proving that Every ring of prime order is commutative

Question

I'm having some trouble with the following question:

Let $R$ be a ring with order $p$, where $p$ is prime. Prove that $R$ is comutative.

Because $(R,+)$ is a group then because of Lagrange's Theorem, $\exists k \in R$ such that: $$R=\{0,k,2k,...,(p-1)k\}$$

So any element of $R$ can be written as: $mk$ for $m\in \{0,...,p-1\}$, but I'm having some trouble continuing the proof from here. How can I prove this?

Answer 1

You're almost there. Now take $p,q\in R$. $p=mk, q=nk, n,m\in\Bbb{Z}$. So: $pq=mknk=nkmk=qp$, where second equality is because elements of $\Bbb{Z}$ commute with every element in $R$ and $k$ commutes with itself.