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Let $R$ denote a commutative ring. $r\in R$ is called irreducible if $r\notin R^{\times}, r\ne 0$, and if $r = ab\Rightarrow a\in R^{\times} \vee b\in R^{\times}$. All right, but isn't every non-unit element $r \in R$ irreducible, since $r=1\cdot r$, and $1\in R^{\times}$?

I guess I am wrong, but don't know where I'm wrong.

Kind regards, MathIsFun

user26857
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  • Counterexample: in $\mathbf Z$, $6=2\cdot 3$ is not irreducible. – Bernard Dec 06 '19 at 11:14
  • You're missing the universal quantifier on the implication, i.e. every factorization of $r$ must be trivial (one factor is a unit), i.e. $r$ has only trivial factorizations. – Bill Dubuque Dec 06 '19 at 15:29
  • I wonder why the explicit $r\ne 0$ is included. Given that $0=0\cdot 0$ and $0\notin R^\times$, that should be covered by the last condition, shouldn't it? – celtschk Dec 07 '19 at 07:07

1 Answers1

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No: $r$ is irreducible iff every decomposition $r=ab$ has the property that $a\in R^*$ or $b\in R^*$.

TonyK
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  • Wow, thanks so much! :-) –  Dec 06 '19 at 11:14
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    In addition to Tony's answer: you are right that every element has at least one such decomposition (in the ring of integers: every number is divisible by 1 and itself) but having the property that every decomposition is of this form is quite special. For instance 6 is not irreducible because it can be written as 23. Of course it can also* be written as $16$ and as $(-1)(-6)$ but the decomposition $6 = 2*3$ is of special interest to us because it shows $6$ is non-irreducible. – Vincent Dec 06 '19 at 11:15
  • Okay, but if I can ask another question: Why do we consider $r\notin R^{\star}$? –  Dec 06 '19 at 11:18
  • @Vincent: Good example! –  Dec 06 '19 at 11:19
  • You are essentially asking (in your second question) 'why is 1 not considered a prime number?' but in the more general context of commuative rings. I think in the specific case of numbers there is some great discussion about that here on MSE. I will try to find a link – Vincent Dec 06 '19 at 11:20
  • Here it is: https://math.stackexchange.com/q/120/101420 – Vincent Dec 06 '19 at 11:21
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    I also really like this answer in the more general, commutative ring case: https://math.stackexchange.com/a/1173808/101420 – Vincent Dec 06 '19 at 11:23
  • I appreciate the last link as well! :-) Thanks again. –  Dec 06 '19 at 16:58
  • PS: I clicked on "automatically move this discussion on chat" and there was an auto-comment stating that "Let us continue discussion in chat.". I deleted the comment now. :-) –  Dec 06 '19 at 16:59