On isomorphism of sheaves
$begingroup$
$underline {Background}$: Suppose $X$ is a scheme and $mathcal F$ is a sheaf of $mathcal O_X$ module.
Suppose we have, $mathcal {F}oplus mathcal {O}_X cong oplus_r mathcal {O}_X$ as sheaves of $mathcal O_X$ modules.
$underline {Question}$: Can we say that $mathcal {F} cong oplus_{r-1} mathcal {O}_X$ ?
$underline {Attempt}$:I tried to define map on all {$SpecA$} which in turn will define a morphism as follows,
If we denote the isomorphism $mathcal {F}oplus mathcal {O}_X cong oplus_r mathcal {O}_X$ as $phi$
then I define $T(specA):mathcal F(specA)to oplus_{r-1}mathcal O(specA)$ as
$T(specA)(a):=phi(SpecA)(a,0)$ with the last coordinate omitted
But,I cannot see why this should be injective and surjective?
Any help from anyone is welcome
algebraic-geometry sheaf-theory
asked Dec 28 '18 at 13:02
HARRYHARRY
889
$endgroup$
add a comment |
$begingroup$
$underline {Background}$: Suppose $X$ is a scheme and $mathcal F$ is a sheaf of $mathcal O_X$ module.
Suppose we have, $mathcal {F}oplus mathcal {O}_X cong oplus_r mathcal {O}_X$ as sheaves of $mathcal O_X$ modules.
$underline {Question}$: Can we say that $mathcal {F} cong oplus_{r-1} mathcal {O}_X$ ?
$underline {Attempt}$:I tried to define map on all {$SpecA$} which in turn will define a morphism as follows,
If we denote the isomorphism $mathcal {F}oplus mathcal {O}_X cong oplus_r mathcal {O}_X$ as $phi$
then I define $T(specA):mathcal F(specA)to oplus_{r-1}mathcal O(specA)$ as
$T(specA)(a):=phi(SpecA)(a,0)$ with the last coordinate omitted
But,I cannot see why this should be injective and surjective?
Any help from anyone is welcome
algebraic-geometry sheaf-theory
asked Dec 28 '18 at 13:02
HARRYHARRY
889
$endgroup$
add a comment |
$begingroup$
$underline {Background}$: Suppose $X$ is a scheme and $mathcal F$ is a sheaf of $mathcal O_X$ module.
Suppose we have, $mathcal {F}oplus mathcal {O}_X cong oplus_r mathcal {O}_X$ as sheaves of $mathcal O_X$ modules.
$underline {Question}$: Can we say that $mathcal {F} cong oplus_{r-1} mathcal {O}_X$ ?
$underline {Attempt}$:I tried to define map on all {$SpecA$} which in turn will define a morphism as follows,
If we denote the isomorphism $mathcal {F}oplus mathcal {O}_X cong oplus_r mathcal {O}_X$ as $phi$
then I define $T(specA):mathcal F(specA)to oplus_{r-1}mathcal O(specA)$ as
$T(specA)(a):=phi(SpecA)(a,0)$ with the last coordinate omitted
But,I cannot see why this should be injective and surjective?
Any help from anyone is welcome
algebraic-geometry sheaf-theory
asked Dec 28 '18 at 13:02
HARRYHARRY
889
$endgroup$
$underline {Background}$: Suppose $X$ is a scheme and $mathcal F$ is a sheaf of $mathcal O_X$ module.
Suppose we have, $mathcal {F}oplus mathcal {O}_X cong oplus_r mathcal {O}_X$ as sheaves of $mathcal O_X$ modules.
$underline {Question}$: Can we say that $mathcal {F} cong oplus_{r-1} mathcal {O}_X$ ?
$underline {Attempt}$:I tried to define map on all {$SpecA$} which in turn will define a morphism as follows,
If we denote the isomorphism $mathcal {F}oplus mathcal {O}_X cong oplus_r mathcal {O}_X$ as $phi$
then I define $T(specA):mathcal F(specA)to oplus_{r-1}mathcal O(specA)$ as
$T(specA)(a):=phi(SpecA)(a,0)$ with the last coordinate omitted
But,I cannot see why this should be injective and surjective?
