# $\mathbb{Q}(\sqrt{d})$ with specific integral basis

I would like some help with the following question.

Ireland and Rosen (ch.13#10)

For which $d$ does $\mathbb{Q}(\sqrt{d})$ have an integral basis of the form $\alpha, \alpha '$ where $\alpha '$ is the conjugate of $\alpha$?

As I understand this, {$a,b\sqrt{d}$} is a basis for $\mathbb{Q}(\sqrt{d})$, so we can set $a+b\sqrt{d}=a_{1}\alpha +b_{2}\alpha '$ where $a,b,a_{1},b_{1}\in \mathbb{Q}$ and $\alpha '$ is the conjugate of $\alpha$. However, I do not see where the restrictions on $d$ come from. Thanks.

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I believe they mean a basis for the ring of integers, and not a basis for the field over $\mathbb{Q}$ consisting of elements that happen to be integral (the latter always exists). – user29743 Apr 25 '12 at 5:36

The ring of integers of a number field $K$ is the set of elements of $K$ that are integral over $\mathbb{Z}$, i.e. they are roots of monic irreducible polynomials with coefficients in $\mathbb{Z}$. As the name implies, they form a subring of $K$. The ring of integers of $K$ is usually denoted $\mathcal{O}_K$.

An integral basis for a number field $K$ is a basis for $\mathcal{O}_K$ as a $\mathbb{Z}$-module (it is a theorem that $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $n$, where $n=[K:\mathbb{Q}]$).

Proposition 13.1.1 in Ireland & Rosen tells you that, for $K=\mathbb{Q}(\sqrt{d})$ with $d$ a squarefree integer, $$\mathcal{O}_K=\begin{cases}\mathbb{Z}[\sqrt{d}] & \text{ if }d\equiv 2,3\bmod 4,\\\\\mathbb{Z}\left[\tfrac{-1+\sqrt{d}}{2}\right] & \text{ if }d\equiv 1\bmod 4.\end{cases}$$

So, given two $\alpha,\beta\in\mathcal{O}_K$, the set $\{\alpha,\beta\}$ is an integral basis for $K$ if every $\gamma\in \mathcal{O}_K$ can be uniquely represented as $\gamma=r\alpha+s\beta$ for some $r,s\in\mathbb{Z}$. Clearly, if $\{\alpha,\alpha'\}$ is to be an integral basis, $\alpha$ cannot be in $\mathbb{Z}$, so we must have that $$\alpha=\begin{cases}h+k\sqrt{d} \text{ for some }h,k\in\mathbb{Z}, k\neq 0 & \text{ if }d\equiv 2,3\bmod 4,\\\\h+k\left(\tfrac{-1+\sqrt{d}}{2}\right) \text{ for some }h,k\in\mathbb{Z}, k\neq 0 & \text{ if }d\equiv 1\bmod 4.\end{cases}$$ Suppose $d\equiv 2,3\bmod 4$, and that $\alpha=h+k\sqrt{d}$. Then $\alpha'=h-k\sqrt{d}$. Can you write every element of $\mathcal{O}_K$ as a $\mathbb{Z}$-linear combination of $\alpha$ and $\alpha'$?

Now try the case of $d\equiv 1\bmod 4$ for yourself :)

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Hint $\rm\ (j\alpha + k\alpha'= 1)'\Rightarrow\: j\alpha'+k\alpha= 1\:\Rightarrow\: j = k\:\Rightarrow\: k\:(\alpha+\alpha') = 1\:\Rightarrow\: tr\:\alpha = \pm 1$

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So, here's a hint: if $d$ is 1 mod 4, there's no problem, because we know the ring of integers has the basis $$1, \frac{1 + \sqrt{d}}{2},$$ but that means that $$\frac{1 + \sqrt{d}}{2}, \frac{1 - \sqrt{d}}{2}$$ is also a basis since the sum of those two elements is 1.

I claim there's at least sometimes a problem when $d$ is not 1 mod 4. Now the known basis is $$1, \sqrt{d}.$$ It's clear that $$-\sqrt{d}, \sqrt{d}$$ won't work (those aren't even linearly independent!), but of course we could try to work with something like $$a + b \sqrt{d}, a - b\sqrt{d},$$ and what you need to show is that you can't always pick $a$ and $b$ to make this work. As a hint, think about the norms of all the elements in the $\mathbb{Z}$ span of this basis.

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There's always a problem if $d \not\equiv 1 \bmod 4$ (for sqfree $d$). One ${\mathbf Z}$-basis of integers of ${\mathbf Q}(\sqrt{d})$ is $\{1,\sqrt{d}\}$, so index of ${\mathbf Z}$-span of $\{a+b\sqrt{d},a-b\sqrt{d}\}$ in ${\mathbf Z}[\sqrt{d}]$ is $|2ab| \geq 2$. A nec. condition for normal integral basis of Galois $K/{\mathbf Q}$ is that $K/{\mathbf Q}$ is tamely ramified, but ${\mathbf Q}(\sqrt{d})/{\mathbf Q}$ is wildly ramified at 2 if $d \not\equiv 1 \bmod 4$. See my answer at math.stackexchange.com/questions/87290/… for more. – KCd Apr 29 '12 at 20:57