# 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. Aug 12, 2017 at 16:08
• @reuns In characteristic 2, is it possible to find an element which is not a square? Oct 21, 2018 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 $$p(t)=(t-\alpha)q(t)$$, forcing $$q(t)$$ to be degree $$1$$. If the characteristic of $$K$$ is not equal to $$2$$ then $$p^\prime(t) = 2t + \cdots \neq 0$$, 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 $$p(x)=x^2-t \in \mathbb{F}_2(t)[x]$$. 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 $$\operatorname{char}(F) \neq 2$$ 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 exists a Galois closure of $$K/F$$, that is, there exists 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 be proven 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 $$K$$ 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.

• hi @Nocturne in the first line of the proof about normal closures it should read "Let .... be a basis of K considered ... " and not "of E" as it says. The thing won't let me edit it so please could you or someone else fix it? Aug 11, 2020 at 14:53