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A divisor of zero in a commutative ring with unity can have no multiplicative inverse.

I don't understand why this statement is true.

So for $a,b$ in the ring, $ab=0$ by zero divisor.

How can we guarantee that there does not exist $c$ such that $ac=1$?

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marked as duplicate by rschwieb abstract-algebra Apr 8 '15 at 9:57

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    $\begingroup$ Dear @nancy : I spotted the duplicate question in the related questions, indicating that it probably appeared in the list that appeared as you typed your question. Please try to do a search before you post a new question. $\endgroup$ – rschwieb Apr 8 '15 at 9:59
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Hint Suppose $a$ has a multiplicative inverse, and multiply both sides of $ab = 0$ by $a^{-1}$.

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  • $\begingroup$ is that a contradiction? Assuming ab=0 for a,b not 0, then if there is inverse of a then cab=b=0 then ab is not zero divisor? $\endgroup$ – Nancy Apr 8 '15 at 3:55
  • $\begingroup$ Yes, by hypothesis $b \neq 0$ but the assumption that $a$ has a multiplicative inverse lead to the conclusion that $b = 0$, which is indeed a contradiction. $\endgroup$ – Travis Apr 8 '15 at 3:57
  • $\begingroup$ Appreciate your concise and exact example $\endgroup$ – Nancy Apr 8 '15 at 4:01
  • $\begingroup$ You're welcome, I'm glad you found it helpful. I wouldn't call it an "example", however: the argument is totally general, in that it applies to every zero divisor in every ring. $\endgroup$ – Travis Apr 8 '15 at 4:02
  • $\begingroup$ @Nancy: If you find Travis' answer useful, you can upvote it (click the up arrow on the left of the answer). Furthermore, if you feel that it has completely answered your question, you can accept it by clicking the tick to the left of the answer. $\endgroup$ – Michael Albanese Apr 8 '15 at 5:12

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