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!