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 $R$.

$\blacksquare$


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