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Let A be a Noetherian with maximal ideal M. let S=A-M. then S contains no zero divisor???

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Arturo, is the notation A-M correct? – Will Jagy Oct 30 2010 at 4:39
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 I can't see a nontrivial question here. (I assume $A-M$ should read $A \setminus M$.) Then, as stated, the answer is no, and one doesn't have to cook up a very complicated ring to see it. (Hint: look for a $2$-dimensional algebra over a field.) On the other hand, if the ring is local, then of course every element outside the maximal ideal is a unit and the answer is yes. – Pete L. Clark Oct 30 2010 at 8:17
Dear Lee, I voted to close, without motivation this looks like a homework question. Here is a hint if you are looking for a counter example: take a ring, say $k[x]$, with at least 2 different maximal ideals. Then kill their product. Suddenly something out of a max. ideal becomes a zero-divisor. – Hailong Dao Oct 30 2010 at 8:41

closed as too localized by Martin Brandenburg, Pete L. Clark, Hailong Dao, Steven Sam, Tom Goodwillie Oct 30 2010 at 12:44

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I do not have the rep to leave a comment, so I must respond here. Perhaps you mean to say that A is a local ring, rather than Noetherian?

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