Element in Ideal represents the norm of ideal. Let $K$ be a number field and $\mathcal{O}_K$ its integer ring. Let $I$ be an ideal of $\mathcal{O}_K$. Is there an element $\alpha\in I$ such that $N(\alpha)=cN(I)$ where c is some integer coprime to $N(I)$?
 A: I believe the below works.
EDIT: Note that in the below, I write $\|\cdot\|$ for the absolute norm in lieu of $N$--it's just my usual notation. Sorry if it was confusing!
Factor $I=\mathfrak{p}_1^{e_1}\cdots \mathfrak{p}_i^{e_i}$. Each prime $\mathfrak{p}_i$ lies above some prime $p_i$ of $\mathbb{Z}$. Up to possible reordering, we may group the primes as $S_1=\{\mathfrak{p}_1,\ldots,\mathfrak{p}_{i_1}\},S_2=\{\mathfrak{p}_{i_1+1},\ldots,\mathfrak{p}_{i_2}\},\ldots$ where $S_i$ is the set of primes lying over $p_i$. Now, for each $\ell$ let $T_\ell$ be the set of primes lying over $p_\ell$ not contained in $S_\ell$. 
Now, by the CRT we may find $\alpha\in\mathcal{O}_K$ such that for each $\ell$
$$v_{\mathfrak{p}_i}(\alpha)=e_i\qquad\text{for all }\mathfrak{p}_i\in S_\ell$$
and
$$v_\mathfrak{p}(\alpha)=0\qquad\text{for all }\mathfrak{p}\in T_\ell$$
Now, consider the factorization of $(\alpha)$. Since $v_{\mathfrak{p}_i}(\alpha)=e_i$ for all $\mathfrak{p}_i\mid I$ we know that $I\mid(\alpha)$ so that $(\alpha)=IJ$. But, note that by construction, we must necessarily have that $J$ is divisible only by primes of $\mathcal{O}_K$ lying over primes $q$ different from each prime $p_i$. In particular, we see that $\|J\|$ is coprime to $\|I\|$. Noting then that 
$$|N_{K/\mathbb{Q}}(\alpha)|=\|(\alpha)\|=\|IJ\|=\|J\|\|I\|$$
finishes the problem.
