# Diagonal Relation on Ring is Ordering Compatible with Ring Structure

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## Theorem

Let $\struct {R, +, \circ, \preceq}$ be a ring whose zero is $0_R$.

Then the diagonal relation $\Delta_R$ on $R$ is an ordering compatible with the ring structure of $R$.

## Proof

From Diagonal Relation is Ordering and Equivalence, we have that $\Delta_R$ is actually an ordering on $R$.

From the definition of the diagonal relation:

- $\tuple {x, y} \in \Delta_R \iff x = y$

Thus:

\(\displaystyle \tuple {x, y}\) | \(\in\) | \(\displaystyle \Delta_R\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x + z\) | \(=\) | \(\displaystyle y + z\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \tuple {x + z, y + z}\) | \(\in\) | \(\displaystyle \Delta_R\) |

Similarly:

\(\displaystyle \tuple {x, y}\) | \(\in\) | \(\displaystyle \Delta_R\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle z + x\) | \(=\) | \(\displaystyle z + y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \tuple {z + x, z + y}\) | \(\in\) | \(\displaystyle \Delta_R\) |

So $\Delta_R$ is compatible with $+$.

Then note that:

\(\displaystyle \tuple {0_R, x}\) | \(\in\) | \(\displaystyle \Delta_R\) | |||||||||||

\(\, \displaystyle \land \, \) | \(\displaystyle \tuple {0_R, y}\) | \(\in\) | \(\displaystyle \Delta_R\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0_R\) | \(=\) | \(\displaystyle x\) | ||||||||||

\(\, \displaystyle \land \, \) | \(\displaystyle 0_R\) | \(=\) | \(\displaystyle y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0_R\) | \(=\) | \(\displaystyle x \circ y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \tuple {0_R, x \circ y}\) | \(\in\) | \(\displaystyle \Delta_R\) |

Hence the result, from the definition of an ordering compatible with the ring structure of $R$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 23$