An exotic ring structure on Z isomorphic to the integers

Detailed Solution

Define $\varphi:\mathbb Z\to\mathbb Z$ by $\varphi(x)=1-x$; this is a bijection (it is its own inverse). Check $\varphi(a\oplus b)=1-(a+b-1)=2-a-b=(1-a)+(1-b)=\varphi(a)+\varphi(b)$, and $\varphi(a\otimes b)=1-(a+b-ab)=1-a-b+ab=(1-a)(1-b)=\varphi(a)\varphi(b)$. So $\varphi$ is a ring isomorphism from $(\mathbb Z,\oplus,\otimes)$ onto the ordinary ring $(\mathbb Z,+,\times)$, which is an integral domain; hence so is $(\mathbb Z,\oplus,\otimes)$, proving (B) and simultaneously ruling out (C) (an integral domain has no zero divisors). Under this correspondence, the $\oplus$-identity is $\varphi^{-1}(0)=1$, not $0$ (ruling out (D)), and the $\otimes$-identity is $\varphi^{-1}(1)=0$, not $1$ (so (A), which swaps the two, is false). Thus the correct option is $\boxed{\mathrm{B}}$.