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 $[L:K]=8$


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

1) Basically, yes.

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.

  • $\begingroup$ thanks. I don't understand how my argument about adjoining roots twice isn't sufficient. can you explain why? what does the positive discriminant imply? $\endgroup$ – animalcroc Dec 11 '13 at 0:27

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