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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?

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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 $\alpha_j = x+3, \, x+2, \, x+1, \, x, \, x-1, \, x-2, \, x-3$.

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Let $K$ be an extension of the rationals of degree four, containing $\sqrt{-1}$. Then $33+4i$, $32+9i$, $31+12i$, and $24+23i$ are distinct algebraic integers in $K$, each of norm $1105^2$, with distinct principal ideals, but they generate a vector space of dimension two over the rationals.

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  • $\begingroup$ That is a nice example. Are there any examples when $K$ is primitive? $\endgroup$ – Stanley Yao Xiao Mar 5 at 3:05
  • $\begingroup$ We can take $K=\mathbb{Q}(\omega)$, where $\omega$ is a primitive 8-th root of unity satisfying $\omega^2=\sqrt{-1}$. $\endgroup$ – eloiprime Mar 5 at 3:11
  • $\begingroup$ Sorry, what does it mean for a field to be primitive? $\endgroup$ – Gerry Myerson Mar 5 at 11:29
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A cubic counterexample: Let $K$ be the first totally real cubic field ${\bf Q}[x] / (x^3+x^2-2x-1)$ (the roots are $2 \cos (2m\pi/7)$ for $m=1,2,3$); and let $a=13$, the first totally split prime in $K$. For each of the three primes above $13$ there are at least three generators in the $2$-dimensional subspace generated by $1$ and $x$. Choose one $\alpha_j$ from each of these three sets $$ \{7x+3, 2x-1, -3x-5\}, \{5x+2, x+3, -3x+4\}, \{-2x+3, -3x-2, -4x-7\}. $$ If you prefer a non-cyclic field, use ${\bf Q}[x] / (x^3-4x-1)$ (discriminant $229$) and $$ \{\alpha_1, \alpha_2, \alpha_3\} = \{ 6x+11, 3x+2, 2x-5 \}, $$ each of which has norm $-37$.

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