diff --git "a/data_all_eng_slimpj/shuffled/split2/finalzzgski" "b/data_all_eng_slimpj/shuffled/split2/finalzzgski" new file mode 100644--- /dev/null +++ "b/data_all_eng_slimpj/shuffled/split2/finalzzgski" @@ -0,0 +1,5 @@ +{"text":"\n\\section{Introduction}\nLet $K$ be a number field, $\\mathcal{O}_K$ be its ring of integers and $\\mathfrak{A}\\subseteq\\mathcal{O}_K$ be a principal ideal.\nOur aim is to investigate the following property of $\\mathfrak{A}$:\n\n\\begin{equation}\\label{eq:property}\\hbox{ $\\mathfrak{A}$ admits a generator $x$ such that $|\\sigma(x)|\\geq 1$ for every embedding $\\sigma: K\\to\\mathbb{C}$}.\n\\end{equation}\n\nWe shall call such a generator a \\emph{large} element of $\\mathcal{O}_K$ (\\emph{strictly large} when the strict inequality holds), and we shall call \\emph{large} every ideal satisfying property \\eqref{eq:property}.\n\nWe shall see that all but finitely many principal ideals $\\mathfrak{A}$ are strictly large; in particular this happens when the \\emph{logarithmic norm} $n(\\mathfrak{A})=\\frac {N(\\mathfrak{A})}{[K:\\mathbb{Q}]}$ exceeds the \\emph{covering radius} of the lattice of units with respect to the $L_\\infty$ norm. Moreover, every non-trivial ideal becomes large in a suitable finite extension of $K$.\n\nIt is possible to relate the notion of largeness of a principal ideal to the Weil height of its generators. Therefore, lower bounds of the Weil height on $K$, as given by the Bogomolov property, may help to prove that some ideals are not large. \nTo this aim, we shall state some inequalities concerning \nthe covering radius, the regulator and the Weil height of systems of multiplicatively independent units of $K$. We shall apply the technique described above in some concrete example.\n\nAs soon as the group of units of $K$ is known, it is relatively easy to decide if a principal ideal is large, and we give the corresponding algorithm. Then, we shall present\nthe results of applying it to some particular ideal in cyclotomic fields.\n\nAs a last application, we shall define the notion of floor function for $K$ relatively to $\\mathfrak{A}$ and show that condition \\eqref{eq:property} allows to explicitly construct a bounded floor function.\nThis turns out to be a good property for the the resulting continued fractions (\\cite{CapuanoMurruTerracini2022}).\n\n\\section{The largeness property and general results}\\label{sect:theproblem}\nLet $K$ be a number field of degree $d$.\nWe denote by $\\calo_K$ the ring of integers of $K$ and $\\calo_K^\\times$ the group of units.\nThe number field $K$ has $r_1$ real embeddings $\\sigma_1,\\dots,\\sigma_{r_1}$\nand $2r_2$ complex embeddings $\\sigma_{r_1+1},\\dots,\\sigma_{r_1+2r_2}$,\nwhere $r_1+2r_2 = d$ and $\\sigma_{r_1+i}$ and $\\sigma_{r_1+r_2+i}$\nare conjugates for $1\\leq i \\leq r_2$. We denote by $\\Sigma$ the whole set of Archimedean embeddings. We shall denote by $|\\cdot |$ the standard complex absolute value.\nAn element $x \\in K$ therefore has\n$r_1+r_2$ Archimedean absolute values, namely $|\\sigma_1(x)|, \\dots, |\\sigma_{r_1+r_2}(x)|$,\nand we have $|\\sigma_{r_1+i}(x)| = |\\sigma_{r_1+r_2+i}(x)|$\nfor all $1\\leq i \\leq r_2$.\nWe put $s=r_1+r_2$, $r=s-1$.\n\nLet\n\\begin{align*} \\iota: K& \\longrightarrow \\mathbb{R}^{r_1}\\times \\mathbb{C}^{r_2}\\\\\n\\lambda &\\longmapsto (\\sigma_1(\\lambda),\\ldots,\\sigma_{r_{1}}(\\lambda),\\sigma_{r_1+1}(\\lambda),\\ldots,\\sigma_{r_1+r_2}(\\lambda))\n\\end{align*} be the canonical embedding of $K$, and\n$$\\ell: K^\\times \\to \\mathbb{R}^{r_1+r_2}$$ be the logarithmic embedding, i.e., the composition $\\mathcal{L}\\circ\\iota$ where \n\\begin{align*} \\mathcal{L}: (\\mathbb{R}^\\times)^{r_1}\\times (\\mathbb{C}^\\times)^{r_2}& \\longrightarrow \\mathbb{R}^{r_1}\\times \\mathbb{R}^{r_2}\\\\\n(x_1,\\ldots, x_{r_1}, y_1,\\ldots, y_{r_2}) &\\longmapsto (\\log|x_1|,\\ldots, \\log|x_{r_1}|, 2\\log|y_1|,\\ldots, 2\\log|y_{r_2}|).\n\\end{align*} \\\\\nFor $\\mathbf{x}=(x_1,\\ldots, x_{r_1},y_1,\\ldots, y_{r_2})\\in \\mathbb{R}^{r_1}\\times \\mathbb{C}^{r_2}$, let us define\n$$N(\\mathbf{x})=\\prod_{i=1}^{r_1}|x_i \n|\\cdot \\prod_{j=1}^{r_2}|y_j|^2;$$ \nthen, $N(\\iota(a))=|N_{K\/\\mathbb{Q}}(a)|$ for every $a\\in K.$\nThe {\\em absolute norm} of $x\\in K$ is equal to the absolute value of the norm of $x$.\\\\\nWe shall denote $\\Lambda_K=\\ell(\\mathcal{O}_K^\\times)$; it is a lattice of rank $r$ in $\\mathbb{R}^s$ by Dirichlet's Unit Theorem. We recall also that the \\emph{regulator} $R_K$ of $K$ is the determinant of any submatrix of order $r$ of the $r\\times (r+1)$ matrix whose rows are $\\ell(u_1),\\ldots, \\ell(u_r)$ \nfor a system $u_1,\\ldots, u_r$ of fundamental units for $K$. If $V_K$ is the volume of a fundamental domain for $\\Lambda_K$ then the relation $V_K=\\sqrt{s}R_K$ holds.\n\n\\begin{definition}\\ \\begin{itemize}\n \\item[a)] We say that $x\\in \\mathcal{O}_K$ is \\emph{large} (resp. \\emph{strictly large}) if\n$|\\sigma(x)|\\geq 1$ (resp. $|\\sigma(x)|> 1$) for every $\\sigma\\in\\Sigma$ . \n\\item[b)] An ideal $\\mathfrak{A}\\subseteq\\mathcal{O}_K$ is \\emph{large} (resp. \\emph{strictly large}) if it principal and has a large (resp. strictly large) generator.\n\\end{itemize} \n\\end{definition}\n\nFor $x \\in \\mathcal{O}_K$\nas in the above definition,\nwe observe that\n$x$ is large in $\\mathcal{O}_K$ if and only if $x$ is large in $\\mathcal{O}_L$ for any finite extension $L$ of $K$.