Direct Product of a Clean Rings

Question

Commutative rings whose elements are a sum of an unit and idempotent by Anderson & Camillo (2002)

Definition 1.
A commutative ring $R$ is a clean ring if every element $x\in R$ can be written in the form of $x=u+i$ where $u\in U(R)$ and i$\in Id(R)$.

Proposition 2.
(1) A quasilocal ring is a clean ring.
(2) A homomorphic image of a clean ring is a clean ring.
(3) A direct product $R=∏R_α $ of rings $\{R_α\}$ is a clean ring if and only if each $R_\alpha$ is a clean ring.
Proof.
(2) This is immediate since the homomorphic image of a unit (resp., idempotent) is a unit (resp., idempotent).
(3) ($\Rightarrow$) This follow from (2).

How does the proof of (3) to the right is follow from (2)? Is there any relationship between direct product and homomorphic image?

Answer 1

If $R=\prod R_\alpha$ is clean, the image of the projection homomorphism $\pi_\alpha:R\to R_\alpha$ is a clean ring by (2). The projection is surjective, so $R_\alpha$ is clean.

Answer 2

Only half of 3 follows "immediately" from 2.

Each $R_\alpha$ is of course the homomorphic image of $R$ by projecting onto the coordinate $\alpha$.

The remaining thing to do is to prove that the product is clean if the $R_\alpha$ are all clean.

But this is easy to see because an idempotent in the product is an element whose entries are idempotent, and a unit in the product is an element whose entries are units.

In fact, knowing that, you don't even need $2$ to show $3$: it is more or less obvious that a clean element of the product is an element whose entries are clean elements of the factors.