Finite Rings whose additive structure is isomorphic to $\mathbb{Z}/(n \mathbb{Z})$ I am having trouble proving the following conjecture: If $R$ is a ring with $1_R$ different from $0_R$ s.t. its additive structure is isomorphic to  $\mathbb{Z}/(n \mathbb{Z})$ for some $n$, must $R$ always be isomorphic to the ring  $\mathbb{Z}/(n \mathbb{Z})$ ? How do we go about defining a ring isomorphism with a proper multiplication on $R$? 
 A: If a finite ring $R$ has cyclic additive group, it is commutative, for it is generated as a ring by any generator of its additive group.
Now, if $n$ is the additive order of $1_R$, then the additive order of every element of $R$ divides $n$. It follows immediately from this that $1_R$ must be an additive generator. It follows that the unique additive group homomorhism $\mathbb Z\to R$ which maps $1_{\mathbb Z}$ to $1_R$ is surjective. It is easy to see that it is a map of rings, and then your desired conclusion follows.
A: Hint: Apply the distributive law. Since any $a,b\in R$ can be written as finite sums of $1_R$, what does $ab$ look like?
Edit: To show that $1_R$ generates the group, all you need to do is show that it has order $n$. Since some $a\in R$ has order $n$ (as the additive structure is that of $\mathbb Z/(n\mathbb Z)$) we have $$0\neq a+\cdots+a=1_ra+\cdots+1_ra=(1_R+\cdots+1_R)a$$ whenever we are adding $m<n$ copies of $a$, thus the order of $1_R$ cannot be less then $n$. Hence it must be $n$.
A: Combine the following general facts:
For any ring $R$, the prime ring (i.e. the subring generated by $1$) is isomorphic to the quotient of $\mathbb Z$ by the annihilator of $R$ in $\mathbb Z$. 
Any cyclic group $R$ is isomorphic to the quotient of $\mathbb Z$ by the annihilator of $R$ in $\mathbb Z$.
(This is Mariano's answer with slightly different words.)
