Ring Product with Zero

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Theorem

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


Then:

$\forall x \in R: 0_R \circ x = 0_R = x \circ 0_R$


That is, the zero is a zero element for the ring product, thereby justifying its name.


Proof

Because $\struct {R, +, \circ}$ is a ring, $\struct {R, +}$ is a group.

Since $0_R$ is the identity in $\struct {R, +}$, we have $0_R + 0_R = 0_R$.

From the Cancellation Laws, all group elements are cancellable, so every element of $\struct {R, +}$ is cancellable for $+$.


Thus:

\(\ds x \circ \paren {0_R + 0_R}\) \(=\) \(\ds x \circ 0_R\) Definition of Ring Zero
\(\ds \leadsto \ \ \) \(\ds \paren {x \circ 0_R} + \paren {x \circ 0_R}\) \(=\) \(\ds x \circ 0_R\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \leadsto \ \ \) \(\ds \paren {x \circ 0_R} + \paren {x \circ 0_R}\) \(=\) \(\ds \paren {x \circ 0_R} + 0_R\) Definition of Ring Zero
\(\ds \leadsto \ \ \) \(\ds x \circ 0_R\) \(=\) \(\ds 0_R\) Cancellation Laws


Next:

\(\ds \paren {0_R + 0_R} \circ x\) \(=\) \(\ds 0_R \circ x\) Definition of Ring Zero
\(\ds \leadsto \ \ \) \(\ds \paren {0_R \circ x} + \paren {0_R \circ x}\) \(=\) \(\ds 0_R \circ x\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \leadsto \ \ \) \(\ds \paren {0_R \circ x} + \paren {0_R \circ x}\) \(=\) \(\ds 0_R + \paren {0_R \circ x}\) Definition of Ring Zero
\(\ds \leadsto \ \ \) \(\ds 0_R \circ x\) \(=\) \(\ds 0_R\) Cancellation Laws

$\blacksquare$


Sources

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