Is the following statement true/false ?
Let $I = \{ f(x) \in \mathbb{Z}[x]: f(1)=f(-1)=0\}$. Then $I$ is a principal ideal of $\mathbb{Z}[x]$.
My attempt: I thinks this is false because I is generated by $ -1$ and $+1$ and we know that for principal ideal it must be generated by single element.
Actually, $I$ is generated by $x^2-1\in \Bbb Z[x]$.
If $f\in I$ then $f$ vanishes at both $+1$ and $-1$. So $(x-1)$ as well as $(x+1)$ are factors of $f$. That's $f$ is divisible by $x^2-1$. Hence $I\subseteq \langle x^2-1\rangle$.
Also for any $f\in \langle x^2-1\rangle $ we have $g\in \Bbb Z[x]$ such that, $f(x)=(x^2-1)g(x)\implies f(\pm 1)=0\implies f\in I$. So $\langle x^2-1\rangle \subseteq I$.
Combining these $I=\langle x^2-1\rangle$.