\nFor the ideal $\\mathfrak{A}$\nit is true that if it is large in $K$, then\nthe ideal $\\mathfrak{A} \\mathcal{O}_L$ is large in $\\mathcal{O}_L$, but the converse is not true.\n\nThen the following definition makes sense and extends the notion of largeness to possibly infinite extensions:\n\\begin{definition}\nLet $L$ be an infinite extension of $K$.\nFor an element $x \\in \\mathcal{O}_K$\nand an ideal $\\mathfrak{A} \\subseteq \\mathcal{O}_K$, we say that:\n\\begin{itemize}\n\\item[a)]\n$x$ is \\emph{large} (resp. \\emph{strictly large}) in $L$ if there is a number field $K'$ with $K \\subseteq K'\\subseteq L$ such that $x$ is large (resp. strictly large) in $K'$.\n\\item[b)]\n$\\mathfrak{A}$ is \\emph{large} (resp. \\emph{strictly large}) in $L$ if there is a number field $K'$ with $K \\subseteq K'\\subseteq L$ such that $\\mathfrak{A} \\mathcal{O}_{K'}$ is large (resp. strictly large) in $K'$.\n\\end{itemize}\n\\end{definition}\n\nSince every unit in $\\mathcal{O}_K$ has norm $\\pm 1$, a unit $x$ is large if and only if $|\\sigma(x)|=1$ for every $\\sigma\\in \\Sigma$; by a theorem of Kronecker \\cite{Kronecker1857}, this happens if and only if $x$ is a root of unity. Therefore no unit can be strictly large. So every strictly large element $x$ in $\\mathcal{O}_K$ must satisfy $|N_{K\/\\mathbb{Q}}(x)|\\geq 2$.\\\\\nWe shall see in Proposition \\ref{prop:normalarge} that almost all principal ideals in $\\mathcal{O}_K$ are (strictly) large. In order to prove this fact we need to recall some terminology from lattice theory. \n\nLet $\\Lambda$ be a lattice in $\\mathbb{R}^n$ of rank $r$ and for a real number $p\\in [1,\\infty)\\cup\\{\\infty\\}$ let $|| \\cdot||_p$ be the norm $L_p$ in $\\mathbb{R}^n$. The \\emph{distance function} relatively to $p$ is by definition\n $$\\rho_p(\\mathbf{v},\\Lambda)=\\min_{\\mathbf{w}\\in\\Lambda} ||\\mathbf{v}-\\mathbf{w}||_p.$$\n The \\emph{covering radius} of $\\Lambda$ with respect to $|| \\cdot||_p$ is\n $$\\rho_p(\\Lambda)=\\sup_{\\mathbf{v}\\in \\mathrm{span}(\\Lambda)} \\rho_p(\\mathbf{v},\\Lambda).$$\n Balls of radius $\\rho_p(\\Lambda)$ centered around all lattice points cover the whole space $\\mathrm{span}(\\Lambda)$.\\\\\n By the well known inequality\n \\begin{equation}\\label{eq:disuglp} ||\\mathbf{v}||_p\\leq ||\\mathbf{v}||_r\\leq n^{\\frac 1 r-\\frac 1 p}||\\mathbf{v}||_p\\quad \\hbox{for } \\infty\\geq p\\geq r,\\end{equation}\n we get \n \\begin{equation}\\label{eq:disugcovrad} \\rho_p(\\Lambda)\\leq \\rho_r(\\Lambda) \\leq n^{\\frac 1 r-\\frac 1 p}\\rho_p(\\Lambda)\\quad \\hbox{ for } \\infty\\geq p\\geq r.\\end{equation} \n If $K$ is a number field we shall write $\\rho_p(K)$ instead of $\\rho_p(\\Lambda_K)$.\n \n For every algebraic number $x\\in\\overline{\\mathbb{Q}}^\\times$ we define the \\emph{logarithmic norm}\n $$n(x)=\\frac{\\log|N_{\\mathbb{Q}(x)\/\\mathbb{Q}}(x)|}{[\\mathbb{Q}(x):\\mathbb{Q}]}.$$\n Analogously, if $\\mathfrak{A}\\subseteq \\mathcal{O}_K$ is any non-zero ideal, we write\n $$n(\\mathfrak{A})=\\frac{\\log|N_{K\/\\mathbb{Q}}(\\mathfrak{A})|}{[K:\\mathbb{Q}]}.$$\n Notice that $$n(x)=\\frac{\\log|N_{K\/\\mathbb{Q}}(x)|}{[K:\\mathbb{Q}]},$$\n for every finite extension $K$ of $\\mathbb{Q}(x)$; moreover\n $$n(x)=n(ux),$$\n for every algebraic unit $u\\in\\overline{\\mathbb{Q}}$. \n Then $n(a)=n(a\\mathcal{O}_K)$\n depends only on the principal ideal generated by $a$ in the ring of integers of every number field containing $a$.\\\\\n We also observe that $n:\\overline{\\QQ}^\\times\\to \\mathbb{R}$ is a morphism; in particular $n(x^k)=kn(x)$ for every $k\\in\\mathbb{N}$.\n We also have $n(x) \\ge 0$ when $x$ is an algebraic integer.\n\\begin{proposition}\\label{prop:normalarge} \n Every principal ideal $\\mathfrak{A}$ of $\\mathcal{O}_K$ such that $n(\\mathfrak{A})>\\rho_\\infty(K)$ is strictly large.\\\\\n Therefore all but finitely many integral principal ideals of $\\mathcal{O}_K$ are strictly large.\n\\end{proposition}\n\\begin{proof}\nLet $x\\in \\mathcal{O}_K$ be a generator of $\\mathfrak{A}$ and put $N=|N_{K\/\\mathbb{Q}}(x)|$. The image of units $\\ell(\\mathcal{O}_K^\\times)$ is a lattice in $\\mathbb{R}^s$ of rank $r=s-1$; it spans the hyperplane $\\mathcal{H}$ of $\\mathbb{R}^s$ with equation $\\sum_{i=1}^{r_1} x_i+\\sum_{i=1}^{r_2}y_i=0$. The vector $$\\mathbf{y}=\\ell(x)-\\frac 1 d \\log(N)(1,\\ldots,1,2,\\ldots,2)$$ lies on $\\mathcal{H}$ . Let $\\rho=\\rho_\\infty(K)$ and assume $n(x)>\\rho$; by definition of covering radius, there exists $u\\in\\mathcal{O}_K^\\times$ such that $||\\mathbf{y}+\\ell(u)||_\\infty \\leq \\rho$. This means that $|\\log|\\sigma(ux)|-n(x)|\\leq \\rho$ for every Archimedean embedding $\\sigma$ of $K$, so that\n$$|\\log|\\sigma(ux)||\\geq n(x)-\\rho> 0.$$\nThe second assertion follows from the fact that the ideals of norm $\\le\\rho$ are finitely many.\n\\end{proof}\n\n\\begin{proposition}\\label{prop:esisteL} Every non-trivial integral ideal $\\mathfrak{A} \\subsetneq \\mathcal{O}_K$ is strictly large in $\\overline{\\mathbb{Q}}$.\n\\end{proposition}\n\\begin{proof} The statement is a consequence of \\cite[Th\u00e9or\u00e8me 5.1]{BergeMartinet1989}, here we present a more direct proof.\nFirst of all, by class field theory, it is well known that $\\mathfrak{A}$ becomes principal in a suitable finite extension $K'$ of $K$.\nWe have $\\mathfrak{A} \\mathcal{O}_{K'} = x \\mathcal{O}_{K'}$ for some $x \\in \\mathcal{O}_{K'}$.