Let $d$ be a positive integer which is not a perfect square. Let $K = \mathbb{Q}(\sqrt{d})$, and let $I$ be a principal ideal in the ring of integers $\mathcal{O}_K$ of $K$. My question is, suppose that I know that $N(I)$ is bounded by some positive number $X$, and I know the regulator $\log \epsilon_d = \log(u_0 + v_0 \sqrt{d})$, where $(u_0, v_0)$ is the smallest positive solution to the equation $x^2 - dy^2 = \pm 4$. Can I bound the size of a generator of $I$, say $u+v\sqrt{d}$? More specifically, is it true that there exists a generatoe $g = u + v\sqrt{d} \in \mathcal{O}_K$ such that $\max\{|u|,|v|\} = O(X \log \epsilon_d)$?

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    $\begingroup$ Yes; write $I = (\alpha)$ and pick among all $\alpha \varepsilon^n$ one with small absolute value. Bound its conjugate using $N(I)$. If both an element and its conjugate have bounded absolute value, you can find bounds on the absolute values of $u$ and $v$. $\endgroup$
    – user23365
    Aug 6, 2016 at 6:09


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