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Exact contravariant endofunctor preserving iso classes of simple objects sends objects of finite length to objects of finite length?
Clash Royale CLAN TAG#URR8PPP
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Suppose you have a module category, and an exact, contravariant endofunctor $F$ that preserves isomorphism classes of simple modules. If $M$ is a module of finite length, why is $F(M)$ also of finite length?
I figure there must be a way to construct a finite composition series of $F(M)$ from that of $M$, since $F$ preserves the isomorphism classes of the composition factors. But since a composition series is not exact and $F$ reverses the arrows, I don't see how you could do this.
(The motivation for my question is that dualization in category $mathcalO$ preserves finitely generated modules.)
abstract-algebra category-theory modules functors
asked Jul 28 at 2:32
Joie Hwang
350210
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up vote
2
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Suppose you have a module category, and an exact, contravariant endofunctor $F$ that preserves isomorphism classes of simple modules. If $M$ is a module of finite length, why is $F(M)$ also of finite length?
I figure there must be a way to construct a finite composition series of $F(M)$ from that of $M$, since $F$ preserves the isomorphism classes of the composition factors. But since a composition series is not exact and $F$ reverses the arrows, I don't see how you could do this.
(The motivation for my question is that dualization in category $mathcalO$ preserves finitely generated modules.)
abstract-algebra category-theory modules functors
asked Jul 28 at 2:32
Joie Hwang
350210
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose you have a module category, and an exact, contravariant endofunctor $F$ that preserves isomorphism classes of simple modules. If $M$ is a module of finite length, why is $F(M)$ also of finite length?
I figure there must be a way to construct a finite composition series of $F(M)$ from that of $M$, since $F$ preserves the isomorphism classes of the composition factors. But since a composition series is not exact and $F$ reverses the arrows, I don't see how you could do this.
(The motivation for my question is that dualization in category $mathcalO$ preserves finitely generated modules.)
abstract-algebra category-theory modules functors
asked Jul 28 at 2:32
Joie Hwang
350210
Suppose you have a module category, and an exact, contravariant endofunctor $F$ that preserves isomorphism classes of simple modules. If $M$ is a module of finite length, why is $F(M)$ also of finite length?
I figure there must be a way to construct a finite composition series of $F(M)$ from that of $M$, since $F$ preserves the isomorphism classes of the composition factors. But since a composition series is not exact and $F$ reverses the arrows, I don't see how you could do this.
(The motivation for my question is that dualization in category $mathcalO$ preserves finitely generated modules.)
abstract-algebra category-theory modules functors
asked Jul 28 at 2:32
Joie Hwang
350210
asked Jul 28 at 2:32
Joie Hwang
350210
asked Jul 28 at 2:32
Joie Hwang
350210
asked Jul 28 at 2:32
Joie Hwang
350210
350210
add a comment |Â
add a comment |Â
1 Answer
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This follows immediately from the fact that length is additive in short exact sequences. So, by induction on $n$, we can show that if $M$ has length $n$ then $F(M)$ has length $n$. The case $n=0$ is trivial, since $M$ must be trivial. If $M$ has length $n>0$, there exists a short exact sequence $$0to Nto Mto Sto 0$$ where $N$ has length $n-1$ and $S$ is simple. Applying $F$, we get a short exact sequence $$0to F(S)to F(M)to F(N)to 0.$$ By assumption $F(S)$ is simple and by induction $F(N)$ has length $n-1$. Thus $F(M)$ has length $n$.
answered Jul 28 at 3:36
Eric Wofsey
162k12188299
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This follows immediately from the fact that length is additive in short exact sequences. So, by induction on $n$, we can show that if $M$ has length $n$ then $F(M)$ has length $n$. The case $n=0$ is trivial, since $M$ must be trivial. If $M$ has length $n>0$, there exists a short exact sequence $$0to Nto Mto Sto 0$$ where $N$ has length $n-1$ and $S$ is simple. Applying $F$, we get a short exact sequence $$0to F(S)to F(M)to F(N)to 0.$$ By assumption $F(S)$ is simple and by induction $F(N)$ has length $n-1$. Thus $F(M)$ has length $n$.
answered Jul 28 at 3:36
Eric Wofsey
162k12188299
add a comment |Â
up vote
2
down vote
accepted
This follows immediately from the fact that length is additive in short exact sequences. So, by induction on $n$, we can show that if $M$ has length $n$ then $F(M)$ has length $n$. The case $n=0$ is trivial, since $M$ must be trivial. If $M$ has length $n>0$, there exists a short exact sequence $$0to Nto Mto Sto 0$$ where $N$ has length $n-1$ and $S$ is simple. Applying $F$, we get a short exact sequence $$0to F(S)to F(M)to F(N)to 0.$$ By assumption $F(S)$ is simple and by induction $F(N)$ has length $n-1$. Thus $F(M)$ has length $n$.
answered Jul 28 at 3:36
Eric Wofsey
162k12188299
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This follows immediately from the fact that length is additive in short exact sequences. So, by induction on $n$, we can show that if $M$ has length $n$ then $F(M)$ has length $n$. The case $n=0$ is trivial, since $M$ must be trivial. If $M$ has length $n>0$, there exists a short exact sequence $$0to Nto Mto Sto 0$$ where $N$ has length $n-1$ and $S$ is simple. Applying $F$, we get a short exact sequence $$0to F(S)to F(M)to F(N)to 0.$$ By assumption $F(S)$ is simple and by induction $F(N)$ has length $n-1$. Thus $F(M)$ has length $n$.
answered Jul 28 at 3:36
Eric Wofsey
162k12188299
This follows immediately from the fact that length is additive in short exact sequences. So, by induction on $n$, we can show that if $M$ has length $n$ then $F(M)$ has length $n$. The case $n=0$ is trivial, since $M$ must be trivial. If $M$ has length $n>0$, there exists a short exact sequence $$0to Nto Mto Sto 0$$ where $N$ has length $n-1$ and $S$ is simple. Applying $F$, we get a short exact sequence $$0to F(S)to F(M)to F(N)to 0.$$ By assumption $F(S)$ is simple and by induction $F(N)$ has length $n-1$. Thus $F(M)$ has length $n$.
answered Jul 28 at 3:36
Eric Wofsey
162k12188299
answered Jul 28 at 3:36
Eric Wofsey
162k12188299
answered Jul 28 at 3:36
Eric Wofsey
162k12188299
answered Jul 28 at 3:36
Eric Wofsey
162k12188299
162k12188299
add a comment |Â
add a comment |Â
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