# Diagonal Relation on Ring is Ordering Compatible with Ring Structure

## 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:

 $\ds \tuple {x, y}$ $\in$ $\ds \Delta_R$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds y$ $\ds \leadsto \ \$ $\ds x + z$ $=$ $\ds y + z$ $\ds \leadsto \ \$ $\ds \tuple {x + z, y + z}$ $\in$ $\ds \Delta_R$

Similarly:

 $\ds \tuple {x, y}$ $\in$ $\ds \Delta_R$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds y$ $\ds \leadsto \ \$ $\ds z + x$ $=$ $\ds z + y$ $\ds \leadsto \ \$ $\ds \tuple {z + x, z + y}$ $\in$ $\ds \Delta_R$

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

Then note that:

 $\ds \tuple {0_R, x}$ $\in$ $\ds \Delta_R$ $\, \ds \land \,$ $\ds \tuple {0_R, y}$ $\in$ $\ds \Delta_R$ $\ds \leadsto \ \$ $\ds 0_R$ $=$ $\ds x$ $\, \ds \land \,$ $\ds 0_R$ $=$ $\ds y$ $\ds \leadsto \ \$ $\ds 0_R$ $=$ $\ds x \circ y$ $\ds \leadsto \ \$ $\ds \tuple {0_R, x \circ y}$ $\in$ $\ds \Delta_R$

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

$\blacksquare$