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To prove that all sub rings of $Q$ Euclidean Domain , what will be the Euclidean Valuation? Can anyone please give me a hint.

My attempt : Can I take $v(a/b) = |a|$?

If I take this how would the following properties would be satisfied? If $x, y \in S$ then $v(x) \le v(xy)$ .

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  • $\begingroup$ Your attempt is Ok. This function satisfies the first axiom of being euclidean valuation, and it does not satisfies the second. But it doesn't have to, because you can redefine it to obtain that the new one satisfies both. You can see the trick here: en.wikipedia.org/wiki/Euclidean_domain $\endgroup$ – SMM Apr 28 '18 at 7:34
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By this answer of mine, any localization of $\Bbb Z$ is a Euclidean domain. Now if $R\subseteq\Bbb Q$ is any subring, it is some localization of $\Bbb Z$, simply take $S$ to be the set of denominators of all completely reduced elements or $R$, then $R=S^{-1}\Bbb Z$ is a Euclidean domain. That old answer also details the construction of the Euclidean degree function.

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