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.

$\blacksquare$