# 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.

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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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Think of the roots of a quadratic polynomial $aX^2 + bX + c$. What happens to the denominator in char $2$.
@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 $2$ –  Andrew Jan 19 '13 at 21:39
oops should be $- b$ –  Andrew 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