\nBy Proposition \\ref{prop:normalarge} there exists a positive integer $j$ such that $x^j$ is strictly large in $K'$. Let $u\\in \\mathcal{O}_{K'}^\\times$ be such that $|\\sigma(u x^j)|>1$ for every embedding $\\sigma:K'\\to\\mathbb{C}$. Let $L=K'(\\omega)$ where $\\omega^j=u$. Let $\\tau:L\\to\\mathbb{C}$ be any embedding and $\\sigma$ be the restriction of $\\tau$ to $K'$. Then\n$$|\\tau(\\omega x)|^j=|\\sigma(ux^j)|>1$$\nso that $|\\tau(\\omega x)|>1$.\n\\end{proof}\n\nBy looking at the proof of Proposition \\ref{prop:esisteL}, we see that a uniform and stronger version holds true.\nFor every number field $K$ and every positive $j\\in\\mathbb{N}$, we denote by $K_j$ the field obtained from $K$ by adding the $j$-th roots of every unit of $K$; it is a finite extension of $K$ by Dirichlet's Unit Theorem.\n\\begin{proposition}\\label{prop:esisteLuniforme}\nLet $K$ be a number field, and let $j> \\frac{\\rho_\\infty(K)[K:\\mathbb{Q}]}{\\log2}$. Every non-trivial principal ideal $\\mathfrak{A} \\subsetneq \\mathcal{O}_K$ is strictly large in $K_j$.\n \\end{proposition}\n \\begin{proof}\n Let $x \\in \\mathcal{O}_K$ be a generator of $\\mathfrak{A}$.\n We have $n(x)\\geq \\frac{\\log 2}{[K:\\mathbb{Q}]}$, so that $n(x^j)=jn(x) >\\rho_\\infty(K)$. Then one can choose $L=K_j$ in the proof of Proposition \\ref{prop:esisteL}.\n \\end{proof}\n\\section{Largeness and Weil height}\\label{sect:height}\nLet $h$ denote the logarithmic Weil height of an algebraic number (see for example \\cite[\\S 1.5.7]{BombieriGubler2006}). For $x\\in K$\n$$h(x)=\\frac 1 d\\sum_{\\sigma\\in \\Sigma} \\max\\{0,\\log|\\sigma(x)|\\}+\\log|a|$$\nwhere $a$ is the leading coefficient of a primitive equation for $x$ over $\\mathbb{Z}$; in particular\nfor an algebraic integer $x$ in $\\mathcal{O}_K$\n$$h(x)=\\frac 1 d\\sum_{\\sigma\\in \\Sigma} \\max\\{0,\\log|\\sigma(x)|\\}.$$\nIt follows that \n\\begin{equation}\\label{eq:h>n} h(x)\\geq \\frac 1 d\\log|N_{K\/\\mathbb{Q}}(x)|=n(x) \\quad \\hbox{ for every non-zero algebraic integer $x$},\\end{equation}\nand equality holds exactly when $x\\mathcal{O}_K$ is large.\\\\\nThen we can draw necessary conditions for largeness of ideals when some explicit minoration for the height of elements in $\\mathcal{O}_K$ is known. Namely, if there is a constant $c>0$ such that \n\\begin{equation}\\label{eq:maggiorazioneperinteri} h(x)>c\\hbox{ for every } x\\in\\mathcal{O}_K\\setminus\\mathcal{O}_K^\\times\\end{equation}\nand $\\mathfrak{A}$ is a principal ideal such that $n(\\mathfrak{A})\\leq c$, then $\\mathfrak{A}$ cannot be large.\\\\\nWe are thus lead to make use of the well known \\emph{Bogomolov property} $\\PB$ and an additional property $\\PS$ defined below.\\\\\nLet $\\mathcal{A}$ be a set of algebraic numbers.\n We put\n \\begin{align*}\n b(\\mathcal{A})&=\\inf\\{h(x)\\ |\\ x\\in \\mathcal{A}, x\\not=0, x\\hbox{ not a root of unity }\\};\\\\\n s(\\mathcal{A})&=\\inf\\{n(x)\\ |\\ x\\in\\mathcal{A}, x\\not=0, N_{\\mathbb{Q}(x)\/\\mathbb{Q}}(x)\\not=\\pm 1\\}.\n \\end{align*}\n\\begin{definition}\\label{def:propBS} We say that a set $\\mathcal{A}$ of algebraic numbers satisfies\n \\begin{itemize} \n \\item[a)] \\emph{property $\\PB$} if $b(\\mathcal{A})>0$;\n\\item[b)] \\emph{property $\\PS$} if $s(\\mathcal{A})>0$.\n\\end{itemize}\n\\end{definition}\n\nIn particular, if $x\\in\\mathcal{O}_L\\setminus \\mathcal{O}_L^\\times $ for an (infinite) extension $L$ with the property $\\PB$, then $h(x)\\geq c_L$ for some $c_L>0$ depending only on $L$ and thus, by \\eqref{eq:h>n}, \n$$x\\mathcal{O}_{\\mathbb{Q}(x)} \\hbox{ large} \\Longrightarrow \nn(x)\\geq c_L.\n$$\nProperty $\\PB$ is known for some special algebraic extensions, as the compositum ${\\mathbb Q}^{\\rm tr}$ of all totally real fields; it is also known for extensions having bounded local degrees at some finite place, and for Abelian extensions of number fields (see~\\cite[Remark 5.2, p.1902]{AmorosoDavidZannier2014}). \\\\\nNote however that property $\\PB$ for the whole field $L$, and even for the ring of integers of $L$, is much stronger that condition \\eqref{eq:maggiorazioneperinteri}, which assumes a lower bound only for the height of algebraic \\emph{integers} which are not units. \n\n\\begin{examples}\\label{exs:esempi}~\n\n\\begin{itemize}\n\\item[a)] Of course $\\PS\\Rightarrow \\PB$ if $\\mathcal{A}$ is a set of algebraic integers containing only a finite number of units. \n \n\\item[b)] On the other hand, there exist sets of algebraic integers satisfying \n$\\PB$ but not $\\PS$: for example the ring $\\mathcal{O}^{\\rm ab}$ of integers of $\\mathbb{Q}^{\\rm ab}$ satisfies property $\\PB$ with $b(\\mathcal{O}^{\\rm ab})\\geq \\frac{\\log 5}{12}$, (see the main theorem in \\cite{AmorosoDvornicich2000}), but \n$$n(1-\\zeta_p)=\\frac{\\log(p)}{p-1}$$ \nfor a prime $p$ and $\\zeta_p$ a primitive $p$-th root of unity; therefore $s(\\mathcal{O}^{\\rm ab})=0$.\n\\item[c)] It is proven in \\cite[Corollary 1]{AmorosoDvornicich2000} that property $\\PS$ holds for the set $\\mathcal{A}$ of algebraic integers $x$ lying in an Abelian extension of $\\mathbb{Q}$ and such that $ x\/{\\overline{x}}$ is not a root of unity. More precisely \n$$n(x)\\geq \\frac{\\log 5 } {12},\\hbox{ for every $x\\in\\mathcal{A}$ }.