I noticed this is true for $\mathbb{Z}$, but I was wondering whether it was true in general. Let $R$ be a Dedekind domain and $P_1, ... , P_s$ maximal ideals. The localized ring $R_{P_i}$ is a discrete valuation ring, and if $\nu_{P_i}(\pi) = 1$, then the unique maximal ideal $PR_{P_i}$ is equal to $\pi R_{P_i}$, where $\nu_{P_i}$ is the discrete valuation on $R$ induced by $P_i$.
My question is, can we pick $\pi$ to be a unit in EVERY other discrete valuation ring $R_P$ for $P \neq P_1$? I know that $\pi$ can only be a nonunit in at most finitely many prime ideal localizations. But ultimately I want to know if we can choose $\pi_1, ... \pi_n \in R$ with $\nu_{P_i}(\pi_j) = \delta_{ij}$.
I was thinking this might be related to the approximation theorem (for the absolute values induced by the discrete valuations). I haven't worked out the details, but I'm thinking it should at least be possible to have $\nu_{P_1}(\pi) = 1$ and $\nu_{P_i}(\pi)$ very close to $0$ for $2 \leq i \leq n$.
Suppose for any prime $P$, there exists an element $\pi = \pi_P$ which has value $1$ at $P$, and value $0$ at all other primes $Q$. Let $I = R \pi$. Then the ideal $I$ factors as a product of prime ideals $$\prod\limits_Q Q^{\nu_Q(I)}$$ But here $\nu_Q(\pi) = \nu_Q(I) = 0$ for all $Q \neq P$, and $\nu_P(\pi) = \nu_P(I) = 1$. Therefore $R\pi = I = P$, which means that $P$ is principal. Since every prime ideal of $R$ is principal, it follows that every ideal of $R$ is principal.