Principal Ideal is Ideal

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Theorem

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 $R$.


Proof

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 $x, y \in \ideal a$.

Then:

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


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 $R$.

$\blacksquare$


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