$$ \n\\item[d)] Recall that ${\\mathbb Q}^{\\rm tr}(i)$ is the compositum of all CM fields, see \\cite[page 1902]{AmorosoDavidZannier2014}; therefore $x\\in {\\mathbb Q}^{\\rm tr}(i)$ if and only if $\\mathbb{Q}(x)$ is either a totally real or a CM field. Since the complex conjugation commutes with all the embeddings of ${\\mathbb Q}^{\\rm tr}(i)$ in $\\mathbb{C}$, \nwe have $|\\sigma(x)| = 1$ for some $\\sigma \\in \\Sigma$ if and only if \n$|\\sigma(x)| = 1$ for all $\\sigma \\in \\Sigma$.\nIn this case, we just write $|x| = 1$.\\\\\nBy a result of Schinzel (apply~\\cite[Corollary 1', p. 386]{Schinzel1973}, to the linear polynomial $P(z)=z-x$), if\n$\\mathcal{A}=\\{x\\in {\\mathbb Q}^{\\rm tr}(i)\\ |\\ |x|\\not=1\\}$ then \n\\begin{equation}\\label{eq:Schinzel} b(\\mathcal{A})\\geq \\frac 1 2\\log \\frac{1+\\sqrt{5}} 2.\\end{equation}\n\\end{itemize}\n\\end{examples}\n\nBy Example \\ref{exs:esempi}.d) we obtain the following\n\\begin{proposition}\\label{prop:notlargeperSchinzel}\nLet $L={\\mathbb Q}^{\\rm tr}(i)$, and let $x\\in\\mathcal{O}_L $ be a non-zero element.\nIf $n(x)< \\frac 1 2 \\log\\frac{1+\\sqrt{5}}2$ then $x \\mathcal{O}_{\\mathbb{Q}(x)}$ is not large in $L$ except if $x$ is a unit.\n\\end{proposition}\n\\begin{proof}\nIf $x$ is a unit, then $x\\mathcal{O}_{\\mathbb{Q}(x)} = \\mathcal{O}_{\\mathbb{Q}(x)}$ is trivially large. If not, we have $|x|\\not=1$, so that Schinzel result \\eqref{eq:Schinzel} implies that $ h(x)\\geq \\frac 1 2\\log \\frac{1+\\sqrt{5}} 2>n(x)$. Then the result follows from \\eqref{eq:h>n}.\n\\end{proof}\n\n\n\n\n\\subsection{Example} Let $p$ be one of the primes for which ${\\QQ}(\\zeta_{p-1})$ has class number one. Note that $p$ splits completely in ${\\QQ}(\\zeta_{p-1})$. Recall (\\cite{MasleyMontgomery1976}) that the cyclotomic field ${\\QQ}(\\zeta_m)$ has class number one if and only if $m$ is one of the following forty-four numbers:\n\\begin{multline*}\n3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28,\\\\ \n30, 32, 33, 34, 35, 36, 38, 40, 42, 44, 45, 48, 50, 54, 60, 66, 70, 84, 90.\n\\end{multline*}\n(which corresponds to twenty-nine distinct cyclotomic fields). Thus the relevant primes are\n\\begin{equation}\n\\label{list}\n5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 61, 67, 71.\n\\end{equation} \n\\begin{question}\n\\label{qu:continued}\nLet $p$ be one of the fifteen primes~\\eqref{list} and let $\\mathfrak{P}$ be a prime ideal over $p$ in the ring of integers of $\\mathbb{Q}(\\zeta_{p-1})$. Is $\\mathfrak{P}$ large?\n\\end{question}\nSince ${\\mathbb Q}^{\\rm ab}\\subseteq{\\mathbb Q}^{\\rm tr}(i)$ and\n$$\np\\leq \\Big(\\frac{1+\\sqrt{5}}2\\Big)^{\\varphi(p-1)\/2}\\hbox{ for }p=41,67\\hbox{ and }71\n$$\nthe answer to Question~\\ref{qu:continued} is negative for these primes, by Proposition \\ref{prop:notlargeperSchinzel}. \n\nNote that we could have tried the same strategy as in Proposition \\ref{prop:notlargeperSchinzel}\nbut using the inequality of Example \\ref{exs:esempi}b. This would work only for the primes $p$ satisfying\n\\begin{equation}\\label{eq:valetutti}\np < 5^{\\varphi(p-1)\/12}.\n\\end{equation}\nHowever, this inequality is satisfied by none of the primes in the list~\\eqref{list}.\n\nIn the subsequent Theorem \\ref{teo:risposta}, Question \\eqref{qu:continued} will receive a complete answer.\n \\subsection{Number theoretic minorations of the covering radius}\n In the light of the propositions \\ref{prop:normalarge} and \\ref{prop:esisteLuniforme}, it is useful to have some quantitative information on the covering radius $\\rho_\\infty(K)$ of a number field $K$ of degree $d$. \\\\\n Let $\\lambda_1,\\ldots,\\lambda_r$ be the successive minima (w.r.t. the Euclidean norm $||\\cdot ||_2$) of the lattice $\\Lambda_K$. \n It is well known (see for example \\cite[Theorem 7.9]{MicciancioGoldwasser2002} that \n\\begin{equation}\\label{eq:disugminsucccovrad} \\lambda_1\\leq\\ldots\\leq \\lambda_r \\leq 2\\rho_2\\leq \\sqrt{s}\\lambda_r.\\end{equation}\nMoreover by \\eqref{eq:disugcovrad} we have\n \\begin{equation}\\label{eq:disugcovrad2}\n \\rho_\\infty(K)\\leq \\rho_2(K)\\leq \\sqrt{s}\\rho_\\infty(K).\\end{equation}\n Therefore \n \\begin{align} \\rho_\\infty(K)&\\geq \\frac 1 {\\sqrt{s}}\\rho_2(K)\\quad\\hbox{from \\eqref{eq:disugcovrad2}}\\nonumber \\\\\\\n& \\geq \\frac 1 {2\\sqrt{s}}\\lambda_r\\quad\\hbox{from \\eqref{eq:disugminsucccovrad}}\\label{eq:keyminoration}\n \\end{align} \n \n \\begin{theorem}\\label{teo:regolatore}~\n \\begin{itemize}\n \\item [a)] Let $K$ be a number field such that $r\\geq 1$. Then $\\rho_\\infty(K)\\geq \\frac 1 2 R_K^{\\frac 1r}$. \n \\item[b)] \n There exists a constant $c>0$ such that $\\rho_\\infty(K)\\geq c$ for every number field $K$ such that $r\\geq1$.\n \\end{itemize}\n \\end{theorem}\n \n \\begin{proof}\n Recall that $V_K$ is the volume of a fundamental domain for $\\Lambda_K$.\n By Minkowski's Second Theorem \\cite[Theorem 1.5]{MicciancioGoldwasser2002}\n $$(\\lambda_1\\cdot\\ldots\\cdot \\lambda_r)^{\\frac 1 r}\\geq \\sqrt{r}\\cdot V_K^{\\frac 1r}= \\sqrt{r}\\cdot (\\sqrt{s}R_K)^{\\frac 1r}.$$\n Then from \\eqref{eq:keyminoration}\n \\begin{align}\\label{eq:covradreg} \\rho_\\infty(K)\\geq \\frac {\\sqrt{r}}{2\\sqrt{s}}(\\sqrt{s}R_K)^{\\frac 1r}\\geq c' R_K^{\\frac 1r},\\end{align}\n for a suitable constant $c'$. Studying the function $\\frac {\\sqrt{r}}{2\\sqrt{s}}(\\sqrt{s})^{\\frac 1 r}$ with $s=r+1$ we see that $c'=\\frac 1 2$. This proves a). Then b) follows from the well known fact that there exist constants $c_0>0$ and $c_1>1$ such that $R_K>c_0\\cdot c_1^d$ (\\cite[\\S 3]{Zimmert1981}, see also \\cite{FriedmanSkoruppa1999}. \\end{proof}\n \n The next results provide some lower bounds for the covering radius involving the Weil height on $K$.