# Splitting field of irreducible polynomial $f(x)=x^4 +ax^2+b$ over field with charactersitic not equal to 2

this is from Seth Warner's Classical Modern Algebra:

Let $K$ be a field whose characteristic is not $2$, let $f=x^4+ax^2+b$ be an irreducible polynomial over $K$, and let $L$ be a splitting field of $f$ over $K$. show:

1) $[L:K]$ is either $4$ or $8$ [hint: consider $g=x^2 +ax+b$]

I did not use the hint in my answer. What I did was factor $f$ into $(x-c_1)(x-c_2)(x+c_1)(x+c_2)$ where $c_1$ and $c_2$ are in terms of $a$ and $b$. I then concluded that if $c_1$ is in $K(c_2)$, $[L:K]=4$. Otherwise $[L:K]=8$ since you will need adjoin roots twice and $4\times2=8$. Is this argument valid?

2) If $K$ is a totally ordered field and $b<0$, then $[L:K]=8$

thanks

• @Carlos Eugenio Thompson. thanks for the editing. I'm new to this so every time you guys edit, I learn :). – animalcroc Dec 10 '13 at 23:23

The hint is essentially the same as observe that if $c_1$ is a root of $f$, then so is $-c_1$, as you did.
The part: if $c_1\notin K(c_2)$ then $[L:K]=8$ 'since you will need to adjoin roots twice' is not a good reasoning. What we need to show for this is that $c_1$ has order $2$ over $K(c_2)$. So, can you express $c_1$ by a root of a quadratic polynomial over $K(c_2)$?
2) If $b<0$ then the discriminant of $g$ (with roots ${c_1}^2$ and ${c_2}^2$) is positive.