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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?

rschwieb
  • 153,510
priti
  • 473
  • Do rings have multiplicative identities for you? – Chris Eagle Feb 08 '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 08 '13 at 10:56
  • The number of (isomorphism classes) of rings with unity and $n$ elements is tabulated at http://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 08 '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 02:26
  • Related: https://math.stackexchange.com/questions/1825661 – Watson Dec 11 '16 at 18:44

1 Answers1

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

Chris Eagle
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