\\\\\nFor $n=1,...,r$ we put, as in \\cite[Page 9]{AmorosoDavid2021}\n$$\\mu_K(n)=\\inf_{\\atop{v_1,\\ldots, v_n\\in\\mathcal{O}_K^\\times}{{\\rm multipl. indep.}}}(h(v_1)\\cdot\\ldots\\cdot h(v_n)),$$\n \n \\begin{theorem}\n \\label{teo:eccolo}\n For $n=1,\\ldots,r$, we have\n $$\\rho_\\infty(K)\\geq \\frac d s\\mu_K(n)^{\\frac 1 n}.$$\n In particular $$\\rho_\\infty(K)\\geq \\frac d s b(\\mathcal{O}_K^\\times).$$\n \\end{theorem}\n \\begin{proof}\nFor every $u\\in\\mathcal{O}_K^\\times$, by \\eqref{eq:disuglp},\n\\begin{equation}\\label{eq:due} 2dh(u)= ||\\ell(u)||_1 \\leq \\sqrt{s}||\\ell(u)||_2.\\end{equation} \n \nThere exist multiplicatively independent $u_1,\\ldots, u_r \\in\\mathcal{O}_K^\\times$ such that $\\lambda_i=||\\ell(u_i)||_2$ (\\cite[Theorem 1.2]{MicciancioGoldwasser2002}). \nWe notice that for $n=1,\\ldots, r$\n\\begin{align*} \\lambda_n & =\\inf_{\\atop{v_1,\\ldots, v_n\\in\\mathcal{O}_K^\\times} {\\rm multipl. indep.}} \\max\\{ ||\\ell(v_1)||_2,\\ldots, ||\\ell(v_n)||_2\\}\\\\\n&\\geq \\frac {2d} {\\sqrt{s}} \\inf_{\\atop{v_1,\\ldots, v_n\\in\\mathcal{O}_K^\\times} {\\rm multipl. indep.}} \\max \\{ h(v_1),\\ldots, h(v_n)\\}\\quad\\hbox{ by \\eqref{eq:due}.}\n\\end{align*}\nIt follows from \\eqref{eq:keyminoration} that\n\\begin{align*} \\rho_\\infty(K) &\\geq \\frac 1 {2\\sqrt{s}} \\lambda_n\\geq \\frac d s\\inf_{\\atop{v_1,\\ldots, v_n\\in\\mathcal{O}_K^\\times} {\\rm multipl. indep.}} \\max \\{ h(v_1),\\ldots, h(v_n)\\}\\geq \\frac d s\\mu_K(n)^{\\frac 1 n}.\\end{align*}\n\\end{proof}\n\nBy inequality \\eqref{eq:covradreg}, it is possible to use known lower bounds for the regulator $R_K$ in order to deduce lower bounds for the covering radius $\\rho_\\infty(K)$. See \\cite[\\S 3.5, 15]{Narkiewicz2004} for an overview on evaluations of the regulator and \\cite[\\S 8]{Narkiewicz2004} for the special case of Abelian extensions. Moreover \\cite[Proposition 3.3]{AmorosoDavid2021} provides a tool allowing to improve the bound from below of extensions $K$ for which $\\mathcal{O}_K^\\times$ has property $\\PB$. In particular \\cite[Corollaire 3.5]{AmorosoDavid2021} deals with the case of totally real and CM field.\n\nSome remarkable results are collected below:\n \n\\begin{itemize}\n\\item[a)] Let $L$ be an infinite extension. Assume that $\\mathcal{O}_L^\\times$ satisfies property $\\PB$ and let $c_L= b(\\mathcal{O}_L^\\times)=\\inf_{u\\in\\mathcal{O}_L^\\times\\setminus\\mathcal{O}_L^{\\times,\\rm{tors}}} h(u) >0$. Then by Theorem \\ref{teo:eccolo}$$ \\rho_\\infty(K)\\geq c_L,$$ for every number field $K \\subseteq L$ such that $r(K) \\geq 1$.\n\\item[b)] In particular, if a number field $K$ is contained in $\\mathbb{Q}^{\\rm{tr}}(i)$, then by Example \\ref{exs:esempi} d),\n$$\\rho_\\infty(K)\\geq \\frac 1 2 \\log \\frac{1+\\sqrt{5}} 2.$$\n\\item[c)] Silverman's theorem \\cite{Silverman1984} allows to construct fields with fixed degree and covering radius arbitrarily large. It suffices to choose a non-CM field of discriminant large enough. \n\\end{itemize}\n\n\n\n\n\n\\section{An algorithm for largeness}\\label{sect:algorithm}\n\n We describe an algorithm that solves the following problem:\n \\begin{problem}\\label{problem1}\n Given a non-zero algebraic number $x\\in K$ and a bound $B>0$,\n find all units $u\\in \\calo_K^\\times$ such that\n $|\\sigma(xu)| \\geq B$ for all $\\sigma \\in \\Sigma$.\n\\end{problem}\nWhen $B=1$ this algorithm detects when a principal ideal is large.\n\\subsection{Basic results}\n\n\\begin{lemma}\\label{upperbound}\n If $u \\in \\calo_K^\\times$ is a solution of Problem \\ref{problem1},\n then for all $S \\subset \\Sigma$, we have\n $$B^{\\# S} \\leq \\prod_{\\sigma \\in S} |\\sigma(xu)| \\leq B^{\\# S - d} N(x)$$\n where $\\# S$ denotes the cardinality of $S$.\n\\end{lemma}\n\\begin{proof}\nLet $u \\in \\mathcal{O}_K^\\times$ be a solution of Problem \\ref{problem1}.\nFor $\\sigma \\in S$, we have the trivial inequality $B \\leq |\\sigma(xu)|$.\nMultiplying these inequalities gives the announced left inequality.\n\n\nFor $\\sigma \\in S$, we have the inequality $|\\sigma(xu)| \\leq |\\sigma(xu)|$, and\nfor $\\sigma \\not\\in S$, we have $B \\leq |\\sigma(xu)|$.\nMultiplying these inequalities gives $B^{d-\\# S} \\prod_{\\sigma \\in S} |\\sigma(xu)| \\leq N(xu)$.\nBut $u$ is a unit, hence $N(xu) = N(x)$, whence the result.\n\\end{proof}\n\n\\begin{proposition}\\label{smallnorm}\n If $N(x) < B^d$, then Problem \\ref{problem1} has no solution.\n\\end{proposition}\n\\begin{proof}\nApply Lemma \\ref{upperbound} with $S = \\Sigma$.\n\\end{proof}\n\n\\begin{proposition}\\label{finitesol}\n Given $x\\in K$ and $B>0$, Problem \\ref{problem1} has only finitely many solutions.\n\\end{proposition}\n\\begin{proof}\nBy Proposition \\ref{smallnorm}, Problem \\ref{problem1} has no solution for $x=0$.\nWe assume now that $x\\neq 0$.\n\nDividing the right inequality of Lemma \\ref{upperbound} by $\\prod_{\\sigma \\in S}|\\sigma(x)| \\neq 0$ gives\n$$ \\prod_{\\sigma\\in S} |\\sigma(u)| \\leqslant B^{\\# S - d} \\prod_{\\sigma \\not\\in S}|\\sigma(x)|\n\\leqslant B^{\\#S-d} \\left( \\frac {||x||_1}{d-\\# S} \\right)^{d-\\# S} \\; .$$\nLet us consider the characteristic polynomial of $u$ for the extension $K\/{\\QQ}$ and\ndenote it by $P_u$.\nSince $u$ is a unit, $P_u \\in \\mathbb{Z}[X]$ and $P_u$ is monic.