{-2}\Delta-9(2a+3)^{-1}Vp^{3s_2-v_p(\Delta)}+V^2p^{6s_2-2v_p(\Delta)}. \end{align*} $$

This expression is divisible by $p^{2s_2-v_p(\Delta )}$ if $p^{2s_2-v_p(\Delta )} \mid \Delta $ , which is valid for ${s_2\leq s-\lfloor {s}/{3}\rfloor }$ . This completes the proof.

We can use the above lemmas to find an explicit basis as in (3.1) for any simplest cubic field K. If $\delta =p_1^{s_1}\cdots p_m^{s_m}$ for primes $p_i$ , then we can take $n_1=p_1^{s_{1,1}}\cdots p_m^{s_{1,m}}$ and $n_2=p_1^{s_{2,1}}\cdots p_m^{s_{2,m}}$ , where $s_{1,i}=\lfloor {s_i}/{3}\rfloor $ and $s_{2,i}=s_i-\lfloor {s_i}/{3}\rfloor $ . To prove that $n_1$ and $n_2$ can be chosen this way, it is enough to check that there exist $j,k,l$ with $0\leq j\leq n_1-1$ and $1\leq k,l\leq n_2-1$ such that $g_2$ and $g_3$ are algebraic integers. For example, for $g_2$ , there exists $j_i$ with $0\leq j_i\leq p_i^{s_{1,i}}-1$ such that ${(j_i+\rho )}/{p^{s_{1,i}}}$ is an algebraic integer for each i, by Lemma 3.2. We will proceed by induction on i. Assume that ${(j_{i+1}+\rho )}/{p^{s_{1,i+1}}}$ and ${(J_i+\rho )}/{p^{s_{1,1}}\cdots p^{s_{1,i}}}$ are algebraic integers, where $1\leq J_i\leq p^{s_{1,1}}\cdots p^{s_{1,i}}-1$ . Since $p^{s_{1,i+1}}$ and $p^{s_{1,1}}\cdots p^{s_{1,i}}$ are coprime, by Bezout’s identity, there exist $u,v\in {\mathbb Z}$ such that $up^{s_{1,i+1}}+vp^{s_{1,1}}\cdots p^{s_{1,i}}=1$ . Then

$$ \begin{align*} v\frac{j_{i+1}+\rho}{p^{s_{1,i+1}}}+u\frac{J_i+\rho}{p^{s_{1,1}}\cdots p^{s_{1,i}}}=\frac{vj_{i+1}p^{s_{1,1}}\cdots p^{s_{1,i}}+uJ_ip^{s_{1,i+1}}+\rho}{p^{s_{1,1}}\cdots p^{s_{1,i+1}}}. \end{align*} $$

Since the summands are algebraic integers, the element on the right-hand side is also an algebraic integer. Moreover, we can take $J_{i+1}$ as $vj_{i+1}p^{s_{1,1}}\cdots p^{s_{1,i}}+uJ_ip^{s_{1,i+1}}$ modulo $p^{s_{1,1}}\cdots p^{s_{1,i+1}}$ . We can use similar arguments for $g_3$ .

4 Lifting problem for universal quadratic forms

In this section, we prove the nonexistence of $\mathbb {Z}$ -forms over a simplest cubic field $K=\mathbb {Q}(\rho )$ . As highlighted in the introduction, we will exhibit a nonunit $\alpha \in \mathcal {O}_K^+$ with minimal trace one and apply [Reference Kala and Yatsyna6, Lemma 4.6]. The elements with minimal trace one are, in particular, indecomposables of K.

