Is it the case that every quotient ring ${F[x]}/{(p(x))}$ is a PID?
Let $F$ be a field and $F[x]$ the univariate polynomial ring over $F$. Is it the case that every quotient ring ${F[x]}/{(p(x))}$, where $p(x) \in F[x]$, is a PID (principal ideal domain)?
This is just a minor issue that's been bothering me for the last couple of hours. I'm aware that when the ideal $(p(x))$ is prime (i.e. $p(x)$ irreducible) then ${F[x]}/{(p(x))}$ is a PID.
If $p(x)$ isn't irreducible (and isn't the zero polynomial), then $F[x]/(p(x))$ isn't an integral domain, much less a PID. (And when $p(x)$ is irreducible, then $F[x]/(p(x))$ is a field.)