Simple definition of irreducible in algebra In commutative ring theory, defining irreducibility often takes several assimptions, for instance :
$x$ is irreducible if


*

*$x \neq 0$

*$x$ is non-invertible

*$x=ab \implies a$ 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.
