Embedding $\mathbb{Z}[\sqrt{-3}]$ in a larger ring

Question

I'm struggling with the following question:

Show how $\mathbb{Z}[\sqrt{-3}]$ may be embedded as a subring in a larger ring R, which is a PID, with the quotient of additive groups having index 2.

I've tried $\mathbb{Z}[\sqrt{-3}]+\mathbb{Z}[\sqrt{3}]$ among other things but not sure if I'm on the right track.

$1,\sqrt3$ and $\sqrt{-3}$ are linearly independent over $\Bbb{Z}$, so your suggestion cannot work. The index of a lower rank free abelian group is automatically infinite. — Jyrki Lahtonen, May 16 '19 at 13:00
@JyrkiLahtonen - I don't know that the OP wanted the answer, or a hint? (This is often an issue, isn't it, in this type of question...) — peter a g, May 16 '19 at 13:00
@peterag That is, indeed, a recurring problem. I only gave it in a comment, for I suspect this question is actually a duplicate (in a wider sense of the word). — Jyrki Lahtonen, May 16 '19 at 13:02
Anyway, the sought after ring is called Eisensteinian integers. Too famous not to have been covered. — Jyrki Lahtonen, May 16 '19 at 13:05
May be a good answer here would be to first explain why this ring is not a PID, and then explain how to fix that :-/ — Jyrki Lahtonen, May 16 '19 at 13:06

score 2 · Accepted Answer · edited May 16 '19 at 13:18

2

The integral closure of $\Bbb Z$ in $\Bbb{Q}\left(\sqrt{-3}\right)$ is the larger ring $\Bbb Z\left[\frac{1 + \sqrt{-3}}{2}\right]$, since $-3 \equiv 1 \bmod{4}$. This is norm-Euclidean and in particular a PID. For $d\equiv 2,3 \bmod 4$, $d$ squarefree the integral closure of $\Bbb Z$ in $\Bbb{Q}(\sqrt{d})$ is equal to $\Bbb Z[\sqrt{d}]$.

Edit: It would be useful to see that $\Bbb Z[\sqrt{-3}]$ is not a PID, see Jyrki's comment. Indeed, this is not even an UFD:

Why is $\mathbb{Z}[\sqrt{-n}], n\ge 3$ not a UFD?

edited May 16 '19 at 13:18

Thomas Andrews

177,126

answered May 16 '19 at 12:59

Dietrich Burde

130,978

it is norm euclidean because $\sqrt{3}/2 < 1$ – reuns May 16 '19 at 13:17
@reuns $\Bbb Z[\omega]$ is norm-Euclidean, but not $\Bbb Z[\sqrt{-3}]$. I assume with "it" you mean the ring of Eisenstein integers? – Dietrich Burde May 16 '19 at 13:22

Embedding $\mathbb{Z}[\sqrt{-3}]$ in a larger ring

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