Show that if the ring $R$ with $1$ has a central nilpotent element then it is not semisimple.
I couldn't find a solution directly but I have a solution. Since any central nilpotent element is contained in the Jacobson Radical hence the Jacobson Radical is non-trivial. Now any semisimple ring's Jacobson radical must be trivial, we get a contradiction. Hence $R$ is not semisimple. Hence proved.
But can anyone give me a better proof of this theorem? If there is any!
On
I'll try to give another answer to your question using Wedderburn's theorem. Please let me know if it is correct.
If $R$ is semisimple then it is a product $$R=\prod_iM_{n_i}(D_i)$$ where $D_i$'s are division rings. Hence $$C(R)=\prod_iC(M_{n_i}(D_i))\cong \prod_i C(D_i) $$
Since $D_i's$ are division rings they can't have nilpotent elements hence the same is true for $C(R)$.
Your proof’s idea is fine, I you have that fact already.
If $x$ is central and nilpotent and nonzero, then $(x)$ is a nonzero nilpotent ideal, and such ideals are contained in the Jacobson radical. The Jacobson radical of a semisimple ring is trivial, so it is not semisimple.
Another way to see it is that for any $r\in R$, $xr$ is nilpotent, hence $1-xr$ is a unit. By a well known characterization of elements of the Jacobson radical, $x\in J(R)$. Or alternatively you can just notice why a nilpotent ideal must certainly annihilate every simple module.
As far as seeing why centrality is necessary to mention, just consider any matrix ring with side more than 1, over any field. It has nilpotent elements, but none of them are central.