If $K$ is an extension field of $\mathbb{Q}$ such that $[K:\mathbb{Q}]=2$, prove that $K=\mathbb{Q}(\sqrt{d})$ for some square free integer $d$ I think I have the later parts of this proof worked out pretty well but what's really stumping me is how to go from knowing $[K:\mathbb{Q}]=2$ to knowing that $K = \mathbb{Q}[x]/a_2x^2 + a_1x + a_0$.
I mean all I know from $[K:\mathbb{Q}]=2$ is that every element of $K$ can be written in the form $bk_1 + ck_2$ for $b,c\in \mathbb{Q}$.  As far as I can tell I don't yet have any theorems at my disposal that say if $[K:\mathbb{Q}]$ is finite than $K$ must be algebraic over $\mathbb{Q}$, or anything like that.  How do I go from this premise about $K$ as a 2-dimensional vector space over $\mathbb{Q}$ to knowing something about elements of $K$ as roots of polynomials in $\mathbb{Q}[x]$?  Thanks.
 A: This is easy enough once you know the primitive element theorem. Since any finite extension of  $\Bbb{Q}$ is separable it follows that $K = \Bbb{Q}(\alpha)$ for some $\alpha\notin \Bbb{Q}$ that has degree 2 over $\Bbb{Q}$. Suppose without loss of generality that the minimal polynomial of $\alpha$ over $\Bbb{Q}$ is 
$$x^2 + bx + c.$$
Then by the quadratic formula, $\alpha = \frac{-b\pm \sqrt{b^2 - 4c}}{2}.$ Now $b^2 - 4c$ cannot be a square in $\Bbb{Q}$ for then this would contradict $\alpha$ not being a rational number. Furthermore because $b$ is a  rational number we have 
$$\Bbb{Q}(\alpha)= \Bbb{Q}(\sqrt{b^2 -4c})$$
so  finally at the end we can say that $K = \Bbb{Q}(\sqrt{e})$ where $e = b^2 - 4c$ that is not a square in $\Bbb{Q}$. Now write $e = \alpha/\beta$ where $\alpha,\beta$ are integers with $\alpha,\beta \neq 0$. We may assume that $e$ is in its lowest terms otherwise we can cancel factors off top and bottom. Then 
$$\sqrt{e} = \sqrt{\alpha}/\sqrt{\beta} = \sqrt{\alpha\beta}/\beta$$
and consequently $K = \Bbb{Q}(\sqrt{e}) = \Bbb{Q}(\sqrt{\alpha\beta})$ because $\beta$ is an integer. Now if we write $d = \alpha\beta$, we have shown that $K = \Bbb{Q}(\sqrt{d})$, where $d$ is not the square of any integer.
$$\hspace{5in} \square$$
A: Let $\alpha \in K - \mathbb{Q}$.
Since $1 , α, α^2$ are linearly dependent over $\mathbb{Q}$, $aα^2 + bα + c = 0$, where $a, b, c \in \mathbb{Q}$ and not all of $a, b, c$ are zero.
By multiplying a suitable nonzero integer, we can assume $a, b, c \in \mathbb{Z}$.
If $a = 0$, we get $bα + c = 0$.
Since $b$ or $c$ is not zero, this can't happen.
Hence $a \neq 0$.
Hence $\mathbb{Q}(\alpha) = \mathbb{Q}((b^2 - 4ac)^{1/2})$.
Since 1$ , α$ are linearly independent over $\mathbb{Q}$, [$\mathbb{Q}(\alpha) : \mathbb{Q}$] = 2. Hence $K = \mathbb{Q}(\alpha)$ and we are done.
A: First, we know that if $K/\Bbb Q$ is quadratic.  Then $K/\Bbb Q$ simple extension. Let $K=\Bbb Q(\alpha).$ Let $f$ be minimal polynomial of $\alpha$. Then $K=\Bbb Q(D(f)).$
