Ring Homomorphism Preserves Zero

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Theorem

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.

Let:

$0_{R_1}$ be the zero of $R_1$
$0_{R_2}$ be the zero of $R_2$.


Then:

$\map \phi {0_{R_1} } = 0_{R_2}$


Proof

By definition, if $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ are rings then $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ are groups.

Again by definition:

the zero of $\struct {R_1, +_1, \circ_1}$ is the identity of $\struct {R_1, +_1}$
the zero of $\struct {R_2, +_2, \circ_2}$ is the identity of $\struct {R_2, +_2}$.


The result follows from Group Homomorphism Preserves Identity.

$\blacksquare$


Sources

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