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Suppose that there are some numbers, or mathematical objects. Suppose that these numbers have operation $+$ and $\cdot$. $\cdot$ and $+$ are commutative - that is $x \cdot y = y \cdot x$ and $x+y = y+x$. There is $0$ which makes $x+0 = 0+x = x$ for every $x$, and $x \cdot 0 = 0$. Also, $x(y+z) = xy+xz$.

If the set $A$ of all numbers has cardinality equal or greater than countably infinite, is there any consistent way to have a set of numbers $B \subset A$ of infinite cardinality that exhibits $x \cdot x = 0$ where $x$ is same $x$, while there is also a set of numbers $C \subset A$of infinite cardinality that exhibits $x \cdot x \neq 0$?

Asaf Karagila
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    You might be interested in the topic of nilpotent infinitesimals, but your Question seems pretty open-ended. – hardmath Aug 04 '14 at 12:39
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    If you are just looking for some example, you can take something like a polynomial ring over a field in the variables $(X_i)_{i \in I}$ indexed by an arbitrary set and take the quotient by the ideal generated by the $X_i^2$. – Matthias Klupsch Aug 04 '14 at 12:43
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    You only need polynomials in 1 variable. In, say, ${\bf Q}[x]/(x^2)$, for every rational $a$ you get $(ax)^2=0$, but for every nonzero rational $a$, you get $a^2\ne0$. – Gerry Myerson Aug 04 '14 at 13:16
  • I'm surprised this was closed, especially considering that the OP's previous attempt at asking this question is still open. The terminology is still a little confused, but a clear question is being asked this time, and Gerry Myerson has provided a correct answer. I've voting to reopen. – Alex Kruckman Aug 08 '14 at 06:31

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