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 $\displaystyle x^4 - 4x - 1$ happens to have Galous group equal to $\displaystyle S_4$ (this can be shown using what your favorite method is). Thus, we require splitting field for $\displaystyle x^4 + ax^2 + b$ to be a degree $\displaystyle 24$ extension. Now, the equation $\displaystyle x^4 + ax^2 + b=0$ solves into $\displaystyle x = \pm \sqrt{ \frac{-a+r}{2}}, \pm \sqrt{ \frac{-a-r}{2}}$ where $\displaystyle r = \sqrt{a^2 - 4b}$. Therefore, the splitting field is $\displaystyle K=\mathbb{Q} \left( \sqrt{ \frac{-a+r}{2}}, \sqrt{\frac{-a+r}{2}} \right)$. We require that $\displaystyle [K:\mathbb{Q}]=24$. However, $\displaystyle [K:\mathbb{Q}] \leq \left[\mathbb{Q}\left( \sqrt{\frac{-a+r}{2}} \right):\mathbb{Q} \right]\left[ \mathbb{Q}\left( \sqrt{\frac{-a-r}{2}} \right) : \mathbb{Q} \right]$. But $\displaystyle \left[\mathbb{Q}\left( \sqrt{\frac{-a+r}{2}} \right):\mathbb{Q} \right] \leq 4$ because $\displaystyle \sqrt{\frac{-a+r}{2}}$ solves $\displaystyle r^2 = (2x^2 + a)^2$. Likewise, $\displaystyle \left[\mathbb{Q}\left( \sqrt{\frac{-a+r}{2}} \right):\mathbb{Q} \right] \leq 4$. Therefore, $\displaystyle [K:\mathbb{Q}] \leq 16 < 24$ and so it seems it is impossible.