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 $s \in \ideal a, x \in R$.
\(\ds s\) | \(\in\) | \(\ds \ideal a, x \in R\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists r \in R: \, \) | \(\ds s\) | \(=\) | \(\ds r \circ a\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ s\) | \(=\) | \(\ds x \circ r \circ a\) | |||||||||||
\(\ds \) | \(\in\) | \(\ds \ideal a\) |
and similarly $s \circ x \in \ideal a$.
Thus by Test for Ideal, $\ideal a$ is an ideal of $R$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 59.$ Principal ideals in a commutative ring with a one: $\S 59.1$