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Let $V$ be a valuation ring. Then any two principal ideals $A_1$ and $A_2$ of $V$ are ordered by inclusion.

I need a proof for this lemma and I don't know how to start.

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closed as off-topic by user26857, Cesareo, Brahadeesh, Leucippus, ancientmathematician Nov 19 '18 at 7:59

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A valuation ring $V$ is an integral domain such that for any $x$ in its field of fractions $F$ either $x$ or $x^{-1}$ is in $V$. Now in your case let $A_1 = (a_1)$ and $A_2 = (a_2)$. Then either $\frac{a_1}{a_2} \in V$ or $\frac{a_2}{a_1} \in V$ . In the first case let $\frac{a_1}{a_2} = b_1$ and note that $(a_1) = (a_2 b_1) \subset (a_2)$. The second case is similar.

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  • $\begingroup$ Thank you very much. It makes sense. $\endgroup$ – Gentiana Nov 15 '18 at 9:58

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