Let \(F \geq K\) be an extension field of degree 2.
If \(K\) has characteristic not 2, show \(F\) is Galois over \(K\).
Give an example where \(F\) is Galois over \(K\) even though \(\operatorname{char} K = 2\).
Give an example where \(F\) is not Galois over \(K\).
Solution
1. If \(\operatorname{char}K\neq2\), every degree-\(2\) extension \(F/K\) is Galois.
Choose \[ \alpha\in F\setminus K. \] Since \([F:K]=2\), one has \[ F=K(\alpha), \] and the minimal polynomial of \(\alpha\) over \(K\) has degree \(2\). Write it as \[ m(T)=T^2+bT+c\in K[T]. \] Because \(m\) is irreducible, \(\alpha\notin K\).
The other root of \(m\) is \[ \beta=-b-\alpha, \] which also belongs to \(F\). Thus \(m\) splits completely over \(F\). Moreover \(m\) has distinct roots. Indeed, if it had a repeated root, that root would be \[ -\frac b2\in K \] because \(2\) is invertible in \(K\), contradicting irreducibility of \(m\). Hence \(m\) is separable.
Therefore \(F\) is the splitting field over \(K\) of the separable polynomial \(m\). So \(F/K\) is normal and separable, hence Galois.
2. There are quadratic Galois extensions in characteristic \(2\).
Take \[ K=\mathbb F_2 \] and \[ F=\mathbb F_4 =\mathbb F_2[\alpha], \] where \(\alpha\) satisfies \[ \alpha^2+\alpha+1=0. \] The polynomial \[ T^2+T+1 \] has no root in \(\mathbb F_2\), so it is irreducible and \[ [F:K]=2. \] Its derivative is \(1\), so it is separable, and its two roots are \[ \alpha \qquad\text{and}\qquad \alpha+1, \] both of which lie in \(F\). Thus \(F\) is its splitting field over \(K\), so \[ \mathbb F_4/\mathbb F_2 \] is Galois. Its nontrivial automorphism is the Frobenius map \[ x\longmapsto x^2. \]
3. There are quadratic extensions in characteristic \(2\) which are not Galois.
Let \(t\) be transcendental over \(\mathbb F_2\), set \[ K=\mathbb F_2(t), \] and let \[ F=K(u), \qquad u^2=t. \] First, \(t\) is not a square in \(K\). Indeed, if \[ t=\left(\frac{a(t)}{b(t)}\right)^2 \] for nonzero coprime polynomials \(a,b\in\mathbb F_2[t]\), then \[ a(t)^2=t\,b(t)^2. \] The exponent of the irreducible factor \(t\) on the left is even, whereas on the right it is odd, a contradiction. Thus \[ T^2-t \] has no root in \(K\) and is irreducible, so \[ [F:K]=2. \]
In characteristic \(2\), \[ \frac{d}{dT}(T^2-t)=0. \] In an algebraic closure, if \(u^2=t\), then \[ T^2-t=(T-u)^2. \] Hence the minimal polynomial of \(u\) is inseparable. Therefore \(F/K\) is not separable and consequently is not Galois.