Let $K$ be a number field with $[K:\mathbb{Q}]=n$ with $n \geq 2$ and let $\mathcal{O}_K$ be its ring of integers. Suppose that $\alpha_1, \cdots, \alpha_n \in \mathcal{O}_K$ are distinct algebraic integers such that $N_{K/\mathbb{Q}}(\alpha_j) = a$ for some fixed rational integer $|a| > 1$ and the principal ideals $(\alpha_j)$ are distinct for $j =1, \cdots, n$. Does it follow that $\alpha_1, \cdots, \alpha_n$ are $\mathbb{Q}$-linearly independent?

When $n = 2$ this is obvious. Indeed, a quadratic algebraic integer is of the form $u = u_1 + \omega u_2$ with $u_1,u_2 \in \mathbb{Z}$ and where $\omega = \sqrt{d}$ for some integer $d$ or $\omega = \frac{1 + \sqrt{d}}{2}$ for some rational integer $d$. Then $u,v$ are $\mathbb{Q}$-linearly dependent in $\mathcal{O}_K$ if and only if there exist rational integers $\ell,m$ such that $\ell u = mv$. Since $u,v$ have the same norm, it follows that $N_{K/\mathbb{Q}}(\ell) = N_{K/\mathbb{Q}}(m)$, and from here we see that $u,v$ are necessarily associates. But then we must have $u = v$.

Is the result above true when $n \geq 3$? If not, what is a simple counterexample to the claim?