Principal Right Ideal is Right Ideal

Theorem

Let $\struct {R, +, \circ}$ be a ring with unity.

Let $a \in R$.

Let $aR$ be the principal right ideal of $R$ generated by $a$.

Then $aR$ is an right ideal of $R$.

Proof

We establish that $aR$ is an right ideal of $R$, by verifying the conditions of Test for Right Ideal.

$aR \ne \O$, as $a \circ 1_R = a \in aR$.

Let $x, y \in ar$.

Then:

\(\ds \exists r, s \in R: \, \)

\(\ds x\)

\(=\)

\(\ds a \circ r, y = a \circ s\)

\(\ds \leadsto \ \ \)

\(\ds x + \paren {-y}\)

\(=\)

\(\ds a \circ r + \paren {a \circ -s}\)

Product with Ring Negative

\(\ds \)

\(=\)

\(\ds a \circ \paren {r + \paren {-s} }\)

\(\ds \leadsto \ \ \)

\(\ds x + \paren {-y}\)

\(\in\)

\(\ds aR\)

Let $s \in aR, x \in R$.

\(\ds s\)

\(\in\)

\(\ds aR, x \in R\)

\(\ds \leadsto \ \ \)

\(\ds \exists r \in R: \, \)

\(\ds s\)

\(=\)

\(\ds a \circ r\)

\(\ds \leadsto \ \ \)

\(\ds s \circ x\)

\(=\)

\(\ds a \circ r \circ x\)

\(\ds \)

\(\in\)

\(\ds aR\)

Thus by Test for Right Ideal, $aR$ is a right ideal of $R$.

$\blacksquare$