Negative of Subring is Negative of Ring

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Theorem

Let $\struct {R, +, \circ}$ be an 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:

\(\displaystyle \mathbin \sim x\) \(=\) \(\displaystyle \map {i_S} {\mathbin \sim x}\) as $\mathbin \sim x \in S$
\(\displaystyle \) \(=\) \(\displaystyle -x\) Group Homomorphism Preserves Inverses

$\blacksquare$