Any help from anyone is welcome
algebraic-geometry sheaf-theory
algebraic-geometry sheaf-theory
asked Dec 28 '18 at 13:02
HARRYHARRY
889
asked Dec 28 '18 at 13:02
HARRYHARRY
889
asked Dec 28 '18 at 13:02
HARRYHARRY
889
asked Dec 28 '18 at 13:02
HARRYHARRY
889
asked Dec 28 '18 at 13:02
HARRYHARRY
889
889
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the case that $r=2$ the answer is yes. The affine case is discussed at this question: $Moplus A cong Aoplus A$ implies $Mcong A$?, and the proof works just as well for all schemes.
If $r geq 3$ the answer is no. It is enough to consider affine schemes. So we have a ring $R$ and an isomorphism of modules $M oplus R cong R^{oplus r}$, the question is whether $M$ is free of rank $r-1$.
You should be aware of the following definition:
Definition. A projective module $P$ is called stably free if $Poplus R^{oplus n} cong R^{oplus m}$ for some $m,n$.
This is a little more general than what you are asking (the extra parameter $n$) but given any stable free module as above then, $M = Poplus R^{oplus n-1}$ will be a counterexample to your question. (Note: your $M$ is necessarily a projective module, so locally it is isomorphic to $R^{oplus r-1}$.)
Examples of stably-free modules which are not free are a little hard to come by, but for instance Keith Conrad has a note with an example over the real coordinate ring of the 2-sphere. For more examples see papers of Swan and of Kumar et al. There is some discussion about this at this question on overflow as well. It mentions the Bass Cancellation theorem, which says that for large enough rank (compared to the Krull dimension of the ring) stably free modules actually are free.
answered Dec 28 '18 at 13:25
BenBen
3,946617
$endgroup$
$begingroup$
,thank you for your answer.I understand that M has to be projective module,but can you explain why locally it is isomorphic to $R^{oplus r-1}$
$endgroup$
– HARRY
Dec 28 '18 at 13:36
1
$begingroup$
Being projective is equivalent to being locally free. Are you asking why its dimension has to be $r-1$? You can just pass to a closed point where you have an isomorphism of vector spaces, and you have additivity of their dimensions.
$endgroup$
– Ben
Dec 28 '18 at 13:39
1
$begingroup$
@HARRY Yes you can derive that. This is the equivalence between locally free and projective modules. Did you want a reference for the proof? You can find it by scouring the standard set of commutative algebra textbooks I think.
$endgroup$
– Ben
Dec 28 '18 at 13:55
1
$begingroup$
@HARRY You know that $Moplus R = R^{oplus n}$. Locally $M cong F$ to a free module $F$. So on an open set $Foplus R = R^{oplus n}$. Now you need to know the rank of $F$ is $n-1$. Let $m subset R$ be a maximal ideal and $R/m = k$ is a field. Then $R^{oplus n} otimes_R k = k^{oplus n}$. In other words the rank is preserved. Then the fact that $dim F = n-1$ is a familiar fact about vector spaces.
$endgroup$
– Ben
Dec 28 '18 at 14:16
1
$begingroup$
@HARRY You could also just use that commutative rings have invariant basis property, if you are familiar with that.
$endgroup$
– Ben
Dec 28 '18 at 14:17
|
show 2 more comments
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1 Answer
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$begingroup$
In the case that $r=2$ the answer is yes. The affine case is discussed at this question: $Moplus A cong Aoplus A$ implies $Mcong A$?, and the proof works just as well for all schemes.
If $r geq 3$ the answer is no. It is enough to consider affine schemes. So we have a ring $R$ and an isomorphism of modules $M oplus R cong R^{oplus r}$, the question is whether $M$ is free of rank $r-1$.
You should be aware of the following definition:
Definition. A projective module $P$ is called stably free if $Poplus R^{oplus n} cong R^{oplus m}$ for some $m,n$.
This is a little more general than what you are asking (the extra parameter $n$) but given any stable free module as above then, $M = Poplus R^{oplus n-1}$ will be a counterexample to your question. (Note: your $M$ is necessarily a projective module, so locally it is isomorphic to $R^{oplus r-1}$.)
Examples of stably-free modules which are not free are a little hard to come by, but for instance Keith Conrad has a note with an example over the real coordinate ring of the 2-sphere. For more examples see papers of Swan and of Kumar et al. There is some discussion about this at this question on overflow as well. It mentions the Bass Cancellation theorem, which says that for large enough rank (compared to the Krull dimension of the ring) stably free modules actually are free.
answered Dec 28 '18 at 13:25
BenBen
3,946617
$endgroup$
$begingroup$
,thank you for your answer.I understand that M has to be projective module,but can you explain why locally it is isomorphic to $R^{oplus r-1}$
$endgroup$
– HARRY
Dec 28 '18 at 13:36
1
$begingroup$
Being projective is equivalent to being locally free. Are you asking why its dimension has to be $r-1$? You can just pass to a closed point where you have an isomorphism of vector spaces, and you have additivity of their dimensions.
