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.

$\blacksquare$


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