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 ?

  • 3
    $\begingroup$ $a$ is a root of a quadratic equations. You can solve quadratic equations by "completing the square" (except in characteristic two). $\endgroup$ – Lord Shark the Unknown Feb 3 at 6:59
  • $\begingroup$ 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 $0$. $\endgroup$ – Christian Singer Feb 3 at 7:11
  • $\begingroup$ * I meant to say $K$ instead of $L$ $\endgroup$ – 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

$\sigma \ne 0, \tag 7$


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

but then

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


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

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

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


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

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

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

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

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