Degree 2 Field extensions 
Are all degree $2$ field extensions Galois?   

I know that this is true over the rationals. But is it true in general?
 A: It is true over any field of characteristic different from $2$. 
If $F$ has characteristic different from $2$, and $K$ is of degree $2$ over $F$, then let $\alpha\in K$, $\alpha\notin F$. Then $K=F(\alpha)$, since $2=[K:F]=[K:F(\alpha)][F(\alpha):F]$, and $[F(\alpha):F]\gt 1$. Let $p(x)$ be the minimal polynomial of $\alpha$ over $K$. Then $p(x) = x^2 + rx+t$ for some $r,t\in F$, and $\alpha = \frac{-r+\sqrt{r^2-4t}}{2}$ or $\alpha=\frac{-r-\sqrt{r^2+4t}}{2}$ (since the characteristic is not $2$). Moreover, the polynomial is irreducible and separable, and $\sqrt{r^2-4t}\notin F$. So $K = F(\sqrt{r^2-4t})$ and $K$ is a splitting field of an irreducible separable polynomial (namely, $x^2 - (r^2-4t)$), hence is Galois over $F$. 
If the characteristic is $2$, then the result is true for perfect fields, but not in general, as the examples by Zev Chonoles and Giovanni De Gaetano show. 
A: No, It's not true in general. 
Here is a counterexample. Let $F$ a field of characteristic 2, and consider the field $k$ of rational functions over $F$, i.e. $k=F(T)$. Now consider the extension of degree 2 obtained adding the roots of the polynomial $x^2-T=0$. This extension is not separable (remember that $F$ so $k$ has characheristic 2) hence not a Galois extension (this is a result of Artin).
A: No - for example, let $K=\mathbb{F}_2(T)$, and let $L=\mathbb{F}_2(\sqrt{T})$. We have that $[L:K]=2$, but the extension $L/K$ is not separable, and therefore not Galois.
However, $[L:K]=2$ implies $L/K$ is Galois when $K$ (and hence $L$) is of any characteristic other than 2. This is because we can take the minimal polynomial $f$ of a primitive element $\alpha\in L$ (i.e., an element such that $L=K(\alpha)$), which will be irreducible of degree 2, and use the quadratic formula (which works because the characteristic isn't 2) to show that the other root of $f$ must be in $L$.
