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 $\mathbb{Z}[\sqrt{m}]$?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.