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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.

Watson
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MyNameIs
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    Do you see that it is enough to show that the characteristic must be $n$? – Tobias Kildetoft Jun 14 '16 at 11:18
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    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? – MyNameIs Jun 14 '16 at 11:22
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    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. – Tobias Kildetoft Jun 14 '16 at 11:23
  • 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? – Tobias Kildetoft Jun 14 '16 at 11:26
  • you also need to verify that this map preserves the multiplicative structure as well – clark Jun 14 '16 at 11:29
  • Dumb question: Why is there an element of order $p$ in the abelian group $(R,+)$? @TobiasKildetoft – MyNameIs Jun 14 '16 at 11:30
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    By Cauchy's theorem. – Tobias Kildetoft Jun 14 '16 at 11:31
  • So if $p$ divides $n$ we get some $x\in R$ with $px=0$. And if $p$ does not divide the char of $R$, then $p$ is nonzero is $R$ (is that true?). So $x=0$. – MyNameIs Jun 14 '16 at 11:38
  • What you need to use here is that because the order is $p$ and $p$ does not divide $n$, you have $nx\neq 0$. – Tobias Kildetoft Jun 14 '16 at 11:38
  • Sorry I am confused, @TobiasKildetoft. What is the logic of the argument we're trying to do? Are we trying to show that any prime dividing $n$ divides the characteristic of $R$? Does this imply $\operatorname{char} R=n$? – MyNameIs Jun 14 '16 at 11:46
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    Yes, if the characteristic is divisible by all primes dividing $n$ then it equals $n$ since $n$ is squarefree. – Tobias Kildetoft Jun 14 '16 at 11:47
  • Ah, OK, and we get that $\operatorname{char} R=n$ because we always have $\operatorname{char} R\leq n$ and if all primes dividing $n$ divide char, then the reverse inequality is true too? – MyNameIs Jun 14 '16 at 11:54
  • (sorry for asking all these silly questions) – MyNameIs Jun 14 '16 at 11:55
  • Another way to prove is to show that every ring of order $n$ with cyclic abelian group is actually $\Bbb Z/n\Bbb Z$. Then your claim follows from this. – Watson Jun 20 '16 at 12:35
  • For the claim in my previous comment, see this. – Watson Dec 27 '16 at 14:08

3 Answers3

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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 $m=n/p^k$.

egreg
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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.

2

We can smuggle the core idea of Cayley's theorem from group theory into ring theory and show that $R$ embeds in the ring of endomorphisms of $R$ as an additive group:

$$ \phi : R \to \text{End}(R,+) \quad \text{as} \quad \phi(r) = [s \mapsto rs] \text.$$

Now, $(R, +)$ is cyclic because $n = |R|$ is squarefree, so $\text{End}(R, +) \cong \text{End}(C_n)$, where $C_n$ is the cyclic group of order $n$.

And $\text{End}(C_n) \cong \mathbb{Z}/n\mathbb{Z}$. (If $n \in \mathbb{Z}$, then $g \mapsto n \cdot g = \sum_{i = 1}^n g$ is an endomorphism of any abelian group; in the case of $C_n$, the image of the generator determines the endomorphism.)

So $R$ embeds in $\mathbb{Z}/n\mathbb{Z}$. The embedding must be an isomorphism since the two rings in question are finite.

jskattt797
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