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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.


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up vote 6 down vote accepted

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.

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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.

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Think of the roots of a quadratic polynomial $aX^2 + bX + c$. What happens to the denominator in char $2$.

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I guess it's not immediately clear to me how failure of the quadratic formula relates to failure of separability. – JSchlather Jan 19 '13 at 21:20
@JacobSchlather My math is probably weak, but I was thinking you can map $X = - a + \sqrt {b^2 - 4ac} /2a$ to its conjugate, but you run into trouble in char $2$ – TheBirdistheWord Jan 19 '13 at 21:39
oops should be $- b$ – TheBirdistheWord Jan 19 '13 at 21:54
In the case that the quadratic formula works, that's the conjugate. But just because the quadratic formula fails doesn't ensure it's not going to work. In any perfect field of characteristic $2$ you're still fine such as any of the finite fields of order $2^n$. – JSchlather Jan 19 '13 at 21:57
Thanks for the insight. – TheBirdistheWord Jan 19 '13 at 22:09

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