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What is the number of non-isomorphic rings of order $135$?

Can anyone tell me how can I able to solve this problem and which theorem or results are helpful? Thanks for your help.

More generally, can anybody tell me how many non-isomorphic rings there are for all orders?

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Do rings have multiplicative identities for you? –  Chris Eagle Feb 8 '13 at 10:54
    
may or may not. I get a example as $Z_{135}$.but how can I find all the non isomorphic rings –  priti Feb 8 '13 at 10:56
    
The number of (isomorphism classes) of rings with unity and $n$ elements is tabulated at oeis.org/A037291 but only up to $n=63$. There are $12$ for $n=27=(135)/5$, and I'd guess there'd be more for $n=135$. –  Gerry Myerson Feb 8 '13 at 11:01
    
priti - FYI: if you find an answer to be helpful, you may "accept it": you can accept one answer per question asked. To accept an answer, click on the $\checkmark$ to the left of the answer you want to accept. I think you have enough rep, too, to upvote helpful answers, (as many as you wish). –  amWhy Feb 23 '13 at 2:26

1 Answer 1

According to the OEIS, the number of rings (up to isomorphism) with $n$ elements is a multiplicative function, so $f(135)=f(5)\cdot f(27)=2\cdot 59=118$.

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can you please explain how did you find the values of $f$ –  priti Feb 8 '13 at 11:08
    
By reading the list in the OEIS. –  Chris Eagle Feb 8 '13 at 11:13
    
Nice - number of rings with unity is evidently not multiplicative. –  Gerry Myerson Feb 8 '13 at 11:14
    
I've never used OEIS: the keyword "mult" is intended to flag multiplicative sequences? –  rschwieb Feb 8 '13 at 14:24
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Hint: The theorem to use is the Basis Theorem For Finitely Generated Abelian Groups, which will give you all possible additive groups of the ring from the prime factorization. –  Barbara Osofsky Feb 9 '13 at 13:37

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