Centralizer of Ring Subset is Subring

From ProofWiki
Jump to: navigation, search

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.

$\blacksquare$


Sources