Artin in "Algebra" mentions the following:
To determine whether or not a given ring is isomorphic to a product ring, one looks for the elements that in a product ring would be $(1,0)$ and $(0,1)$. They are idempotent elements.
So I don't think I am understanding this. What does he mean and could someone provide example of using idempotent elements in demonstrating that $\mathbb{Z}/(6)$ is isomorphic to $\mathbb{Z}/(2) \times \mathbb{Z}/(3)$. I could explicitly list our the isomorphism, but I am trying to understand where the idempotent elements are coming in?
For a commutative ring $R$ to be isomorphic to a product of rings, there has to exist a non-trivial idempotent $e \in R$ and then one has an isomorphism $R \cong Re \times R (1 - e) $ given by $r \mapsto (r e, r (1-e) )$.
So let us have a look at the idempotents of $R = \mathbb{Z}/6\mathbb{Z}$. The nontrivial ones are given by the cosets $e_1 = 3 + 6\mathbb{Z}$ (as $3^2 = 9 \equiv 3$ mod $6$ ) and $e_2 = 4 + 6\mathbb{Z}$ (as $4^2 = 16 \equiv 4$ mod $6$). Also $e_1 + e_2 = 1$ in $\mathbb{Z}/6\mathbb{Z}$. We thus have an isomorphism $R \cong R e_1 \times R e_2$. Note that we have $$R e_1 = \{0 + 6 \mathbb{Z}, 3 + 6\mathbb{Z}\} \cong \mathbb{Z}/2 \mathbb{Z}$$ and $$R e_2 = \{0 + 6\mathbb{Z}, 2 + 6 \mathbb{Z}, 4 + 6\mathbb{Z}\} \cong \mathbb{Z}/3 \mathbb{Z}.$$