Mixing
Finitely generated projective modules are reflexive
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How to show $i_P$ is an isomorphism ? We need to show that it is injective and surjective.
abstract-algebra ring-theory modules projective-module
edited Feb 26 at 21:56
user26857
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asked Jan 23 at 9:43
maths studentmaths student
6251521
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$begingroup$
How to show $i_P$ is an isomorphism ? We need to show that it is injective and surjective.
abstract-algebra ring-theory modules projective-module
edited Feb 26 at 21:56
user26857
39.4k124183
asked Jan 23 at 9:43
maths studentmaths student
6251521
$endgroup$
add a comment |
$begingroup$
How to show $i_P$ is an isomorphism ? We need to show that it is injective and surjective.
abstract-algebra ring-theory modules projective-module
edited Feb 26 at 21:56
user26857
39.4k124183
asked Jan 23 at 9:43
maths studentmaths student
6251521
$endgroup$
How to show $i_P$ is an isomorphism ? We need to show that it is injective and surjective.
abstract-algebra ring-theory modules projective-module
abstract-algebra ring-theory modules projective-module
edited Feb 26 at 21:56
user26857
39.4k124183
asked Jan 23 at 9:43
maths studentmaths student
6251521
edited Feb 26 at 21:56
user26857
39.4k124183
asked Jan 23 at 9:43
maths studentmaths student
6251521
edited Feb 26 at 21:56
user26857
39.4k124183
edited Feb 26 at 21:56
user26857
39.4k124183
edited Feb 26 at 21:56
user26857
39.4k124183
39.4k124183
asked Jan 23 at 9:43
maths studentmaths student
6251521
asked Jan 23 at 9:43
maths studentmaths student
6251521
asked Jan 23 at 9:43
maths studentmaths student
6251521
6251521
add a comment |
add a comment |
1 Answer
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$begingroup$
Let's show that $i_P$ is injective. If $i_P x=0$ then for all $fin P^*$ we have $f(x)=0$ so $x=0$. Hence $i_P$ is injective. Let $F=Poplus Q$ be a finitely generated free module (which exists since $P$ is projective). Then $i_F:Frightarrow F^{**}$ is an isomorphism (probably this is covered in the previous page of the book you are citing). Since $i_F$ maps $P$ to $P^{**}$, we deduce that $i_P=i_Fmid_P$ is surjective.
answered Jan 23 at 10:12
LeventLevent
2,729925
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1 Answer
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$begingroup$
Let's show that $i_P$ is injective. If $i_P x=0$ then for all $fin P^*$ we have $f(x)=0$ so $x=0$. Hence $i_P$ is injective. Let $F=Poplus Q$ be a finitely generated free module (which exists since $P$ is projective). Then $i_F:Frightarrow F^{**}$ is an isomorphism (probably this is covered in the previous page of the book you are citing). Since $i_F$ maps $P$ to $P^{**}$, we deduce that $i_P=i_Fmid_P$ is surjective.
answered Jan 23 at 10:12
LeventLevent
2,729925
$endgroup$
add a comment |
$begingroup$
Let's show that $i_P$ is injective. If $i_P x=0$ then for all $fin P^*$ we have $f(x)=0$ so $x=0$. Hence $i_P$ is injective. Let $F=Poplus Q$ be a finitely generated free module (which exists since $P$ is projective). Then $i_F:Frightarrow F^{**}$ is an isomorphism (probably this is covered in the previous page of the book you are citing). Since $i_F$ maps $P$ to $P^{**}$, we deduce that $i_P=i_Fmid_P$ is surjective.
answered Jan 23 at 10:12
LeventLevent
2,729925
$endgroup$
add a comment |
$begingroup$
Let's show that $i_P$ is injective. If $i_P x=0$ then for all $fin P^*$ we have $f(x)=0$ so $x=0$. Hence $i_P$ is injective. Let $F=Poplus Q$ be a finitely generated free module (which exists since $P$ is projective). Then $i_F:Frightarrow F^{**}$ is an isomorphism (probably this is covered in the previous page of the book you are citing). Since $i_F$ maps $P$ to $P^{**}$, we deduce that $i_P=i_Fmid_P$ is surjective.
answered Jan 23 at 10:12
LeventLevent
2,729925
$endgroup$
Let's show that $i_P$ is injective. If $i_P x=0$ then for all $fin P^*$ we have $f(x)=0$ so $x=0$. Hence $i_P$ is injective. Let $F=Poplus Q$ be a finitely generated free module (which exists since $P$ is projective). Then $i_F:Frightarrow F^{**}$ is an isomorphism (probably this is covered in the previous page of the book you are citing). Since $i_F$ maps $P$ to $P^{**}$, we deduce that $i_P=i_Fmid_P$ is surjective.
answered Jan 23 at 10:12
LeventLevent
2,729925
answered Jan 23 at 10:12
LeventLevent
2,729925
answered Jan 23 at 10:12
LeventLevent
2,729925
answered Jan 23 at 10:12
LeventLevent
2,729925
2,729925
add a comment |
add a comment |
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