# For a ring $A$ with $A^*\cong \Bbb{Z}$ show that $1+1=0$

Let $A$ be a ring and denote $A^*:=\{x\in A|\exists y:xy=1\}$. Suppose $A^*\cong \Bbb{Z}$, show that $1+1=0$.

I will assume I am to show that $1_{A}+1_A=0$ but I am not quite sure what assumption I may make based on the isomorphism (A groups one I presume).

$A^*$ is a group, there is not doubt about it. $\Bbb{Z}$ is an additive group. I guess that by the definition of isomorphism, there exists $\phi:\Bbb{Z}\to A^*$ such that $\phi(0)=1_A$ (Is it true? How can I formally justify that?). If it is true, then $\phi(0+0)=\phi(0)=1_A+1_A=1_A$ meaning $1_A=0$ (I suppose). I am quite new to Ring Theory. Is there a point in my approach? How should I approach it?

• First, $A^\ast$ is a group with respect to multiplication. So your penultimate line should read $\phi(0+0) = \phi(0) = 1_A\cdot 1_A = 1_A$. Now, for the actual problem here is a hint: What is multiplicative order of $-1$? What is the order of a non-identity element in $\mathbb{Z}$? – Jason DeVito Apr 15 '17 at 23:45
• An infinite one? Are you suggesting to look at the integers as a multiplicative group? – Meitar Apr 15 '17 at 23:46
• No, look at $A^\ast$ a multiplicative group and $\mathbb{Z}$ as an additive group. – Jason DeVito Apr 15 '17 at 23:47
• Oh you are correct, thank you. – Meitar Apr 15 '17 at 23:47

If $A$ is any (unital) ring, there is a unique ring homomorphism $\mathbb{Z} \to A$ sending $1_{\mathbb Z} \mapsto 1_A$. If the kernel of this map is $n \mathbb{Z}$ then the universal property of quotients gives an embedding $\mathbb Z/n\mathbb Z \hookrightarrow A$. It follows that there is a group embedding $(\mathbb Z/n \mathbb Z)^* \hookrightarrow A^*$. First of all, if $n = 1$ then $1_A = 0_A$ so we can rule out that case. Otherwise, if $n \ne 2$ you can show that it is impossible to embed $(\mathbb Z/n\mathbb Z)^*$ into $\mathbb Z$ since the former has torsion and the latter is torsion-free.