0/0 exists in the 0 ring

{0} is a ring under multiplication. 0 is the multiplicative identity and additive identity.

we have 0 x 0 = 0
so 0/0 = 0 and everythign works fine!

>>7
I didn't say that the definition implies that 0 != 1, I said it demands that 0 != 1. That's just stated as part of the definition of an integral domain. If you ignore that part of the definition though, then yes, the trivial ring is a field in which you can divide by zero.