# When is $\mathbb{G}_m(R)$ enough to determine $R$?

Say I have a ring, $R$, with 1 which I consider my universe, and I know its group of units $G=\mathbb{G}_m(R)$. Then given a subgroup, $H\le G$, can I determine if there is there a subring $S_H$ such that $\mathbb{G}_m(S)=H$? If so, is $S_H$ unique with this group of units? If so, is there in fact a--canonical in the sense above--1-1 correspondence between subgroups of $G$ and subrings of $R$ with 1? Preliminary attempts at a solution don't indicate any problems with the truth of the statement, but naturally one should be skeptical of limited data especially in a subject with so many intricacies as groups and rings.

The motivating example is $\overline{\mathbb{Q}}/\mathbb{Q}$, due to some interesting number theory that could come out of such a correspondence.

In the case of fields the question is supposed to collapse into the question "Can I add 0 to a subgroup of the group of units of some big field and get a subfield without doing anything else?"

There is no possibility for general rings, but are there assumptions on $R$ or $G$ which can ensure existence or uniqueness? And it is also fine to induce assumptions on what kind of $S$ we are allowed to have as well, fields instead of just rings for example.

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Sorry, the answer to your last question (in the case of fields) seems to be an obvious "no", maybe I misunderstood the question ? Take for example $R=\mathbb{C}$, take the subgroup of $R^\times$ composed of nth roots of 1 (for a fixed n), this is not the group of units of any subring. Or take any subgroup of $R^\times$ that does not contain $-1$ (defined in this case as "a square root of 1 distinct from 1", if you want to use only the multiplicative structure), this cannot be the group of units of a subring. As for the uniqueness part of the question, it also seems to fail. For example... –  Alex Nov 1 '11 at 0:33
...still taking $R=\mathbb{C}$, the subgroup $\{\pm 1\}$ of $R^\times$ is the group of units of many subrings of $R$, including most rings of integers of imaginary quadratic extensions of $\mathbb{Q}$. –  Alex Nov 1 '11 at 0:35
Adam: Why do you believe there are any right hypotheses here? It really doesn't seem like this is a line of inquiry where there is something nonsubtle one can work out. –  KConrad Nov 1 '11 at 1:57
Actually, no, I don't think it works for finite fields. Take $R=\mathbb{F}_p$, where $p$ is an prime number $\geq 5$. Then $R^\times$ is a cyclic group of order $p-1$, so it has nontrivial subgroups. But $R$ does not have any nontrivial subrings. –  Alex Nov 1 '11 at 2:34
@Adam, usually such widely open questions («come up with hypothesis on anything you want for the following very stange property to hold, and the only motivation for it to hold is that it will prove something interesting that I will not elaborate upon») are not a good way of making use of the potential of MO. –  Mariano Suárez-Alvarez Nov 1 '11 at 11:45

Uniqueness is hopeless; let $k$ be any reduced ring and $R=k[x]$. Then $k\subset R$ and $R\subset R$ have the same group of units.

Existence is also hopeless in general: Let $S=Z/5Z$, and let $H$ be the subgroup conisting of 1 and 4.

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That's great for general rings, and I thought I said this, but I apparently did not: can I include assumptions on $R$ which will make it hold true. I'll edit the original question. Thanks! –  Adam Hughes Nov 1 '11 at 0:34
@Adam: This is just my gut feeling, but I don't think many interesting hypotheses on your ring could circumvent Steven's counterexample to Uniqueness... –  M Turgeon Nov 1 '11 at 1:38
M Turgeon: if R is required to be a field that circumvents the problem with $k[x]$, so there is hope in that direction. –  Adam Hughes Nov 1 '11 at 1:45
If S is required to be a subfield, then of course you get uniqueness, but I still don't see a natural condition that would guarantee existence. Besides, you wrote that your motivating example if the absolute Galois group of $\mathbb{Q}$, but you would need your subgroup to be commutative if it has to be the group of units in a field. Maybe I totally misunderstood, but I don't really see what you're trying to do. –  Alex Nov 1 '11 at 2:31
Surely we need that k has no nontrivial nilpotents? –  Harry Altman Nov 1 '11 at 3:59
It is also worth to recall that if $F$ is a field then a free algebra $F[X]$ has $F^\times$ as group of units for any set $X$...