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How many associative rings(with unity) of order 35 are there upto isomorphism? I think the answer is one. This is because there is just one abelian group of order 35. Is my reasoning right. Note that I have read this question but it did not serve my purpose as no proof was given there. Thanks beforehand

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vidyarthi
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    Do your rings have a unity? If so, there is only one ring of order $pq$, but your reasoning is not adequate—yes, we know the additive structure, but why does that tell us the multiplicative structure? (also, there are three other non-unital examples) – Andrew Dudzik Dec 11 '16 at 17:27
  • There is the ring where every product evaluates to $0$, for instance. – Arthur Dec 11 '16 at 17:29
  • @Slade yes the rings consist of unity. edited the question. – vidyarthi Dec 11 '16 at 17:30
  • @Arthur so are there many rings without the unity? – vidyarthi Dec 11 '16 at 17:33

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Let $R$ be a ring of order $pq$, with $p$ and $q$ distinct primes. As you have observed, $(R,+) \cong (\mathbb{Z}/pq,+)$ as abelian groups.

To show uniqueness of the ring structure, it is enough to show that $1\in R$ is an additive generator, as then the morphism $\mathbb{Z}\to R$ is surjective, so $R\cong \mathbb{Z}/pq$ as rings.

But there exists an element $x\in R$ of additive order $pq$. If $1+\cdots + 1 = 0$, then $(1+\cdots + 1)x = x+\cdots + x = 0$, so $1$ has order $pq$ as well.

More generally, for non-unital rings, we can write $R\cong pR \oplus qR$ by the Chinese Remainder Theorem, and consider the four cases where each of $pR$ and $qR$ are either cyclic or have trivial multiplication.

Andrew Dudzik
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