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My question is the following:

If we take the set theoretic complement of the set of zero divisors in a commutative ring with identity will the resulting set be a field? If not is there a process to take out more elements so that at we arrive at a field.

user11937
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  • Have you tried any examples? – Eric Wofsey Jul 25 '18 at 04:13
  • @Eric Wofsey I thought about this question when reading this question https://math.stackexchange.com/questions/60969/every-nonzero-element-in-a-finite-ring-is-either-a-unit-or-a-zero-divisor – user11937 Jul 25 '18 at 04:17

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If you remove any set including the zero divisors, the result won't be a field (using the same operations + and ×) because you'll no longer have a $0$ element!

Sambo
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Sambo has given a clear answer. Assuming we modify your definition to put back zero, after removing zero-divisors, we can still find interesting counter-examples which are not trivial.

Take $F$ to be any field (say real numbers, if you prefer). The in the commutative ring $R=F\times F$ the zero divisors are elements of the form $(a,0)$ and $(0,b)$.

The resulting set is closed under multiplication, but not under addition, so not a subgroup: $(2,3) + (-2,3) = (0,3)$ which is a zero-divisor.

P Vanchinathan
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