Inclusion Mapping on Subring is Homomorphism
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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.
$\blacksquare$