Direct Product iff Nontrivial Idempotent
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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 if and only if $R$ contains an idempotent element not equal to $0_R$ or $1_R$.
Proof
Necessary Condition
Suppose $R_1$ and $R_2$ are non-trivial rings, such that $R = R_1 \times R_2$.
Then:
| \(\ds \tuple {1_{R_1}, 0_{R_2} }^2\) | \(=\) | \(\ds \tuple {1_{R_1}^2, 0_{R_2}^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \tuple {1_{R_1}, 0_{R_2} }\) |
Thus $\tuple {1_{R_1}, 0_{R_2} }$ is an idempotent element of $R$ not equal to $0_R$ or $1_R$.
$\Box$
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:
| \(\ds \map \phi {e c + \paren {1 - e} d}\) | \(=\) | \(\ds \tuple {e \paren {e c + \paren {1 - e} d}, \paren {1 - e} \paren {e c + \paren {1 - e} d} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \tuple {e c, \paren {1 - e} d}\) |
Thus $\phi$ is surjective.
Hence $\phi$ is a ring isomorphism.
$\blacksquare$