For $m\geq2$, is $\Bbb{F}[x]/(p(x)^m)$ a field?

Question

Suppose $\Bbb F$ is a field and $p(x)$ and $p(x)$ are irreducible in $\Bbb F[x]$. Then $\Bbb F[x]/(p(x))$ is a field.

On the other hand $\Bbb F[x]/(p(x)^mq(x)^n)$ is not a field since $$\Bbb F[x]/(p(x)^mq(x)^n)\cong \Bbb F[x]/(p(x)^m)\times \Bbb F[x]/(q(x)^n),$$ and even if both the rings on the right are fields, their direct product is always not a field.

But what about $\Bbb F[x]/(p(x)^m)$ ($m\ge2$)? Is it always/never/sometimes a field?

Note that if it is never a field then (guess) $\Bbb F[x]/(f(x))$ is a field iff $f(x)$ is irreducible in $\Bbb F[x]$ and it should (guess) not be always a field or the theorem in textbook will talk about $\Bbb F[x]/(p(x)^m)$ instead $\Bbb F[x]/(p(x))$

Answer 1

If $f,g\in\Bbb{F}[x]$ are nonconstant then $f,g\in\Bbb{F}[x]/(fg)$ are nonzero with $fg=0$.