We now summarise the method of [Reference Kala and Tinková5] to locate indecomposables of K. We write $B=B_{n_1,n_2}(j,k,l)=\{g_1,g_2,g_3\}$ for the integral basis of K, as above, and embed $K\hookrightarrow \mathbb {R}^3$ by means of this integral basis: that is,

$$ \begin{align*}\begin{array}{ccl} K & \longrightarrow & \mathbb{R}^3, \\ \sum_{i=1}^3x_ig_i & \mapsto & (x_1,x_2,x_3). \end{array}\end{align*} $$

The parallelepiped with nodes $\ell _1,\ell _2,\ell _3\in \mathbb {Z}^3$ will be the set

$$ \begin{align*}\mathcal{D}(\ell_1,\ell_2,\ell_3)=[0,1]\ell_1+[0,1]\ell_2+[0,1]\ell_3.\end{align*} $$

In the case of a simplest cubic field K, $(\rho ^2,1+2\rho +\rho ^2)$ is a proper pair of totally positive units; therefore, we are left with the parallelepipeds

$$ \begin{align*} \mathcal{D}(1,\rho^2,1+2\rho+\rho^2),\quad\mathcal{D}(1,\rho^2,\rho^2\rho^{\prime\prime 2}). \end{align*} $$

It can be checked that $\rho ^2\rho ^{\prime \prime 2} =-1-a-(a^2+3a+3)\rho +(a+2)\rho ^2$ .

An element $\alpha \in \mathcal {O}_K^+$ has minimal trace one if there is $\delta \in \mathcal {O}_K^{\vee ,+}$ such that $\mathrm {Tr}(\delta \alpha )=1$ . By [Reference Narkiewicz11, Proposition 4.14], the elements $\varphi _1,\varphi _2,\varphi _3\in \mathcal {O}_K$ , the coordinates of which with respect to B are the columns of the inverse of the matrix $(\mathrm {Tr}(g_rg_s))_{r,s=1}^3$ , form a $\mathbb {Z}$ -basis of $\mathcal {O}_K^{\vee }$ such that

$$ \begin{align*}\mathrm{Tr}\bigg(\bigg(\sum_{r=1}^3a_rg_r\bigg)\bigg(\sum_{s=1}^3b_s\varphi_s\bigg)\bigg)=\sum_{r=1}^3a_rb_r\end{align*} $$

for every $a_r,b_s\in \mathbb {Z}$ . In our case,

$$ \begin{align*} \varphi_1 & =\frac{1}{\Delta}(( a^2+5a+11 - (2 a^2+8a+4) j + 2( a^2+a+1) j^2 + 2 (a+4) k - 2(2 a+1) j k \\ &\quad + 2 k^2 - 2( a+4) j l + 2(2 a+1) j^2 l - 4 j k l + 2 j^2 l^2)g_1 \\ &\quad +( a^2+ 4 a +2 - 2( a^2+a+1) j + (2 a+1) k + (a+4) l - 2(2 a+1) j l + 2 k l \\ &\qquad - 2 j l^2 )n_1g_2 -( a+4 - (2 a+1) j + 2 k - 2 j l )n_2g_3), \\ \varphi_2 & =\frac{1}{\Delta}((a^2 +4a+2 - 2(a^2+a+1)j + (2 a+1) k + (a+4) l - 2(2 a+1) j l + 2 k l \\ &\qquad - 2 j l^2 )n_1g_1+(2 a^2 +2a+2 + 2 l + 4 a l + 2 l^2 )n_1^2g_2-( 2 a + 2 l +1)n_1n_2g_3), \\ \varphi_3 & =\frac{1}{\Delta}(-( a +4 - j - 2 a j + 2 k - 2 j l )n_2g_1-( 2 a + 2 l +1)n_1n_2g_2+ 2 n_2^2 g_3). \end{align*} $$

4.1 The case $j=0$

We will start with the case when $j=0$ . This implies that ${n_1=1}$ . Indeed, since ${\rho }/{n_1}$ is an algebraic integer, it is a root of some monic polynomial ${g\in \mathbb {Z}[x]}$ . Now, $\rho $ is a root of the polynomial $h(x)=n_1^3g({x}/{n_1})$ , and since this is monic, $h(x)=f(x)$ is the minimal polynomial of $\rho $ . But then $n_1^3\mid f(0)=-1$ , so $n_1=1$ .

