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Definitions of prime element:

$(1)$ We say $p$ is prime if $p|ab$ it implies $p|a$ or $p|b$ (I don't need definition of unity here)

$(2)$ We say $p$ is prime if $p=ab$ it implies $p|a$ or $p|b$ (I don't need definition of unity here)

Are these two definitions equivalent?

Note: $p = ab$ it may not imply $p|ab$ using definition $(1)$ (as much as I can see because there is no unity)

Motivation: ring without unity with an prime element $p$ such that $ab=p$ but $p$ does not divide $a$ nor it divides $b$ (if possible)

Adam Hughes
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Sushil
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  • If $p|a$ or $p|b$ then WLOG say $p|a$ i.e. $a=pr$, then $ab=p(rb)$ so certainly with your definition $p=ab$ implies $p|ab$ – Adam Hughes Feb 14 '15 at 09:09
  • @AdamHughes but you start with assumption p/a or p/b. What if we don't know intially p/a or p/b – Sushil Feb 14 '15 at 09:12
  • That's because you said that was your definition of a prime element. – Adam Hughes Feb 14 '15 at 09:17
  • @AdamHughes Yes if p/ab but p=ab does not imply p/ab – Sushil Feb 14 '15 at 09:20
  • Yes it does, according to your definition, if $p$ is prime, then $p|a$ or $p|b$, hence $p|ab$ by my argument above – Adam Hughes Feb 14 '15 at 09:22
  • @AdamHughes I am not saying you are wrong. But I want a and b such that p = ab but p doesn't divide a nor b(if possible) You are doing converse of statement as much as I can see – Sushil Feb 14 '15 at 09:26
  • Sushil, that's impossible. The one inviolate thing in all mathematics is the definition. You are asking for something which is simultaneously prime and not prime. By the law of the excluded middle, this is impossible. – Adam Hughes Feb 14 '15 at 09:32
  • I don't think I am finding example which is prime as well as not prime. I just have two different definition. And I am looking whether these are equivalent or not. There is no violation of law of excluded middle – Sushil Feb 14 '15 at 09:34
  • @AdamHughes I have modified the question, may be this will help you in explaining things to me – Sushil Feb 14 '15 at 09:37

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An example where (1) and (2) are not equivalent is the Rng $2\mathbb{Z}$.

If we take definition (1), then 2 is not prime since if $a=b=6$, we have $ab=36$ and $2|36$ since $36=2\times 18$. However 2 does not divide 6 in $2\mathbb{Z}$ as we cannot write $6=xy$ where $x,y\in 2\mathbb{Z}$.

On the other hand 2 is prime according to (2) since the condition holds vacuously.

Robin Balean
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  • Very correct thanks but may I ask one more question. p=ab where p prime but ring without unity would it imply p/a or p/b – Sushil Feb 16 '15 at 18:53
  • @Sushil - I assume by prime, here you mean your definition (1). In that case the answer is yes. If $p=ab$ then in particular $p|(ab)(ab)$ and hence $p|ab$ by definition (1) since $p$ is assumed to be prime. Again, since $p$ is prime we then have $p|a$ or $p|b$. – Robin Balean Feb 16 '15 at 20:53