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

[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!

 

[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.