Does there exist a unital ring whose underlying abelian group is $\mathbb{Q}^*$?

Let $\mathbb{Q}^*$ be the group of units of the rational numbers.

Does there exist a unital ring whose underlying additive group is $\mathbb{Q}^*$?

I don't really have a gut feeling yea or nea.

Hope I'm not overlooking something obvious.

Thanks!

• Can you be more precise about the condition "if this group admits a unital ring structure"? If this is no constraint at all, then you can identify $\mathbb{Q}^*$ as a set with the underlying set of any countably infinite ring, as Pedro points out in his answer. – Travis Willse Dec 21 '14 at 4:41
• Maybe you had in mind some structure respecting the order on $\mathbb Q$? – Lubin Dec 21 '14 at 4:42
• Be more precise, because otherwise answers can be uninteresting (just at the level of countable sets with no attention to the group structure). Are you asking for a ring structure on $\mathbf Q^\times$ for which the ring addition is ordinary multiplication in $\mathbf Q^\times$? – KCd Dec 21 '14 at 4:46
• As a multiplicative group, $\mathbf Q^\times$ is isomorphic to the additive group $\mathbf Z/(2) \oplus \bigoplus_{p} \mathbf Z$, where $p$ runs over the primes. That comes from looking at the sign of the number and the exponents in its prime factorization. So you are asking if this additive group (infinite direct sum) has a multiplication making it a ring. – KCd Dec 21 '14 at 4:50
• What do you mean by compatible? Seriously, just say what you mean! – anon Dec 21 '14 at 4:53


But, there is a non-obvious multiplication that works! The additive group of the polynomial ring $\Z[x]$ is isomorphic to a countable direct sum of copies of $\Z$, and so (upon choosing a bijection between the natural numbers and the set of prime numbers), the additive group of $(\Z/2\Z) \times \Z[x]$ is isomorphic to the additive group $(\Z/2\Z) \times \bigoplus_{p \text{ prime}} \Z$.

Composing these isomorphisms, we see that the group of units $\Q^\times$ is isomorphic to the additive group of the ring $(\Z/2\Z) \times \Z[x]$.

Here's a simple example: let $\star$ be the operation defined by

• $2 \star 2 = 2$
• $2 \star p = p$
• $p \star p = 1$
• $p \star q = 1$
• $2 \star -1 = -1$
• $p \star -1 = 1$
• $-1 \star -1 = 1$

and extended to all nonzero rational numbers by distributivity: e.g.

\begin{align}6 \star 8 &= (2 \cdot 3) \star (2 \cdot 2 \cdot 2) \\&= (2 \star 2) \cdot (2 \star 2) \cdot (2 \star 2) \cdot (3 \star 2) \cdot (3 \star 2) \cdot (3 \star 2) \\&= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \\&= 216 \end{align}

This would be more evident by noting the positive rationals are the free abelian group whose basis is the primes. The subring of positive rationals described above is constructed as the polynomial ring on countably many indeterminate varaibles (one for each odd prime), modulo the relation that the product of any two variables is the additive identity, which ensures that the additive group of this ring is the free abelian group generated by $1$ and each of the indeterminate variables.

I've taken $2$ as the multiplicative identity in this ring.

Adding $-1$ into this ring is essentially the same, except I add the extra identity that multiplying $-1$ by itself gives zero.

There is a much simpler construction if we just look at the abstract structure: the additive group of $\mathbf{Z}[x]$ is already a free abelian group on countably many elements. The additive group of $\mathbf{Z}[x,y] / (2y, xy, y^2)$ has precisely the structure we are looking for.

Take a bijection $\mathbf Q\to \mathbf Q^\times$ and transport the ring structure. You can even take a bijection from $\mathbf Z$. Any countably infinite ring works.

• Any countable infinite ring...and with characteristic zero, I think. – Timbuc Dec 21 '14 at 4:42
• I meant for the ring structure to be compatible with the multiplicative group structure of $\mathbb {Q}^*$. – Doug Dec 21 '14 at 4:45
• I wonder if the OP wants to use the multiplicative group structure as the new additive structure in a ring. If so, I believe there is no such ring structure – Rachmaninoff Dec 21 '14 at 4:46
• Yes, Rachmaninoff, that is what I intended. – Doug Dec 21 '14 at 4:48