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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.5$