Let F be a field, and $p(x) \in F[x]$ an irreducible polynomial. Consider $n \in \mathbb{N}$ and let's define an ideal $J = F[x]p(x)^n$ in polynomials algebra $F[x]$, and now we shall define the quotient algebra $A = F[x]/J$.
Prove that $A$ is semi-simple iff $n=1$. (when we define $A$ to be semi-simple if it can be written as a sum of minimal left ideals - which is equivalent to the definition that Jacobson radical is 0).
Now, I believe that i have an argument to the first part: if $A$ is semi-simple means $A = \sum_{\alpha \in \Gamma}J_{\alpha}$ (where $J_{\alpha}$ are left minimal ideals) we can assume by contradiction that $n>1$. let's define $I_k:= ${$[q(x)] \in A: p(x)^k | q(x)$} for $k=1,...,n-1$ - which are all left ideals ($I_n = ${$[0]$} and $I_{n-1}$ is minimal and also $I_{n-1} \subseteq I_1$). now since we know that there exists $\alpha_0 \in \Gamma$ such that $p(x) \in J_{\alpha _ 0}$ we deduce $I_1 \subseteq J_{\alpha _0}$ which means contradiction to minimalism of $J_{\alpha _0}$ or $I_{n-1} = J_{\alpha_0}$ which means $I_{n-1} = I_{1}$ also a contradiction.
I'm not sure in my solution and also i seek for a proof in the other direction.
thanks ahead
In the reverse direction, if n $>$ 1, then there exists a non-zero polynomial, namely $p(x)$ that is a nilpotent. The nil radical has a non-zero element therefore, so does the Jacobson radical.