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Ring with order the square of a prime

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

#1

Post by Alkesk » Thu Jun 09, 2016 9:56 am

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

#2

Post by Tsakanikas Nickos » Thu Jun 09, 2016 9:58 am

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.
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