Question about Galois Theory. Extension of a field of odd characteristic.

Question

Let $F$ be a field of characteristic $\ne 2$, and let $K$ be an extension of $F$ with $[K:F]=2$. Show that $K = F(\sqrt{a})$ for $a \in F$; that is, $K = F(\alpha)$ with $\alpha^{2}=a$. Moreover, show that $K$ is Galois over $F$.

My doubt is in the first part.

Take $\alpha \in K\setminus F$. Then $\lbrace 1, \alpha, \alpha^{2} \rbrace$ are LD over $F$, so $\alpha^{2} + p\alpha + q = 0$ with $p,q \in F$ ($1$ and $\alpha$ are LI over $F$, since $\alpha \not\in F$). Completing the squares, we have $$\left(\alpha + \frac{p}{2}\right)^{2} = \frac{p^{2}}{4} - q$$ because char$(F) \neq 2$. Let $\displaystyle a = \frac{p^{2}}{4} - q$, so $\displaystyle \sqrt{a}=\alpha + \frac{p}{2}$ and $\sqrt{a} \not\in F$.

Here is my doubt: I know $K/F$ is finite, then $K$ is finitely generated, but why $\sqrt{a} \not\in F$ ensure that $K=F(\sqrt{a})$?

Answer 1

$$F\subsetneq F[\sqrt{a}]\subseteq K$$ What can you say about the degrees of extensions of $F[\sqrt{a}]/F$, $K/F[\sqrt{a}]$ and $K/F$?

Answer 2

The quickest way to see this that I know is to note that $\sqrt a \notin F$ implies $1$ and $\sqrt a$ are linearly independent over $F$, since for $\alpha, \beta \in F$, not both zero,

$\alpha + \beta \sqrt a = 0 \Longrightarrow \sqrt a = -\dfrac{\alpha}{\beta} \in F \Rightarrow \Leftarrow \sqrt a \notin F; \tag 1$

therefore $F(\sqrt a)$ is a two-dimensional subspace of $K$; since $[K:F] = 2$, we are done.

Answer 3

We can also consider the following for $\mathbb{Q}$ to get the hang of it. To show: every quadratic extension of $\mathbb{Q}$ is of the shape $\mathbb{Q}(\sqrt{d}), d\in\mathbb{Z}$. Let $K/\mathbb{Q}$ be a quadratic extension of $\mathbb{Q}$, i.e. $[K:Q]=2$. First we will prove that $K$ is a simple extension; i.e., $K$ is an extension of $\mathbb{Q}$ that is generated by one single element, also called a primitive extension or simple extension. Take $\alpha\in K\backslash \mathbb{Q}$ and the tower of field extensions $\mathbb{Q}\subset\mathbb{Q}(\alpha)\subset K$. By the tower rule for field extensions we have $[K:\mathbb{Q}]=[K:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=2$, but $[\mathbb{Q}(\alpha):\mathbb{Q}]>1$ since deg $f_{\mathbb{Q}}^{\alpha} \neq 1$ because $\alpha \notin \mathbb{Q}$; this implies $[K:\mathbb{Q}(\alpha)]=1$ and $K=\mathbb{Q}(\alpha)$. Hence, we conclude that $K$ is an extension of $\mathbb{Q}$ by just one element (simple extension). We say that K is a quadratic extension iff $\deg(f_{\mathbb{Q}}^{\sqrt{d}})=2$. We take this minimum polynomial as the following: $f_{\mathbb{Q}}^{\sqrt{d}}=x^2+bx+c,\ b,c\in\mathbb{Q}$. The general solution in $\mathbb{Q}$ to such an equation is $x=\frac{-b\pm\sqrt{b^2-4c}}{2}$. If $\sqrt{b^2-4c}\in\mathbb{Q}$, then $x\in\mathbb{Q}$ and $\deg(f_{\mathbb{Q}}^{\sqrt{d}})=1$. $ \Rightarrow\Leftarrow$. If $\sqrt{b^2-4c}\notin\mathbb{Q}$, then $x\in\mathbb{Q}(\sqrt{d})$ with $d\in\mathbb{Z}$. We verify that $d$ is an element of $\mathbb{Z}$ by saying we could write $f_{\mathbb{Q}}^{\sqrt{d}}$ with integer coefficients; so $b^2-4c$ then becomes an element in $\mathbb{Z}$. In fact, $d=b^2-4c$ now. $\Box$