Unique representations in quadratic integer rings $\mathbb{Z}[\omega]$

Question

Let $D$ be a squarefree integer and consider the ring $\mathbb{Z}[\omega]=\{a+b\omega: a,b \in \mathbb{Z}\}$ with

$$\omega = \begin{cases}\sqrt{D} \quad D \equiv_4 2,3\\ 1/2(1+\sqrt{D}) \quad D \equiv_4 1\end{cases}$$

I'm trying to show that every element in this ring has a unique representation of the form $a+b\omega$. For this, it sufficies to show that

$$a+b\omega = 0 \implies a = 0 = b$$

Here is my proof:

Suppose $b \neq 0$ and thus also $a \neq 0$. By dividing out common factors, we may assume that $(a,b) = 1$ and $b \geq 1$. Let's consider the case $D \equiv_4 2,3$.

We then see that $a= -b\sqrt{D}$ and hence $D = (a/b)^2$. It is not possible that $b = 1$, or this would imply that $D$ is not squarefree. Thus $b > 1$ and $a/b$ is a real fraction (we can't divide out the denominator because $a$ and $b$ have no common factors). Hence, $(a/b)^2$ is a real fraction (squaring does not introduce new factors that can be canceled out). This contradicts the assumption that $D$ is an integer.

Is this correct? It looks like I'm overcomplicating things.

Let's now look at the case that $D \equiv_4 1$. Then we have

$$2a= -b(1+ \sqrt{D}) = -b -b\sqrt{D}\implies \left(\frac{2a+b}{b}\right)^2 = D$$

and then I'm stuck. Any ideas?

Answer 1

More simply, in both cases $\,\omega\,$ is a rational root of a monic quadratic $\,f(x)\in\Bbb Z[x]\,$ so $\,\omega\in\Bbb Z\,$ by the Rational Root Test. This implies $\,\sqrt D\in\Bbb Z,\,$ contra hypothesis.

Remark $ $ Generally an algebraic number is defined to be an algebraic integer if it is a root of a polynomial $\,f(x)\in\Bbb Z[x]\,$ that is monic (lead coef $= 1).\,$ So applying RRT as above we deduce that

$$ \text{ a rational } r\, \text{ is an algebraic integer} \iff r\ \text{ is an integer}\qquad $$

See here for some motivation of the definition of an algebraic integer.

Answer 2

In either case, you have shown that $D$ is a rational square that is assumed to be an integer, so it is an integer square.

Answer 3

The key point is that $\omega$ is irrational (*). Once you know that, $a+b\omega=0$ implies $a=b=0$ because $b\ne0$ implies that $\omega$ is rational.

(*) This boils down to $\sqrt D$ being irrational, which follows from $D$ being a squarefree integer.