Are all degree $2$ field extensions Galois?
I know that this is true over the rationals. But is it true in general?
Are all degree $2$ field extensions Galois?
I know that this is true over the rationals. But is it true in general?
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 $L$.
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.
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).