Negative of Subring is Negative of Ring
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Theorem
Let $\struct {R, +, \circ}$ be a ring.
For each $x \in R$ let $-x$ denote the ring negative of $x$ in $R$.
Let $\struct {S, + {\restriction_S}, \circ {\restriction_S}}$ be a subring of $R$.
For each $x \in S$ let $\mathbin \sim x$ denote the ring negative of $x$ in $S$.
Then:
- $\forall x \in S: \mathbin \sim x = -x$
Proof
Let $i_S: S \to R$ be the inclusion mapping from $S$ to $R$.
By Inclusion Mapping on Subring is Homomorphism, $i_S$ is a ring homomorphism.
By Ring Homomorphism of Addition is Group Homomorphism, $i_S$ is a group homomorphism on ring addition $+$.
Let $x \in S$.
Then:
\(\ds \mathbin \sim x\) | \(=\) | \(\ds \map {i_S} {\mathbin \sim x}\) | as $\mathbin \sim x \in S$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -x\) | Group Homomorphism Preserves Inverses |
$\blacksquare$