# Let $[L:K]=2$ and char $K\neq 2$ show that $L/K$ is a simple 2-radical extension.

I want to show that

$$[L:K]=2$$ and char $$K\neq 2$$ $$\Rightarrow$$ $$L/K$$ is a simple 2-radical extension.

I know that, since $$[L:K]$$ is prime, the extension is simple. I also concluded that, since $$[L:K]=[K(a):K]=[a:K]$$ for all $$a\in L/K$$, the minimal polynomial of $$a$$ has to have degree 2. But how do I proceed from here ? And how does char$$K\neq 2$$ help me ?

• $a$ is a root of a quadratic equations. You can solve quadratic equations by "completing the square" (except in characteristic two). – Lord Shark the Unknown Feb 3 at 6:59
• So if I'm correct I just have to write down the "abc-formular", square it and hence get a representation of $a^{2}$ with coefficients from $L$. This would then show that $a^{2}\in L$. The char$K\neq2$ is needed to gurantee that the denominator of the abc formular is not $0$. – Christian Singer Feb 3 at 7:11
• * I meant to say $K$ instead of $L$ – Christian Singer Feb 3 at 7:18

Choose any

$$a \in L \setminus K; \tag 1$$

then since

$$[L:K] = 2, \tag 2$$

it follows that

$$\{1, a \} \subset L \tag 3$$

forms a basis for $$L$$ over $$K$$; thus the set

$$\{1, a, a^2 \} \tag 4$$

is linearly dependent over $$K$$; hence we have

$$\sigma, \tau, \rho \in K, \tag 5$$

not all zero, such that

$$\sigma a^2 + \tau a + \rho = \sigma a^2 + \tau a + \rho 1 = 0; \tag 6$$

we note that

$$\sigma \ne 0, \tag 7$$

lest

$$\tau a + \rho = 0; \tag 8$$

but then

$$\tau \ne 0 \Longrightarrow a = -\dfrac{\rho}{\tau} \in K \Rightarrow \Leftarrow a \in L \setminus K, \tag 9$$

whereas

$$\tau = 0 \Longrightarrow \rho = 0 \Longrightarrow \sigma, \tau, \rho = 0, \tag{10}$$

contradicting our hypothesis that there is a non-zero element amongst $$\sigma$$, $$\tau$$, $$\rho$$; thus (7) binds and setting

$$\alpha = \dfrac{\tau}{\sigma}, \; \beta = \dfrac{\rho}{\sigma}, \tag{11}$$

we write (6) in the form

$$a^2 + \alpha a + \beta = 0, \tag{12}$$

that is, $$a$$ satisfies the monic quadratic polynomial

$$x^2 + \alpha x+ \beta \in K[x]; \tag{13}$$

(12) yields

$$a^2 + \alpha a = -\beta, \tag{14}$$

whence we have, since $$\text{char}(K) \ne 2$$,

$$\left (a + \dfrac{\alpha}{2} \right )^2 = a^2 + \alpha a + \dfrac{\alpha}{4} = \dfrac{\alpha}{4} - \beta \in K; \tag{15}$$

we thus see that

$$b = a + \dfrac{\alpha}{2} \in L \setminus K \tag{16}$$

with

$$b^2 \in K. \tag{17}$$

It is now manifestly evident that

$$L = K(a) = K \left (a + \dfrac{\alpha}{2} \right ) = K(b) \tag{18}$$

is a simple extension of $$K$$ (since it is generated by adjoining either of the single elements $$a$$ or $$b$$ to $$K$$), and by virtue of (17), is what our OP Christian Singer refers to a a "2-radical extension", since we may write

$$b = \sqrt { \dfrac{\alpha}{4} - \beta }. \tag{19}$$

• Thanks alot for the in depth answer! – Christian Singer Feb 3 at 20:04
• My pleasure sir, and thanks for the "acceptance"! – Robert Lewis Feb 3 at 20:05