# 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 $a\in F$

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 $F$.

• It’s the Quadratic Formula. – Lubin Sep 5 '17 at 0:37

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 $\alpha^2\in F$.
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.
• 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 $x+\frac a2$ – Lubin Sep 5 '17 at 0:36