Let $f=x^3+x^2-2x-1\in \Bbb Q[x]$. Let $\alpha\in \Bbb R$ be a zero of $f$.
$\Bbb Q[x]/(f)$ is isomorphic to the subring $R=\{c_0+c_1\alpha + c_2\alpha^2:c_i\in \Bbb R\}$ of $\Bbb R$.
The map $\Bbb Q[x]\to R$ given by $g\mapsto g(\alpha)$ decends to a map on $\Bbb Q[x]/(f)$ because $f(\alpha)=0$. It is obviously surjective and $(f)$ is contained in the kernel.
Question Why is the kernel of this map precisely $(f)$?
For what it's worth I formerly showed that $f$ is irreducible.
Theorem: $F[x]$ is an Euclidean domain for every field $F$. the kernel of the map namely K is an ideal of the ring $Q[x]$ .As $Q[x]$ is a PID , $K=(p(x))$ , for any monic polynomial $p(x)$ of lowest degree contained in $K$. By rational root theorem $f(x)$ is irreducible in $Q[x]$. Also it is of lowesr degree monic irreducible polynomial in $k$. Because if $p(x)$ is of lower degree and contained in $K$ then it would divide $f(x)$ leaving a monic factor of degree one , disproving that it is irreducibility .Hence $K$ is generated by $f$.