Centralizer of Ring Subset is Subring

Theorem

Let $S$ be a subset of a ring $\struct {R, +, \circ}$

Then $\map {C_R} S$, the centralizer of $S$ in $R$, is a subring of $R$.

Proof

Certainly $0_R \in \map {C_R} S$ as $0_R$ commutes (trivially) with all elements of $R$.

Suppose $x, y \in \map {C_R} S$.

Then:

 $\displaystyle \forall s \in S: s \circ \paren {x + \paren {-y} }$ $=$ $\displaystyle s \circ x + s \circ \paren {-y}$ $\quad$ Distributivity of $\circ$ over $+$ $\quad$ $\displaystyle$ $=$ $\displaystyle x \circ s + \paren {-y} \circ s$ $\quad$ $x$ and $y$ are in $\map {C_R} S$ $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {x + \paren {-y} } \circ s$ $\quad$ Distributivity of $\circ$ over $+$ $\quad$

So:

$x + \paren {-y} \in \map {C_R} s$

Suppose $x, y \in \map {C_R} S$ again.

$x \circ y \in \map {C_R} S$

Thus all the conditions are fulfilled for Subring Test, and $\map {C_R} S$ is a subring of $R$.

$\blacksquare$