Principal Ideal is Ideal

From ProofWiki
Jump to navigation Jump to search


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

Let $a \in R$.

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

Then $\ideal a$ is an ideal of $R$.


First we establish that $\ideal a$ is an ideal of $R$, by verifying the conditions of Test for Ideal.

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

Let $x, y \in \ideal a$.


\(\ds \exists r, s \in R: \ \ \) \(\ds x\) \(=\) \(\ds r \circ a, y = s \circ a\)
\(\ds \leadsto \ \ \) \(\ds x + \paren {-y}\) \(=\) \(\ds r \circ a + \paren {-s \circ a}\)
\(\ds \) \(=\) \(\ds \paren {r + \paren {-s} } \circ a\)
\(\ds \leadsto \ \ \) \(\ds x + \paren {-y}\) \(\in\) \(\ds \ideal a\)

Let $s \in \ideal a, x \in R$.

\(\ds s\) \(\in\) \(\ds \ideal a, x \in R\)
\(\ds \leadsto \ \ \) \(\ds \exists r \in R: s\) \(=\) \(\ds r \circ a\)
\(\ds \leadsto \ \ \) \(\ds x \circ s\) \(=\) \(\ds x \circ r \circ a\)
\(\ds \) \(\in\) \(\ds \ideal a\)

and similarly $s \circ x \in \ideal a$.

Thus by Test for Ideal, $\ideal a$ is an ideal of $R$.