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If $K = \mathbb{Q}(\sqrt{d})$ is a real quadratic field, then any unit $u \in \mathcal{O}_K^\times$ with $u > 1$ must not be too small: indeed, such a $u = u_1 + u_2 \sqrt{d}$ with $u_1, u_2 > 0$ must satisfy $u_1^2 - d u_2^2 = \pm 1$, so $u_1 \gg \sqrt{d}$, say. Thus the gap between the smallest unit $u \in \mathcal{O}_K^\times$ and $1$ must be quite large.

This seems a particular quirk of fields whose unit group is 1-generated. For real cubic fields, whose unit group is 2-generated, there seems to be substantial possibility for cancellation and even possible for the smallest unit $u > 1$ to be arbitrarily close to $1$ as the fields vary.

My question is this: let $K$ be a cyclic cubic field (in particular, necessarily totally real) having discriminant $d_K = c_K^2$. Let $u_K \in \mathcal{O}_K^\times$ be the smallest (in terms of usual archimedean valuation) element satisfying $u_K > 1$. Write

$$\displaystyle u_K = 1 + \kappa_K.$$

Can one effectively bound $\kappa_K$ in terms of $d_K$ (or equivalently, $c_K$)?

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2 Answers 2

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Asking about a smallest unit bigger than $1$ in a unit group of rank greater than $1$ feels like the wrong question, sort of like asking for a smallest algebraic integer of absolute value greater than $1$ in a number field (inside $\mathbf C$) of degree greater than $1$. The ring of integers is discrete when you use all Archimedean embeddings, but not fewer Archimedean embeddings (e.g., $\mathbf Z[\sqrt{2}]$ is dense in $\mathbf R$ but its image in $\mathbf R^2$ using the Euclidean embedding is a lattice). Likewise, the unit group of that ring is discrete using the logarithm mapping to a hyperplane in $\mathbf R^{r_1+r_2}$.

Use the regulator as a measure of the size of the unit group. For a unit group of rank $1$ with a real embedding, taking logarithms shows that comparing the size of the smallest unit greater than $1$ with a power of the absolute value of the discriminant is like comparing the regulator of that unit group with a multiple of the logarithm of the absolute value of discriminant. For cubic fields of rank $1$, Artin showed its regulator $R$ and (negative) discriminant $D$ satisfy $R > \log(|D|/4 - 6)$ except for the unique cubic field of discriminant $-23$. For totally real cubic fields, the unit rank is $2$ and Cusick showed the regulator $R$ and (positive) discriminant $D$ for such fields satisfy $R > (1/16)(\log(D/4))^2$. See Theorem 5.8 here.

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There will not be a smallest one. As you say, $O_K^\times$ has rank $2$, meaning it has two multiplicatively independent generators $u_1,u_2$, which we may take to be positive. This is equivalent to saying $\log u_1,\log u_2$ are linearly independent over $\mathbb Q$, and that implies the linear combinations of them will be dense in $\mathbb R$. Hence the subgroup of $O_K^\times$ generated by $u_1,u_2$ will be dense in $(0,\infty)$, and thus $O_K^\times$ will be dense in $\mathbb R$.

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  • $\begingroup$ I do believe the set $\{\log u : u \in \mathcal{O}_K^\times\}$ is a rank-2 $\mathbb{Z}$-lattice, so it is discrete. In particular we do not allow $\mathbb{Q}$-combinations in general and only $\mathbb{Z}$-combinations. This is manifested in the fact that $\mathcal{O}_K^\times$ is not closed under taking arbitrary radicals, for example $\endgroup$ Commented Jul 9, 2022 at 14:46
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    $\begingroup$ @StanleyYaoXiao That set will be a subgroup isomorphic to $\mathbb Z^2$, but it is a subgroup of $\mathbb R$, not of $\mathbb R^2$. Such a subgroup is necessarily dense, even though we allow only integer combinations. $\endgroup$
    – Wojowu
    Commented Jul 9, 2022 at 14:54

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