Mixing
Inclusion of closed subschemes.
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Let $X$ be a scheme and $Y$ and $Z$ closed subschemes. What does it mean for $Y$ to be contained in $Z$?
This question adresses the same question, but it covers only the affine case and it uses a different definition of a closed subscheme. So for the non-affine case, let $(Y, mathcal{O}_Y)$ and $(Z, mathcal{O}_Z)$ be two closed subschemas of a scheme $(X, mathcal{O}_X)$. Then $mathcal{O}_Y$ and $mathcal{O}_Z$ are quotients of the structure sheaf $mathcal{O}_X$ by a quasi-coherent sheaf of ideals, that is,
$$mathcal{O}_Y = mathcal{O}_X/mathcal{I}$$
and
$$mathcal{O}_Z = mathcal{O}_X/mathcal{J} , .$$
Then, to say that $Y subset Z$, we would like to say that $|Y| subset |X|$ as topological spaces, and that $mathcal{O}_Y$ is a sub sheaf of $mathcal{O}_Z$. I believe we can sharpen it a bit more by saying that the quasi-coherent sheaf $mathcal{J}$ is a subsheaf of $mathcal{I}$. Would this suffice?
algebraic-geometry sheaf-theory schemes
asked 2 days ago
user313212
282520
add a comment |
up vote
1
down vote
favorite
Let $X$ be a scheme and $Y$ and $Z$ closed subschemes. What does it mean for $Y$ to be contained in $Z$?
This question adresses the same question, but it covers only the affine case and it uses a different definition of a closed subscheme. So for the non-affine case, let $(Y, mathcal{O}_Y)$ and $(Z, mathcal{O}_Z)$ be two closed subschemas of a scheme $(X, mathcal{O}_X)$. Then $mathcal{O}_Y$ and $mathcal{O}_Z$ are quotients of the structure sheaf $mathcal{O}_X$ by a quasi-coherent sheaf of ideals, that is,
$$mathcal{O}_Y = mathcal{O}_X/mathcal{I}$$
and
$$mathcal{O}_Z = mathcal{O}_X/mathcal{J} , .$$
Then, to say that $Y subset Z$, we would like to say that $|Y| subset |X|$ as topological spaces, and that $mathcal{O}_Y$ is a sub sheaf of $mathcal{O}_Z$. I believe we can sharpen it a bit more by saying that the quasi-coherent sheaf $mathcal{J}$ is a subsheaf of $mathcal{I}$. Would this suffice?
algebraic-geometry sheaf-theory schemes
asked 2 days ago
user313212
282520
1
Does this question answer your question? math.stackexchange.com/questions/2352513/…
– Alfred Yerger
2 days ago
@AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
– user313212
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ be a scheme and $Y$ and $Z$ closed subschemes. What does it mean for $Y$ to be contained in $Z$?
This question adresses the same question, but it covers only the affine case and it uses a different definition of a closed subscheme. So for the non-affine case, let $(Y, mathcal{O}_Y)$ and $(Z, mathcal{O}_Z)$ be two closed subschemas of a scheme $(X, mathcal{O}_X)$. Then $mathcal{O}_Y$ and $mathcal{O}_Z$ are quotients of the structure sheaf $mathcal{O}_X$ by a quasi-coherent sheaf of ideals, that is,
$$mathcal{O}_Y = mathcal{O}_X/mathcal{I}$$
and
$$mathcal{O}_Z = mathcal{O}_X/mathcal{J} , .$$
Then, to say that $Y subset Z$, we would like to say that $|Y| subset |X|$ as topological spaces, and that $mathcal{O}_Y$ is a sub sheaf of $mathcal{O}_Z$. I believe we can sharpen it a bit more by saying that the quasi-coherent sheaf $mathcal{J}$ is a subsheaf of $mathcal{I}$. Would this suffice?
algebraic-geometry sheaf-theory schemes
asked 2 days ago
user313212
282520
Let $X$ be a scheme and $Y$ and $Z$ closed subschemes. What does it mean for $Y$ to be contained in $Z$?
This question adresses the same question, but it covers only the affine case and it uses a different definition of a closed subscheme. So for the non-affine case, let $(Y, mathcal{O}_Y)$ and $(Z, mathcal{O}_Z)$ be two closed subschemas of a scheme $(X, mathcal{O}_X)$. Then $mathcal{O}_Y$ and $mathcal{O}_Z$ are quotients of the structure sheaf $mathcal{O}_X$ by a quasi-coherent sheaf of ideals, that is,
$$mathcal{O}_Y = mathcal{O}_X/mathcal{I}$$
and
$$mathcal{O}_Z = mathcal{O}_X/mathcal{J} , .$$
Then, to say that $Y subset Z$, we would like to say that $|Y| subset |X|$ as topological spaces, and that $mathcal{O}_Y$ is a sub sheaf of $mathcal{O}_Z$. I believe we can sharpen it a bit more by saying that the quasi-coherent sheaf $mathcal{J}$ is a subsheaf of $mathcal{I}$. Would this suffice?
algebraic-geometry sheaf-theory schemes
algebraic-geometry sheaf-theory schemes
asked 2 days ago
user313212
282520
asked 2 days ago
user313212
282520
asked 2 days ago
user313212
282520
asked 2 days ago
user313212
282520
asked 2 days ago
user313212
282520
282520
1
Does this question answer your question? math.stackexchange.com/questions/2352513/…
– Alfred Yerger
2 days ago
@AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
– user313212
2 days ago
add a comment |
1
Does this question answer your question? math.stackexchange.com/questions/2352513/…
– Alfred Yerger
2 days ago
@AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
– user313212
2 days ago
1
1
Does this question answer your question? math.stackexchange.com/questions/2352513/…
– Alfred Yerger
2 days ago
Does this question answer your question? math.stackexchange.com/questions/2352513/…
– Alfred Yerger
2 days ago
@AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
– user313212
2 days ago
@AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
– user313212
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).
Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.
The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.
And then the answer to your question is given by the biblical simple equivalence:
$$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$
Edit: a simple but edifying example.
Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.
a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.
These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.
Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.
b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.
This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.
answered 2 days ago
Georges Elencwajg
118k7179322
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).
Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.
The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.
And then the answer to your question is given by the biblical simple equivalence:
$$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$
Edit: a simple but edifying example.
Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.
a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.
These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.
Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.
b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.
This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.
answered 2 days ago
Georges Elencwajg
118k7179322
add a comment |
up vote
3
down vote
accepted
For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).
Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.
The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.
And then the answer to your question is given by the biblical simple equivalence:
$$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$
Edit: a simple but edifying example.
Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.
a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.
These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.
Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.
b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.
This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.
answered 2 days ago
Georges Elencwajg
118k7179322
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).
Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.
The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.
And then the answer to your question is given by the biblical simple equivalence:
$$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$
Edit: a simple but edifying example.
Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.
a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.
These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.
Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.
b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.
This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.
answered 2 days ago
Georges Elencwajg
118k7179322
For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).
Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.
The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.
And then the answer to your question is given by the biblical simple equivalence:
$$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$
Edit: a simple but edifying example.
Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.
a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.
These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.
Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.
b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.
This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.
answered 2 days ago
Georges Elencwajg
118k7179322
edited yesterday
answered 2 days ago
Georges Elencwajg
118k7179322
answered 2 days ago
Georges Elencwajg
118k7179322
answered 2 days ago
Georges Elencwajg
118k7179322
118k7179322
add a comment |
add a comment |
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1
Does this question answer your question? math.stackexchange.com/questions/2352513/…
– Alfred Yerger
2 days ago
@AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
– user313212
2 days ago