Find the Galois group of $x^3-2x-1$ over $\mathbb{Q}$ and over $\mathbb{Q}(\sqrt{5})$
$x^3-2x-1$ has $-1$ as root, so $(x^3-2x-1)/(x+1) = x^2-x-1$ which has $x = \frac{1}{2}\pm \frac{\sqrt{5}}{2}$ as roots. I must find the splitting fields of these roots over $\mathbb{Q}$ and over $\mathbb{Q}(\sqrt{5})$.
$\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)$ is the splitting field of the polynomial over $\mathbb{Q}$. We have $3$ roots and a polynomial of degree $3$ that contains these roots, so $\left[\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right):\mathbb{Q}\right] = 3$ and $\left|Gal\left(\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)/\mathbb{Q}\right)\right| = 3$, so $Gal\left(\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)/\mathbb{Q}\right) = \{p_1, p_2, p_3\}$, where $p_1$ is the identity automorphism, but what are $p_2$ and $p_3$?
Now, for $Gal\left(\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)/\mathbb{Q}(\sqrt{5})\right)$, there is only one root and its polynomial is $x+1$, so the degree is one and $\left|Gal\left(\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)/\mathbb{Q}(\sqrt{5})\right)\right| = 1$, so the Galois group is the trivial one. Is this right?