$\bar{\mathbb{Z}}\cap \mathbb{Q}\left[\sqrt{-3}\right] = \mathbb{Z}\left[\omega\right]$

Question

How do you go about proving $\bar{\mathbb{Z}}\cap \mathbb{Q}\left[\sqrt{-3}\right] = \mathbb{Z}\left[\omega\right]$, where $\omega$ is $\frac{-1+\sqrt{-3}}{2}$? I have tried to approach it number theoretically, but no luck so far. I have so far only proven that $\bar{\mathbb{Z}}\cap \mathbb{Q}\left[\sqrt{-3}\right]$ is a ring.

Answer 1

One of the containments is easy, since $\mathbb Z [\omega]$ is a subring of both $\bar{\mathbb Z}$ and $\mathbb Q [\sqrt {-3}]$. So you want to prove the other identity.

Now, an element of $\bar{\mathbb Z}$ is a number satisfying a monic polynomial equation, i.e. one of the form (where the $a_i$ are integer coefficients) $$\alpha^n+ a_{n-1}\alpha^{n-1} + \cdots +a_1\alpha + a_0=0.$$

And an element of $\mathbb Q[\sqrt {-3}]$ is something of the form $p+q\sqrt{-3}$ where $p,q$ are rational numbers. Now it is easy to calculate its minimal polynomial to be (assuming $ q \neq 0$) $$\alpha^2-2p\alpha + p^2+3q^2=0.$$

The problem is now to determine conditions on $p,q$ such that $2p$ and $p^2+3q^2$ are both integers. The first condition is easy: $p$ must be an element of $\frac 12 \mathbb Z$. The other condition is trickier. It translates into the condition that $$\frac {a^2}{4}+\frac{3b^2}{c^2} \in \mathbb Z$$ where we can assume that $b$ and $c$ have no common factors. We do two cases. If $2 \mid a$, then the left term is already an integer. So the condition is that $c^2 \mid 3b^2$. But $3$ is not a square, so we must have $c \mid b$, but they are relatively prime, so $c= 1$.

Now assume $2 \not \mid a$. Then we must have $4c^2 \mid (ac)^2+12b^2$. This implies that $c^2 \mid 12$, and again, $3$ is not a square, so $c \mid 2$, and in fact $c=2$.

In conclusion, an element of $\bar {\mathbb{Z}} \cap \mathbb Q [\sqrt {-3}]$ must be either of the form $a+b\sqrt{-3}$ or $\frac{a}{2}+\frac{b}{2}\sqrt{-3}$, where $a,b$ are integers. Now it is easy to check that both of these is contained in the right-hand side.