Extension field of degree 2 contains an element whose square is in the field Little question.
Let $F$ be a field with $\operatorname{char}(F) \neq 2$. Let $E$ be a field extension such that $[E:F]=2$
I need to show there exists $y \in E$ such that $y^2 \in F$ and $E=F[y]$.
Let $x \in E$ and not in $F$. If $[E:F]=2$, then $\deg(m_x(T))=2$ and so $m_x(T)=T^2+bT+c$ for some $b,c \in F$. So for all $x$ not in $F$: $x^2+bx+c=0$ (still for some $b(x),c(x) \in F$ )
The idea now is to set $y=x-\frac{b}{2}$ which
$\textbf{if} y \notin F$ leads to $y^2=\frac{b^2}{4}-c \in F$ and clearly $y$ satisfies $E=F[y]$
My questions is: why (if true) may I assume $y$ is not in $F$?
Thank you.
 A: Since $m_x(T)$ has degree $2$, $\{1,x\}$ is an $F$-basis for $F[x]$, that is, $F[x] = \{a+bx : a,b \in F\}$. So, an element of $F[x]$ is in $F$ if and only if can be written as $a+bx$ with $b=0$. In particular $y \notin F$.
A: Let's see . . .
$[E:F] = 2 \Longrightarrow F \subsetneq E \Longrightarrow \exists x \in E \setminus F; \tag 1$
$x \in E \setminus F \Longrightarrow F \subsetneq F(x) \subset E \Longrightarrow [E:F(x)][F(x):F] = [E:F] = 2 \Longrightarrow [F(x):F] \mid 2; \tag 2$
$[F(x):F] \mid 2 \Longrightarrow [[F(x):F] = 1] \vee [[F(x):F] = 2]; \tag 3$
$[F(x):F] = 1 \Longrightarrow x \in F \Rightarrow \Leftarrow x \in E \setminus F; \tag 4$
therefore,
$[F(x):F] = 2, \tag 5$
which implies that there exist
$a, b, c \in F, \text{not all zero}, \tag{6}$
with
$ax^2 + bx + c = 0; \tag 7$
we have
$a \ne 0, \tag 8$
lest
$bx + c = 0, \tag 9$
which prohibits
$b = 0, \tag{10}$
for this forces
$c = 0; \tag{11}$
but then
$a, b, c = 0, \tag{12}$
in contradiction to (6); we thus infer that (8) binds and hence we may write (7) in the form
$x^2 + \alpha x + \beta = 0, \tag{13}$
where
$\alpha = \dfrac{b}{a}, \beta = \dfrac{c}{a}; \tag{14}$
now, from (13) we find
$x^2 + \alpha x = -\beta, \tag{15}$
and since $\text{char} \; F \ne 2, \tag{16}$
$\dfrac{\alpha}{2} \in F, \tag{17}$
whence
$\left (x + \dfrac{\alpha}{2} \right)^2 = x^2 + \alpha x + \dfrac{\alpha^2}{4} = \dfrac{\alpha^2}{4} -\beta \in F; \tag{18}$
setting
$y = x + \dfrac{\alpha}{2} \in E \setminus F, \tag{19}$
we see that
$y^2 \in F, \tag{20}$
and that
$F(y) = F[y] = F[x] = F(x) = E, \tag{21}$
which binds by virtue of the fact that
$F[y] = F(y), F[x] = F(x) \tag{22}$
when $x$ and $y$ are algebraic over $F$, and that (19) implies
$F(y) = F(x). \tag{23}$
$OE\Delta$.
