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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 $R$?

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3 Answers 3

up vote 6 down vote accepted

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.

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I knew that if I could show $1 = 1_R$, I would be done. But I can't convince myself that this is the case. Why must $n$ be the additive order of $1_R$? Assuming otherwise, I have not found the contradiction. –  Peter Feb 26 '12 at 6:40
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Well, if the additive order of $1_R$ is not $n$, say it is $m \neq n$. Note $m|n$, so we have $m < n$. Then for all $a \in R$, $ma = (m1_R)a = 0$. This means that the additive order of every element of $R$ is less than or equal to $m$ which is itself less than $n$.But, $R$ is isomorphic as additive group to $\mathbb{Z}/n\mathbb{Z}$, so some element of $R$ must have order $n$, which contradicts what we just proved. –  Rankeya Feb 26 '12 at 6:46
    
@Peter, there is no $1$, really, and I did not prove that «$1=1_R$». A ring structure on the abelian group $\mathbb Z/n\mathbb Z$ need not have the class $1+n\mathbb Z$ as unit element! –  Mariano Suárez-Alvarez Feb 26 '12 at 8:53

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 $n$.

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(You do need to prove that $1_R$ is a generator of the additive group of $R$ for that, though) –  Mariano Suárez-Alvarez Feb 26 '12 at 6:30

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.)

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