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

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!

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 $$1, \sqrt{-3}, \frac{1 \pm \sqrt{-3}}{2}, \frac{3 \pm \sqrt{-3}}{2}$$.

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.