I am looking for a nice proof for the following lemma, which will later help prove that a ring is artinian if and only if it is noetherian and all prime ideals are maximal.
Let $A$ be an artinian ring. Then $\mathfrak{R}$ is nilpotent, id est, there exists an $n \in \mathbb{N}$ such that \begin{equation} \mathfrak{R}^n = (0) \; , \end{equation} where $\mathfrak{R} = \operatorname{Jac}A$ denotes the Jacobson radical of $A$.
We shall not use that artinian rings are noetherian, since, as stated above, this is what I want to prove after having shown this statement. My ansatz is the following:
Since $A$ is artinian, we have $\mathfrak{R}^{n+1} = \mathfrak{R}^n$ for some $n\in \mathbb{N}$. Let $x \in \mathfrak{R}^n$. I would like to find a finitely generated $A$-submodule (that is, a finitely generated ideal contained in $\mathfrak{R}^n$) $x \in I \unlhd \mathfrak{R}^n$ such that we have $\mathfrak{R}I = I$, and we could thus use Nakayama's lemma to obtain $I=\{0\}$ and therefore $x = 0$, which in conclusion would show $\mathfrak{R}^n = (0)$. Do we already have $\mathfrak{R}(x) = (x)$? I don't think so. Maybe we need to use Zorn's lemma to find such a submodule.
What I know is that $\operatorname{Max} A$, the set of all maximal ideals of $A$, is finite, and that we have $\operatorname{Spec}A = \operatorname{Max}A$, that is, every prime ideal is maximal. Also, for a short exact sequence of $A$-modules $0 \rightarrow M^\prime \rightarrow M \rightarrow M^{\prime\prime} \rightarrow 0$, $M $ is artinian if and only if $M^\prime$ and $M^{\prime\prime}$ are artinian. I don't think this will help, though. Rather, these facts together with this lemma are useful to show that a ring is artinian if and only if it is noetherian and every prime ideal is maximal.
Assume that $\newcommand{\R}{\mathfrak{R}}\R^{n+1}=\R^n\ne0$, and let $P = \R^n$. Then, there exist left ideals $I$ which are contained in $P$ for which $IP\neq 0$, since $P^2 = P \neq 0$.
There is (by the Artinian condition) a minimal left ideal with $IP \ne0$ and $I\subseteq P$. There must exist some $x\in I$ for which $Px\neq 0$ so $I = Ax$ is principal by minimality. Choose $z\in P$ for which $zx =x$. Then $(1-z)x=0$, but $1-z$ is invertible, so $x=0$ which is a contradiction.