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I'm studying the ring $(\mathbb{Z}_2^s, \oplus, \odot)$, where $s$ is arbitrary, $\oplus$ is the sum modulo $2$, and $\odot$ is the AND.

Does it have a name? Even for a certain fixed $s>1$? Does anyone know of a book that studies its properties? Thanks

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  • $\begingroup$ I know that addition modulo 2 is used to sum games in game theory... $\endgroup$ – Guillermo Mosse Jul 11 '18 at 17:24
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    $\begingroup$ $\mathbb Z_{2^s}$? How do you do AND with elements from that ring? Do you mean $\mathbb Z_2^s$? $\endgroup$ – rschwieb Jul 11 '18 at 17:33
  • $\begingroup$ @rschwieb, to do an AND in my original ring, just express the numbers in binary. What I had written wasn't a typo but you are right anyway: it's clearer this way. $\endgroup$ – Guillermo Mosse Jul 11 '18 at 18:28
  • $\begingroup$ $\mathbb Z_{2^s}$ suggests you are talking about $\mathbb Z/2^s\mathbb Z$ which does not have an AND operation (afaik) and doesn't have binary addition. So, many people will think it is a typo. $\endgroup$ – rschwieb Jul 11 '18 at 20:14
  • $\begingroup$ you are right. In fact I'm adding a new operation to $\mathbb{Z}_{2^s}$ via the homomorphism with the other ring $\endgroup$ – Guillermo Mosse Jul 16 '18 at 13:03
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This is a special case of Boolean ring. Each element in this ring is a Boolean vector of length $s$.

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  • $\begingroup$ I guess you answered first because your answer appears above the other one? $\endgroup$ – Guillermo Mosse Jul 11 '18 at 18:29
  • $\begingroup$ @GuillermoMosse The order of answers depends on the sort you have selected on your tab. Our answers were within a few seconds of each other. $\endgroup$ – rschwieb Jul 11 '18 at 20:13
  • $\begingroup$ OK, thanks. I ordered by time and I see yours was first. I'm accepting yours and upvoting theirs :-) $\endgroup$ – Guillermo Mosse Jul 11 '18 at 20:39
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It sure looks to me like you are talking about a finite Boolean ring. Every finite Boolean ring has this structure.

When the product is infinite, you still have a Boolean ring, but some Boolean rings are not just such a product. They are all subrings of such a product, though.

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