Any ring of prime order commutative ?

Is any ring of prime order commutative ?

Solution 1:

Not only it is true that any ring of prime order $p$ (i.e. number of elements is $p$) is commutative, what's more, there are exactly two non-isomorphic rings, one is the ring $\mathbb{Z}/p\mathbb{Z}$ with usual ring structure, and the other one is $(\mathbb{Z}/p\mathbb{Z}, +, \times)$ where $+$ is the usual addition and $a \times b = 0, \forall a, b, \in \mathbb{Z}/p\mathbb{Z}.$

Let $R$ be a ring with $p$ number of elements which is not isomorphic to the ring of the second type as written above. Note that as a group $(\mathbb{Z}/p\mathbb{Z}, +)$ and $(R, +)$ are isomorphic. Let $\overline{a} \in \mathbb{Z}/p\mathbb{Z}$ is a generator. Then $\overline{a}^2 = n\overline{a},$ for some $n \in \mathbb{N}.$ Choose $m \in \mathbb{N}$ such that $mn \equiv 1 \pmod p.$ Let $\overline{b} = m\overline{a}.$ If $\phi : (\mathbb{Z}/p\mathbb{Z}, +) \rightarrow (R, +)$ is a group isomorphism, then define a map $\psi: \mathbb{Z/p\mathbb{Z}} \rightarrow R$ by $\overline{x} \mapsto x\phi{(\overline{b})}.$ (here both $\mathbb{Z/p\mathbb{Z}}$ and $R$ are considered as rings.) Now one can check that $\psi$ is a ring isomorphism.