Centralizer of Ring Subset is Subring

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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:

\(\ds \forall s \in S: \, \) \(\ds s \circ \paren {x + \paren {-y} }\) \(=\) \(\ds s \circ x + s \circ \paren {-y}\) Distributivity of $\circ$ over $+$
\(\ds \) \(=\) \(\ds x \circ s + \paren {-y} \circ s\) $x$ and $y$ are in $\map {C_R} S$
\(\ds \) \(=\) \(\ds \paren {x + \paren {-y} } \circ s\) Distributivity of $\circ$ over $+$


So:

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


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

Then from Element Commutes with Product of Commuting Elements:

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

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

$\blacksquare$


Sources

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