Consider the ring $R=\Bbb Z[x]/(x^3+x+1)$. It is easily seen that $(x^3+x+1)$ is a prime ideal, so that $R$ is an integral domain. Consider its field of fractions $F$. I am asked to describe the group of automorphisms of $F$ fixing $\Bbb Z$.
My attempt: Let $y=x+(x^3+x+1) \in R$. Since $R$ contains $\Bbb Z$ as a subring, $F$ must contain $\Bbb Q$. Thus $F\supset \Bbb Q(y)$. On the other hand, $\Bbb Q(y)$ is a field containing $\Bbb Z$ and $y$, so it must contain the quotient field of $\Bbb Z[y]=R$, and hence $\Bbb Q(y)=F$. Any automorphism of $F$ must send $1$ to $1$, and hence fixes $\Bbb Z$, so we don't need to think about the condition fixing $\Bbb Z$. Note that any automorphism $\sigma$ of $\Bbb Q(y)$ is determined by its value on $y$. Since $x^3+x+1$ is a polynomial in $\Bbb Z[x]$ having $y$ as a root, $\sigma(y)$ must be a root of $x^3+x+1$ also. Moreover since $[\Bbb Q(y):\Bbb Q]=\deg(x^3+x+1)=3$, the group of automorphisms of $\Bbb Q(y)$ has at most $3$ elements, and hence must be either $\{id\}, \Bbb Z_2$ or $\Bbb Z_3$, but I got stuck here.
Hint: $F \cong \mathbb Q[x]/(x^3+x+1) \cong \mathbb Q(\alpha) \subseteq \mathbb R$, where $\alpha$ is the real root of $x^3+x+1$.