◀ ▲ ▶Branches / Algebra / Proof

Proof

(related to Lemma: Divisibility of Principal Ideals)

"$\Rightarrow$"

• By assumption, \((i)=Ri\) and \((j)=Rj\) are principal ideals for some \(i,j\in R\), where $(R, + ,\cdot)$ is an integral domain.
• If \(i \mid j\), then following the definition of divisibility in the ring there exists an element \(x\in R\) with \(xi=j\).
• It follows that $(j)=Rj=R(xi)\subseteq Ri=(i).$
• Following the definition of divisibility of ideals $(j)\subseteq (i)$ is equivalent to $(i)\mid (j).$

"$\Leftarrow$"

• If, vice versa, \((i)\mid (j)\), we have the set inclusion \((j)\subseteq (i)\).
• Because \(j\in (j)\), it follows that \(j\in (i)\).
• Therefore, there exists an element \(x\in R\) with \(j=xi\).
• Thus, \(i\mid j\).
  ∎

Thank you to the contributors under CC BY-SA 4.0!

Github:

References

Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013