Let $R$ be a ring with $10$ elements, show that $R$ is commutative.
$R$ is a ring which contains $10$ elements and doesn't have to include $1$.
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To check: are you thinking about something like the even integers modulo $20$? But non-commutative? – Mark Bennet Apr 07 '14 at 16:55
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The question i'm facing doesn't specify what sort of ring it is, only that when one has 10 elements it is commutative. – Leijten Apr 07 '14 at 16:57
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This answer might help you - http://math.stackexchange.com/questions/110271/ring-with-10-elements-is-isomorphic-to-mathbbz-10-mathbbz – The very fluffy Panda Apr 07 '14 at 17:02
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Thanks a lot, I'll check it out – Leijten Apr 07 '14 at 17:04
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Related: https://math.stackexchange.com/questions/1825661 – Watson Dec 11 '16 at 18:42
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The additive group is cyclic, let $g$ be a generator. Then for arbitrary elements $a=ng$, $b=mg$, we have $$a\cdot b = ng\cdot mg=nm(g\cdot g)= mn(g\cdot g)=mg\cdot ng=b\cdot a.$$
Hagen von Eitzen
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And note that $g\cdot g$ defines the multiplication, so the rings of order $10$ can be enumerated quite easily. – Mark Bennet Apr 07 '14 at 17:11
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@drhab Here is one exaplanation: the ring is isomorphic to $\mathbb{Z}/10 \mathbb{Z}$ which has an additive generator: $2$. Sets of generators are invariant under isomorphism. – Rustyn Apr 08 '14 at 19:59
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@Rustyn The OP says that the ring has not necessarily an identity. That gives more possibilities. – drhab Apr 09 '14 at 09:57
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@drhab You are right, i didn't see that in the statement of the problem – Rustyn Apr 09 '14 at 17:44
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@Rustyn Later I realized that there is only one abelian group having exactly $10$ elements. And it is indeed cyclic. – drhab Apr 09 '14 at 18:53