Describe the set of quadratic integers α in Q[sqrt−3] for which α ̄ and α are associates.

Question

I was working through some textbook problems for my Number Theory class and needed some help with the following question:

Describe the set of quadratic integers α in Q[sqrt−3] for which α ̄ and α are associates.

I'd really like some help with this problem. Thank you!

Answer 1

First note that if the field is $K = \mathbb{Q}(\sqrt{-3})$ then the ring of integers is $\mathcal{O}_K = \mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right] = \mathbb{Z}[\zeta_6]$ (where $\zeta_6$ denotes the primitive sixth root of unity $e^{i\pi/3}$).

The units in $\mathbb{Z}[\zeta_6]$ are those of norm $\pm 1$, and the norm is given by $N_{K/\mathbb{Q}}(a + b\zeta_6) = (a + b\zeta_6)(a+b\zeta_6^*) = a^2 + ab + b^2$.

The equation $a^2 + ab + b^2 = \pm 1$ has finitely many solutions in the integers, as can be seen by rewriting it as $(a+b/2)^2+3b^2/4 = \pm 1$ after completing the square. In particular the quadratic form is always positive, so only solutions of norm $1$ are possible, and the only possible solutions are those where $b = 0, 1$ or $-1$ (by bounding the second term).

We therefore find all solutions to be $(a, b) = (1, 0), (-1, 0), (0, 1), (-1, 1), (0, -1)$ and $(1, -1)$. These correspond to the values $1, -1, \pm\zeta_6, \pm(1-\zeta_6)$, which is just the group of units generated by the powers of $\zeta_6$. Also note that $\zeta_6^* = 1 - \zeta_6$, so the conjugate of $a + b\zeta_6$ is $(a + b) - b\zeta_6$.

Now the question asks about which elements $\alpha \in \mathcal{O}_K$ are associated with their conjugate. By the previous result, we can check for each case using the general form of $\alpha = a + b\zeta_6$, and using the linear independence of $\{1, \zeta_6\}$:

1. $(a + b) - b\zeta_6 = a + b\zeta_6$
2. $(a + b) - b\zeta_6 = -a - b\zeta_6$
3. $(a + b) - b\zeta_6 = -b + (a + b)\zeta_6$
4. $(a + b) - b\zeta_6 = b - (a + b)\zeta_6$
5. $(a + b) - b\zeta_6 = (a + b) - a\zeta_6$
6. $(a + b) - b\zeta_6 = -(a + b) + a\zeta_6$

Each of these ends up in a system of equations for $a$ and $b$ that has a one-dimensional solution space, and we find the following solutions corresponding to each case: $c(1, 0), c(1, -2), c(-2, 1), c(0, 1), c(1, 1)$ and $c(1, -1)$ for arbitrary integer $c$.

This corresponds to the 6 distinct sets spanned by $1, 1 - 2\zeta_6, -2 + \zeta_6, \zeta_6, 1 + \zeta_6$ and $1 - \zeta_6$.

In other words any integer multiple of $1, \sqrt{-3}, \frac{1 \pm \sqrt{-3}}{2}, \frac{3 \pm \sqrt{-3}}{2}$.

Answer 2

Let $\omega$ be a primitive 3-rd root of unity, $K=\mathbf Q(\omega)=\mathbf Q(\sqrt{-3})$. Using minimal information on the arithmetic of $K$, you can practically avoid any further calculation. It is known (thanks to the discriminant) that the ring of integers of $K$ is $O_K=\mathbf Z [\omega]$, the so called Eisenstein ring, which a PID. The group of units $U_K$ is also known to be equal to the group $\mu_6$ of $6$-th roots of unity (a particular case of Dirichlet's theorem).

To determine the integers $\alpha\in\ O_K$ s.t. $\bar \alpha=u.\alpha, u\in U_K$, just write down the prime decomposition of $\alpha$ in $O_K$. The primes $\pi$ of $O_K$ above the primes $p$ of $\mathbf Z$ are of three possible types : 1) $p$ is inert, i.e. remains prime in $O_K$ ; 2) $p$ splits, i.e. has the form $p=\pi \bar {\pi}$ ; 3) $p=3$ is totally ramified, precisely $-3=(i\sqrt 3)^2$. So we can obviously write $\alpha = \epsilon .a.(i\sqrt 3)^n$, with $\epsilon \in U_K, a\in \mathbf Z, n=0,1$ and $N(\alpha)=3^n.a^2$. Taking conjugates gives $\bar \alpha=\bar\epsilon.a.(-i\sqrt 3)^n$, the condition $\bar \alpha=u.\alpha$ is equivalent to $\bar\epsilon=\pm u.\epsilon$, and we recover the solutions given by Tob Ernack.