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With only the additional axiom that $0\neq 1$, I think I have been able to formally construct the a subset $n$ of the ring $(R,+,*,0,1)$ using only a subset axiom (specification in ZF).

Informally, $n=\{1, 1+1, 1+1+1, ...\}$

I have shown the that equivalent of Peano's Axioms (including induction) holds on this subset. Can this be true? Would this not imply that all rings with $0\neq 1$ are infinite?

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Maybe you could show the construction? – nonpop Dec 2 '12 at 7:29
up vote 1 down vote accepted

As you pointed out, this cannot be true, because there exist finite rings. The Peano Axioms that is not satisfied in this case is:

For every natural number n, S(n) = 0 is false. That is, there is no natural number whose successor is 0.

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My mistake. I was, however, able to establish the equivalent of the other Peano Axioms (including the induction principle) plus $x\neq S(x)$ or $x\neq x+1$. – Dan Christensen Dec 1 '12 at 19:18
Is it then correct to say that finite rings must "loop around" like modular arithmetic? – Dan Christensen Dec 1 '12 at 19:28

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