\nThe roots of $P_u$ in $\\mathbb{C}$ are the real or complex numbers $\\sigma(u)$.\nUsing the above inequality and expressing the coefficient $a_k$ of $X^k$ in $P_u$ in terms of the roots of $P_u$,\nwe deduce that $|a_k|$ is bounded independently of $u$.\nFor example we have\n$|a_d| = |a_0|=1$ since $u$ is a unit\nand \n$$|a_k| \\leq \\binom{d}{k} \\left( \\frac {||x||_1}{kB} \\right)^k$$\nfor the other values of $k$.\nSince this bound does not depend on $u$, there are only finitely many\npossibilities for $P_u$, hence for $u$.\n\\end{proof}\n\n\\noindent{\\bf Remark.}\nWe could turn the proof of Proposition \\ref{finitesol} into an algorithm\nthat tests all polynomials with coefficients within some bounds depending on $B$ and $x$.\nExplicitly, using the bounds given during the proof,\nwe see that, for a given number field of fixed degree $d$,\nthe number of polynomials that need to be tested, is proportional to\n$\\displaystyle \\left( \\frac {||x||_1}{B} \\right)^\\alpha$\nwith $\\displaystyle \\alpha = \\sum_{k=1}^{d-1} k = \\frac {d(d-1)}2$.\nThe number of polynomials that need to be tested in the algorithm\nis therefore exponential in the input $x$, hence very large,\nand the resulting algorithm is very slow.\n\nWe will give another algorithm in the next section.\n\n\\subsection{An algorithm to solve Problem \\ref{problem1}}\nIf $A=(a_{i,j})$ is a matrix (or a vector) with real entries,\nwe write $A \\geq 0$ to indicate that $a_{i,j} \\geq 0$ for all $i$ and $j$.\nWe also write $A\\geq B$ if $A-B \\geq 0$.\nWe will use the fact that, if $A\\geq 0$ and $B\\geq 0$, then $AB \\geq 0$.\n\nWe recall that, for a number field $K$ of degree $d$, with $r_1$ real embeddings and $r_2$ complex embeddings, we have set $s = r_1+r_2$ and $r = s-1$.\nIf $u_1,\\dots,u_r$ are generators of $\\calo_K^\\times$ modulo torsion,\nwe define the matrix $L$ of size $r\\times s$ by\n$$ L_{i,j} = \\log | \\sigma_j(u_i)|$$\nif $\\sigma_j$ is real and \n$$ L_{i,j} = 2 \\log | \\sigma_j(u_i)|$$\nif $\\sigma_j$ is complex.\nThe $i$-th row of $L$ is equal to $\\ell(u_i)$.\nAt last, we define the column vector $V = (1,\\dots,1)^t \\in \\mathbb{Z}^s$.\n\n\\medskip\nUsing logarithms, we can reformulate our Problem \\ref{problem1} as:\n\n\\begin{problem}\\label{problem2}\n Given a non-zero algebraic number $x\\in K^\\times$ and a bound $B>0$,\n find all rows $U \\in \\mathbb{Z}^r$ such that\n $$ UL + X \\geq 0 $$\n where $X = \\ell(x)-\\mathcal{L}(B)$.\n\\end{problem}\n\nFormulated in this way, we see that Problem \\ref{problem2} can be solved by integer linear programming. However, the situation is not generic here, and a simpler algorithm is given below.\n\n\\begin{algo}\\label{algo}~\n \n Input : $x\\in K$, $x\\neq 0$, and $B>0$.\n\n Output : all solutions $U \\in \\mathbb{Z}^r$ of Problem \\ref{problem2}.\n \n {\\tt\n \\begin{enumerate}\n \\item Compute the matrix $L$ of size $r\\times s$\n and the column vector $V$ of size $s$ as in the above definition.\n \\item Remove from $L$ its last column and call $M$ the inverse of this matrix.\n Concatenate $M$ with the row vector of size $r$ whose all entries are $0$\n and obtain a new matrix $M$ of size $s\\times r$.\n \n For the next steps, we use the notation $N_{,j}$ for the $j$-th column of a matrix $N$.\n \\item Define the matrix $N^+$ of size $s\\times r$, such that, for $1\\leq j \\leq r$, \n $N^+_{,j} = M_{,j} - \\min_i\\{M_{i,j}\\} V$.\n \\item Define the matrix $N^-$ of size $s\\times r$, such that, for $1\\leq j \\leq r$, \n $N^-_{,j} = M_{,j} - \\max_i\\{M_{i,j}\\} V$.\n \n\n \\item Compute the row vector $X = \\ell(x) - \\mathcal{L}(B)$.\n \\item For all row vector $U \\in \\mathbb{Z}^r$ in the range\n $ -XN^+ \\leq U \\leq -XN^-$, test if $UL+X \\geq 0$.\n If this is the case, output $U$.\n \\end{enumerate}\n }\n\\end{algo}\n\n\\begin{proposition}\\label{proofalgo}\n Algorithm \\ref{algo} is correct.\n\n Furthermore, when the number field $K$ is fixed,\n the number of $U$ that need to be tested during step 6\n is at most proportional to $( \\log N(x)- d\\log B +1)^r$.\n\n\\end{proposition}\n\\begin{proof}\nWe follow the algorithm step by step.\n\\begin{enumerate}\n\\item By Dirichlet's Unit Theorem, the matrix $L$ constructed in step 1 has rank $r$.\n Since the absolute norm of a unit is equal to $1$, we have $LV = 0$,\n hence $V$ is in the right kernel of $L$.\n\\item By Dirichlet's Unit Theorem,\n when we remove any column of $L$, the determinant of the remaining square matrix\n is always the same and equals the regulator of $K$, which is not $0$.\n This matrix of size $r \\times r$ is invertible.\n By construction, we have $LM = I_r$.\n\\item For all columns of $N^+$, we have $N^+_{,j} = M_{,j} - \\min_i\\{M_{i,j}\\} V$.\n Let $i(j)$ be the index such that $\\min_i\\{M_{i,j}\\} = M_{i(j),j}$.\n We have $N^+_{i,j} = M_{i,j} - M_{i(j),j} \\geq 0$ by minimality.\n Hence $N^+ \\geq 0$.\n Because $V$ is in the right kernel of $L$, we deduce that\n $LN^+ = LM = I_r$.\n\\item Using a similar argument, we can prove that\n $N^- \\leq 0$ and $LN^- = I_r$.\n\\item There is nothing to say here.\n\\item If $U$ is a solution of Problem \\ref{problem2},\n then $UL+X \\geq 0$.\n But $N^+ \\geq 0$, hence $ULN^+ + XN^+ \\geq 0$.\n By the relation $LN^+ = I_r$, we deduce $U \\geq -XN^+$.\n By $N^- \\leq 0$, we deduce $ULN^- + XN^- \\leq 0$, and $U \\leq -XN^-$.\n\\end{enumerate}\n\nIn order to bound the number of $U$ tested in step $6$,\nwe observe that $-XN^+ \\leq U \\leq -XN^-$.