Ring is Commutative iff Opposite Ring is Itself

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Theorem

Let $\struct {R, +, \times}$ be a ring.

Let $\struct {R, +, *}$ be the opposite ring of $\struct {R, +, \times}$.


Then $\struct {R, +, \times}$ is a commutative ring if and only if:

$\struct {R, +, \times} = \struct {R, +, *}$


Proof

By definition of the opposite ring:

$\forall x, y \in R: x * y = y \times x$


Necessary Condition

Let $\struct {R, +, \times}$ be a commutative ring.

Then:

\(\ds \forall x, y \in R: \, \) \(\ds x \times y\) \(=\) \(\ds y \times x\)
\(\ds \leadsto \ \ \) \(\ds x \times y\) \(=\) \(\ds x * y\) Definition of opposite ring
\(\ds \leadsto \ \ \) \(\ds \struct {R, +, \times}\) \(=\) \(\ds \struct {R, +, *}\) Equality of Algebraic Structures

$\blacksquare$


Sufficient Condition

Let $\struct {R, +, \times} = \struct {R, +, *}$.

Then:

\(\ds \forall x, y \in R: \, \) \(\ds x \times y\) \(=\) \(\ds x * y\) Equality of Algebraic Structures
\(\ds \leadsto \ \ \) \(\ds x \times y\) \(=\) \(\ds y \times x\) Definition of opposite ring

Thus by definition $\struct {R, +, \times}$ is a commutative ring.

$\blacksquare$


Sources

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