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Given the number field $K=\mathbb{Q}[\sqrt{m}]$ for some $m\equiv 2$or$3$ mod$4$ we take $b$ to be the minimum integer such that $mb^{2} \pm1 = a^{2}$. Then $a + b\sqrt{m}$ is a unit in $\mathbb{Z}[\sqrt{m}]$.

I need to show that $a + b\sqrt{m}$ is actually a fundamental unit. Now I think I'm able to assume $a + b\sqrt{m}$ is a power of the fundamental unit that generates the multiplicative unit group, $\mathcal{O}_{K}^{*}$ because we are dealing with a real quadratic field. So I want to show that power can't be greater than $1$. So would the correct approach involve showing $a + b\sqrt{m}$ can't be factored in $\mathbb{Z}[\sqrt{m}]$?

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  • $\begingroup$ If $a + \sqrt{m} b = (c + \sqrt{m} d )^e $, then also $a - \sqrt{m} b = (c - \sqrt{m} b)^e$? If so, $a^2 - m b^2 = ( c^2 - m d^2 ) ^e$. $ l := c^2 - m d^2 \in \mathbb{Z}$ and $l = \pm 1$. Then $b$ was not minimal. $\endgroup$ – MPB94 Nov 25 '17 at 18:26
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Let $\epsilon$ be a fundamental unit, and let $\eta = \pm \epsilon^k$ be any other unit. Your question is how to tell if $\eta = \pm \epsilon$, or whether $k = \pm 1$. You already have the key idea: if $\eta = a + b \sqrt{m}$, then the size if $\eta$ is related to the size of $b$. Hence the smallest $b$ can only occur for $k = \pm 1$.

Let $\epsilon = r + s \sqrt{m}$, and $\eta = a + b \sqrt{m}$. Without loss of generality, we may assume that $r,s \ge 1$ and $a,b \ge 1$. Thus if $\eta = \pm \epsilon^k$, then $\eta = \epsilon^k$ for $k \ge 1$.

Lemma: $s$ is the closest integer to $\displaystyle{\frac{\epsilon}{2 \sqrt{m}}}$ and $b$ is the closest integer to $\displaystyle{\frac{\eta}{2 \sqrt{m}} = \frac{\epsilon^k}{2 \sqrt{m}}}$.

In particular, if $m > 1$, then $b > s$. Since $ms^2 \pm 1 = r^2 $ and $m b^2 \pm 1 = a^2$, this answers your question.

Proof: The argument for both is the same, so let's just do $\eta = a + b \sqrt{m}$. We have $\eta > \sqrt{m}$. The conjugate of $\eta$ is is $\sigma \eta = a - b \sqrt{m}$. The assumption that $\eta$ is a unit implies that $N(\eta):=\eta \cdot (\sigma \eta) = a^2 - m b^2 = \pm 1$. This means that the conjugate $\sigma \eta$ is very small, less than $1/\sqrt{m}$. It follows that:

$$b = \frac{1}{2 \sqrt{m}} \left(\eta + \sigma \eta\right) = \frac{1}{2 \sqrt{m}} \left(\eta + \frac{N(\eta)}{\eta}\right).$$

In particular,

$$\left| b - \frac{1}{2 \sqrt{m}} \eta \right| \le \frac{1}{2 \sqrt{m} |\eta|} \le \frac{1}{2m} < \frac{1}{2},$$

and so $b$ is the closest integer to $\displaystyle{\frac{\eta}{2 \sqrt{m}}}$, as claimed.

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  • $\begingroup$ right after the lemma it should $ms^2 \pm 1 = r^2$, right? and the exponent of the $\epsilon$ in the lemma should be $k$, not $m$, I think. $\endgroup$ – MPB94 Nov 26 '17 at 16:55

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