\nFor the $j$-th entry, this is explicitly\n$-XN^+_{,j} \\leq U_j \\leq -XN^-_{,j}$\nhence the number of $U_j$ that need to be tested is at most equal to\n$-XN^-_{,j} + XN^+_{,j}+1$. But we have\n\\begin{align*} -XN^-_{,j} + XN^+_{,j} & = X(M_{,j}-\\min_i\\{M_{i,j}\\}V - M_{,j}+\\max_i\\{M_{i,j}\\}V)\\\\\n& = (\\max\\{M_{,j}\\} - \\min\\{M_{,j}\\}) XV\n\\end{align*}\nWe also have $XV = \\log N(x)- d\\log B$.\nIf $\\log N(x) -d\\log B < 0$, we have seen in Proposition \\ref{smallnorm} that the problem has no solution. When\n$\\log N(x) -d\\log B \\geq 0$, we have\n$$-XN^-_{,j} + XN^+_{,j}+1\n\\leq \n(\\max\\{M_{,j}\\} - \\min\\{M_{,j}\\}+1) (\\log N(x) - d \\log B + 1)$$\nwhence a bound for the number of $U$ by\n$$ (\\log N(x) - d\\log B+1)^r \\times \\prod_j (\\max\\{M_{,j}\\} - \\min\\{M_{,j}\\}+1)$$\n\n\n\\end{proof}\n\n\\section{A complete example}\\label{sect:cyclotomic}\n\nIn this section, we shall give a detailed execution of Algorithm \\ref{algo}, which answers Question \\ref{qu:continued} for $p=17$.\nAll computations were done using PARI\/gp \\cite{PARI2}.\n\n\\medskip\nLet us consider the $16$-th cyclotomic field $K$ equal to ${\\QQ}(\\zeta) = {\\QQ}[X]\/\\Phi_{16}(X)$,\nwhere $\\Phi_{16}(X) = X^8+1$.\n\nIn this field, we consider $x = -\\zeta^7 - \\zeta^3 + \\zeta^2$.\nWe have $N(x) = 17$, hence $x$ is a generator of a principal prime ideal above $17$.\nWe are looking for another generator $x'$ of this principal ideal\nsuch that $|\\sigma(x')| \\geq 1$ for all $\\sigma\\in\\Sigma$. We need to solve Problem \\ref{problem2} with $B = 1$.\n\nWe follow here the steps of Algorithm \\ref{algo}.\n\n\\begin{enumerate}\n\\item\n For this field, we have $d=8$, $r_1 = 0$ and $r_2 = 4$.\n In this case, we have $s = r_1+r_2 = 4$ and $r = 3$.\n \n The units of $K$ are generated by\n $u_0 = \\zeta$, $u_1 = -\\zeta^6 + \\zeta^2 - 1$, $u_2 = \\zeta^2+\\zeta+1$ and $u_3=-\\zeta^6+\\zeta^3-\\zeta$,\n where $u_0$ generates the torsion part and $u_1,u_2,u_3$ generate the free part.\n The matrix $L$ is equal to\n $$L =\n \\left(\n \\begin{array}{rrrr}\n -1.76274&-1.76274& 1.76274& 1.76274\\\\\n -0.33031& 2.09306&-2.89946& 1.13671\\\\\n 1.13671&-2.89946&-0.33031& 2.09306\n \\end{array} \\right)\n $$\n We easily check that $L\\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = 0$.\n\\item\n We have\n $$M = \n \\left( \\begin{array}{rrrr}\n -0.46575&-0.29144& 0.07276\\\\\n -0.17430&-0.07276&-0.29144\\\\\n -0.07276&-0.36421&-0.21868\\\\\n 0 & 0 & 0\n \\end{array} \\right)\n $$\n We can check that $LM = I_3$, the identity matrix of order $3$.\n\\item We have\n $\\min(M_{,1}) = -0.46575$,\n $\\min(M_{,2}) = -0.36421$,\n $\\min(M_{,3}) = -0.29144$.\n This gives\n $$N^+ = \\left( \\begin{array}{rrr}\n 0 & 0.07276 & 0.36421 \\\\\n 0.29144 & 0.29144 & 0 \\\\\n 0.39298 & 0 & 0.07276 \\\\\n 0.46575 & 0.36421 & 0.29144 \\\\\n \\end{array}\\right)$$\n We can check that $N^+ \\geq 0$ and $LN^+ = I_3$.\n\\item We have\n $\\max(M_{,1}) = 0$,\n $\\max(M_{,2}) = 0$,\n $\\max(M_{,3}) = 0.07276$. This gives\n $$N^- = \\left( \\begin{array}{rrr}\n -0.46575 & -0.29144 & 0 \\\\\n -0.17430 & -0.07276 & -0.36421\\\\\n -0.07276 & -0.36421 & -0.29144\\\\\n 0 & 0 & -0.07276\\\\\n \\end{array}\\right)$$\n We can check that $N^- \\leq 0$ and $LN^- = I_3$.\n\\item Since $B=1$, we have $\\mathcal{L}(B) = 0$.\n For $x = -\\zeta^7 - \\zeta^3 + \\zeta^2$, we have\n $$X = \\ell(x) = (1.40668, 0.65107, 1.72510, -0.94965)$$\n\\item\n We compute\n $$-XN^+ = (-0.42539, 0.05376, -0.36109)$$\n $$-XN^- = ( 0.89419, 1.08567, 0.67081)$$\n In this example, the only $U \\in \\mathbb{Z}^3$ within the bounds is\n $U = (0,1,0)$.\n However, for this $U$, we have\n $$UL+X = (1.07636, 2.74414, -1.17435, 0.18706)$$\n hence this is not a solution.\n\n\n\\end{enumerate}\n This computation shows that, in this example, Problem \\ref{problem2} has no solution.\\\\\nAnalogous computations applied to all prime $p$ in the list \\eqref{list} allow to give a complete answer to Question \\ref{qu:continued}:\n\\begin{theorem}\\label{teo:risposta}\nLet $p$ be one of the primes for which $\\mathbb{Q}(\\zeta_{p-1})$ has class number 1, as listed in \\eqref{list}, and let $\\mathfrak{P}$ be a prime ideal over $p$ in the ring of integers of $\\mathbb{Q}(\\zeta_{p-1})$. Then $\\mathfrak{P}$ is large if and only if\n\\begin{equation}\\label{eq:listalarge} p\\in \\{5,7,11,13,19,31\\}.\\end{equation}\n\\end{theorem}\nMore precisely, for each primes in the list \\eqref{eq:listalarge} the following table gives (up to Galois conjugation and multiplication by a root of unity) the elements $\\pi$ of absolute norm $p$ having all the components $\\geq 1$ in the canonical embedding ($\\zeta=\\zeta_{p-1}$ in each case): \n\n\\begin{table}[h]\n\\def1.5{1.5}\n\\begin{tabular}{l|l} $p$ & $\\pi$\\\\ \\hline $5$ & $2\\zeta+1$\\\\\n $7$ & $\\zeta-3$\\\\\n $11$ & $2\\zeta^3-1$, \\quad $2\\zeta^2-\\zeta+1$\\\\\n $13$ & $-2\\zeta^3-\\zeta^2$, \\quad $\\zeta^3-\\zeta^2+2$\\\\\n $19$ & $-\\zeta^4-\\zeta^3+\\zeta^2+\\zeta+1$\\\\\n $31$ & $-\\zeta^7-\\zeta^3-\\zeta$,\\quad $ -\\zeta^6-\\zeta^5+\\zeta^3+\\zeta^2+\\zeta-1$\\\\\n &\n \\end{tabular}\n \n \\caption{Strictly large integers of absolute norm $p$ in $\\mathbb{Q}(\\zeta_{p-1})$} \\label{tavola} \n\\end{table}\n\n\n\n\n\\section{Another application: floor functions and types}\\label{sect:floorfunctions}\n\nLet $K$ be a number field of degree $d$ over $\\mathbb{Q}$, and let $\\mathcal{O}_K$ be its ring of integers. We fix an ideal $\\mathfrak{A}$ of $\\mathcal{O}_K$.