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Let $R$ be a ring of square-free order $n$.

If $p \mid n$ then $\Bbb Z/p\Bbb Z\to R/pR$ is a well-defined isomorphism.

I'm really unsure how to approach this problem. So we need to show that if $p \mid n$ then the kernel of $\Bbb Z\to R/pR$ is $p\Bbb Z$ and moreover the map is surjective. I can't figure out either of those and I don't really know how to get started on them.

I would really be happy about help.

MyNameIs
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  • But you already know precisely what the ring is by a previous question. – Tobias Kildetoft Jun 15 '16 at 07:42
  • Yes but I am trying to follow the hint given by Hurkyl here http://math.stackexchange.com/questions/1825661/ring-of-order-n-is-isomorphic-to-bbb-z-n-bbb-z-with-n-square-free#1825688 in order to prove the assertion from that previous question =) – MyNameIs Jun 15 '16 at 07:48

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$R$ should be assumed to be commutative with unit here. Consider the natural map $\mathbb Z \to R$ and the composition $\mathbb Z \to R \to R/pR$. Clearly $p$ is mapped into $pR$, hence the composition factors over $\mathbb Z/p\mathbb Z$ and we get a map $\mathbb Z/p\mathbb Z \to R/pR$, which is injective because $\mathbb Z/p\mathbb Z$ is a field.

Thus $R/pR$ is a vector-space over $\mathbb Z/p\mathbb Z$, hence has $p^e$ elements. Now you should use the squarefree assumption to finish the proof.

MooS
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