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I have a problem and the problem is:

Let $F$ be a field and char $F$=2.Let $$E=F[x]/(x^2+bx+c)~~b,c\in F.$$

When $b$ and $c$ satisfy what condition, $E$ is a Galois extension with $[E:F]=2$.

My opinion:

Firstly $x^2+bx+c$ should an irreducible polynomial in $F[x]$.

Then in $E$ , $x^2+bx+c$ should have two different roots.

$E$ is splitting field over $F$ with a separable polynomial, so $E/F$ is a Galois extension with $[E:F]=2$

But how to find the specific conditions which $b$ and $c$ should satisfy?

Thanks!

fusheng
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    Separability is equivalent to $b\neq 0$. For irreducibility I see no better condition than $F$ having no element $y$ satisfying $y^2+y +cb^{-2}=0$. (Make the variable change $x=by$ in $x^2+bx+c$.) – Claudius Dec 06 '22 at 15:56
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    https://math.stackexchange.com/questions/290003 – Anne Bauval Dec 06 '22 at 16:05
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    See this old answer of mine for an exposition of how solvability of a quadratic is determined over a finite field of characteristic two. At heart it is the idea in @Claudius's comment. But in the case of finite fields we can describe the solvability condition using the trace function, which is often handy. This is also a case of Artin-Schreier theory in action. – Jyrki Lahtonen Dec 06 '22 at 16:27

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