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Using standard notations, let $K$ be a number field and $S = \left\{p_{1}, ..., p_{n}\right\}$ a finite set of non-zero prime ideals of $K$. Let $a$ be a non-zero fractional ideal of $K$. Prove that there exists $\alpha \in K^{\times}$ such that $b := \alpha a$ has prime factorization $b = \prod_{p \in S}{p^{n_{p}(b)}},$ with $n_{p}(b)=0$ for all $p_i$'s in $S$.

I know this should follow from the unique factorization directly somehow, but I'm traving trouble picking the $\alpha$...

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If we write $\mathfrak a = \prod_{i = 1}^n \mathfrak p_i^{r_i} \prod_{\mathfrak p \notin S} \mathfrak p^{s_\mathfrak p}$ then the goal is to find an $\alpha \in K^*$ such that $v_i(\alpha) = -r_i$ for all $i$. Now, fix an $i$. Pick an $a_i \in \mathfrak p_i - \mathfrak p_i^2$, so that $v_{\mathfrak p_i}(a_i) = 1$. By the Chinese remainder theorem, you can pick a $b_i \in \mathscr O_K$ such that $b_i \equiv a_i \mod{\mathfrak p_i^2}$, and such that $b_i \equiv 1 \mod{\mathfrak p_j}$ for $j \neq i$. Does this help?