# Extensions of degree two are Galois Extensions.

This question from Allan Clark's "Elements of Abstract Algebra"

Show that an extension of degree 2 is Galois except possibly when the characteristic is 2. What is the case when the characteristic is 2?

Tips are helpful, a solution is ideal.

Thanks.

• (in characteristic $2$) If $a$ is not a square in $F$ then $x^2-a = (x+\sqrt{a})^2$ is irreducible and $F(\sqrt{a})/F$ is inseparable. Now $\mathbb{F}_4 = \mathbb{F}_2[x]/(x^2+x+1) = \mathbb{F}_2(\zeta_3)$ is Galois of degree $2$, so some degree $2$ extensions are Galois, some other are not. – reuns Aug 12 '17 at 16:08
• @reuns In characteristic 2, is it possible to find an element which is not a square? – roi_saumon Oct 21 '18 at 14:28

Let $L/K$ be a field extension of degree $2$. If $\alpha \in L \setminus K$ then $p(t)=\min_K(\alpha,t)$ has degree $2$. In particular $p(t)$ must split over $L$ since $p(t)=(x-\alpha)q(t)$, forcing $q(t)$ to be degree $1$. If the characteristic of $K$ is not equal to $2$ then $p^\prime(t) \neq 0$, so $\alpha$ is separable over $L$. Thereby $L/K$ is Galois.

For a counterexample in the case of characteristic $2$ consider the splitting field of $p(x)=x^2-t \in \mathbb F_2(t)$. It's not hard to see that $p(x)=(x+\sqrt{t})^2$ so the extension is purely inseparable and not Galois.

You can also use Galois theory to prove the statement. Suppose $K/F$ is an extension of degree $2$. In particular, it is finite and $char(F) \neq 2$ implies that it is separable (every $\alpha \in K/F$ has minimal polynomial of degree $2$ whose derivative is non-zero).

We know that if $K/F$ is finite and separable, there exist a Galois closure of $K/F$, that is, a field $E$ containing $K$ such that $E/F$ is Galois (see below for a proof of this theorem).

The field inclusions I will use are $F \subseteq K \subseteq E$.

Let $E$ be a Galois closure of $K$ over $F$ and let $H \le G=Gal(E/F)$ be the subgroup corresponding to $K$ (the group that fixes $K$ pointwise). By the fundamental theorem of Galois theory, we know that the index of $H$ in $G$, $[G:H]$, equals the degree of the extension $K/F$, i.e. the degree of the fixed field of $H$ over the base field.

But, $[K:F]=2$ by hypothesis, so our $H$, being of index $2$ is normal in $G$. Again, Galois theory tells us that $K/F$ is Galois.

It remains to prove that for every finite and separable extension $K/F$ there exist a Galois closure of $K$ over $F$.

Let $\alpha_1, \alpha_2, ... , \alpha_n$ be a basis of $E$ considered as a vector space over $F$ and let $p_i(X) \in F[X]$ be the minimal polynomial of $\alpha_i$ over $F$, $1\le i \le n$. By hypothesis, $K/F$ is separable, so the minimal polynomials $p_i(X)$ are all separable. If we denote by $K_i$ the splitting field of $p_i(X)$ over $F$, then $K_i/F$ will be a Galois extension (as the splitting field of a separable polynomial). Since every $K_i/F$ is Galois, the composite $K_1K_2 \cdots K_n/F$ is also Galois which contains $K$ and proves the existence. Note that by taking the intersection of all Galois extensions of $F$ containing $K$, we can obtain a minimal Galois extension $E_{min}$ in the sense that every Galois extension of $F$ containing $K$ contains $E_{min}$. When we say the Galois closure of $K$ over $F$, we actually refer to this minimal Galois extension.

Of course, in your question, the Galois closure $E$ is actually $K$ itself and $H$ is the trivial subgroup, but we didn't know it a priori.