Mixing
About dual of finitely generated projective module
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Let $M$ be a finitely generated projective module, $x in M$ and $x neq 0$. Then is it true that there is $g in M^*$ such that $gx neq 0$?
If yes how to prove it? For vector space dual this result is true, but what about projective module?
Also if $f in M^*$, $f neq 0$ then $fy neq 0$ is also true or not ?
linear-algebra modules projective-module free-modules
edited Feb 26 at 21:59
user26857
39.5k124283
asked Jan 29 at 17:22
maths studentmaths student
6321521
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add a comment |
$begingroup$
Let $M$ be a finitely generated projective module, $x in M$ and $x neq 0$. Then is it true that there is $g in M^*$ such that $gx neq 0$?
If yes how to prove it? For vector space dual this result is true, but what about projective module?
Also if $f in M^*$, $f neq 0$ then $fy neq 0$ is also true or not ?
linear-algebra modules projective-module free-modules
edited Feb 26 at 21:59
user26857
39.5k124283
asked Jan 29 at 17:22
maths studentmaths student
6321521
$endgroup$
add a comment |
$begingroup$
Let $M$ be a finitely generated projective module, $x in M$ and $x neq 0$. Then is it true that there is $g in M^*$ such that $gx neq 0$?
If yes how to prove it? For vector space dual this result is true, but what about projective module?
Also if $f in M^*$, $f neq 0$ then $fy neq 0$ is also true or not ?
linear-algebra modules projective-module free-modules
edited Feb 26 at 21:59
user26857
39.5k124283
asked Jan 29 at 17:22
maths studentmaths student
6321521
$endgroup$
Let $M$ be a finitely generated projective module, $x in M$ and $x neq 0$. Then is it true that there is $g in M^*$ such that $gx neq 0$?
If yes how to prove it? For vector space dual this result is true, but what about projective module?
Also if $f in M^*$, $f neq 0$ then $fy neq 0$ is also true or not ?
linear-algebra modules projective-module free-modules
linear-algebra modules projective-module free-modules
edited Feb 26 at 21:59
user26857
39.5k124283
asked Jan 29 at 17:22
maths studentmaths student
6321521
edited Feb 26 at 21:59
user26857
39.5k124283
asked Jan 29 at 17:22
maths studentmaths student
6321521
edited Feb 26 at 21:59
user26857
39.5k124283
edited Feb 26 at 21:59
user26857
39.5k124283
edited Feb 26 at 21:59
user26857
39.5k124283
39.5k124283
asked Jan 29 at 17:22
maths studentmaths student
6321521
asked Jan 29 at 17:22
maths studentmaths student
6321521
asked Jan 29 at 17:22
maths studentmaths student
6321521
6321521
add a comment |
add a comment |
1 Answer
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$begingroup$
The answer to your first question is yes, since finitely generated projective modules are reflexive (see here), meaning that the canonical injective map $Mto M^{**}$ is an isomorphism. Hence if $xneq 0$, then $xneq 0in M^{**}$, so there is $gin M^*$ such that $gx=langle g, xrangleneq 0$.
I don't know what is $y$ in your second question. If you means there exists such an $y$, then it's true by definition of $fneq 0$.
edited Feb 26 at 22:00
user26857
39.5k124283
answered Feb 14 at 10:27
Lao-tzuLao-tzu
1,0611024
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer to your first question is yes, since finitely generated projective modules are reflexive (see here), meaning that the canonical injective map $Mto M^{**}$ is an isomorphism. Hence if $xneq 0$, then $xneq 0in M^{**}$, so there is $gin M^*$ such that $gx=langle g, xrangleneq 0$.
I don't know what is $y$ in your second question. If you means there exists such an $y$, then it's true by definition of $fneq 0$.
edited Feb 26 at 22:00
user26857
39.5k124283
answered Feb 14 at 10:27
Lao-tzuLao-tzu
1,0611024
$endgroup$
add a comment |
$begingroup$
The answer to your first question is yes, since finitely generated projective modules are reflexive (see here), meaning that the canonical injective map $Mto M^{**}$ is an isomorphism. Hence if $xneq 0$, then $xneq 0in M^{**}$, so there is $gin M^*$ such that $gx=langle g, xrangleneq 0$.
I don't know what is $y$ in your second question. If you means there exists such an $y$, then it's true by definition of $fneq 0$.
edited Feb 26 at 22:00
user26857
39.5k124283
answered Feb 14 at 10:27
Lao-tzuLao-tzu
1,0611024
$endgroup$
add a comment |
$begingroup$
The answer to your first question is yes, since finitely generated projective modules are reflexive (see here), meaning that the canonical injective map $Mto M^{**}$ is an isomorphism. Hence if $xneq 0$, then $xneq 0in M^{**}$, so there is $gin M^*$ such that $gx=langle g, xrangleneq 0$.
I don't know what is $y$ in your second question. If you means there exists such an $y$, then it's true by definition of $fneq 0$.
edited Feb 26 at 22:00
user26857
39.5k124283
answered Feb 14 at 10:27
Lao-tzuLao-tzu
1,0611024
$endgroup$
The answer to your first question is yes, since finitely generated projective modules are reflexive (see here), meaning that the canonical injective map $Mto M^{**}$ is an isomorphism. Hence if $xneq 0$, then $xneq 0in M^{**}$, so there is $gin M^*$ such that $gx=langle g, xrangleneq 0$.
I don't know what is $y$ in your second question. If you means there exists such an $y$, then it's true by definition of $fneq 0$.
edited Feb 26 at 22:00
user26857
39.5k124283
answered Feb 14 at 10:27
Lao-tzuLao-tzu
1,0611024
edited Feb 26 at 22:00
user26857
39.5k124283
edited Feb 26 at 22:00
user26857
39.5k124283
edited Feb 26 at 22:00
user26857
39.5k124283
39.5k124283
answered Feb 14 at 10:27
Lao-tzuLao-tzu
1,0611024
answered Feb 14 at 10:27
Lao-tzuLao-tzu
1,0611024
answered Feb 14 at 10:27
Lao-tzuLao-tzu
1,0611024
1,0611024
add a comment |
add a comment |
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