I have to prove the following: If $R$ is a finite-dimensional algebra over a field $F$, then $J(R)$ is nilpotent.

Because $R$ is in particular a finite dimensional vector space, there is a composition series $0=R_0\subset R_1\subset\dots\subset R_k=R$ of VECTOR SPACES. Do I know that this is a composition series of $R$-modules? If yes, why? Or how can I find some?

If that is the case, I would proceed as follows: $J(R)R_n/R_{n-1}\subseteq R_n/R_{n-1}$. Now, I would like to use, that the quotient is simple, so if the above is a strict inclusion (how to show that?), then $J(R)R_n/R_{n-1}=0$. Now, by induction it is easy to see that $J(R)^k=\{0\}$.

• Commutative version is here although I think most of the logic has noncommutative translation. Commented Aug 3, 2018 at 13:38

We can begin by saying $J=J(R)$ is finitely generated. The descending sequence of $J\supseteq J^2\supseteq \ldots$ must stabilize eventually, so at some point we have $J^n=J^{n+1}$.

But the noncommutative formulation of Nakayama's Lemma says that if $M$ is a finitely generated right $R$ module $I$ is an ideal contained in $J(R)$ and $MI=M$, then $M=\{0\}$.

Applying that to this case, we have that $J^nJ=J^n$, so $J^n=\{0\}$.

I would like to use, that the quotient is simple, so if the above is a strict inclusion (how to show that?)

You have difficulties right away because it is unclear what $J(R)R_n/R_{n-1}$ means. It is not obvious that there should be a module structure on the vector space $R_n/R_{n-1}$.

But you do not have to use only vector spaces: a finite dimensional algebra is also right and left Artinian, so you could consider $R_n$'s to be right $R$ modules, and that $R_n/R_{n-1}$ is a simple right $R$ module.
At any rate, $R_{n-1}/R_{n-1}=J(R)(R_n/R_{n-1})\subsetneq R_n/R_{n-1}$ is a proper containment because $J(R)$ annihilates simple $R$ modules, including $R_n/R_{n-1}$.

• I have a question to your solution: I understand that $J(R)$ is finitely generated as a vecor space and I know that it is a right ideal, i.e. a right module over $R$, so also $J(R)^n$ is. For Nakayama's Lemma you need that $J(R)^n$ is finitely generated as an $R$-module, how do I know that? Commented Aug 3, 2018 at 14:05
• And for your improvement of my try: I know that I need the $R_n$'s to be modules, but how do I know that such a chain exist? Commented Aug 3, 2018 at 14:08
• @mathstackuser Because a finite dimensional algebra is Artinian and Noetherian, and such a module always has a composition series. You can even simply argue that a composition series of $R$ ideals must exist because of the dimensionality directly. Commented Aug 3, 2018 at 14:38
• @mathstackuser You know that all submodules of $R$ are finitely generated because of finite dimensionality. Suppose you had a right ideal that wasn't finitely generated, and construct an infinitely ascending chain of f.g. right ideals inside of it. Those are all subspaces, so they'd have to have strictly increasing dimensions. Contradiction. Commented Aug 3, 2018 at 14:40
• I understand that a finite dimensional algebra is Artinian and Noetherian over the FIELD $F$ over which it is defined, but you want it Artinian and Noetherian as an $R$-module right? Commented Aug 3, 2018 at 14:41