Let $R$ be a ring of order $n$ and suppose $n$ has no square in its prime decomposition. How do I see that $R$ is isomorphic to $\Bbb Z/n\Bbb Z$?

I bet that the map $\Bbb Z \to R, \, 1\mapsto 1_R$ descends to an iso $\Bbb Z/n\Bbb Z \to R$ but I don't see how $n$ having no squares implies the desired descent.

  • 1
    $\begingroup$ Do you see that it is enough to show that the characteristic must be $n$? $\endgroup$ – Tobias Kildetoft Jun 14 '16 at 11:18
  • $\begingroup$ The characteristic is the integer $k$ with such that the kernel of the map above is $k \Bbb Z$. So if the char is $n$, the map descends to $\Bbb Z/n$. The map is also injective and therefore surjective (finiteness) and therefore an iso. That's why it suffices to show that the characteristic in $n$ right? $\endgroup$ – MyNameIs Jun 14 '16 at 11:22
  • $\begingroup$ That is one way to think about it. Or it is the size of the subgroup generated by $1$, which you are trying to show is the entire group. $\endgroup$ – Tobias Kildetoft Jun 14 '16 at 11:23
  • $\begingroup$ Now, if $p$ is some prime which divides $n$ but not the characteristic, what happens if you consider an element of order $p$ in the abelian group underlying the ring? $\endgroup$ – Tobias Kildetoft Jun 14 '16 at 11:26
  • $\begingroup$ you also need to verify that this map preserves the multiplicative structure as well $\endgroup$ – clark Jun 14 '16 at 11:29

Hint: If $n$ is square-free, then the additive group of $R$ is isomorphic to a direct sum of cyclic groups of prime order and its characteristic is then equal to $n$.

If $n$ is not square free, say $p^k\mid n$ and $p^{k+1}\nmid n$, with $p$ a prime and $k>1$, then there are two nonisomorphic rings with order $n$, namely $$ \mathbb{Z}/p^k\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z} $$ and $$ \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p^{k-1}\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z} $$ where $m=n/p^k$.


Reduce the problem to the following lemma

If $p \mid n$, $\mathbb{Z} / p \mathbb{Z} \to R / pR$ is well-defined an isomorphism

Since all of the ideals $pR$ and $qR$ are coprime, the Chinese remainder theorem asserts

$$ R = R / nR \cong \prod_{p \mid n} \mathbb{Z} / p \mathbb{Z} \cong \mathbb{Z} / n \mathbb{Z} $$

If $n$ is not squarefree, there are examples where $\mathbb{Z} / p^2 \mathbb{Z} \to R / p^2 R $ is not an isomorphism.


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