Inclusion Mapping on Subring is Homomorphism

Theorem

Let $\struct {R, +, \circ}$ be a ring.

Let $\struct {S, +{\restriction_S}, \circ {\restriction_S}}$ be a subring of $R$.

Let $i_S: S \to R$ be the inclusion mapping from $S$ to $R$.

Then ${i_S}$ is a ring homomorphism.

Proof

Let $x, y \in S$.

Then:

\(\ds \map {i_S} x + \map {i_S} y\)

\(=\)

\(\ds x + y\)

Definition of Inclusion Mapping

\(\ds \)

\(=\)

\(\ds x \mathbin{ + {\restriction_S} } y\)

as $x, y \in S$

\(\ds \)

\(=\)

\(\ds \map {i_S} {x \mathbin{ + {\restriction_S} } y}\)

as $x \mathbin{ + {\restriction_S} } y \in S$

and:

\(\ds \map {i_S} x \circ \map {i_S} y\)

\(=\)

\(\ds x \circ y\)

Definition of Inclusion Mapping

\(\ds \)

\(=\)

\(\ds x \mathbin{\circ {\restriction_S} } y\)

as $x, y \in S$

\(\ds \)

\(=\)

\(\ds \map {i_S} {x \mathbin{ \circ {\restriction_S} } y}\)

as $x \mathbin{ \circ {\restriction_S} } y \in S$

Hence ${i_S}$ is a ring homomorphism by definition.

$\blacksquare$