In commutative ring theory, defining irreducibility often takes several assimptions, for instance : $x$ is irreducible if

  1. $x \neq 0$

  2. $x$ is non-invertible

  3. $x=ab \implies a$ invertible and $b$ non invertible or vice-versa

I think it's safe to say that an irreducible element can be defined as a non-invertible which isn't the product of several non-invertibles. But since I've nevet actually read this anywhere I'm a bit skeptical.

Is this definition equivalent to the three listed above ? If so, is it less convenient to state it that way ?


Note that in this form, 3. implies both 1. and 2.:
1. Take $x=a=b=0$, which is clearly non-invertible, yet $x=ab$. So $x=0$ doesn't satisfy 3.
2. If $x$ was invertible (also called 'unit' in ring theory), we could write $x=x\cdot 1$ where both factors are invertible.

Now it remains to prove that 3. is equivalent to your definition:
If a noninvertible $x$ can't be written as $x=ab$ for $a,b$ noninvertibles, it means that at least one of $a,b$ is invertible whenever $x=ab$. But can't be both, as $x=ab$ is noninvertible.

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    $\begingroup$ I hadn't spotted that 3 $\implies $ 1 and 2, indeed. I don't know I'm just under the impression I never find it stated that way and it got me suspicious. $\endgroup$ – James Well Jun 9 '17 at 12:44

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