Direct Product iff Nontrivial Idempotent

Theorem
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Then $R$ is the direct product of two non-trivial rings $R$ contains an idempotent element not equal to $0_R$ or $1_R$.

Necessary Condition
Suppose $R_1$ and $R_2$ are non-trivial rings, such that $R = R_1 \times R_2$.

Then:

Thus $\tuple {1_{R_1}, 0_{R_2} }$ is an idempotent element of $R$ not equal to $0_R$ or $1_R$.

Sufficient Condition
Suppose there exists an idempotent element $e \in R$ not equal to $0_R$ or $1_R$.

That is:
 * $0_R, 1_R \ne e$
 * $e^2 = e$

Note that:
 * $\paren {1 - e}^2 = 1 - e$

Let $R_1 = \ideal e$, the ideal generated by $e$.

Let $R_2 = \ideal {1 - e}$, the ideal generated by $1 - e$.

By Ring by Idempotent $\struct {R_1, +, \circ}$ and $\struct {R_2, +, \circ}$ are commutative rings with unity.

Define $R_1 \times R_2$ to be the direct product ring of $\struct {R_1, +, \circ}$ and $\struct {R_2, +, \circ}$.

We define a ring homomorphism $\phi : R \to R_1 \times R_2$ by:


 * $\map \phi a := \tuple {e a, \paren {1 - e} a}$

By Ring Homomorphism by Idempotent $\phi$ is a ring homomorphism.

Let $a \in \map \ker \phi$.

Then:
 * $e a = \paren {1 - e} a = 0$.

It follows that:
 * $a = e a + \paren {1 - e} a = 0_R$

So:
 * $\map \ker \phi = 0$

Thus $\phi$ is injective.

Let $\struct {e c, \paren {1 - e} d} \in R_1 \times R_2$ for arbitrary $c, d \in R$.

Then:

Thus $\phi$ is surjective.

Hence $\phi$ is a ring isomorphism.