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 $\mathfrak o \subset \mathfrak 0_v$. 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 $v_{\mathfrak p_v}(z)=0$. Then we can write $z=a^\prime/b^\prime$ with $a^{\prime},b^{\prime}\in \mathfrak o$ and $v_{\mathfrak p_v}(a^\prime)=v_{\mathfrak p_v}(b^\prime)=l$, say. Next, choose $\pi\in \mathfrak p_v^{-l}-\mathfrak p_v^{-l+1}$ and set $a=\pi a^\prime$, $b=\pi b^\prime$.
The book then states that $a,b \in \mathfrak o-\mathfrak p_v$, and so $v(z)=0$.
Can someone tell me why $a$ and $b$ should be in $\mathfrak o$? Can we assume that $l\le 0$?
Many thanks for your help.
To check that $a \in \mathfrak o$, it suffices to check that for every discrete valuation $v$ of $K$, $v(a)\geq 0$.
By construction, $v_{\mathfrak p_v}(a) = 0$, since $\pi$ was chosen to have valuation exactly $-l = -v(a')$ at $\mathfrak p$.
At any other prime $\mathfrak q$, we must show that $v_{\mathfrak q}(\pi ) \geq 0$. Indeed, this follows from taking $v_\mathfrak q$ valuations on both sides of $\mathfrak p^l \pi \in \mathcal o - \mathfrak p$ (using the fact that $v_\mathfrak q(\mathfrak p^l) = 0$).