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In a non-commutative ring with unity without zero divisors find a prime element which is not irreducible (if possible).

$p$ is prime iff $p|ab$ implies that $p|a$ or $p|b$, and $x$ is irreducible iff $x = ab$ implies that either $a$ or $b$ is a unit.

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If we deal with a commutative ring:

Let $p$ be a prime with $p=ab$. Then some $r$ exists with $a=rp\vee b=rp$. If $a=rp$ then we have $p=rpb$ leading to $p(1-rb)=0$ hence $1=rb$ showing that $b$ is a unit. Conclusion: $p$ is irreducible. Of course $b=rp$ leads to the same conclusion.

You could say that this doesn't have to work if the ring is not commutative because we also need $1=br$. Normally prime and irreducible elements are defined in commutative rings.


addendum concerning non-commutative ring (with thanks to @rschwieb).

$1=rb$ implies that $r(1-br)=0$ hence also $br=1$. Proved is now that $b$ is a unit.

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  • $\begingroup$ What will counterexample in case of non-commutative ring with unity without zero divisor $\endgroup$ – Sushil Feb 14 '15 at 7:48
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    $\begingroup$ I must admit that I have no idea. In your edited question you now speak of non-commutative ring. So after a short while I will delete this answer. $\endgroup$ – drhab Feb 14 '15 at 8:18
  • $\begingroup$ What if I remove condition 'no zero divisor' also. Motivation as mentioned in question. $\endgroup$ – Sushil Feb 15 '15 at 13:44
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    $\begingroup$ It is not a good thing to add new questions to an old one. This especially if answers are allready given. You should just ask a new question and let it eventually be accompanied by a reference to this question. The edit of this question should be withdrawn in this context. $\endgroup$ – drhab Feb 15 '15 at 13:51
  • $\begingroup$ okay i'll do it $\endgroup$ – Sushil Feb 15 '15 at 14:03

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