Determining Galois extension based on degree of extension

[PsychonautQQ]

Homework Statement

If Char(K) = 0 and [L:K]=2, is L:K a galois extension?

Homework Equations

The Attempt at a Solution

My gut is saying yes because if [L:K]=2 then it seems that any polynomial in K[x] with a root in L should split in L[x]. Something about how some hypothetical minimal polynomial of some element m call it m(x) where L[x] is isomorphic to the quotient field K[x]/m(x), then K(u) is isomorphic to L where given the conditions that [L:K] =2 then deg(m(x)) = 2 and so if L has one of the roots of m(x) then that means that m(x) factors linearly because it only has degree 2 and thus all the roots are in L thus the extension is normal. Am I leaving anything important out?

 

[fresh_42]

I've thought along the same lines. More formal we can choose an element $u \in L-K$. Then $\{1,u\}$ is $K-$linear independent and $\{1,u,u^2\}$ is not. This means we can write $u^2=\alpha \cdot 1 + \beta \cdot u$ which means $u$ is a root of $x^2- \beta x - \alpha$.
Now we can write down both solutions $u$ and $v$ and see that $v \in K(u)$, i.e. our minimal polynomial splits and is separable. (Why?)

 

[PsychonautQQ]

I've thought along the same lines. More formal we can choose an element $u \in L-K$. Then $\{1,u\}$ is $K-$linear independent and $\{1,u,u^2\}$ is not. This means we can write $u^2=\alpha \cdot 1 + \beta \cdot u$ which means $u$ is a root of $x^2- \beta x - \alpha$.
Now we can write down both solutions $u$ and $v$ and see that $v \in K(u)$, i.e. our minimal polynomial splits and is separable. (Why?)

Our minimal polynomial is separable because K is given to have Characteristic 0, and our minimal polynomial splits because both of it's roots are in K(u). Thanks!

 

[fresh_42]

Our minimal polynomial is separable because K is given to have Characteristic 0, and our minimal polynomial splits because both of it's roots are in K(u). Thanks!

One can also see it directly: If $u$ were a double root, then $u=-\frac{\beta}{2} \in K$ which we ruled out. And with $u=-\frac{\beta}{2}-\sqrt{sth.}\, , \,v=-\frac{\beta}{2}+\sqrt{sth.}$ we get $v=-u-\beta \in K(u)$.