All objects in $bf{FinVect}$ have finite length
I want to show that every object $X$ of $bf{FinVect}$ has finite length. i.e. there is a sequence of monos
$$
0=X_0 hookrightarrow X_1 hookrightarrow ... hookrightarrow X_{n-1} hookrightarrow X_n = X
$$
such that $forall i,{X_i}/{X_{i-1}}$ is simple.
I think this is supposed to be a trivial case, but I do not see how. If we use just the zero map $0xrightarrow{0} X$ it does not work, since $X/{0}cong X$ need not be simple.
vector-spaces category-theory abelian-categories
edited Nov 24 at 12:36
user 170039
10.4k42465
asked Nov 24 at 12:32
Soap
1,027615
add a comment |
I want to show that every object $X$ of $bf{FinVect}$ has finite length. i.e. there is a sequence of monos
$$
0=X_0 hookrightarrow X_1 hookrightarrow ... hookrightarrow X_{n-1} hookrightarrow X_n = X
$$
such that $forall i,{X_i}/{X_{i-1}}$ is simple.
I think this is supposed to be a trivial case, but I do not see how. If we use just the zero map $0xrightarrow{0} X$ it does not work, since $X/{0}cong X$ need not be simple.
vector-spaces category-theory abelian-categories
edited Nov 24 at 12:36
user 170039
10.4k42465
asked Nov 24 at 12:32
Soap
1,027615
2
What are simple objects in your category? Can you show that any object $X$ in this category is of length $dim(X)$?
– Batominovski
Nov 24 at 12:35
@Batominovski Thanks, that hint was enough - it feels quite obvious now
– Soap
Nov 25 at 10:45
add a comment |
I want to show that every object $X$ of $bf{FinVect}$ has finite length. i.e. there is a sequence of monos
$$
0=X_0 hookrightarrow X_1 hookrightarrow ... hookrightarrow X_{n-1} hookrightarrow X_n = X
$$
such that $forall i,{X_i}/{X_{i-1}}$ is simple.
I think this is supposed to be a trivial case, but I do not see how. If we use just the zero map $0xrightarrow{0} X$ it does not work, since $X/{0}cong X$ need not be simple.
vector-spaces category-theory abelian-categories
edited Nov 24 at 12:36
user 170039
10.4k42465
asked Nov 24 at 12:32
Soap
1,027615
I want to show that every object $X$ of $bf{FinVect}$ has finite length. i.e. there is a sequence of monos
$$
0=X_0 hookrightarrow X_1 hookrightarrow ... hookrightarrow X_{n-1} hookrightarrow X_n = X
$$
such that $forall i,{X_i}/{X_{i-1}}$ is simple.
I think this is supposed to be a trivial case, but I do not see how. If we use just the zero map $0xrightarrow{0} X$ it does not work, since $X/{0}cong X$ need not be simple.
vector-spaces category-theory abelian-categories
vector-spaces category-theory abelian-categories
edited Nov 24 at 12:36
user 170039
10.4k42465
asked Nov 24 at 12:32
Soap
1,027615
edited Nov 24 at 12:36
user 170039
10.4k42465
asked Nov 24 at 12:32
Soap
1,027615
edited Nov 24 at 12:36
user 170039
10.4k42465
edited Nov 24 at 12:36
user 170039
10.4k42465
edited Nov 24 at 12:36
user 170039
10.4k42465
10.4k42465
asked Nov 24 at 12:32
Soap
1,027615
asked Nov 24 at 12:32
Soap
1,027615
asked Nov 24 at 12:32
Soap
1,027615
1,027615
2
What are simple objects in your category? Can you show that any object $X$ in this category is of length $dim(X)$?
– Batominovski
Nov 24 at 12:35
@Batominovski Thanks, that hint was enough - it feels quite obvious now
– Soap
Nov 25 at 10:45
add a comment |
2
What are simple objects in your category? Can you show that any object $X$ in this category is of length $dim(X)$?
– Batominovski
Nov 24 at 12:35
@Batominovski Thanks, that hint was enough - it feels quite obvious now
– Soap
Nov 25 at 10:45
2
2
What are simple objects in your category? Can you show that any object $X$ in this category is of length $dim(X)$?
– Batominovski
Nov 24 at 12:35
What are simple objects in your category? Can you show that any object $X$ in this category is of length $dim(X)$?
