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:

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

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 $R$.

$\blacksquare$


Sources

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