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Let $F$ be a field of characteristic not $2$, and let $K$ be an extension of $F$ with $[K: F] = 2$. Show that $K = F (\sqrt{a})$ for some $a\in F$; that is, show that $K = F(\alpha)$ with $\alpha^2= a\in F$. Moreover, show that $K$ is Galois over $F$.

I do not know how to try the first part, could someone help me please? For the second part I am using the following theorem:

Let $K$ be a finite extension of $F$. Then $K/F$ is Galois if and only if $|Gal(K/F)| = [K:F]$

Because $Gal(K/F)=\left \{ id,\sigma \right \} $ where $\sigma$ is such that $\sigma(\sqrt{a})=-\sqrt{a}$, then $K$ is Galois over $F$.

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    $\begingroup$ It’s the Quadratic Formula. $\endgroup$
    – Lubin
    Commented Sep 5, 2017 at 0:37

2 Answers 2

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Pick $\beta \in K\setminus F$. Then $1,\beta,\beta^2$ are $F$-linear dependent, but we know that $1,\beta$ are not. Hence $\beta^2+p\beta+q=0$ with $p,q\in F$. Let $\alpha=\beta+\frac12\cdot p$ (this is where the characteristic is used: otherwise we are not allowed to divide by $2$). Now verify that $\alpha^2\in F$.

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  • $\begingroup$ I do not understand why you make that substitution, could you explain better? Why is the characteristic used there? $\endgroup$
    – user425181
    Commented Sep 4, 2017 at 22:04
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Consider the minimal polynomial $x^2+ax+b$, then translate by $a/2$. We are allowed to do so since the characteristic is not 2, then the minimal polynomial becomes $x^2-d$, hence $F=K(\sqrt{d})$ as wanted.

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    $\begingroup$ That is the minimal polynomial of what? $\endgroup$
    – user425181
    Commented Sep 4, 2017 at 22:07
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    $\begingroup$ Of your degree 2 extension. $\endgroup$
    – SC30
    Commented Sep 4, 2017 at 22:09
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    $\begingroup$ I don’t think that extensions have minimal polynomials; rather an element of the extension has a minimal polynomial. In this case, it’s minimal polynomial of $x+\frac a2$ $\endgroup$
    – Lubin
    Commented Sep 5, 2017 at 0:36

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