Associated to the quadratic field $K = \mathbb{Q}[\sqrt{D}]$ (for $D$ square-free) is a number, denoted $\omega$ in all the discussions I’ve seen, defined by $$\omega = \begin{cases} \frac{-1 + \sqrt{D}}{2} & \text{if}\ D \equiv 1 \pmod{4} \\ \sqrt{D} & \textrm{otherwise.} \end{cases}$$

(It’s used, for instance, in the standard description of the algebraic integers of K as $\mathbb{Z} + \omega \mathbb{Z}$.)

Is there a standard term for this number, other than just $\omega$?

  • 2
    $\begingroup$ I don’t think that that notation is all that standard. For me (and maybe for me only) $\omega$ is that constant when $D=-3$. $\endgroup$
    – Lubin
    Commented Jun 13, 2013 at 22:25
  • $\begingroup$ @Lubin: In e.g. a complex-analysis context, I’d quite agree that $\omega$ conventionally means $(\sqrt{3}i - 1)/2$. In algebraic number theory, though, this seems quite common — I’m very unfamiliar with the field, but the first few sources I met it in all used $\omega$ for this: Jozsa’s Notes on Hallgren’s algotithm, these notes of Keith Conrad, and Wikipedia. $\endgroup$ Commented Jun 13, 2013 at 23:33

1 Answer 1


I just came across this post, almost $10$ years too late. Since a comment by the OP includes a link to something I wrote, let me first point out that when $D \equiv 1 \bmod 4$, I did not use $\omega$ at the link to mean $(-1 + \sqrt{D})/2$: I wrote $\omega$ for $(1+\sqrt{D})/2$, and it is $(1+\sqrt{D})/2$, not $(-1+\sqrt{D})/2$, that is used universally in the case $D \equiv 1 \bmod 4$. The Wikipedia page on quadratic integers that is mentioned in the question also uses $(1+\sqrt{D})/2$. Perhaps the OP made a sign error.

The only time when I'd expect to find anyone singling out $(-1 + \sqrt{D})/2$ as being worthy of having a special label is the case mentioned in Lubin's comment: $D = -3$, since $(-1+\sqrt{-3})/2$ is a nontrivial cube root of unity $\zeta_3$ (while $(1+\sqrt{-3})/2$ is a primitive $6$th root of unity $\zeta_6$). The numbers $(1+\sqrt{D})/2$ and $(-1+\sqrt{D})/2$ differ by $1$, so they generate the same ring over $\mathbf Z$, and it's simply a long-standing tradition when $D \equiv 1 \bmod 4$ to prefer $(1+\sqrt{D})/2$ as the generator of the ring of integers of $\mathbf Q(\sqrt{D})$.

To answer the question that was asked: there is no standard term. It's just called $\omega$.

Something worth noting is that there is a unified way to write the ring of integers of all quadratic fields without having to make up a notation $\omega$ for $\sqrt{D}$ or $(1+\sqrt{D})/2$ depending on $D \bmod 4$: if the quadratic field $K$ has discriminant $\Delta$, then $\Delta = D$ if $D \equiv 1 \bmod 4$ and $\Delta = 4D$ if $D \not\equiv 1 \bmod 4$, and you can check in both cases that $$ \mathcal O_K = \mathbf Z\left[\frac{\Delta + \sqrt{\Delta}}{2}\right] = \mathbf Z + \mathbf Z\frac{\Delta + \sqrt{\Delta}}{2}. $$

The first book treatment of "general" algebraic number theory is Supplement XI in Dirichlet-Dedekind's Vorlesungen über Zahlentheorie, where quadratic fields are the topic of Sections 186 and 187 (at least in the 4th edition). The only basis used there for the ring of integers is $\{1,(\Delta + \sqrt{\Delta})/2\}$. The notation $\omega$ is used there for an algebraic number in $K$, not necessarily an algebraic integer, and the reason for using $\omega$ is that the notation in the book for a general number field is $\Omega$. A number field is called a Schaar = flock/herd -- how picturesque!

  • $\begingroup$ I haven’t worked on this topic for most of the decade since I posted the question, but it’s great to get such a thorough answer! The background, incidentally, was implementing Hallgren’s algorithm in Quipper — not yet public when I posted. Apologies for misquoting your notes; I’d been sloppy unifying conventions between yours and Josza’s (see the question’s edit history…). But it still seems odd if none of these widely-used choices of generating algebraic integer has a distinctive name! $\endgroup$ Commented Jan 16, 2023 at 10:06

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