# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Cesareo, Brahadeesh, Leucippus, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.

## 1 Answer

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.

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