A question regarding the definition of a quadratic field

Question

A quadratic field $F$ is a finite field extension of $\mathbb{Q}$ such that $F$ has degree $2$ over $\mathbb{Q}$.

Many sources claim that any quadratic field has the form $\mathbb{Q}(\sqrt{d})$, where $d$ is a squarefree integer and $d \neq 0,1$. I have two questions about this that I'd like to resolve :

• $\mathbb{Q}(\omega)$ has degree $2$ over $\mathbb{Q}$, where $\omega$ is a primitive cube root of unity. But, can this be written in the above form ? If not, why is that not considered a quadratic field ?
• Why do we have to require $d$ to be squarefree ? For example, $8$ is not squarefree, but if I'm not mistaken, $\mathbb{Q}(\sqrt{8})$ has degree $2$ over $\mathbb{Q}$, so it should still be considered a quadratic field.

Thanks!

Answer 1

• The primitive cube roots of $1$ are $-\frac12+\frac i2\sqrt3$ and $-\frac12-\frac i2\sqrt3$, therefore $\Bbb Q(\omega)=\Bbb Q\left(\sqrt{-3}\right)$.

• Having $d$ be squarefree is not a necessity: you could say that the extensions of degree $2$ of $\Bbb Q$ are the ones in the form $\Bbb Q(\sqrt{n})$, where $n$ is an integer which is not a square. However, one does not lose generality in writing $n=k^2d$, where $d$ is squarefree and $\ne1$, and then notice that $\Bbb Q\left(\sqrt{k^2d}\right)=\Bbb Q\left(\lvert k\rvert\sqrt{d}\right)=\Bbb Q\left(\sqrt d\right)$. For instance, if $n=8$, then $\Bbb Q\left(\sqrt 8\right)=\Bbb Q\left(2\sqrt2\right)=\Bbb Q\left(\sqrt2\right)$. It is also more convienient, because with the restriction of $d\ne 1$ and squarefree the map $d\mapsto \Bbb Q\left(\sqrt d\right)$ is injective.

Answer 2

Take a quadratic equation with rational coefficients but with no rational roots. Take the root $\alpha$ of such an equation, then the set $\{a+b\alpha\mid a,b\in\mathbf{Q}\}$ is a field (subfield of complex numbers) and is a quadratic field.

There are so many choices of such quadratic equations. Though they all yield quadratic fields for two such different equations the corresponding fields could be identical: for example the set of numbers $\{a + b(-7+3\sqrt{5})\mid a,b \in \mathbf{Q}\}$ is the same as the set $\{a + b\sqrt{5}\mid a,b \in \mathbf{Q}\} $

The first set is from the quadratic equation $x^2+14x-4=0$ whereas the second is from $x^2-5=0$.

A careful but routine analysis of these situation leads to the fact that $x^2-d=0$ for various squarefree $d$ covers all the quadratic fields.