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## Ring with order the square of a prime

Groups, Rings, Domains, Modules, etc, Galois theory
Alkesk
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### Ring with order the square of a prime

Let $\displaystyle{R}$ an associative ring with unit and order $\displaystyle{p^{2}}$ , where $\displaystyle{p}$ is a prime number.

Prove that $\displaystyle{R}$ is commutative.

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Tsakanikas Nickos
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### Re: Ring with order the square of a prime

Since $\displaystyle o(R) = p^2$, $\displaystyle \forall r \in R : o(r) | o(R) = p^2$Therefore, if $\displaystyle r \in R$, then $\displaystyle o(R) = 1 \vee o(R) = p \vee o(r) = p^2$.

If $\displaystyle \exists r \in R : o(r) = p^2 = o(R)$
then $\displaystyle R = <r>$ and therefore $\displaystyle R$ is commutative. If it is not the case, then consider the cyclic group $\displaystyle <1_{R}>$. Since $\displaystyle 1_{R} \neq 0_{R}$, $\displaystyle o(1_{R})=p$. Let $\displaystyle r \notin <1_{R}>$. Then $\displaystyle r \neq 0_{R}$ and $\displaystyle o(r) = p$. Observe that $\displaystyle <1_{R}> \bigcap <r> = \{ 0_{R} \}$Hence the sum $\displaystyle <1_{R}> + <r>$ is direct and therefore $\displaystyle o\left( <1_{R}> + <r> \right) = p^2 = o(R)$, which means that $\displaystyle R = <1_{R}> + <r>$. Now, let
$\displaystyle x,y \in R$. Then
$\displaystyle x = \kappa 1_{R} + \lambda r$ $\displaystyle y = \mu 1_{R} + \nu r$ so \begin{align*}
xy &= (\kappa 1_{R} + \lambda r)(\mu 1_{R} + \nu r) \\
&= \kappa \mu 1_{R} + \kappa \nu r + \lambda \mu r + \lambda \nu r^2 \\
&= \mu \kappa 1_{R} + \nu \kappa r + \mu \lambda r + \nu \lambda r^2 \\
&= (\mu 1_{R} + \nu r)(\kappa 1_{R} + \lambda r) \\
&= yx
\end{align*} Hence $\displaystyle R$ is commutative.