Let $U\subseteq\mathbb{A}^1$ be an open set in affine $1$-space. Why is $U$ necessarily a principal open set?
Since $U$ is the complement of a closed set, I write $U=\mathbb{A}^1\setminus V(S)$ for some $S\subseteq F[X]$. I want to show $U=P_f=\{a\in P:f(a)\neq 0\}$ for some polynomial $f$. I think $P_f=\mathbb{A}^1\setminus V(\langle f\rangle)$. Is $S=\langle f\rangle$, or at least $V(S)=V(\langle f\rangle)$ to get the conclusion?
Since $F[X]$ is a PID, and thus the ideal generated by $S$ is generated by a single polynomial $f$, you have that $V(S)=V(f)$, and so $U$ is principal open.