$endgroup$
– Ben
Dec 28 '18 at 13:39
1
$begingroup$
@HARRY Yes you can derive that. This is the equivalence between locally free and projective modules. Did you want a reference for the proof? You can find it by scouring the standard set of commutative algebra textbooks I think.
$endgroup$
– Ben
Dec 28 '18 at 13:55
1
$begingroup$
@HARRY You know that $Moplus R = R^{oplus n}$. Locally $M cong F$ to a free module $F$. So on an open set $Foplus R = R^{oplus n}$. Now you need to know the rank of $F$ is $n-1$. Let $m subset R$ be a maximal ideal and $R/m = k$ is a field. Then $R^{oplus n} otimes_R k = k^{oplus n}$. In other words the rank is preserved. Then the fact that $dim F = n-1$ is a familiar fact about vector spaces.
$endgroup$
– Ben
Dec 28 '18 at 14:16
1
$begingroup$
@HARRY You could also just use that commutative rings have invariant basis property, if you are familiar with that.
$endgroup$
– Ben
Dec 28 '18 at 14:17
|
show 2 more comments
$begingroup$
In the case that $r=2$ the answer is yes. The affine case is discussed at this question: $Moplus A cong Aoplus A$ implies $Mcong A$?, and the proof works just as well for all schemes.
If $r geq 3$ the answer is no. It is enough to consider affine schemes. So we have a ring $R$ and an isomorphism of modules $M oplus R cong R^{oplus r}$, the question is whether $M$ is free of rank $r-1$.
You should be aware of the following definition:
Definition. A projective module $P$ is called stably free if $Poplus R^{oplus n} cong R^{oplus m}$ for some $m,n$.
This is a little more general than what you are asking (the extra parameter $n$) but given any stable free module as above then, $M = Poplus R^{oplus n-1}$ will be a counterexample to your question. (Note: your $M$ is necessarily a projective module, so locally it is isomorphic to $R^{oplus r-1}$.)
Examples of stably-free modules which are not free are a little hard to come by, but for instance Keith Conrad has a note with an example over the real coordinate ring of the 2-sphere. For more examples see papers of Swan and of Kumar et al. There is some discussion about this at this question on overflow as well. It mentions the Bass Cancellation theorem, which says that for large enough rank (compared to the Krull dimension of the ring) stably free modules actually are free.
answered Dec 28 '18 at 13:25
BenBen
3,946617
$endgroup$
$begingroup$
,thank you for your answer.I understand that M has to be projective module,but can you explain why locally it is isomorphic to $R^{oplus r-1}$
$endgroup$
– HARRY
Dec 28 '18 at 13:36
1
$begingroup$
Being projective is equivalent to being locally free. Are you asking why its dimension has to be $r-1$? You can just pass to a closed point where you have an isomorphism of vector spaces, and you have additivity of their dimensions.
$endgroup$
– Ben
Dec 28 '18 at 13:39
1
$begingroup$
@HARRY Yes you can derive that. This is the equivalence between locally free and projective modules. Did you want a reference for the proof? You can find it by scouring the standard set of commutative algebra textbooks I think.
$endgroup$
– Ben
Dec 28 '18 at 13:55
1
$begingroup$
@HARRY You know that $Moplus R = R^{oplus n}$. Locally $M cong F$ to a free module $F$. So on an open set $Foplus R = R^{oplus n}$. Now you need to know the rank of $F$ is $n-1$. Let $m subset R$ be a maximal ideal and $R/m = k$ is a field. Then $R^{oplus n} otimes_R k = k^{oplus n}$. In other words the rank is preserved. Then the fact that $dim F = n-1$ is a familiar fact about vector spaces.
$endgroup$
– Ben
Dec 28 '18 at 14:16
1
$begingroup$
@HARRY You could also just use that commutative rings have invariant basis property, if you are familiar with that.
$endgroup$
– Ben
Dec 28 '18 at 14:17
|
show 2 more comments
$begingroup$
In the case that $r=2$ the answer is yes. The affine case is discussed at this question: $Moplus A cong Aoplus A$ implies $Mcong A$?, and the proof works just as well for all schemes.