\nThe aim of this section is to apply largeness (when possible) in order to define complete sets of representatives of $K\/\\mathfrak{A}$ (which will be called \\emph{types}) satisfying some integrality properties and having all the Archimedean embeddings bounded in a controlled way. \\\\ Types associated to a prime ideal $\\mathfrak{P}$ of a number field were introduced in \\cite{CapuanoMurruTerracini2022} with the aim of constructing a general notion of $\\mathfrak{P}$-adic continued fractions and studying their finiteness and periodicity properties.\n\nLet $\\mathcal{M}_K^0$ be a set of representatives for the non-Archimedean places of $K$. For every rational prime $p$ and every $v\\in\\mathcal{M}_K^0$ above $p$ let $K_{v}\\subseteq \\overline{\\mathbb{Q}}_p$ be the completion of $K$ w.r.t. the $v$-adic valuation and $\\mathcal{O}_v$ be its valuation ring; we put $d_v=[K_v:\\mathbb{Q}_p]$. Let $|\\cdot |_v=|N_{K_v\/\\mathbb{Q}_p}(\\cdot)|_p^{\\frac 1{d_v}}$ be the unique extension of $|\\cdot |_p$ to $K_v$. \nLet $\\widetilde{K}=\\prod_{v \\mid \\mathfrak{A}} K_v$ be the $\\mathfrak{A}$-adic completion of $K$, with $K$ diagonally embedded, and $\\widetilde{\\mathcal{O}}=\\prod_{v \\mid \\mathfrak{A}}\\mathcal{O}_v$.\\\\\nLet $S_0=\\{v\\in \\mathcal{M}_K^0 \\mid v \\mid \\mathfrak{A}\\}$.\n\\begin{definition}\nAn \\emph{$\\mathfrak{A}$-adic floor function} for $K$ is a function $s:\\widetilde K\\to K$ such that\n\\begin{itemize} \n\\item[a)] $\\alpha-s(\\alpha)\\in \\mathfrak{A}\\widetilde\\mathcal{O}$ for every $\\alpha\\in \\widetilde K$;\n\\item[b)] $|s(\\alpha)|_{v}\\leq 1$ for every $v\\in \\mathcal{M}_K^0\\setminus S_0$;\n\\item[c)] $s(0)=0$;\n\\item[d)] $s(\\alpha)=s(\\beta)$ if $\\alpha-\\beta\\in\\mathfrak{A}\\widetilde\\mathcal{O}$.\n\\end{itemize}\n\\end{definition}\n\n\n \nBy the Strong Approximation Theorem in number fields (see for example \\cite[Theorem 4.1]{Cassels}), $\\mathfrak{A}$-adic floor functions always exist, and there are infinitely many. \\\\\nWe define the ring of $S_0$-integers\n\\[\\mathcal{O}_{K,S_0}=\\{\\alpha\\in K\\ |\\ |\\alpha|_v\\leq 1\\hbox{ for every } v \\in \\mathcal{M}_K^0 \\setminus S_0\\}.\n\\]\nThen, we can regard an $\\mathfrak{A}$-adic floor function as a map $s: \\widetilde K\/\\mathfrak{A}\\widetilde\\mathcal{O}\\to \\mathcal{O}_{K,S_0}$ such that $s(\\mathfrak{A}\\widetilde\\mathcal{O})=0$ and which is a section of the projection map $\\widetilde K\\to \\widetilde K\/\\mathfrak{A}\\widetilde\\mathcal{O}$.\nTherefore the choice of an $\\mathfrak{A}$-adic floor function amounts to choose a set $\\mathcal{Y}$ of representatives of the cosets of $\\mathfrak{A}\\widetilde\\mathcal{O}$ in $\\widetilde K$ containing $0$ and contained in\n$\\mathcal{O}_{K,S_0}$.\\\\\nWe shall call the data $\\tau=(K,\\mathfrak{A},s)$ (or $(K,\\mathfrak{A},\\mathcal{Y})$) a \\emph{type}. \n\\begin{remark} The absolute Galois group $\\mathrm{Gal}(\\overline{\\QQ}\/\\mathbb{Q})$ acts on the set of types; indeed, if $\\tau=(K,\\mathfrak{A}, s)$ is a type, then $\\sigma\\in \\mathrm{Gal}(\\overline{\\QQ}\/\\mathbb{Q})$ induces a continuous map $\\widetilde K \\to \\widetilde{K^\\sigma}$, where $\\widetilde{K^\\sigma}$ is the completion of $K^\\sigma$ with respect to the ideal $\\mathfrak{A}^\\sigma$. Then $\\tau^\\sigma=(K^\\sigma,\\mathfrak{A}^\\sigma, s^\\sigma)$ is also a type, where $s^\\sigma=\\sigma\\circ s\\circ \\sigma^{-1}$. In particular, if $K\/\\mathbb{Q}$ is a Galois extension and $\\sigma$ belongs to the decomposition group \n$$D_\\mathfrak{A}=\\{\\sigma\\in\\mathrm{Gal}(\\overline{\\QQ}\/\\mathbb{Q})\\ |\\ \\mathfrak{A}^\\sigma=\\mathfrak{A}\\},$$\n then $\\tau^\\sigma=(K,\\mathfrak{A}, s^\\sigma)$ is again an $\\mathfrak{A}$-adic type.\n\\end{remark}\n\\subsection{Types arising from generators of $\\mathfrak{A}$} \\label{sec:special_type}\nIn the case $\\mathfrak{A}$ is principal, there is a natural way of defining an $\\mathfrak{A}$-adic floor function.\nIndeed, let $\\pi\\in\\mathfrak{A}$ be generator and let $\\mathcal{R}$ be a complete set of representatives of $\\mathcal{O}_K\/\\mathfrak{A}$ containing $0$. Then, every $\\alpha\\in \\widetilde K$ can be expressed uniquely as a Laurent series $\\alpha=\\sum_{j=-n}^\\infty c_j\\pi^j$, where $c_j\\in\\mathcal{R}$ for every $j$. It is possible to define an $\\mathfrak{A}$-adic floor function by\n$$s(\\alpha)=\\sum_{j=-n}^0c_j\\pi^j\\in K. $$\nWe shall denote the types $\\tau=(K,\\mathfrak{A},s)$ obtained in this way by $\\tau=(K,\\pi,\\mathcal{R})$, and we will usually call them \\emph{special types}.\n\n\\begin{example}[Browkin and Ruban types over $\\mathbb{Q}$]\nWhen $K=\\mathbb{Q}$ and $\\pi=p$ odd prime, two main special types have been studied in the literature:\n\\begin{itemize}\n\\item the \\emph{Browkin type} $\\tau_B=(\\mathbb{Q},p,\\mathcal{R}_B)$ where $\\mathcal{R}_B=\\{- \\frac {p-1} 2,\\ldots, \\frac {p-1} 2\\}$ (see \\cite{Browkin1978, Bedocchi1988, Bedocchi1989, Bedocchi1990, Browkin2000, CapuanoMurruTerracini2020});\n\\item the \\emph{Ruban type} $\\tau_R=(\\mathbb{Q},p,\\mathcal{R}_R)$ where $\\mathcal{R}_R=\\{0,\\ldots, p-1\\}$ (see \\cite{Ruban1970, Laohakosol1985, Wang1985, CapuanoVenezianoZannier2019}).\n\\end{itemize}\n\\end{example}\n\n\\subsection{Bounded types} We say that a type $\\tau=(K,\\mathfrak{A},s)$ is \\emph{bounded} if there exists a real number $C>0$ such that $|\\sigma(s(\\alpha))|