– Batominovski
Nov 24 at 12:35
@Batominovski Thanks, that hint was enough - it feels quite obvious now
– Soap
Nov 25 at 10:45
@Batominovski Thanks, that hint was enough - it feels quite obvious now
– Soap
Nov 25 at 10:45
add a comment |
2 Answers
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The simple objects in $mathbf{FinVect}$ are exactly the 1-dimensional vector spaces. Take any ordered basis $(a_1,ldots,a_n)$ of $X$ (which exists, since $X$ is a finite dimensional vector space), and take $X_k = mathop{mathrm{Span}}{a_1,ldots,a_k}$.
answered Nov 24 at 12:44
user3482749
2,411414
add a comment |
Hints. Let $mathbb{K}$ be the base field. The simple objects of $mathbf{C}:=mathbf{FinVect}(mathbb{K})$ are $1$-dimensional $mathbb{K}$-vector spaces, i.e., they are all isomorphic to $mathbb{K}$. This shows that, if $Xin mathbf{C}$ has a filtration
$$0=X_0hookrightarrow X_1 hookrightarrow X_2hookrightarrow ldots hookrightarrow X_{l-1}hookrightarrow X_l=Xtag{*}$$
in which $X_i/X_{i-1}$ are simple (whence $1$-dimensional), then $$dim_mathbb{K}(X)=dim_mathbb{K}(X_0)+sum_{i=1}^l,dim_mathbb{K}left(X_i/X_{i-1}right)=0+sum_{i=1}^l,1=l,$$
For convenience, we call (*) a composition series of $X$. Ergo, you have the following statement: "if $Xinmathbb{C}$ has a composition series of length $l$, then $dim_mathbb{K}(X)=l$." How would you show that the converse of this statement is also true? That is, show that any composition series of $Xinmathbf{C}$ exists and is of length $dim_mathbb{K}(X)$.
Try induction on $dim_mathbb{K}(X)$.
answered Nov 24 at 12:44
Batominovski
33.7k33292
add a comment |
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2 Answers
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2 Answers
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The simple objects in $mathbf{FinVect}$ are exactly the 1-dimensional vector spaces. Take any ordered basis $(a_1,ldots,a_n)$ of $X$ (which exists, since $X$ is a finite dimensional vector space), and take $X_k = mathop{mathrm{Span}}{a_1,ldots,a_k}$.
answered Nov 24 at 12:44
user3482749
2,411414
add a comment |
The simple objects in $mathbf{FinVect}$ are exactly the 1-dimensional vector spaces. Take any ordered basis $(a_1,ldots,a_n)$ of $X$ (which exists, since $X$ is a finite dimensional vector space), and take $X_k = mathop{mathrm{Span}}{a_1,ldots,a_k}$.
answered Nov 24 at 12:44
user3482749
2,411414
add a comment |
The simple objects in $mathbf{FinVect}$ are exactly the 1-dimensional vector spaces. Take any ordered basis $(a_1,ldots,a_n)$ of $X$ (which exists, since $X$ is a finite dimensional vector space), and take $X_k = mathop{mathrm{Span}}{a_1,ldots,a_k}$.
answered Nov 24 at 12:44
user3482749
2,411414
The simple objects in $mathbf{FinVect}$ are exactly the 1-dimensional vector spaces. Take any ordered basis $(a_1,ldots,a_n)$ of $X$ (which exists, since $X$ is a finite dimensional vector space), and take $X_k = mathop{mathrm{Span}}{a_1,ldots,a_k}$.
answered Nov 24 at 12:44
user3482749
2,411414
answered Nov 24 at 12:44
user3482749
2,411414
answered Nov 24 at 12:44
user3482749
2,411414
answered Nov 24 at 12:44
user3482749
2,411414
2,411414
add a comment |
add a comment |
Hints. Let $mathbb{K}$ be the base field. The simple objects of $mathbf{C}:=mathbf{FinVect}(mathbb{K})$ are $1$-dimensional $mathbb{K}$-vector spaces, i.e., they are all isomorphic to $mathbb{K}$. This shows that, if $Xin mathbf{C}$ has a filtration
$$0=X_0hookrightarrow X_1 hookrightarrow X_2hookrightarrow ldots hookrightarrow X_{l-1}hookrightarrow X_l=Xtag{*}$$
in which $X_i/X_{i-1}$ are simple (whence $1$-dimensional), then $$dim_mathbb{K}(X)=dim_mathbb{K}(X_0)+sum_{i=1}^l,dim_mathbb{K}left(X_i/X_{i-1}right)=0+sum_{i=1}^l,1=l,$$
For convenience, we call (*) a composition series of $X$. Ergo, you have the following statement: "if $Xinmathbb{C}$ has a composition series of length $l$, then $dim_mathbb{K}(X)=l$." How would you show that the converse of this statement is also true? That is, show that any composition series of $Xinmathbf{C}$ exists and is of length $dim_mathbb{K}(X)$.