If $r geq 3$ the answer is no. It is enough to consider affine schemes. So we have a ring $R$ and an isomorphism of modules $M oplus R cong R^{oplus r}$, the question is whether $M$ is free of rank $r-1$.
You should be aware of the following definition:
Definition. A projective module $P$ is called stably free if $Poplus R^{oplus n} cong R^{oplus m}$ for some $m,n$.
This is a little more general than what you are asking (the extra parameter $n$) but given any stable free module as above then, $M = Poplus R^{oplus n-1}$ will be a counterexample to your question. (Note: your $M$ is necessarily a projective module, so locally it is isomorphic to $R^{oplus r-1}$.)
Examples of stably-free modules which are not free are a little hard to come by, but for instance Keith Conrad has a note with an example over the real coordinate ring of the 2-sphere. For more examples see papers of Swan and of Kumar et al. There is some discussion about this at this question on overflow as well. It mentions the Bass Cancellation theorem, which says that for large enough rank (compared to the Krull dimension of the ring) stably free modules actually are free.
answered Dec 28 '18 at 13:25
BenBen
3,946617
$endgroup$
In the case that $r=2$ the answer is yes. The affine case is discussed at this question: $Moplus A cong Aoplus A$ implies $Mcong A$?, and the proof works just as well for all schemes.
If $r geq 3$ the answer is no. It is enough to consider affine schemes. So we have a ring $R$ and an isomorphism of modules $M oplus R cong R^{oplus r}$, the question is whether $M$ is free of rank $r-1$.
You should be aware of the following definition:
Definition. A projective module $P$ is called stably free if $Poplus R^{oplus n} cong R^{oplus m}$ for some $m,n$.
This is a little more general than what you are asking (the extra parameter $n$) but given any stable free module as above then, $M = Poplus R^{oplus n-1}$ will be a counterexample to your question. (Note: your $M$ is necessarily a projective module, so locally it is isomorphic to $R^{oplus r-1}$.)
Examples of stably-free modules which are not free are a little hard to come by, but for instance Keith Conrad has a note with an example over the real coordinate ring of the 2-sphere. For more examples see papers of Swan and of Kumar et al. There is some discussion about this at this question on overflow as well. It mentions the Bass Cancellation theorem, which says that for large enough rank (compared to the Krull dimension of the ring) stably free modules actually are free.
answered Dec 28 '18 at 13:25
BenBen
3,946617
edited Dec 28 '18 at 13:41
answered Dec 28 '18 at 13:25
BenBen
3,946617
answered Dec 28 '18 at 13:25
BenBen
3,946617
answered Dec 28 '18 at 13:25
BenBen
3,946617
3,946617
$begingroup$
,thank you for your answer.I understand that M has to be projective module,but can you explain why locally it is isomorphic to $R^{oplus r-1}$
$endgroup$
– HARRY
Dec 28 '18 at 13:36
1
$begingroup$
Being projective is equivalent to being locally free. Are you asking why its dimension has to be $r-1$? You can just pass to a closed point where you have an isomorphism of vector spaces, and you have additivity of their dimensions.
$endgroup$
– Ben
Dec 28 '18 at 13:39
1
$begingroup$
@HARRY Yes you can derive that. This is the equivalence between locally free and projective modules. Did you want a reference for the proof? You can find it by scouring the standard set of commutative algebra textbooks I think.
$endgroup$
– Ben
Dec 28 '18 at 13:55
1
$begingroup$
@HARRY You know that $Moplus R = R^{oplus n}$. Locally $M cong F$ to a free module $F$. So on an open set $Foplus R = R^{oplus n}$. Now you need to know the rank of $F$ is $n-1$. Let $m subset R$ be a maximal ideal and $R/m = k$ is a field. Then $R^{oplus n} otimes_R k = k^{oplus n}$. In other words the rank is preserved. Then the fact that $dim F = n-1$ is a familiar fact about vector spaces.
$endgroup$
– Ben
Dec 28 '18 at 14:16
1
$begingroup$
@HARRY You could also just use that commutative rings have invariant basis property, if you are familiar with that.
