If $[K : \mathbb{Q}] = 2$, then $K = \mathbb{Q}(\sqrt{d})$ This isn't for homework, but I would just a like a hint please.  The question asks

If $K$ is an extension field of $\mathbb{Q}$ and $[K : \mathbb{Q}] = 2$ (the dimension of $K$ over $\mathbb{Q}$), then $K = \mathbb{Q}(\sqrt{d})$ for some square-free integer $d$.

I started by considering the linearly independent set $\{ 1 \}$ in $K$.  Now since $[K : \mathbb{Q}] = 2$, I can extend this set to a basis $\{ 1, v \}$ of $K$ over $\mathbb{Q}$, where $v \notin \text{span} \{ 1 \}$.  I see that $v^2 \in \text{span} \{ 1, v \}$, so that $v^2 = a_0 + a_1 v$ for some $a_0, a_1 \in \mathbb{Q}$.  Now $a_0 \neq 0$ for otherwise $a_1 = v$ and $v \in \text{span} \{ 1 \}$.  However, I'm not sure how to conclude that $v$ is a square-free integer.  I feel like I am on the right track (hopefully), and would greatly appreciate a hint please!

$v$ satisfies a quadratic equation $ax^2 + bx + c = 0$, where $a \ne 0, b, c \in \mathbb{Z}$.
Solve this equation and deduce $K = \mathbb{Q}(\sqrt D)$, where $D = b^2 -4ac$.
