# Linear independence of algebraic integers of equal norm

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?

## 3 Answers

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

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.

• That is a nice example. Are there any examples when $K$ is primitive? – Stanley Yao Xiao Mar 5 at 3:05
• We can take $K=\mathbb{Q}(\omega)$, where $\omega$ is a primitive 8-th root of unity satisfying $\omega^2=\sqrt{-1}$. – eloiprime Mar 5 at 3:11
• Sorry, what does it mean for a field to be primitive? – Gerry Myerson Mar 5 at 11:29

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 $$-37$$.