Every Ring of Order $p^2$ is Commutative
Problem 501
Let $R$ be a ring with unit $1$. Suppose that the order of $R$ is $|R|=p^2$ for some prime number $p$.
Then prove that $R$ is a commutative ring.
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Proof.
Let us consider the subset
\[Z:=\{z\in R \mid zr=rz \text{ for any } r\in R\}.\]
(This is called the center of the ring $R$.)
This is a subgroup of the additive group $R$.
In fact, if $z, z’\in Z$, then we have for any $r\in R$,
\begin{align*}
(z-z’)r=zr-z’r=rz-rz’=r(z-z’).
\end{align*}
It follows that $z-z’\in Z$, and thus $Z$ is a subgroup of $R$.
Note that $0, 1 \in Z$, hence $Z$ is not a trivial subgroup.
Thus, we have either $|Z|=p, p^2$ since $R$ is a group of order $p^2$.
If $|Z|=p^2$, then we have $Z=R$.
By definition of $Z$, this implies that $R$ is commutative.
It remains to show that $|Z|\neq p$.
Assume that $|Z|=p$.
Then $R/Z$ is a cyclic group of order $p$.
Let $\alpha$ be a generator of $R/Z$.
Since $Z\neq R$, there exist $r, s\in R$ such that $rs\neq sr$.
Write
\[r=m\alpha+z \text{ and } s=n\alpha+z’\]
for some $m, n\in \Z$, $z, z’\in Z$.
Then we have
\begin{align*}
rs&=(m\alpha+z)(n\alpha+z’)\\
&=(m\alpha)(n\alpha)+m\alpha z’ + n z\alpha +z z’\\
&=(n\alpha)(m\alpha)+m z’ \alpha +n \alpha z +z’ z\\
&=(n\alpha+z’)(m\alpha+z)\\
&=sr.
\end{align*}
This contradicts $rs\neq sr$, and we conclude that $|Z|\neq p$.
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