Field extension with degree=2 Let $K:F$ be a field extensions with $|K:F|=2$
I want to show that $K=F(\alpha)$ for $\alpha \in K$ and $\alpha ^2 \in F$ given that char $F\neq2$
Then that when F=$\mathbb{Q}$ you can choose $\alpha$ with $\alpha ^2 \in \mathbb{Z}$
I have set $K=a+b\alpha$ with $a,b \in F$ but I don't know where to go from here 
 A: Let
$\beta \in K \setminus F; \tag 1$
then, since $[K:F] = 2$, there must exist a linear dependence over $F$ 'twixt
$1$, $\beta$, and $\beta^2$; that is, there are $a, b, c \in F$, not all zero, such that
$a \beta^2 + b\beta + c = 0; \tag 2$
if now $a = 0$, 
$b\beta + c = 0; \tag 3$
then $b = 0$ forces $c = 0$, so $a = b = c = 0$, contrary to our assumption;
if $b \ne 0$, then (3) implies $\beta \in F$, another contradiction; thus we may rule out the case $a = 0$.
With $a \ne 0$, (2) yields
$\beta^2 + d \beta + e = 0, \tag 4$
where $d = a^{-1}b$ and $e = a^{-1}c$; then we may write
$\beta^2 + d \beta = -e, \tag 5$
whence, with $\text{char}(F) \ne 2$,
$(\beta + \dfrac{d}{2})^2 = \beta^2 + d \beta+ \dfrac{d^2}{4} = \dfrac{d^2}{4} - e \in F; \tag 6$
we note that $4 = 2^2 \ne 0$ in $F$; otherwise $2^2 = 0 \Longrightarrow 2 = 0$, and $\text{char}(F) = 2$; now setting
$\alpha = \beta + \dfrac{d}{2}, \tag 7$
we see from (6) that $\alpha^2 \in F$; also,
$\alpha \in K \setminus F, \tag 8$
thus $[F(\alpha):F] = 2$, so 
$K = F(\alpha), \tag 9$
and we have produced the requisite $\alpha$.
If $F = \Bbb Q$, then 
$\alpha^2 = \dfrac{p}{q} \tag{10}$
with $p, q \in \Bbb Z$; note $q \ne 0$; then
$q \alpha^2 = p, \tag{11}$
so
$(q \alpha)^2 = q^2 \alpha^2 = pq \in \Bbb Z; \tag{12}$
finally,
$K = F(\alpha) = F(q\alpha). \tag{13}$
A: First remember that a finite field extension is algebraic. Then there exists $\alpha\in K$ with $\min(\alpha,F)\in F[x]$ a degree 2 polynomial. 
Now we write the minimal polynomial: $\min(\alpha,F) = x^2+bx+c$ with $b,c\in F$. We know that the roots of this polynomial are $\frac{-b+\sqrt{b^2-4c}}{2},\frac{-b-\sqrt{b^2-4c}}{2}$. 
The primitive element theorem guarantees the existence of a primitive element since we know the roots for $\alpha$'s minimal polynomial are different, which means it's separable, and $K|F$ is finite.
With either of these roots, since we know $b,c\in F$, we can see that the part which is not in $F$ is $\sqrt{b^2-4c}$. We can take this element as our primitive element for the extension. 
Therefore, $K=F(\sqrt{b^2-4c})$, the splitting field for the polynomial $x^2-(b^2-4c)$. 
