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For $R = \mathbb Z_2 \times \mathbb Z_2$ we can define a product by the following table:

\begin{matrix} \cdot & a & b & c & d \\ a & a & a & a & a \\ b & a & a & d & d \\ c & a & a & c & c \\ d & a & a & b & b \end{matrix}

where $a = (0,0), b = (0,1), c= (1,0), d= (1,1). $ We can verify that $R$ is indeed a ring. I've checked that this product is not commutative: $bc=d\neq a = cb,$ is not associative: $(db)c = ac = a \neq b = dd = d(bc)$ and clearly does not have a unit element.

My question is: Can I generalize this result? Is there a result concerning those kind of products (non-associative, non-commutative and non-unital) for a ring $R$ with more than 4 elements?

Observation: I am asking this because my professor posed us a question to define such a product on $\mathbb Z_2\times \mathbb Z_2$. At the end of the exercise, he asked: Can you generalize for $R$ with $\geq4 $ elements?

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    $\begingroup$ Why call it a ring if it has none of those nice properties? I suppose in general, that if you define a "product" at random, then you shouldn't expect it to be commutative, you shouldn't expect it to be associative and shouldn't expect it to be unital. $\endgroup$ – Lord Shark the Unknown Sep 8 '18 at 1:06
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    $\begingroup$ This operation does not distribute over addition: $b+c = d$ and $a(b+c) = ad = a$ while $ab + ac = a + a = 0$. Also $(b+c)a = da = a$ while $ba+ca = a+a = 0$. Therefore it makes absolutely no sense how "We can verify that $R$ is indeed a ring." What is your definition of a ring? Having a binary operation on an abelian group is not enough to call the system a ring under all conventions I have ever heard. $\endgroup$ – KCd Sep 8 '18 at 1:40
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    $\begingroup$ @KCd why is it wrong? we have $a=0$, as it is the identity of the sum. For me you have just verified that $a(b+c) =a+ a = a = ab+ac$ $\endgroup$ – math.h Sep 8 '18 at 2:01
  • $\begingroup$ Ah, whoops! Sorry. I wasn't even paying attention to what $a$ actually was there. If your multiplication is genuinely distributive over addition but lacks any other decent properties (esp. being nonassociative and no identity) I would not consider it to be a ring at all. Call it a Z-algebra, but definitely not a ring. Please read arxiv.org/abs/1404.0135. $\endgroup$ – KCd Sep 8 '18 at 3:10
  • $\begingroup$ What do we want to "generalize", that there exist a structure $*$ on a given abelian group which does not have all properties in a list? (Which abelian groups shall we consider, all of the shape $\Bbb Z/m\times \Bbb Z/m$? Which is the list of properties that should not hold?) Please try to state a clear proposition, that should be generalized. $\endgroup$ – dan_fulea Sep 8 '18 at 4:23

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