Nilpotent Jacobson radical

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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 $0=R_0subset R_1subsetdotssubset R_k=R$ 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: $J(R)R_n/R_n-1subseteq R_n/R_n-1$. 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 $J(R)^k=0$.

asked Aug 3 at 13:31

mathstackuser

1497

Commutative version is here although I think most of the logic has noncommutative translation.
â€“Â rschwieb
Aug 3 at 13:38

add a commentÂ |Â

up vote
1
down vote

favorite

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:

asked Aug 3 at 13:31

mathstackuser

1497

Commutative version is here although I think most of the logic has noncommutative translation.
â€“Â rschwieb
Aug 3 at 13:38

add a commentÂ |Â

up vote
1
down vote

favorite

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:

asked Aug 3 at 13:31

mathstackuser

1497

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:

asked Aug 3 at 13:31

mathstackuser

1497

asked Aug 3 at 13:31

mathstackuser

1497

asked Aug 3 at 13:31

mathstackuser

1497

asked Aug 3 at 13:31

mathstackuser

1497

Commutative version is here although I think most of the logic has noncommutative translation.
â€“Â rschwieb
Aug 3 at 13:38

add a commentÂ |Â

Commutative version is here although I think most of the logic has noncommutative translation.
â€“Â rschwieb
Aug 3 at 13:38

Commutative version is here although I think most of the logic has noncommutative translation.
â€“Â rschwieb
Aug 3 at 13:38

add a commentÂ |Â

1 Answer
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We can begin by saying $J=J(R)$ is finitely generated. The descending sequence of $Jsupseteq J^2supseteq ldots$ 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 $R_n/R_n-1$.

edited Aug 3 at 13:49

answered Aug 3 at 13:43

rschwieb

99.5k1190225

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?
â€“Â mathstackuser
Aug 3 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?
â€“Â mathstackuser
Aug 3 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.
â€“Â rschwieb
Aug 3 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.
â€“Â rschwieb
Aug 3 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?
â€“Â mathstackuser
Aug 3 at 14:41

Â |Â
show 1 more comment

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1 Answer
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1 Answer
1

active

oldest

votes

up vote
1
down vote

We can begin by saying $J=J(R)$ is finitely generated. The descending sequence of $Jsupseteq J^2supseteq ldots$ 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 $R_n/R_n-1$.

edited Aug 3 at 13:49

answered Aug 3 at 13:43

rschwieb

99.5k1190225

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?
â€“Â mathstackuser
Aug 3 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?
â€“Â mathstackuser
Aug 3 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.
â€“Â rschwieb
Aug 3 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.
â€“Â rschwieb
Aug 3 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?
â€“Â mathstackuser
Aug 3 at 14:41

Â |Â
show 1 more comment

up vote
1
down vote

We can begin by saying $J=J(R)$ is finitely generated. The descending sequence of $Jsupseteq J^2supseteq ldots$ 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 $R_n/R_n-1$.

edited Aug 3 at 13:49

answered Aug 3 at 13:43

rschwieb

99.5k1190225

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?
â€“Â mathstackuser
Aug 3 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?
â€“Â mathstackuser
Aug 3 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.
â€“Â rschwieb
Aug 3 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.
â€“Â rschwieb
Aug 3 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?
â€“Â mathstackuser
Aug 3 at 14:41

Â |Â
show 1 more comment

up vote
1
down vote

We can begin by saying $J=J(R)$ is finitely generated. The descending sequence of $Jsupseteq J^2supseteq ldots$ 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 $R_n/R_n-1$.

edited Aug 3 at 13:49

answered Aug 3 at 13:43

rschwieb

99.5k1190225

We can begin by saying $J=J(R)$ is finitely generated. The descending sequence of $Jsupseteq J^2supseteq ldots$ 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 $R_n/R_n-1$.

edited Aug 3 at 13:49

answered Aug 3 at 13:43

rschwieb

99.5k1190225

edited Aug 3 at 13:49

answered Aug 3 at 13:43

rschwieb

99.5k1190225

answered Aug 3 at 13:43

rschwieb

99.5k1190225

answered Aug 3 at 13:43

rschwieb

99.5k1190225

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?
â€“Â mathstackuser
Aug 3 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?
â€“Â mathstackuser
Aug 3 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.
â€“Â rschwieb
Aug 3 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.
â€“Â rschwieb
Aug 3 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?
â€“Â mathstackuser
Aug 3 at 14:41

Â |Â
show 1 more comment

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?
â€“Â mathstackuser
Aug 3 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?
â€“Â mathstackuser
Aug 3 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.
â€“Â rschwieb
Aug 3 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.
â€“Â rschwieb
Aug 3 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?
â€“Â mathstackuser
Aug 3 at 14:41

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?
â€“Â mathstackuser
Aug 3 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?
â€“Â mathstackuser
Aug 3 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.
â€“Â rschwieb
Aug 3 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.
â€“Â rschwieb
Aug 3 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?
â€“Â mathstackuser
Aug 3 at 14:41

Â |Â
show 1 more comment

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