Let $A$ be an Artinian ring whose nilradical is zero. Show that $A$ has only finitely many ideals. Let $A$ be an Artinian ring whose nilradical is zero. Show that $A$ has only finitely many ideals.
We know for Artianian ring nilradical=jacobson radical= $0$.
Again it has finitely many maximal ideal.
Now what conclusion can be drawn from here? Please help
 A: Let $M_1, \dotsc, M_n$ be the finitely many maximal ideals of $A$. By the Chinese remainder theorem, we have that 
$$A/\operatorname{nil}(R) \cong A/M_1 \times \dotsb \times A/M_n$$ is a finite direct product of fields, which has only finitely many ideals.
So $A/\operatorname{nil}(R)$ has only finitely many ideals for any artinian ring.
A: For the sake of completeness let me show that an artinian ring with non-zero nilradical can very well have infinitely many ideals.
The simplest such example is $Q[x,y]:=\mathbb Q[X,Y]/\langle X^2,XY,Y^2\rangle $.
This $3$-dimensional $\mathbb Q$-algebra has the infinitely many $1$-dimensional subspaces $$ I_\lambda=\mathbb Q(x+\lambda y) \;(\lambda\in \mathbb Q)$$ as distinct ideals.
A: If the nilradical $R$ of $A$ is zero, then $A$ is a semisimple ring (in general $A/R$ is semisimple for any Artinian ring $A$).
Now use the Artin-Wedderburn structure theorem for semisimple rings, i.e. $A$ is a finite product of matrix rings $M_{n_i}(D_i)$ over skew fields $D_i$
$$A = M_{n_1}(D_1) \times ... \times M_{n_s}(D_s)$$
This has only finitely many ideals.
I guess that there is a much simpler proof without Artin-Wedderburn though...
