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Let $R$ be a ring and let $I$ be a proper ideal of $R$.

If $R/I$ has no zero divisors, then is it true that $R$ has no zero divisors?

My attempt: Suppose $R$ has zero divisors, say $ab=0$ for some $a,b\in R^*$. Then $(a+I)(b+I)=ab+I=I$. However, I cannot exclude the case where $a,b\in I$.

Any ideas?

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    $\begingroup$ Hint: every nontrivial ring has a maximal ideal $\endgroup$ Feb 23, 2014 at 8:41
  • $\begingroup$ So what when $I$ is not the maximal ideal? still confused... $\endgroup$
    – user130893
    Feb 23, 2014 at 9:50
  • $\begingroup$ When $I$ is a maximal ideal, what do we know about the quotient? $\endgroup$ Feb 23, 2014 at 9:51
  • $\begingroup$ Then the quotient is a simple ring. $\endgroup$
    – user130893
    Feb 23, 2014 at 9:55
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    $\begingroup$ It is also a field, hence it has no zero divisors. We can thus take arbitrary ring with zero divisors, and create a quotient which has none. $\endgroup$ Feb 23, 2014 at 9:58

2 Answers 2

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If you try to prove something and hit a wall, then you should switch to finding a counterexample for a while. Then if that doesn't work, try proving again. Back and forth you go.

Really you should find a counterexample right away. You could, for example, take a look at $\Bbb Z/\Bbb Z4$.

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No. Z6/<2> is an integral domain as Z6 is a CRU and <2> is a prime ideal but Z6 is not a integral domain.

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