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

I thought about this, but there are some gaps:

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 $J(R)^k=\{0\}$.

Algebra
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1 Answers1

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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 $R_n/R_{n-1}$.

rschwieb
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  • 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? – Algebra Aug 03 '18 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? – Algebra Aug 03 '18 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. – rschwieb Aug 03 '18 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. – rschwieb Aug 03 '18 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? – Algebra Aug 03 '18 at 14:41
  • @mathstackuser And it is. Every right ideal is also a right subspace. If chains of the latter things are bound by the ACC, chains of the former thing are bound by the ACC. Ditto for DCC. – rschwieb Aug 03 '18 at 14:44
  • I missed something in my original question. My case is when $R$ is just Artinian, not a finite dim algebra. So how do we know $J(R)$ is finitely generated? This should only be true if $J(R)$ is semi-simple right? – Lemon Apr 26 '21 at 02:04
  • @Hawk no. A right Artinian ring is right Noetherian, hence all its right ideals are fg. – rschwieb Apr 26 '21 at 02:58