Let $D$ be a squarefree integer, and let $\mathcal{O}$ be the ring of integers in the quadratic field $\mathbb{Q}(\sqrt{D})$. For positive integer $f$ define the order of conductor $f$, $\mathcal{O}_f$ by $$ \mathcal{O}_f = \mathbb{Z}[f\omega] = \{ a + bf\omega \mid a,b \in \mathbb{Z} \}, $$


$$\omega = \begin{cases} \sqrt{D} & \text{if } D \equiv 2 \text{ or } 3 \pmod 4 \\ \frac{1+\sqrt{D}}{2} & \text{if } D \equiv 1 \pmod 4. \end{cases}$$

In this case $\mathcal{O}_f$ is a subring of $\mathcal{O}$. I would like to prove $[ \mathcal{O}: \mathcal{O}_f] = f$ (where the index is as additive abelian groups).

I'm not sure how to go about this. A first step that seems logical is to write $\varphi: \mathcal{O} \to \mathbb{Z}$ defined by $\varphi (a+b\omega) = b$. This map is then a surjective group homomorphism. I'm not entirely sure how to proceed. Any detailed solution would be welcome since I'm a bit lost on this section of the textbook I am working on.

  • $\begingroup$ Presumably $\omega$ is an algebraic integer such that $\mathcal{O}=\Bbb{Z}[\omega]$? In that case you can study the quotient group $\mathcal{O}/\mathcal{O}_f$ directly using their respective bases $\{1,\omega\}$ and $\{1,f\omega\}$. It looks just like the quotient of two free abelian groups: $\Bbb{Z}^2/M$, where $M$ is spanned by $(1,0)$ and $(0,f)$. You may have seen stacked/aligned basis theorem when studying the structure of f.g. modules over a PID? $\endgroup$ Dec 10, 2015 at 19:01
  • $\begingroup$ @JyrkiLahtonen Yes, $\omega$ is an algebraic integer as you say, I have added that into the question. Unfortunately, I don't understand your suggestion since I don't have the notion of a base, and have not gotten to modules over a PID etc. I've been following Dummit and Foote and this is a piece of an exercise in the introductory chapter on rings (the book goes groups, ring, then modules and vector spaces). $\endgroup$
    – Bamboo
    Dec 10, 2015 at 19:06
  • $\begingroup$ Ok. Let's try something else then. Can you show directly that $\{0,\omega,2\omega,\ldots,(f-1)\omega\}$ is a set of representatives of cosets of $\mathcal{O}_f$ inside $\mathcal{O}$? $\endgroup$ Dec 10, 2015 at 19:17
  • $\begingroup$ Or, building upon your idea, can you show that $$f:\mathcal{O}\to\Bbb{Z}/f\Bbb{Z}, a+b\omega\mapsto \overline{b}$$ is a surjective homomorphism of additive groups with kernel $\mathcal{O}_f$? $\endgroup$ Dec 10, 2015 at 19:19
  • $\begingroup$ @JyrkiLahtonen Yes, I can! Thanks. To finish up, once we have that it is a surjective group homomorphism we can apply the first isomorphism theorem (or a corollary thereof) to conclude that the index of $\mathcal{O}_f$ in $\mathcal{O}$ is equal to the order of the image of the homomorphism. Since the homomorphism is surjective onto $\mathbb{Z}/f\mathbb{Z}$ we know this image has order $f$. Thanks! $\endgroup$
    – Bamboo
    Dec 12, 2015 at 16:52

1 Answer 1


Note that $f\omega\in\mathcal{O}_{f}$. Let $c$ be the smallest positive integer such that $c\omega\in\mathcal{O}_{f}$. By division algorithm, there exists $q,r\in\mathbb{Z}$ such that $f=cq+r,0\leq r <c$. Since $\mathcal{O}_{f}$ is a ring, $r\omega=(f-cq)\omega\in\mathcal{O}_{f}$. Thus $r$ must be $0$, i.e., $f$ is a multiple of $c$. Write $c\omega=a+bf\omega$. Since $1$ and $\omega$ are $\mathbb{Z}$-linearly independent, we have $a=0,c=bf$, in particular, $c$ is a multiple of $f$. Since $c$ and $f$ are positive integers, $c=f$.

Now define $\phi : \mathcal{O}\longrightarrow\mathbb{Z}/f\mathbb{Z}$ as follows:

Let $\alpha\in\mathcal{O}$ and write $\alpha=a+b\omega$. By division algorithm, $\exists ! q,r$ such that $b=fq+r,0\leq r <f$. Set $\phi(\alpha)=r\pmod{f}$. Then one can show that $\phi$ is a group homomorphism. Let $r\pmod{f}\in\mathbb{Z}/f\mathbb{Z}$ Then $\alpha:=r\omega\in\mathcal{O}$ and $\phi(\alpha)=r\pmod{f}$. Suppose that $\phi(a+b\omega)=0$ in $\mathbb{Z}/f\mathbb{Z}$. By construction, this is equivalent to $b$ is a multiple of $f$. Hence, $\ker(\phi)=\mathcal{O}_f$. This proves that $[\mathcal{O}:\mathcal{O}_{f}]=f$.


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