On the other hand, under this condition, we must have $n_2^2\mid \Delta $ , which is the consequence of the following arguments. Let $p\mid n_2$ be a prime. If $p^{s}\mid \Delta $ for some $s\geq 3$ , then either $j\neq 0$ , or $p=3$ , $27\mid \Delta $ but $9\nmid n_2$ (see Lemmas 3.2 and 3.3). Therefore, for $p\neq 3$ , $s=1,2$ . If $s=1$ , then $p\nmid \delta $ , and thus, $p\nmid n_2$ . Thus, $s=2$ and, by the formula for the conductor, $p^2\nmid \delta $ . In summary, we must have $n_2^2\mid \Delta $ .

In the following proof, we use a simple consequence of this fact. If $n_2^2\mid \Delta $ , then, clearly, for $a> 5$ ,

$$ \begin{align*} n_2^2\leq \Delta = a^2+3a+9<a^2+4a+4=(a+2)^2, \end{align*} $$

which gives $n_2\leq a+1$ . For $a\leq 5$ (or $a\leq 7$ , needed later), we obtain only monogenic simplest cubic fields, which are covered by the result of Kala and Yatsyna [Reference Kala and Yatsyna6].

Proposition 4.3. Let K be a nonmonogenic simplest cubic field with $a>5$ and $j=0$ . Then $-g_2+g_3$ is a nonunit indecomposable integer in $\mathcal {O}_K$ with minimal trace one.

Proof. First, we show that $-g_2+g_3$ is totally positive. It will be enough to check that it lies in the parallelepiped $\mathcal {D}(1,\rho ^2,\rho ^2\rho ^{\prime \prime 2})$ . It is easily verified that $-g_2+g_3$ equals

$$ \begin{align*} \frac{ka^2+(3k-l+n_2)(a+1)}{n_2(a^2+3a+3)}+\frac{a^2+3a+3+(l-n_2)(a+2)}{n_2(a^2+3a+3)}\rho^2 +\frac{n_2-l}{n_2(a^2+3a+3)}\rho^2\rho^{\prime\prime 2}. \end{align*} $$

The first and the third terms are clearly positive since $n_2>l$ . The numerator of the coefficient of $\rho ^2$ is

$$ \begin{align*} \geq a^2+3a+3-n_2(a+2)\geq a^2+3a+3-(a+1)(a+2)=1>0. \end{align*} $$

Here, we use the facts that $l\geq 0$ and $n_2\leq a+1$ , which we derived above. Next, we show that all of these coefficients are bounded above by $1$ . For the first one, the inequality

$$ \begin{align*}ka^2+(3k-l+n_2)(a+1)< n_2(a^2+3a+3)\end{align*} $$

is equivalent to

$$ \begin{align*}(n_2-k)a^2+(2n_2+l-3k)(a+1)>0.\end{align*} $$

Now, the inequalities $1\leq k,l\leq n_2-1$ give $l-3k\geq 1-3(n_2-1)=4-3n_2$ , so

$$ \begin{align*}(n_2-k)a^2+(2n_2+l-3k)(a+1)\geq a^2-(n_2-4)(a+1).\end{align*} $$

Since $n_2\leq a+1$ for $a>5$ ,

$$ \begin{align*}a^2-(n_2-4)(a+1)\geq a^2-(a-3)(a+1)=a^2-(a^2-2a-3)=2a+3>0.\end{align*} $$

Therefore, the original inequality is true. For the second coefficient,

$$ \begin{align*}a^2+(l+3-n_2)a+2l-2n_2+3< n_2(a^2+3a+3)\end{align*} $$

if and only if

$$ \begin{align*}(n_2-1)a^2+(4n_2-l-3)a+5n_2-2l-3>0,\end{align*} $$

which holds trivially. Finally, for the third coefficient, clearly, $n_2-l< n_2(a^2+3a+3)$ .

Therefore, $-g_2+g_3$ is totally positive. Since it lies in the interior of the parallelepiped $\mathcal {D}(1,\rho ^2,\rho ^2\rho ^{\prime \prime 2})$ and this is contained in the fundamental domain for the action of totally positive units on elements of K, it follows that $-g_2+g_3$ is not a unit.

