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April 28th, 2017, 02:18 PM   #1
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Rings with four elements

Good morning.
Recently I've been doing some research on a ring theory. I found out that there 11 rings that have four elements (is the number of them correct or are they less or more?)? Here comes my question: why are there only 11 of them? Can someone explain this in a language of abstract algebra?
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April 28th, 2017, 02:42 PM   #2
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Is that right? Great factoid. I've played with the field with four elements but I never thought about this question for rings. Are these commutative? With or without identity?

ps -- Today I learned! These ring are not necessarily commutative, and need not contain a multiplicative identity.

It's a fact that there are exactly 11 such rings of order $p^2$ if $p$ is a prime.

I found a nice Stackexchange thread.

One answer refers to a program alg which lets you input a bunch of axioms and cranks out all the models of the axioms. If only Russell and Hilbert had software like this!

I also found this paper that proves the result that there are 11 rings of order $p^2$.

I'm afraid I'm completely unable to shed the slightest bit of light on this beyond Googling.
Thanks from agentredlum

Last edited by Maschke; April 28th, 2017 at 02:53 PM.
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April 28th, 2017, 03:37 PM   #3
Joined: Apr 2017
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Well, since you have googled the topic, it seems like I don't need to explain anything. :P But I did the same when I came up with my 11 rings, but I wasn't able to find the paper you mentioned at the end. It is really interesting. Thank you kindly, but I'd like to invite others to discuss the problem. Maybe it is not the only way to justify this. Source of my knowledge limits me a bit, because I can say I started dealing with ring theory not long time ago.
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