I'm having a problem with a section of Niven's book the Theory Of Numbers. I am trying to show:

If an integer $\alpha \in \mathbb{Q}(\sqrt{m})$ is neither zero nor a unit, prove that $|N(\alpha)|>1$.

An element of $\mathbb{Q}(\sqrt{m})$ would look like $a+b\sqrt{m}$ where $a,b$ are rational integers $0, \pm1, \pm2,...$ The norm is defined as $N(\alpha)=\alpha\bar{\alpha}$, where $\bar{\alpha}$ is defined as the conjugate of $\alpha$ ($\alpha=a+b\sqrt{m}, \bar{\alpha}=a-b\sqrt{m}$).

So since $\alpha$ can not be zero or a unit, we have the two cases where 1). $a\neq 0, b\neq 0$. 2). $a\neq \pm 1, b\neq 0$. Therefore $a\ge 2, b\ge 1$. Now

$N(\alpha)=\alpha\bar{\alpha}=(a+b\sqrt{m})(a-b\sqrt{m})=a^2 -b^2m$. By our given conditions, we see that $N(\alpha)\neq 0$ or $1$ Now I have to show that $|N(\alpha)|>1.$ I was thinking contradiction, so suppose that there exist an $a,b$ such that $|a+b\sqrt{m}|\le 1$. Then $$-1\le a^2+b^2m\le 1$$ However, with $m\ge 2$ this is impossible unless either $a=b=0$ or $a=1, b=0$. But these cannot be since we have restricted these conditions in the initial problem. Therefore $|N(\alpha)|>1$.

Is this the correct way of solving this particuluar problem? ( I think I worked it out while I was typing, but any verification would be helpful....)

  • $\begingroup$ Looks good to me. $\endgroup$ – Hubble Jul 30 '14 at 6:11
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    $\begingroup$ They are not necessarily of the form $a+b\sqrt{m}$ with $a$ and $b$ integers, example $m=5$. But the norm of an agebraic integer is always an integer. So if it has absolute value not equal to $0$ or $1$, it must have absolute value greater than $1$. $\endgroup$ – André Nicolas Jul 30 '14 at 6:11
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    $\begingroup$ Oh, I think I found it, because $5\equiv 1\pmod{4}$....So should I be doing the proof by cases then to accommodate the different forms of $m$? (i.e., $1\pmod{4}, 3\pmod{4},...$) $\endgroup$ – Lalaloopsy Jul 30 '14 at 6:26
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    $\begingroup$ @Adam, I don't think Niven's book discusses such things as the index of an ideal. $\endgroup$ – Gerry Myerson Jul 30 '14 at 9:30
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    $\begingroup$ @GerryMyerson thanks again for the heads up. The answer is undeleted. $\endgroup$ – Adam Hughes Jul 31 '14 at 1:09

This is fairly straightforward. For elements of the field, $k$, we have that


Iff the index in the ring is $1$ is $\alpha$ a unit when $\alpha$ is an integer of $k$.

Proof of the formula $(*)$:

We know that the map $m_\alpha:x\to\alpha x$ is a linear transformation of the field $K=\Bbb Q(\sqrt{m})$ to itself which has determinant $N(\alpha)$ by definition of the norm down to $\Bbb Q$. Since $\alpha$ is an algebraic integer, $m_\alpha$ takes the integer ring, $\mathcal{O}_k$ into itself as a $\Bbb Z$-module of rank $2$, and indeed the image is just $\alpha\mathcal{O}_k=(\alpha)$. However the index of that ideal is exactly the determinant of the matrix because $\mathcal{O}_k$ is a finitely generated, free $\Bbb Z$-module, hence $|N(\alpha)|=1\iff (\alpha)=\mathcal{O}_k$. That $N(\alpha)\in\Bbb Z$ for all algebraic integers is a byproduct of the fact that $m_\alpha\in GL_{[k:\Bbb Q]}(\Bbb Z)$ is a matrix with integer coefficients, when $\alpha\in\mathcal{O}_k$.

As a side note, it's straightforward to show in fact, that $N^L_{\Bbb Q}(\alpha)=N_{\Bbb Q}^k(\alpha)^{[L:k]}$ from the definition of the norm, whenever $\alpha\in k\subseteq L$, and that the charateristic polynomial of $m_\alpha$ in $L$ is just the $[L:k]$th power of the characteristic polynomial in $k$, so that you can really just assume that the polynomial for $m_\alpha$ is the minimal one, since $|x|^n\le 1\iff |x|\le 1$ as far as sizes are concerned. I mention it because--in your case--it's easier to see more concretely why the constant term in the characteristic polynomial of that matrix exactly matches the ad hoc definition as $a^2-mb^2$.


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