Now, the statement will be established as soon as we prove that the element of the codifferent $\varphi _1+\varphi _3$ is totally positive. Indeed, in that case, we have $\varphi _1+\varphi _3\in \mathcal {O}_K^{\vee ,+}$ and

$$ \begin{align*}\mathrm{Tr}((\varphi_1+\varphi_3)(-g_2+g_3))=1.\end{align*} $$

We prove that $\varphi _1+\varphi _3$ is totally positive. We use the coefficients of the minimal polynomial of $(a^2+3a+9)(\varphi _1+\varphi _3)$ and prove that they are all positive. Now,

$$ \begin{align*} (a^2+3a+9)(\varphi_1+\varphi_3) & =(a^2+5a+11+2k^2+2ak+8k-n_2(a+2k+4))g_1\\ &\quad +(a^2+4a+2+k(2a+2l+1)+l(a+4) \\ &\quad -n_2(2a+2l+1))g_2 +(2n_2^2-n_2(a+2k+4))g_3. \end{align*} $$

Thus,

$$ \begin{align*} \text{Tr}((a^2+3a+9)(\varphi_1+\varphi_3))=a^2+3a+9>0. \end{align*} $$

The second coefficient of the minimal polynomial is

$$ \begin{align*} &(a^2+3a+9)(-1-4k-k^2+4n_2+2kn_2-n_2^2+(n_2-k-1)a)\\&\quad=(a^2+3a+9)g(a,n_2,k), \end{align*} $$

where $g(a,n_2,k)$ is a linear polynomial in a with a positive leading coefficient if $k\neq n_2-1$ . If $k=n_2-1$ , then we obtain $g(a,n_2,n_2-1)=2>0$ . If $k\neq n_2-1$ , then $g(a,n_2,k)$ is positive if

$$ \begin{align*} a>\frac{n_2^2-2kn_2-4n_2+k^2+4k+1}{n_2-k-1}=n_2-k-1+\frac{2(k-n_2)}{n_2-k-1}. \end{align*} $$

This is clearly true if $a\geq n_2-k-1$ , that is, if $a+1\geq n_2-k$ , which is valid for ${n_2\leq a+1}$ and $k\geq 0$ .

We will proceed with the norm, $N((a^2+3a+9)(\varphi _1+\varphi _3))$ , which is equal to

$$ \begin{align*} (a^2+3a & +9)(-1+3k^2+k^3-6kn_2-3k^2n_2+3n_2^2+3kn_2^2-n_2^3 \\ &\quad +(n_2^2-2kn_2-n_2+k^2+k)a) =(a^2+3a+9)h(a,n_2,k). \end{align*} $$

Again, $h(a,n_2,k)$ is a linear polynomial in a, and we now study its leading coefficient. The roots of $n_2^2-2kn_2-n_2+k^2+k$ as a polynomial in k are $n_2-1$ and $n_2$ . Thus, this coefficient is positive if $k\neq n_2-1$ . If $k=n_2-1$ , then we obtain ${h(a,n_2,n_2-1)=1>0}$ . If $k\neq n_2-1$ , then $h(a,n_2,k)$ is positive if a is greater than

$$ \begin{align*} \frac{1 - 3 k^2 - k^3 + 6 k n_2 + 3 k^2 n_2 - 3 n_2^2 - 3 k n_2^2\! + \! n_2^3}{n_2^2-2kn_2-n_2+k^2+k}\!=\!n_2-k-2+\frac{2(k-n_2)+1}{n_2^2-2kn_2-n_2+k^2+k}, \end{align*} $$

which is true because $n_2\leq a+1$ .

Therefore, $\varphi _1+\varphi _3$ is totally positive and gives trace one for $-g_2+g_3$ .

