# Valuation ring's principal ideals [closed]

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.

## closed as off-topic by user26857, Cesareo, Brahadeesh, Leucippus, ancientmathematicianNov 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.