Mixing
Show that $(V/U)/(W/U)cong V/W$
$begingroup$
$K$ is a field,$V$ is a $K$-vectorspace and $Usubseteq Wsubseteq V$.
I know that I have to show that the left and the right side are both vectorspaces which bases with same dimensions. But I don't know how do to that.
Support would be highly appreciated.
linear-algebra
asked Jan 14 at 14:57
RM777RM777
39112
$endgroup$
add a comment |
$begingroup$
$K$ is a field,$V$ is a $K$-vectorspace and $Usubseteq Wsubseteq V$.
I know that I have to show that the left and the right side are both vectorspaces which bases with same dimensions. But I don't know how do to that.
Support would be highly appreciated.
linear-algebra
asked Jan 14 at 14:57
RM777RM777
39112
$endgroup$
1
$begingroup$
"I know that I have to show that [...] which bases with same dimensions" Not necessarily. Just demonstrating the existence of an isomorphism is enough. And once you get to grips with what the elements of each vector space look like, the answer might be obvious.
$endgroup$
– Arthur
Jan 14 at 15:01
$begingroup$
Let $phi:V/U to V/W$ be the function $phi(v+U) = v+W$. So that is a well defined correspondence, surjective and linear... whats $kerphi$?(I guess $U,W$ are subspaces of $V$)
$endgroup$
– dem0nakos
Jan 14 at 15:16
add a comment |
$begingroup$
$K$ is a field,$V$ is a $K$-vectorspace and $Usubseteq Wsubseteq V$.
I know that I have to show that the left and the right side are both vectorspaces which bases with same dimensions. But I don't know how do to that.
Support would be highly appreciated.
linear-algebra
asked Jan 14 at 14:57
RM777RM777
39112
$endgroup$
$K$ is a field,$V$ is a $K$-vectorspace and $Usubseteq Wsubseteq V$.
I know that I have to show that the left and the right side are both vectorspaces which bases with same dimensions. But I don't know how do to that.
Support would be highly appreciated.
linear-algebra
linear-algebra
asked Jan 14 at 14:57
RM777RM777
39112
asked Jan 14 at 14:57
RM777RM777
39112
asked Jan 14 at 14:57
RM777RM777
39112
asked Jan 14 at 14:57
RM777RM777
39112
asked Jan 14 at 14:57
RM777RM777
39112
39112
1
$begingroup$
"I know that I have to show that [...] which bases with same dimensions" Not necessarily. Just demonstrating the existence of an isomorphism is enough. And once you get to grips with what the elements of each vector space look like, the answer might be obvious.
$endgroup$
– Arthur
Jan 14 at 15:01
$begingroup$
Let $phi:V/U to V/W$ be the function $phi(v+U) = v+W$. So that is a well defined correspondence, surjective and linear... whats $kerphi$?(I guess $U,W$ are subspaces of $V$)
$endgroup$
– dem0nakos
Jan 14 at 15:16
add a comment |
1
$begingroup$
"I know that I have to show that [...] which bases with same dimensions" Not necessarily. Just demonstrating the existence of an isomorphism is enough. And once you get to grips with what the elements of each vector space look like, the answer might be obvious.
$endgroup$
– Arthur
Jan 14 at 15:01
$begingroup$
Let $phi:V/U to V/W$ be the function $phi(v+U) = v+W$. So that is a well defined correspondence, surjective and linear... whats $kerphi$?(I guess $U,W$ are subspaces of $V$)
$endgroup$
– dem0nakos
Jan 14 at 15:16
1
1
$begingroup$
"I know that I have to show that [...] which bases with same dimensions" Not necessarily. Just demonstrating the existence of an isomorphism is enough. And once you get to grips with what the elements of each vector space look like, the answer might be obvious.
$endgroup$
– Arthur
Jan 14 at 15:01
$begingroup$
"I know that I have to show that [...] which bases with same dimensions" Not necessarily. Just demonstrating the existence of an isomorphism is enough. And once you get to grips with what the elements of each vector space look like, the answer might be obvious.
$endgroup$
– Arthur
Jan 14 at 15:01
$begingroup$
Let $phi:V/U to V/W$ be the function $phi(v+U) = v+W$. So that is a well defined correspondence, surjective and linear... whats $kerphi$?(I guess $U,W$ are subspaces of $V$)
$endgroup$
– dem0nakos
Jan 14 at 15:16
$begingroup$
Let $phi:V/U to V/W$ be the function $phi(v+U) = v+W$. So that is a well defined correspondence, surjective and linear... whats $kerphi$?(I guess $U,W$ are subspaces of $V$)
$endgroup$
– dem0nakos
Jan 14 at 15:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the map $f:frac{V/U}{W/U} rightarrow V/W$ defined by $f((v+U)+W/U)=v+W$. Note that $(v+U)+W/U$ is a coset in $V/U$. The verification that this is indeed an isomorphism is left as an exercise.
