# Centralizer of Ring Subset is Subring

## Theorem

Let $S$ be a subset of a ring $\left({R, +, \circ}\right)$

Then $C_R \left({S}\right)$, the centralizer of $S$ in $R$, is a subring of $R$.

If a unit $u \in R$ such that $u \in C_R \left({S}\right)$, then $u^{-1} \in C_R \left({S}\right)$.

## Proof

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

Suppose $x, y \in C_R \left({S}\right)$. Then:

 $\displaystyle \forall s \in S: s \circ \left({x + \left({-y}\right)}\right)$ $=$ $\displaystyle s \circ x + s \circ \left({-y}\right)$ $\quad$ Distributivity of $\circ$ over $+$ $\quad$ $\displaystyle$ $=$ $\displaystyle x \circ s + \left({-y}\right) \circ s$ $\quad$ $x$ and $y$ are in $C_R \left({S}\right)$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left({x + \left({-y}\right)}\right) \circ s$ $\quad$ Distributivity of $\circ$ over $+$ $\quad$

So $x + \left({-y}\right) \in C_R \left({S}\right)$.

Suppose $x, y \in C_R \left({S}\right)$ again. Then: $x \circ y \in C_R \left({S}\right)$ from Element Commutes with Product of Commuting Elements.

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

Let $u \in U_R$.

If $u \in C_R \left({S}\right)$, then $u^{-1} \in C_R \left({S}\right)$ from Commutation with Inverse in Monoid.

$\blacksquare$