## Splitting field of f(r)=r^3+r^2-2r-1 over Q

$\mathbf{Problem:}$ Let $E=\mathbb{Q}(r)$ where $r^3+r^2-2r-1=0$. Verify that $r'=r^2-2$ is also a root of $x^3+x^2-2x-1=0$. Determined $\text{Gal }E/\mathbb{Q}$. Show that $E$ is normal over $\mathbb{Q}$.
$\mathbf{Proof:}$ $r^3+r^2-2r-1=0$ means $(r+1)(r^2-2)=-1$. So we only need to prove $\displaystyle \frac{-1}{r+1}$ is another root of $f(x)=x^3+x^2-2x-1=0$.
Let $y=r+1$, $f(r)=0$ means $y^3-2y^2-y+1=0$.
Let $\displaystyle z=\frac{-1}{y}$, then $\displaystyle\frac {1}{z^3}(z^3+z^2-2z-1)=0$, obviously $z\neq 0$, thus $\displaystyle z=\frac{-1}{r+1}$ is a root of $f(x)=0$.
$f(x)$ is monic and has no root in $\mathbb{Z}$, so it is irreducible in $\mathbb{Q}$. $f(x)$ is minimal polynomial of $r$. So $[E:\mathbb{Q}]=3$. $\text{Gal } E/\mathbb{Q}$ must be the cyclic group of order 3. By theorem 4.7 chapter 4 in Jacoboson’s book, $E$ is normal over $\mathbb{Q}$.
$\mathbf{Remark:}$ Jacobson, Algebra I. p243.