Principal Ideal from Element in Center of Ring

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Theorem

Let $\left({R, +, \circ}\right)$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $b \in R$ be in the center of $R$.


Then:

$\left({b}\right) = R \circ b = \left\{{x \circ b: x \in R}\right\}$

where $\left({b}\right)$ is the principal ideal generated by $b$.


Proof

Let $J = R \circ b$

The center of $R$ is defined as:

$Z \left({R}\right) = \left\{{x \in R: \forall s \in R: s \circ x = x \circ s}\right\}$

Therefore:

$R \circ b = b \circ R = \left\{{x \circ b: x \in R}\right\} = \left\{{b \circ x: x \in R}\right\}$

and so:

$x \in J \implies x \circ b \in J \land b \circ x \in J$


Thus $J$ is an ideal of $R$ and so $J = \left({b}\right)$.

$\blacksquare$


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