Principal Ideal in Integral Domain generated by Power Plus One is Subset of Principal Ideal generated by Power
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Theorem
Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $a \in D$ be a proper element of $D$.
Then:
- $\forall n \in \Z_{\ge 0}: \ideal {a^{n + 1} } \subsetneq \ideal {a_n}$
where $\ideal x$ denotes the principal ideal of $D$ generated by $x$.
Proof
We have:
\(\ds x\) | \(\in\) | \(\ds \ideal {a^{n + 1} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists r \in D: \, \) | \(\ds x\) | \(=\) | \(\ds r \circ a^{n + 1}\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \paren {r \circ a} \circ a^n\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \ideal {a^n}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \ideal {a^{n + 1} }\) | \(\subseteq\) | \(\ds \ideal {a^n}\) |
It remains to be shown that $\ideal {a^{n + 1} } \ne \ideal {a^n}$.
Aiming for a contradiction, suppose $\ideal {a^{n + 1} } = \ideal {a^n}$.
Then:
\(\ds \ideal {a^{n + 1} }\) | \(=\) | \(\ds \ideal {a^n}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^n\) | \(\in\) | \(\ds \ideal {a^{n + 1} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in D: \, \) | \(\ds a^n\) | \(=\) | \(\ds x \circ a^{n + 1}\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1_D\) | \(=\) | \(\ds x \circ a\) | Cancellation Law for Ring Product of Integral Domain |
That is, $a$ is a unit of $D$.
This contradicts the assertion that $a$ is a proper element of $D$.
The result follows by Proof by Contradiction.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $7$