Try induction on $dim_mathbb{K}(X)$.
answered Nov 24 at 12:44
Batominovski
33.7k33292
add a comment |
Hints. Let $mathbb{K}$ be the base field. The simple objects of $mathbf{C}:=mathbf{FinVect}(mathbb{K})$ are $1$-dimensional $mathbb{K}$-vector spaces, i.e., they are all isomorphic to $mathbb{K}$. This shows that, if $Xin mathbf{C}$ has a filtration
$$0=X_0hookrightarrow X_1 hookrightarrow X_2hookrightarrow ldots hookrightarrow X_{l-1}hookrightarrow X_l=Xtag{*}$$
in which $X_i/X_{i-1}$ are simple (whence $1$-dimensional), then $$dim_mathbb{K}(X)=dim_mathbb{K}(X_0)+sum_{i=1}^l,dim_mathbb{K}left(X_i/X_{i-1}right)=0+sum_{i=1}^l,1=l,$$
For convenience, we call (*) a composition series of $X$. Ergo, you have the following statement: "if $Xinmathbb{C}$ has a composition series of length $l$, then $dim_mathbb{K}(X)=l$." How would you show that the converse of this statement is also true? That is, show that any composition series of $Xinmathbf{C}$ exists and is of length $dim_mathbb{K}(X)$.
Try induction on $dim_mathbb{K}(X)$.
answered Nov 24 at 12:44
Batominovski
33.7k33292
add a comment |
Hints. Let $mathbb{K}$ be the base field. The simple objects of $mathbf{C}:=mathbf{FinVect}(mathbb{K})$ are $1$-dimensional $mathbb{K}$-vector spaces, i.e., they are all isomorphic to $mathbb{K}$. This shows that, if $Xin mathbf{C}$ has a filtration
$$0=X_0hookrightarrow X_1 hookrightarrow X_2hookrightarrow ldots hookrightarrow X_{l-1}hookrightarrow X_l=Xtag{*}$$
in which $X_i/X_{i-1}$ are simple (whence $1$-dimensional), then $$dim_mathbb{K}(X)=dim_mathbb{K}(X_0)+sum_{i=1}^l,dim_mathbb{K}left(X_i/X_{i-1}right)=0+sum_{i=1}^l,1=l,$$
For convenience, we call (*) a composition series of $X$. Ergo, you have the following statement: "if $Xinmathbb{C}$ has a composition series of length $l$, then $dim_mathbb{K}(X)=l$." How would you show that the converse of this statement is also true? That is, show that any composition series of $Xinmathbf{C}$ exists and is of length $dim_mathbb{K}(X)$.
Try induction on $dim_mathbb{K}(X)$.
answered Nov 24 at 12:44
Batominovski
33.7k33292
Hints. Let $mathbb{K}$ be the base field. The simple objects of $mathbf{C}:=mathbf{FinVect}(mathbb{K})$ are $1$-dimensional $mathbb{K}$-vector spaces, i.e., they are all isomorphic to $mathbb{K}$. This shows that, if $Xin mathbf{C}$ has a filtration
$$0=X_0hookrightarrow X_1 hookrightarrow X_2hookrightarrow ldots hookrightarrow X_{l-1}hookrightarrow X_l=Xtag{*}$$
in which $X_i/X_{i-1}$ are simple (whence $1$-dimensional), then $$dim_mathbb{K}(X)=dim_mathbb{K}(X_0)+sum_{i=1}^l,dim_mathbb{K}left(X_i/X_{i-1}right)=0+sum_{i=1}^l,1=l,$$
For convenience, we call (*) a composition series of $X$. Ergo, you have the following statement: "if $Xinmathbb{C}$ has a composition series of length $l$, then $dim_mathbb{K}(X)=l$." How would you show that the converse of this statement is also true? That is, show that any composition series of $Xinmathbf{C}$ exists and is of length $dim_mathbb{K}(X)$.
Try induction on $dim_mathbb{K}(X)$.
answered Nov 24 at 12:44
Batominovski
33.7k33292
edited Nov 24 at 12:49
answered Nov 24 at 12:44
Batominovski
33.7k33292
answered Nov 24 at 12:44
Batominovski
33.7k33292
answered Nov 24 at 12:44
Batominovski
33.7k33292
33.7k33292
add a comment |
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2
What are simple objects in your category? Can you show that any object $X$ in this category is of length $dim(X)$?
– Batominovski
Nov 24 at 12:35
@Batominovski Thanks, that hint was enough - it feels quite obvious now
– Soap
Nov 25 at 10:45