Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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 :)

share|cite|improve this answer

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 $

share|cite|improve this answer

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.

share|cite|improve this answer
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… for more. – KCd Apr 29 '12 at 20:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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