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Let $K \subset L$ be an extension of degree 2. If $\operatorname{char}(K)=2$ then there exists $a \in K$ such that $L$ is the splitting field over $K$ of a polynomial of the form $X^2-a$ or $X^2-X-a$.

My attempt, Since the extension is of degree 2, $L=K(\alpha)$ and the minimal polynomial of $\alpha$ over $K$ has the form $X^2+bX+c$ Now if $b=0$ the result follows, how to show the other part?

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You know $\alpha$ is a root of $X^{2} + b X + c$, with $b \ne 0$. Write $\alpha = d \gamma$, with $d \in K$, and determine a monic polynomial $f$ in $K[X]$ of degree $2$ of which $\gamma$ is a root. Choosing an appropriate $d$, can you make the coefficient of $X$ in $f$ equal $1$?

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what is $\gamma$? can you please elaborate it a little more. –  Mathematician Feb 11 '13 at 12:37
    
Simply, choose any $d \in K^{\star}$, and take $\gamma = d^{-1} \alpha$. Now substitute $\alpha = d \gamma$ in $X^2 + b X + c$. –  Andreas Caranti Feb 11 '13 at 12:38
    
Well, I am sorry but i just don't understand why $\alpha=d \gamma$ and if it is a root then the monic polynomial for which $\gamma$ is a root is $X^2+bd^{-1}X+cd^{-1}$ –  Mathematician Feb 11 '13 at 12:45
    
@WaqasAliAzhar You're there! Just choose $d = -b^{-1}$, write $a = - cd^{-1} = c b$, and note that $L = K(\alpha) = K(\gamma)$. –  Andreas Caranti Feb 11 '13 at 12:49
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@WaqasAliAzhar, it's just a substitution, to get from the polynomial $X^2 + bX + c$, with $b \ne 0$, to a polynomial of the required form $X^2 - X - a$. You have probably seen similar substitutions in the form $X \mapsto u X + v$ to get from one polynomial to another. –  Andreas Caranti Feb 11 '13 at 12:55
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