1. ## galois Theory

Determine, if there are rational numbers c,d so that the polynomails x^4 -4x-1 and x^4+ax^2+b have splitting field over Q(rational numbers), If yes find a and b.

2. it should be same splitting field

soory I forgor to write

3. Originally Posted by peteryellow
Determine, if there are rational numbers c,d so that the polynomails x^4 -4x-1 and x^4+ax^2+b have splitting field over Q(rational numbers), If yes find a and b.
I think it is impossible. This is just an idea that I have that might work. We make a trivial observation, if they have the same splitting field then the polynomials definitely have the same Galois group. The polynomial $x^4 - 4x - 1$ happens to have Galous group equal to $S_4$ (this can be shown using what your favorite method is). Thus, we require splitting field for $x^4 + ax^2 + b$ to be a degree $24$ extension. Now, the equation $x^4 + ax^2 + b=0$ solves into $x = \pm \sqrt{ \frac{-a+r}{2}}, \pm \sqrt{ \frac{-a-r}{2}}$ where $r = \sqrt{a^2 - 4b}$. Therefore, the splitting field is $K=\mathbb{Q} \left( \sqrt{ \frac{-a+r}{2}}, \sqrt{\frac{-a+r}{2}} \right)$. We require that $[K:\mathbb{Q}]=24$. However, $[K:\mathbb{Q}] \leq \left[\mathbb{Q}\left( \sqrt{\frac{-a+r}{2}} \right):\mathbb{Q} \right]\left[ \mathbb{Q}\left( \sqrt{\frac{-a-r}{2}} \right) : \mathbb{Q} \right]$. But $\left[\mathbb{Q}\left( \sqrt{\frac{-a+r}{2}} \right):\mathbb{Q} \right] \leq 4$ because $\sqrt{\frac{-a+r}{2}}$ solves $r^2 = (2x^2 + a)^2$. Likewise, $\left[\mathbb{Q}\left( \sqrt{\frac{-a+r}{2}} \right):\mathbb{Q} \right] \leq 4$. Therefore, $[K:\mathbb{Q}] \leq 16 < 24$ and so it seems it is impossible.

4. I think it is a good exercise to actually find the conditions on $a,b$ that determine what Galois group $x^4 + ax^2 + b$ actually is.
This should not be too bad since it appears as an exercise problem in my book.

5. which book are you using?

6. Originally Posted by peteryellow
which book are you using?
This one.