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A question from Frohlich and Taylor's book 'Algebraic Number Theory', page 60.

Let $\mathfrak o$ be a Dedekind domain with quotient field $K$ and let $v$ be a discrete valuation on $K$. Let $\mathfrak o_v$ be the valuation ring of $v$ in $K$ and $\mathcal P_v$ be the valuation ideal of $v$ in $K$. Assume that $\mathfrak o\subset \mathfrak o_v$. Set $\mathfrak p_v=\mathcal P_v\cap\mathfrak o$ which is a non-zero prime ideal of $\mathfrak o$ and define $v_{\mathfrak p_v}$ to be the valuation on $K$ induced by $\mathfrak p_v$.

Suppose that $x\in \mathfrak o$, $x\ne 0$, with $v_{\mathfrak p_v}(x)=l$ so that $x\mathfrak o= \mathfrak p_v^l\mathfrak a$ with $(\mathfrak a,\mathfrak p_v)=1$.

Why then is $\mathfrak a \cdot \mathfrak o_v=\mathfrak o_v$ as claimed on p.60?

Many thanks for your help.

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If $a$ is comprime to $p_v$, then $v(a)=0$, it is a unit

Note that $o_v = \{ v \geq 0 \}$ and $v: K_v^\times \rightarrow \mathbb{Z}$ is a group homomorphism.

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