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?



Linear independence of algebraic integers of equal norm
We adapt an idea from a now-deleted answer by **Kenny Lau**
to construct examples for any $n>2$ with the $\alpha\_j$ all contained in
2-dimensional space. Let $a$ be prime, and choose distinct integers
$x\_1,\ldots,x\_n$ that remain different mod $a$ for which
$$
P(x) := \left[\prod\_{j=1}^n (x-x\_j)\right] - a
$$
is irreducible. (By Hilbert's irreducibility theorem,
most $x\_j$ will satisfy the last condition.) Then take $\alpha\_j = x - x\_j$
in the field ${\bf Q}[x] / (P(x))$. Each of these has norm $\pm a$,
and they generate distinct ideals because the ideals above $a$
correspond to factors of $P \bmod a$.


For example, when $n=7$ we may take $a=11$ and $\alpha\_j = j-4$
($j=1,\ldots,7$) to make $P(x) = x^7 - 14 x^5 + 49 x^3 - 36 x - 11$
with $\alpha\_j = x+3, \, x+2, \, x+1, \, x, \, x-1, \, x-2, \, x-3$.


