# A commutative ring with group of invertible elements isomorphic to $\mathbb{Z}$ has characteristic 2

Let $R$ be a commutative ring. Denote by $R^*$ the group of invertible elements (this is a group w.r.t multiplication.) Suppose $R^*\cong \mathbb{Z}$. I need to show that $1+1=0$ in $R$.

I have no clue about why such statement should be true. I don't even have an example for a ring that satisfies these assumptions, so I'd be glad to see one.

Hints (or partial solutions) will be welcomed. Thank you!

• Others have already given the answers. It might be instructive to know some near-misses. Take the ring $\{a+b\sqrt2\mid a,b\in\mathbf{Z}\}$. The unit group is almost $\mathbf{Z}$. It is a direct product of $\mathbf{Z}$ with a group of order 2. The torsion group is generated by $-1$ and the free part is generated by $3+2\sqrt2$, called a fundamental unit. – P Vanchinathan Mar 21 '17 at 11:14

Hint 1: $-1$ is a unit and so is a power of $u$, where $u$ is a generator of $R^\times$.

Hint 2: $(-1)^2=1$. What are the elements of finite order in $\mathbb Z$ ?

• Oh, I see! Thank you. $-1=u^a$ so $1=u^{2a}$ and therefore $a=0$. Do you have an example for such ring? – 35T41 Mar 21 '17 at 10:16
• $\mathbb F_2[X,X^{-1}]$ – MooS Mar 21 '17 at 10:17

Hint: If $R^{\times}\cong{\mathbb Z}$, then in particular it has no nontrivial torsion; on the other hand, there's $-1\in R^{\times}$.

• I don't know what a torsion of a ring is... – 35T41 Mar 21 '17 at 10:20
• @35T41 An element $x\in A$ in an abelian group $A$ is torsion if there is some $n>0$ such that $na=0$. In the case of $R^{\times}$, it coincides with roots of unity / elements of finite order. Anyways, after lhf has edited his answer (+1), my hint is the same as his second one. – Hanno Mar 21 '17 at 10:23
• @35T41 Also note that Hanno did just mean the torsion of the group $R^*$. – MooS Mar 21 '17 at 10:25
• Oh, you're right, I did hear about torsions of groups before – 35T41 Mar 21 '17 at 10:27