Factor ring of polynomial $F[x]$ is a polynomial ring over a certain field $F$. $J$ is an ideal of $F$, $J = (f(x))$. I need to prove that if the polynomial $f(x)$ has a multiple root the factor ring $F[x]/J$ is not a field.
I am trying to show that $J$ is not a maximal ideal. But got stuck…
 A: Of course you got stuck : the result you are trying to prove is false!
For example if $F=\mathbb F_p(t)$ (where $p$ is aprime and $t$ is an indeterminate), then  the polynomial $x^p-t\in F[x]$ has multiple roots but $J=(x^p-t)$ is maximal and  $F[x]/(x^p-t)$ is a field.
Edit
At Bill's request, I'll flesh out my answer a little.
The condition that a polynomial $f(x)\in F[x]$ have a multiple root is a bit ambiguous: a multiple root where ?
The answer is: in any field $K$ in which the polynomial splits into degree one factors, for example in an algebraic  closure  $K=\bar F$of $F$.
A polynomial without multiple factors is then called separable.
We  have the pleasant result that $f(x)$ is separable if and only if the polynomial $f(x)$ and its formal derivative $f'(x)$ are relatively prime in $F[x]$, that is over the ground field $F$: a condition easy to test algorithmically.
An irreducible polynomial in characteristic zero is automatically separable, but the example above shows that in characteristic $p$ an irreducible polynomial like $x^p-t\in F[x]$ needn't be separable: its $p$  roots in an algebraic closure of $F$ coincide and are denoted $\sqrt [p]t$.
A: Hint: As Georges has pointed out the result is false in positive characteristic.  If $F$ has characteristic $0$ then show that the derivative $f'$ is not contained in $J$ and $\langle f, f'\rangle$ is a proper ideal in $F[x]$.
You'll be using the fact that a repeated root of $f$ is also a root of $f'$.
