# Why $0$ is not irreducible?

We took the following definition of irreducible:

If $$R$$ is a commutative ring, $$r \in R$$ is irreducible if $$r = ab$$ for $$a,b \in R$$ implies $$a \in R^{*}$$ or $$b \in R^{*}.$$

Then my professor said: "so $$0$$ is not irreducible" but I do not understand that as I know that $$0 = 1.0$$ and so we have $$1$$ is a unit and so $$0$$ can be irreducible. Could anyone explain that to me, please?

• But also $0=0\cdot0$, and none of the factors on the right are units. The implication has to hold for all possible factorizations, not just one you cherry-picked. Nov 15, 2020 at 10:22
• Ok thank you very much!@Vercassivelaunos
– user778657
Nov 15, 2020 at 10:27
• related although the question here stands on its own as a different one. Nov 15, 2020 at 12:01

## 2 Answers

In any ring, $$0\cdot0 = 0$$, so the given implication does not hold (since 0 is not invertible).

It's worth emphasis that is not the definition of irreducible that is used by many ring theorists studying factorization theory and related matters in general commutative rings. As described in the paper excerpted below, common (integral) domain notions like "associate" and "irreducible" bifurcate into a few inequivalent notions in the presence of zero-divisors. Motivated by extending results on unique factorization to non-domains, this gives more insight for the most useful notion of irreducible and associate, including whether or not $$0$$ should be considered irreducible. It turns out that using such definitions yields that $$0$$ is irreducible iff the ring is a domain (for all notions of associate and irreducible). In particular this means that $$p$$ prime $$\Rightarrow$$ $$p$$ irreducible is true for all $$p$$ (including $$p = 0)$$. Below are the definitions in use.

Excerpted from: Factorization in Commutative Rings with Zero-divisors.
D.D. Anderson, Silvia Valdes-Leon.
Rocky Mountain J. Math. Volume 28, Number 2 (1996), 439-480