4.2 The case $j\neq 0$

Now, let us discuss the element $g_2={(j+\rho )}/{n_1}$ with $j\neq 0$ . Recall that its norm is

$$ \begin{align*} \frac{1-(a+3)j+aj^2+j^3}{n_1^3}. \end{align*} $$

For $j=1$ , we obtain $-{1}/{n_1^3}$ , which is not a rational integer for $n_1\geq 3$ . Therefore, $j\geq 2$ , and $g_2$ is totally positive since the smallest negative conjugate of $\rho $ satisfies ${-2<\rho '<-1}$ . In the following, we prove that $g_2$ has minimal trace one.

We can easily verify that ${(j+\rho )}/{n_1}$ is not a unit:

$$ \begin{align*} \frac{j+\rho}{n_1}\rho^2\rho^{\prime\prime 2} &=\frac{1}{n_1}(a+2-j(a+1)\\ &\quad +(a^2+4a+5-j(a^2+3a+3))\rho-(a+3-j(a+2))\rho^2) \end{align*} $$

and also

$$ \begin{align*} =\frac{1}{n_1(a^2+3a+3)}+\frac{a+1}{n_1(a^2+3a+3)}\rho^2+\frac{j(a^2+3a+3)-(a^2+4a+5)}{n_1(a^2+3a+3)}\rho^2\rho^{\prime\prime 2}. \end{align*} $$

We see that all of the coefficients are positive and less than one for $2\leq j\leq n_1-1$ . Therefore, $({(j+\rho )}/{n_1})\rho ^2\rho ^{\prime \prime 2}$ lies inside $\mathcal {D}(1,\rho ^2,\rho ^2\rho ^{\prime \prime 2})$ and cannot be a unit.

Proposition 4.4. Let K be a simplest cubic field with $a\geq 7$ and $j\neq 0$ . If $a\geq n_1+n_2-j$ , then there exists $W\in {\mathbb N}$ such that the element $\varphi _1+\varphi _2+W\varphi _3$ is totally positive.

Proof. First, it is not difficult to check that the elements of the basis of the codifferent can be rewritten as $\varphi _1=(k-(a+l)j)\psi _1-j\psi _2+\psi _3$ , $\varphi _2=(a+l)n_1\psi _1+n_1\psi _2$ and ${\varphi _3=-n_2\psi _1}$ , where

$$ \begin{align*} \psi_1&=\frac{1}{\Delta}(a+4+(2a+1)\rho-2\rho^2),\\ \psi_2&=\frac{1}{\Delta}(2+(a+2)\rho-\rho^2),\\ \psi_3&=\frac{1}{\Delta}(a^2+5a+11+(a^2+4a+2)\rho-(a+4)\rho^2). \end{align*} $$

The advantage of the elements $\psi _i$ is that they depend only on a. Thus, it is not difficult to find good estimates on their conjugates, which we do in the following steps.

The element $\psi _1$ is a root of the polynomial

$$ \begin{align*} x^3-\frac{1}{\Delta}x-\frac{1}{\Delta^2}. \end{align*} $$

This polynomial has roots satisfying

$$ \begin{align*} -\frac{1}{a^2+3a+7}<\psi_1<-\frac{1}{a^2+3a+8},\ \ -\frac{1}{a+2}<\psi_1'<-\frac{1}{a+3},\ \ \frac{1}{a+2}<\psi_1"<\frac{1}{a+1}. \end{align*} $$

This can be verified by checking the values of the polynomial in the above intervals and by considering estimates on conjugates of $\rho $ . Similarly, $\psi _2$ is a root of

$$ \begin{align*} x^3-\frac{1}{\Delta}x+\frac{1}{\Delta^2}, \end{align*} $$

and we have

$$ \begin{align*} \frac{1}{a+3}<\psi_2<\frac{1}{a+2},\quad -\frac{1}{a+1}<\psi_2'<-\frac{1}{a+2},\quad \frac{1}{a^2+3a+8}<\psi_2"<\frac{1}{a^2+3a+7}. \end{align*} $$

