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Let $\mathcal{O}$ be the ring of integers in an algebraic number field. Define $R \subset \mathcal{O}$ to be the set of all $\mathbb{Z}$-linear combinations of units. Since the product of two units is a unit, the set $R$ is a ring.

Question : Under what circumstances do we have $R = \mathcal{O}$?

Of course, this holds for $\mathcal{O} = \mathbb{Z}$. It is also clearly false for imaginary quadratic number rings aside from the Gaussian integers. But it is true for all the real quadratic number rings I have played with. Maybe it holds whenever $\mathcal{O}$ has infinitely many units? This might be too optimistic...

In case the above has a wild or overly complicated answer, the following question might be easier.

Question : Under what circumstances is $R$ finite-index in $\mathcal{O}$ as an abelian group?

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Huh? It appears to be true for the Eisenstein integers. –  Qiaochu Yuan Oct 23 '13 at 4:35
    
@QiaochuYuan : Whoops, you're right, I missed that one! –  Susan Oct 23 '13 at 4:37
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At mathoverflow.net/questions/27380/ Robin Chapman shows that any totally real field has infinitely many CM extensions where the unit group doesn't increase, providing circumstances where your second question answers in the negative. Actually these should be the only examples, as the subfield generated by the units can only be proper if the original field is CM. –  Peter McNamara Oct 23 '13 at 5:40
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@PeterMcNamara - what if the subring generated by the units is not integrally closed? Then the subfield generated by the units is not as relevant. –  Will Sawin Oct 23 '13 at 5:46
    
Stephane Louboutin has worked on related problems for cubic fields; see e.g. Proc. Amer. Math. Soc. 140 (2012), 429-436. –  Franz Lemmermeyer Oct 23 '13 at 11:53

1 Answer 1

As was already pointed out, the only imaginary quadratic fields with $R=\mathcal O$ are $\mathbb Q(i)$ and $\mathbb Q(\zeta_3)$. This follows pretty easily from the structure of $\mathcal O^\times$. On the other hand, for real quadratic fields $\mathbb Q(\sqrt{d})$, with $d \in \mathbb Z$ square-free, one has $R=\mathcal O$ in the following cases only:

  1. $d\not\equiv 1 \pmod{4}$ and either $d+1$ or $d-1$ is a perfect square.

  2. $d\equiv 1 \pmod{4}$ and either $d+4$ or $d-4$ is a perfect square.

This is was first proved by Belcher, Integers expressible as sums of distinct units, Bull. Lond. Math. Soc. 6 (1974), 66–68.

There are similar results for certain cubic and quartic fields but the problem for general number fields seems to be wide open. Nonetheless, an interesting positive result in this direction was obtained by Frei, On rings of integers generated by their units, Bull. London Math. Soc. 44 (2012), 167–182:

For every number field $K$ there exists a number field $L$ containing $K$ such that $\mathcal O_L = R_L$.

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