$endgroup$
– Ben
Dec 28 '18 at 14:17
|
show 2 more comments
$begingroup$
,thank you for your answer.I understand that M has to be projective module,but can you explain why locally it is isomorphic to $R^{oplus r-1}$
$endgroup$
– HARRY
Dec 28 '18 at 13:36
1
$begingroup$
Being projective is equivalent to being locally free. Are you asking why its dimension has to be $r-1$? You can just pass to a closed point where you have an isomorphism of vector spaces, and you have additivity of their dimensions.
$endgroup$
– Ben
Dec 28 '18 at 13:39
1
$begingroup$
@HARRY Yes you can derive that. This is the equivalence between locally free and projective modules. Did you want a reference for the proof? You can find it by scouring the standard set of commutative algebra textbooks I think.
$endgroup$
– Ben
Dec 28 '18 at 13:55
1
$begingroup$
@HARRY You know that $Moplus R = R^{oplus n}$. Locally $M cong F$ to a free module $F$. So on an open set $Foplus R = R^{oplus n}$. Now you need to know the rank of $F$ is $n-1$. Let $m subset R$ be a maximal ideal and $R/m = k$ is a field. Then $R^{oplus n} otimes_R k = k^{oplus n}$. In other words the rank is preserved. Then the fact that $dim F = n-1$ is a familiar fact about vector spaces.
$endgroup$
– Ben
Dec 28 '18 at 14:16
1
$begingroup$
@HARRY You could also just use that commutative rings have invariant basis property, if you are familiar with that.
$endgroup$
– Ben
Dec 28 '18 at 14:17
$begingroup$
,thank you for your answer.I understand that M has to be projective module,but can you explain why locally it is isomorphic to $R^{oplus r-1}$
$endgroup$
– HARRY
Dec 28 '18 at 13:36
$begingroup$
,thank you for your answer.I understand that M has to be projective module,but can you explain why locally it is isomorphic to $R^{oplus r-1}$
$endgroup$
– HARRY
Dec 28 '18 at 13:36
1
1
$begingroup$
Being projective is equivalent to being locally free. Are you asking why its dimension has to be $r-1$? You can just pass to a closed point where you have an isomorphism of vector spaces, and you have additivity of their dimensions.
$endgroup$
– Ben
Dec 28 '18 at 13:39
$begingroup$
Being projective is equivalent to being locally free. Are you asking why its dimension has to be $r-1$? You can just pass to a closed point where you have an isomorphism of vector spaces, and you have additivity of their dimensions.
$endgroup$
– Ben
Dec 28 '18 at 13:39
1
1
$begingroup$
@HARRY Yes you can derive that. This is the equivalence between locally free and projective modules. Did you want a reference for the proof? You can find it by scouring the standard set of commutative algebra textbooks I think.
$endgroup$
– Ben
Dec 28 '18 at 13:55
$begingroup$
@HARRY Yes you can derive that. This is the equivalence between locally free and projective modules. Did you want a reference for the proof? You can find it by scouring the standard set of commutative algebra textbooks I think.
$endgroup$
– Ben
Dec 28 '18 at 13:55
1
1
$begingroup$
@HARRY You know that $Moplus R = R^{oplus n}$. Locally $M cong F$ to a free module $F$. So on an open set $Foplus R = R^{oplus n}$. Now you need to know the rank of $F$ is $n-1$. Let $m subset R$ be a maximal ideal and $R/m = k$ is a field. Then $R^{oplus n} otimes_R k = k^{oplus n}$. In other words the rank is preserved. Then the fact that $dim F = n-1$ is a familiar fact about vector spaces.
$endgroup$
– Ben
Dec 28 '18 at 14:16
$begingroup$
@HARRY You know that $Moplus R = R^{oplus n}$. Locally $M cong F$ to a free module $F$. So on an open set $Foplus R = R^{oplus n}$. Now you need to know the rank of $F$ is $n-1$. Let $m subset R$ be a maximal ideal and $R/m = k$ is a field. Then $R^{oplus n} otimes_R k = k^{oplus n}$. In other words the rank is preserved. Then the fact that $dim F = n-1$ is a familiar fact about vector spaces.
$endgroup$
– Ben
Dec 28 '18 at 14:16
1
1
$begingroup$
@HARRY You could also just use that commutative rings have invariant basis property, if you are familiar with that.
$endgroup$
– Ben
Dec 28 '18 at 14:17
$begingroup$
@HARRY You could also just use that commutative rings have invariant basis property, if you are familiar with that.
$endgroup$
– Ben
Dec 28 '18 at 14:17
|
show 2 more comments
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