Vector space of finite dimension then $Tor_{K[x]}V=V$
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Alright guys, I have a doubt. Let $K$ be a field and $V$ a vector space of dimesion $n$. Because of this we know $V$ is a finitely generated free module. The goal is to show that $Tor_{K[x]}V=V$ where $V$ has a $K[x]$ module structure induced by $T in End_k(V)$, $xcdot v=T(v)$. So I tried working with the generators and try to show that there is only a finite number of options where they can be send to by the endomorphism to prove that indeed for all $vin V$ there is a polinomyal that is going to kill it, but I'm not very confident about it. Can you help me out? Thanks.
abstract-algebra vector-spaces modules
edited Dec 29 '18 at 11:29
Javi
2,7132827
asked Dec 29 '18 at 11:24
Pedro SantosPedro Santos
1458
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add a comment |
$begingroup$
Alright guys, I have a doubt. Let $K$ be a field and $V$ a vector space of dimesion $n$. Because of this we know $V$ is a finitely generated free module. The goal is to show that $Tor_{K[x]}V=V$ where $V$ has a $K[x]$ module structure induced by $T in End_k(V)$, $xcdot v=T(v)$. So I tried working with the generators and try to show that there is only a finite number of options where they can be send to by the endomorphism to prove that indeed for all $vin V$ there is a polinomyal that is going to kill it, but I'm not very confident about it. Can you help me out? Thanks.
abstract-algebra vector-spaces modules
edited Dec 29 '18 at 11:29
Javi
2,7132827
asked Dec 29 '18 at 11:24
Pedro SantosPedro Santos
1458
$endgroup$
add a comment |
$begingroup$
Alright guys, I have a doubt. Let $K$ be a field and $V$ a vector space of dimesion $n$. Because of this we know $V$ is a finitely generated free module. The goal is to show that $Tor_{K[x]}V=V$ where $V$ has a $K[x]$ module structure induced by $T in End_k(V)$, $xcdot v=T(v)$. So I tried working with the generators and try to show that there is only a finite number of options where they can be send to by the endomorphism to prove that indeed for all $vin V$ there is a polinomyal that is going to kill it, but I'm not very confident about it. Can you help me out? Thanks.
abstract-algebra vector-spaces modules
edited Dec 29 '18 at 11:29
Javi
2,7132827
asked Dec 29 '18 at 11:24
Pedro SantosPedro Santos
1458
$endgroup$
Alright guys, I have a doubt. Let $K$ be a field and $V$ a vector space of dimesion $n$. Because of this we know $V$ is a finitely generated free module. The goal is to show that $Tor_{K[x]}V=V$ where $V$ has a $K[x]$ module structure induced by $T in End_k(V)$, $xcdot v=T(v)$. So I tried working with the generators and try to show that there is only a finite number of options where they can be send to by the endomorphism to prove that indeed for all $vin V$ there is a polinomyal that is going to kill it, but I'm not very confident about it. Can you help me out? Thanks.
abstract-algebra vector-spaces modules
abstract-algebra vector-spaces modules
edited Dec 29 '18 at 11:29
Javi
2,7132827
asked Dec 29 '18 at 11:24
Pedro SantosPedro Santos
1458
edited Dec 29 '18 at 11:29
Javi
2,7132827
asked Dec 29 '18 at 11:24
Pedro SantosPedro Santos
1458
edited Dec 29 '18 at 11:29
Javi
2,7132827
edited Dec 29 '18 at 11:29
Javi
2,7132827
edited Dec 29 '18 at 11:29
Javi
2,7132827
2,7132827
asked Dec 29 '18 at 11:24
Pedro SantosPedro Santos
1458
asked Dec 29 '18 at 11:24
Pedro SantosPedro Santos
1458
asked Dec 29 '18 at 11:24
Pedro SantosPedro Santos
1458
1458
add a comment |
add a comment |
1 Answer
1
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Hint: If $V$ has dimension $n$ and $v in V$, then $v, Tv, T^2v, dots, T^n v$ are linearly dependent.
answered Dec 29 '18 at 11:28
lhflhf
165k10171396
$endgroup$
$begingroup$
See also math.stackexchange.com/a/36662/589
$endgroup$
– lhf
Dec 29 '18 at 11:30
$begingroup$
Im not quite seeing it , can u elaborate pls? sorry
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:36
$begingroup$
@PedroSantos, there are $n+1$ vectors in that list. A linear dependence relation for them is the same as the action of a polynomial in $T$ on $v$.
