# To show a ring of order p (prime) is isomorphic to the integers mod p.

1. Mar 18, 2012

### rachellcb

If R is a finite ring of of order p where p is prime, show that either R is isomorphic to Z/pZ or that xy=0 for all x,y in R

I know that both R and Z/pZ have the same number of elements (up to equivalence) and that R isomorphic to Z/pZ implies R must be cyclic (I think) but am otherwise lost on where to start. Help much appreciated!

2. Mar 18, 2012

### jgens

Suppose that $(R,+,\cdot)$ is not a trivial ring. Notice that $(R,+) \cong (\mathbb{Z}/p\mathbb{Z},+)$ by the Lagrange Theorem.

Let $r \in R$ be an element generating the additive group of $R$. If $a,b \in R$ are non-zero, then it follows that $a = r + \cdots + r$ (n times) and $b = r + \cdots + r$ (m times) where neither $n,m$ divides $p$; and in particular it follows that $ab = r^2 + \cdots + r^2 \; (mn \; \mathrm{times}) = ba$. This proves two things:
1. If $r^2 = 0$, then $(R,+,\cdot)$ would be a trivial ring. This means that $r^2 \neq 0$.
2. Since $p$ is prime and $r^2 \neq 0$, this means that $ab \neq 0$. This shows that $R$ contains no zero-divisors.
Using the fact that $R$ contains no zero-divisors, you should pretty easily be able to show that $R$ is a unital ring; and therefore, $R$ is an integral domain. From there, it should be pretty simply to construct the desired isomorphism.