Principal Left Ideal is Left Ideal

Theorem

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

Let $a \in R$.

Let $Ra$ be the principal left ideal of $R$ generated by $a$.

Then $Ra$ is an left ideal of $R$.

Proof

We establish that $Ra$ is an left ideal of $R$, by verifying the conditions of Test for Left Ideal.

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

Let $x, y \in Ra$.

Then:

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

\(\ds x\)

\(=\)

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

\(\ds \leadsto \ \ \)

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

\(=\)

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

Product with Ring Negative

\(\ds \)

\(=\)

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

\(\ds \leadsto \ \ \)

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

\(\in\)

\(\ds Ra\)

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

\(\ds s\)

\(\in\)

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

\(\ds \leadsto \ \ \)

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

\(\ds s\)

\(=\)

\(\ds r \circ a\)

\(\ds \leadsto \ \ \)

\(\ds x \circ s\)

\(=\)

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

\(\ds \)

\(\in\)

\(\ds Ra\)

Thus by Test for Left Ideal, $Ra$ is a left ideal of $R$.

$\blacksquare$

Also see

• Principal Right Ideal is Right Ideal where it is shown that a principal right ideal is a right ideal