Noncommutative ring of order $np^2$

Question

Could anyone help me to prove this theorem, please?

Let $R_1$ be a ring of order $p^2$ which is the direct product of $C_p$ with itself and a minimal generating system for $R_1$ is $[(a,0),(0,a)]$, where $a$ is a generator of $C_p$. The multiplication is defined as

$$(j_1a,k_1a)(j_2a,k_2a)=(j_2+k_2)(j_1a,k_1a).$$ Let $R_2$ be any ring of order $n$, then the ring $R=R_1\oplus R_2$ (the direct sum of $R_1$ and $R_2$) is a noncommutative ring of order $np^2$.

Which part gives you trouble? What do you mean by order of a ring? — tomasz, Jul 24 '13 at 14:29
I do not know how to start. The order of a ring means the number of elements in this ring. — Monika, Jul 24 '13 at 14:38
$R_1=C_p\times C_p$ is commutative. If $R_2$ is commutative, then $R$ is commutative as well. — Dietrich Burde, Jul 24 '13 at 14:46
There is nothing to prove: the cardinality is of course $n\cdot p\cdot p$, because of counting of sets (has nothing to do with rings). — Dietrich Burde, Jul 24 '13 at 14:49
@DietrichBurde: Monika meant that the additive structure is that of a direct product of cyclic groups, multiplicative structure is different. You're right about the second point, though. — tomasz, Jul 24 '13 at 14:52

score 2 · Answer 1 · answered Aug 14 '13 at 14:21

Clearly $R_2$ plays no role at all in the noncommutativity of $R$, since when $R_2$ is commutative, $R$ is commutative iff $R_1$ is. So proving that $R_1$ isn't commutative is the only thing at stake, and the only way $R_2$ comes into play is in determining the order of $R$.

But it is simple to come up with counterexamples to commutativity in $R_1$, here.

For example, $(0a,1a)(1 a,0a)=(0a,1a)$ but $(1a,0a)(0a,1a)=(1a,0a)$.

Noncommutative ring of order $np^2$

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