This result is known as the third isomorphism theorem.
Claim: $f$ is an isomorphism.
$f((v+U)+W/U+(v'+U)+W/U)=v+W+v'+W=(v+v')+W=f(((v+v')+U)+W/U)$. Hence, $f$ is a homomorphism. Also, $f$ is clearly bijective. Hence, $f$ is an isomorphism.
answered Jan 14 at 15:19
toric_actionstoric_actions
1058
$endgroup$
$begingroup$
Can you please give me the solution?
$endgroup$
– RM777
Jan 14 at 17:30
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Consider the map $f:frac{V/U}{W/U} rightarrow V/W$ defined by $f((v+U)+W/U)=v+W$. Note that $(v+U)+W/U$ is a coset in $V/U$. The verification that this is indeed an isomorphism is left as an exercise.
This result is known as the third isomorphism theorem.
Claim: $f$ is an isomorphism.
$f((v+U)+W/U+(v'+U)+W/U)=v+W+v'+W=(v+v')+W=f(((v+v')+U)+W/U)$. Hence, $f$ is a homomorphism. Also, $f$ is clearly bijective. Hence, $f$ is an isomorphism.
answered Jan 14 at 15:19
toric_actionstoric_actions
1058
$endgroup$
$begingroup$
Can you please give me the solution?
$endgroup$
– RM777
Jan 14 at 17:30
add a comment |
$begingroup$
Consider the map $f:frac{V/U}{W/U} rightarrow V/W$ defined by $f((v+U)+W/U)=v+W$. Note that $(v+U)+W/U$ is a coset in $V/U$. The verification that this is indeed an isomorphism is left as an exercise.
This result is known as the third isomorphism theorem.
Claim: $f$ is an isomorphism.
$f((v+U)+W/U+(v'+U)+W/U)=v+W+v'+W=(v+v')+W=f(((v+v')+U)+W/U)$. Hence, $f$ is a homomorphism. Also, $f$ is clearly bijective. Hence, $f$ is an isomorphism.
answered Jan 14 at 15:19
toric_actionstoric_actions
1058
$endgroup$
$begingroup$
Can you please give me the solution?
$endgroup$
– RM777
Jan 14 at 17:30
add a comment |
$begingroup$
Consider the map $f:frac{V/U}{W/U} rightarrow V/W$ defined by $f((v+U)+W/U)=v+W$. Note that $(v+U)+W/U$ is a coset in $V/U$. The verification that this is indeed an isomorphism is left as an exercise.
This result is known as the third isomorphism theorem.
Claim: $f$ is an isomorphism.
$f((v+U)+W/U+(v'+U)+W/U)=v+W+v'+W=(v+v')+W=f(((v+v')+U)+W/U)$. Hence, $f$ is a homomorphism. Also, $f$ is clearly bijective. Hence, $f$ is an isomorphism.
answered Jan 14 at 15:19
toric_actionstoric_actions
1058
$endgroup$
Consider the map $f:frac{V/U}{W/U} rightarrow V/W$ defined by $f((v+U)+W/U)=v+W$. Note that $(v+U)+W/U$ is a coset in $V/U$. The verification that this is indeed an isomorphism is left as an exercise.
This result is known as the third isomorphism theorem.
Claim: $f$ is an isomorphism.
$f((v+U)+W/U+(v'+U)+W/U)=v+W+v'+W=(v+v')+W=f(((v+v')+U)+W/U)$. Hence, $f$ is a homomorphism. Also, $f$ is clearly bijective. Hence, $f$ is an isomorphism.
answered Jan 14 at 15:19
toric_actionstoric_actions
1058
edited Jan 15 at 7:26
answered Jan 14 at 15:19
toric_actionstoric_actions
1058
answered Jan 14 at 15:19
toric_actionstoric_actions
1058
answered Jan 14 at 15:19
toric_actionstoric_actions
1058
1058
$begingroup$
Can you please give me the solution?
$endgroup$
– RM777
Jan 14 at 17:30
add a comment |
$begingroup$
Can you please give me the solution?
$endgroup$
– RM777
Jan 14 at 17:30
$begingroup$
Can you please give me the solution?
$endgroup$
– RM777
Jan 14 at 17:30
$begingroup$
Can you please give me the solution?
$endgroup$
– RM777
Jan 14 at 17:30
add a comment |
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1
$begingroup$
"I know that I have to show that [...] which bases with same dimensions" Not necessarily. Just demonstrating the existence of an isomorphism is enough. And once you get to grips with what the elements of each vector space look like, the answer might be obvious.
$endgroup$
– Arthur
Jan 14 at 15:01
$begingroup$
Let $phi:V/U to V/W$ be the function $phi(v+U) = v+W$. So that is a well defined correspondence, surjective and linear... whats $kerphi$?(I guess $U,W$ are subspaces of $V$)
$endgroup$
– dem0nakos
Jan 14 at 15:16