$endgroup$
– lhf
Dec 29 '18 at 11:41
$begingroup$
oh yeah right , thx
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:42
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Hint: If $V$ has dimension $n$ and $v in V$, then $v, Tv, T^2v, dots, T^n v$ are linearly dependent.
answered Dec 29 '18 at 11:28
lhflhf
165k10171396
$endgroup$
$begingroup$
See also math.stackexchange.com/a/36662/589
$endgroup$
– lhf
Dec 29 '18 at 11:30
$begingroup$
Im not quite seeing it , can u elaborate pls? sorry
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:36
$begingroup$
@PedroSantos, there are $n+1$ vectors in that list. A linear dependence relation for them is the same as the action of a polynomial in $T$ on $v$.
$endgroup$
– lhf
Dec 29 '18 at 11:41
$begingroup$
oh yeah right , thx
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:42
add a comment |
$begingroup$
Hint: If $V$ has dimension $n$ and $v in V$, then $v, Tv, T^2v, dots, T^n v$ are linearly dependent.
answered Dec 29 '18 at 11:28
lhflhf
165k10171396
$endgroup$
$begingroup$
See also math.stackexchange.com/a/36662/589
$endgroup$
– lhf
Dec 29 '18 at 11:30
$begingroup$
Im not quite seeing it , can u elaborate pls? sorry
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:36
$begingroup$
@PedroSantos, there are $n+1$ vectors in that list. A linear dependence relation for them is the same as the action of a polynomial in $T$ on $v$.
$endgroup$
– lhf
Dec 29 '18 at 11:41
$begingroup$
oh yeah right , thx
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:42
add a comment |
$begingroup$
Hint: If $V$ has dimension $n$ and $v in V$, then $v, Tv, T^2v, dots, T^n v$ are linearly dependent.
answered Dec 29 '18 at 11:28
lhflhf
165k10171396
$endgroup$
Hint: If $V$ has dimension $n$ and $v in V$, then $v, Tv, T^2v, dots, T^n v$ are linearly dependent.
answered Dec 29 '18 at 11:28
lhflhf
165k10171396
answered Dec 29 '18 at 11:28
lhflhf
165k10171396
answered Dec 29 '18 at 11:28
lhflhf
165k10171396
answered Dec 29 '18 at 11:28
lhflhf
165k10171396
165k10171396
$begingroup$
See also math.stackexchange.com/a/36662/589
$endgroup$
– lhf
Dec 29 '18 at 11:30
$begingroup$
Im not quite seeing it , can u elaborate pls? sorry
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:36
$begingroup$
@PedroSantos, there are $n+1$ vectors in that list. A linear dependence relation for them is the same as the action of a polynomial in $T$ on $v$.
$endgroup$
– lhf
Dec 29 '18 at 11:41
$begingroup$
oh yeah right , thx
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:42
add a comment |
$begingroup$
See also math.stackexchange.com/a/36662/589
$endgroup$
– lhf
Dec 29 '18 at 11:30
$begingroup$
Im not quite seeing it , can u elaborate pls? sorry
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:36
$begingroup$
@PedroSantos, there are $n+1$ vectors in that list. A linear dependence relation for them is the same as the action of a polynomial in $T$ on $v$.
$endgroup$
– lhf
Dec 29 '18 at 11:41
$begingroup$
oh yeah right , thx
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:42
$begingroup$
See also math.stackexchange.com/a/36662/589
$endgroup$
– lhf
Dec 29 '18 at 11:30
$begingroup$
See also math.stackexchange.com/a/36662/589
$endgroup$
– lhf
Dec 29 '18 at 11:30
$begingroup$
Im not quite seeing it , can u elaborate pls? sorry
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:36
$begingroup$
Im not quite seeing it , can u elaborate pls? sorry
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:36
$begingroup$
@PedroSantos, there are $n+1$ vectors in that list. A linear dependence relation for them is the same as the action of a polynomial in $T$ on $v$.
$endgroup$
– lhf
Dec 29 '18 at 11:41
$begingroup$
@PedroSantos, there are $n+1$ vectors in that list. A linear dependence relation for them is the same as the action of a polynomial in $T$ on $v$.
$endgroup$
– lhf
Dec 29 '18 at 11:41
$begingroup$
oh yeah right , thx
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:42
$begingroup$
oh yeah right , thx
$endgroup$
– Pedro Santos
Dec 29 '18 at 11:42
add a comment |
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