Ring with non-trivial idempotent splitting as product of two rings

If a commutative ring $T$ has a non-trivial idempotent $e$, then it is easy to show that $f:T \rightarrow R\times S$ defined by $f(t)=(et,(1-e)t)$ is a ring isomorphism. My question is how to prove the converse, viz., if a commutative ring splits as the product of two rings, then it contains a non-trivial idempotent.

Solution 1:

The converse is easier: Suppose $R \approx A \times B$ as rings. Then $R$ is a direct product of two other rings, each with their own identity, $1_A$ and $1_B$. Both of these are non-trivial idempotents: $$1_A^2 = (1_A,0) \cdot (1_A,0) = (1_A^2,0) =(1_A,0)=1_A.$$

In particular, $1_A$ and $1_B$ are orthogonal idempotents, meaning that also $1_A 1_B = 1_B 1_A = 0$.

See also this Tricki page.

Added I think I see where the confusion lies. Okay, here's how to define $R \times S$. Given an idempotent $e \in T$ (in your notation), you get another idempotent $1-e$ automatically ($s := (1-e)^2=1-2e+e=1-e$). Then let $R$ be the subring of $T$ generated by $e$ and $S$ the subring generated by $s$. Then we define a map $f:T \to R \times S$ by $g \mapsto (eg, sg)$. This is surjective (easy), and it is injective: for suppose $eg=0$ and $(1-e)g=0$. Then it follows trivially that $g=0$. So $f$ is an isomorphism. Thus, all of $R \times S$ used, there are no "subsets".