I know that if $F[x]$ is a PID then $F$ is a field. Now $F[x]$ being a field implies that $F[x]$ is a PID, so $F$ is a field. Anyway, I tried to prove that $F$ is a field right away and the following argument came to mind:
Let $A[x]$ be a field, and let $p(x)= a + bx$, then there exists $q(x) = d x^n + ... + r$ where $b$ and $d$ are nonzero, such that $p(x)q(x) = 1.$ Carrying out the multiplication and equating the coefficient of $x^{n+1}$ to zero yields $bd = 0.$ Hence b and d are zero divisors and A is not even an integral domain.
What is wrong with this argument? And is $R[x]$ never a field?