# 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!

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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.