# Rank these notations from most valid to least valid

$$\mathbb{Z}[\sqrt{-3}]$$

$$\mathbb{Z}\left[\frac{i \sqrt{3} - 1}{2}\right]$$

$${\bf{A}}(-3)$$

$$\mathbb{Z}[\omega]$$

$$\mathbb{Z}\left[-\frac{1}{2} + \frac{\sqrt{-3}}{2}\right]$$

$$\mathbb{Z}[\sqrt{-27}]$$

$$\mathbb{Z}\left[\frac{\sqrt{-3}}{2}\right]$$

I don't know how to number equations without a preamble, so please cut me a little slack.

Of course $\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2}$, right?

The notation with the bold A comes from a really old book found in some public libraries and used bookstores.

• What do you mean by "most valid"? Notation is notation. – user61527 Jun 28 '14 at 17:15
• "Valid" is a weird word. Some of these refer to different rings than others. And I've never seen $\mathbf{A}(-3)$ before. – Qiaochu Yuan Jun 28 '14 at 17:22
• @T.Bongers I guess that I mean "most efficient according to professional mathematicians" and "palatable to Math.StackExchange users." – Robert Soupe Jun 28 '14 at 17:23
• @QiaochuYuan If I recall correctly, the choice of the letter A was explained as "algebraic integers." That book is available in a slightly more recent Dover reprint, I believe. – Robert Soupe Jun 28 '14 at 17:25
• It refers to the ring $\{ a + b \sqrt{-27} : a, b \in \mathbb{Z} \}$, which is neither the Eisenstein integers $\mathbb{Z}[\omega]$ nor $\mathbb{Z}[\sqrt{-3}]$ nor $\mathbb{Z} \left[ \frac{\sqrt{-3}}{2} \right]$ (this ring isn't even a ring of algebraic integers). Those are the four rings you're referring to. – Qiaochu Yuan Jun 28 '14 at 17:30

It seems that not all of these notations refer to the same object. For example, $\mathbb{Z}[\sqrt{-3}] \neq \mathbb{Z}[\sqrt{-27}]$. The former set is $\{a+b\sqrt{-3}|a,b\in\mathbb{Z}\}$, while the latter is $\{a+b\sqrt{-27}|a,b\in\mathbb{Z}\}$, which doesn't contain $\sqrt{-3}$. In fact, we have the following:

$\mathbb{Z}[\sqrt{-27}] \subset \mathbb{Z}[\sqrt{-3}] \subset \mathbb{Z}\left[-\frac{1}{2} + \frac{\sqrt{-3}}{2}\right]$

If you're trying to describe the ring of integers of the field $\mathbb{Q}(\sqrt{-3})$, then you want the last of these three, not one of the first two, because they don't contain all the integers in the field. They are sub-rings of the ring of integers, and if that's all you need them to be, then there's nothing wrong with their notation.

The notation $\mathbb{Z}\left[\frac{\sqrt{-3}}{2}\right]$ is kind of surprising. If you form all integer polynomials in $\frac{\sqrt{-3}}{2}$, you get a strange ring that contains the ring of integers, but isn't a finitely generated $\mathbb{Z}$-module at all. If this is supposed to be a notation for the ring of integers, it's kind of misleading.

Other remarks: lots of algebraic number theorists seem to often eschew $\sqrt{3}i$ in favor of $\sqrt{-3}$. The notation $\bf{A}(-3)$ is probably unfamiliar to lots of modern readers. You should be prepared to explain it if you're going to use it. The $\mathbb{Z}[\omega]$ notation is great, as long as you've clearly defined $\omega$.

Finally, if I were talking about the ring of integers of the field $\mathbb{Q}(\sqrt{-3})$, I'd be inclined to say: "Let $K=\mathbb{Q}(\sqrt{-3})$ and consider the ring of integers $\mathcal{O}_K$." I think lots of number theory people would find that very comfortable and familiar. If I needed to be more precise, I might define $\omega$ and use $\mathbb{Z}[\omega]$.

I would add $\mathcal{O}_{\mathbb{Q}(\sqrt{-3})}$ to the list and rank them thus:

1. $\mathcal{O}_{\mathbb{Q}(\sqrt{-3})}$
2. $\mathbb{Z}[\omega]$ (assuming $\omega$ has been defined)
3. $\mathbb{Z}\left[-\frac{1}{2} + \frac{\sqrt{-3}}{2}\right]$
4. $\mathbb{Z}\left[\frac{i \sqrt{3} - 1}{2}\right]$ (I wouldn't call myself an algebraic number theorist, but I don't like this one either, I guess I would prefer to limit $i$ to discussion of Gaussian integers).
5. ${\bf A}(-3)$ (Ethan Bolker used this one in a book that's available as a Dover reprint).

As others have already mentioned, neither $\mathbb{Z}[\sqrt{-3}]$ nor $\mathbb{Z}[\sqrt{-27}]$ are integrally closed.

Even though $\frac{\sqrt{-3}}{2} = \frac{1}{2} + \omega$, I don't know quite what to make of $\mathbb{Z}\left[\frac{\sqrt{-3}}{2}\right]$.