Finally, the element $\psi _3$ is a root of

$$ \begin{align*} x^3-x^2-\frac{a+1}{\Delta}x+\frac{1}{\Delta^2}, \end{align*} $$

which gives

$$ \begin{align*} \frac{1}{\Delta(a+1)+a+3}<\psi_3<\frac{1}{\Delta(a+1)+a+2},\quad & -\frac{1}{a+3}<\psi_3'<-\frac{1}{a+4}, \\ & 1+\frac{1}{a+4}<\psi_3"<1+\frac{1}{a+3}. \end{align*} $$

The element $\delta _W=\varphi _1+\varphi _2+W\varphi _3$ can be rewritten as

$$ \begin{align*} \varphi_1+\varphi_2+W\varphi_3=(M-n_2W)\psi_1+(n_1-j)\psi_2+\psi_3, \end{align*} $$

where . Clearly, the coefficient $n_1-j$ is positive, and the same is true for M. Since $\psi _1',\psi _2',\psi _3'<0$ , the necessary condition to get $\delta _W'>0$ is that ${n_2W>M}$ . If this is the case, then, obviously, $\delta _W>0$ since $\psi _1<0$ and $\psi _2,\psi _3>0$ . Thus, it remains to discuss $\delta _W'$ and $\delta _W"$ .

Using the above estimates, we conclude that

$$ \begin{align*} \delta_W'>\frac{n_2W-M}{a+3}-\frac{n_1-j}{a+1}-\frac{1}{a+3}. \end{align*} $$

The expression on the right-hand side is positive if

$$ \begin{align*} n_2W>M+n_1-j+1+\frac{2}{a+1}(n_1-j). \end{align*} $$

Note that this also implies that $n_2W>M$ . On the other hand,

$$ \begin{align*} \delta_W">\frac{M-n_2W}{a+1}+\frac{n_1-j}{a^2+3a+8}+1+\frac{1}{a+4}. \end{align*} $$

Again, the right-hand side is positive if

$$ \begin{align*} n_2W<M+a+1+\frac{a+1}{a+4}+\frac{(a+1)(n_1-j)}{a^2+3a+8}. \end{align*} $$

The difference between the upper and lower bound is

$$ \begin{align*}a-n_1+j+\frac{a+1}{a+4}+\frac{(a+1)(n_1-j)}{a^2+3a+8}-\frac{2}{a+1}(n_1-j).\end{align*} $$

Moreover, $(a+2)^2>a^2+3a+9\geq n_1n_2\geq n_1^3$ , so $n_1<(a+2)^{{2}/{3}}$ and

$$ \begin{align*} \frac{a+1}{a+4}+\frac{(a+1)(n_1-j)}{a^2+3a+8}-\frac{2(n_1-j)}{a+1}>\frac{a+1}{a+4}+\frac{a+1}{a^2+3a+8}-\frac{2(a+2)^{{2}/{3}}}{a+1}>0 \end{align*} $$

for $a\geq 13$ , which we have checked by computer. For $-1\leq a\leq 12$ , the corresponding simplest cubic fields are monogenic. Now, the hypothesis gives $a-n_1+j\geq n_2$ , so the difference above is clearly bounded below by $n_2$ . Therefore, we obtain the existence of a suitable W.

Corollary 4.5. Let K be a simplest cubic field with $a\geq n_1+n_2-j$ and $j\neq 0$ . Then ${(j+\rho )}/{n_1}$ is a nonunit indecomposable integer with minimal trace one.

Proof. As we have deduced, ${(j+\rho )}/{n_1}$ is a nonunit totally positive algebraic integer. Together with the totally positive element of codifferent $\varphi _1+\varphi _2+W\varphi _3$ from Proposition 4.4, it gives the minimal trace one, as required.

Therefore, if $n_1$ and $n_2$ are fixed and $j\neq 0$ , there are at most finitely many simplest cubic fields for which ${(j+\rho )}/{n_1}$ does not necessarily have minimal trace one. However, simplest cubic fields with $a< n_1+n_2-j$ exist, and there could be infinitely many of them. For example, $a=101471$ , $182451$ and $18128865$ , which are exceptional cases in the paper of Kashio and Sekigawa [Reference Kashio and Sekigawa7, page 7], do not satisfy this condition.