# 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:

\(\displaystyle \forall s \in S: s \circ \paren {x + \paren {-y} }\) | \(=\) | \(\displaystyle s \circ x + s \circ \paren {-y}\) | Distributivity of $\circ$ over $+$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x \circ s + \paren {-y} \circ s\) | $x$ and $y$ are in $\map {C_R} S$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {x + \paren {-y} } \circ s\) | 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

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 21$: Theorem $21.5$