Locally ringed space with noetherian stalks and a non-coherent structural sheaf
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I am looking for a locally ringed space the stalks of which are noetherian and such that the structural sheaf is not coherent over itself. Can you provide me an example of this?
Notice that one may not find such an example which is a scheme, since schemes with noetherian stalks are locally noetherian and any locally noetherian scheme has a coherent structural sheaf (over itself).
sheaf-theory coherent-sheaves locally-ringed-spaces
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I am looking for a locally ringed space the stalks of which are noetherian and such that the structural sheaf is not coherent over itself. Can you provide me an example of this?
Notice that one may not find such an example which is a scheme, since schemes with noetherian stalks are locally noetherian and any locally noetherian scheme has a coherent structural sheaf (over itself).
sheaf-theory coherent-sheaves locally-ringed-spaces
New contributor
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I am looking for a locally ringed space the stalks of which are noetherian and such that the structural sheaf is not coherent over itself. Can you provide me an example of this?
Notice that one may not find such an example which is a scheme, since schemes with noetherian stalks are locally noetherian and any locally noetherian scheme has a coherent structural sheaf (over itself).
sheaf-theory coherent-sheaves locally-ringed-spaces
New contributor
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I am looking for a locally ringed space the stalks of which are noetherian and such that the structural sheaf is not coherent over itself. Can you provide me an example of this?
Notice that one may not find such an example which is a scheme, since schemes with noetherian stalks are locally noetherian and any locally noetherian scheme has a coherent structural sheaf (over itself).
sheaf-theory coherent-sheaves locally-ringed-spaces
sheaf-theory coherent-sheaves locally-ringed-spaces
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GaussianGaussian
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Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.
Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.
Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.
Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$
For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.
In the special case that $Y$ equals $text{Spec} k[t]_{langle trangle}$ and $Z$ equals the singleton of the closed point, then also $mathcal{O}_X(Y)$ equals $k[t]_{langle t rangle} oplus k[t]_{langle t rangle}cdot epsilon$, which is Noetherian. Also $mathcal{O}_X(Ysetminus Z)$ equals $k(t)$, which is Noetherian. Thus, also the ring of sections is Noetherian for every open subset.
There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.
If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.
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A non-coherent ring whose stalks are noetherian, constructed by Harris and Nagata, is described on p. 51 in S. Glaz, Commutative coherent rings, Lecture Notes in Math. 1371, Springer, Berlin, 1989.
(Noetherianness of the stalks of the structure sheaf of a scheme does not imply local noetherianness of the scheme, not even in the affine case.)
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1
$begingroup$
Those example are even simpler. Let $R$ be $mathbb{Z}oplus Iepsilon$ where $I$ is the $mathbb{Z}$-module equal to the direct sum of $mathbb{Z}cdot b_0$ and $mathbb{Z}/pmathbb{Z}cdot b_p$ for every prime integer $p$, where the elements $b_0$ and $b_p$ are placeholders. Every local ring of $R$ is Noetherian, either $mathbb{Q}oplus mathbb{Q}cdot b_0epsilon$ or $mathbb{Z}/pmathbb{Z}oplus (mathbb{Z}/pmathbb{Z}cdot b_0 oplus mathbb{Z}/pmathbb{Z}cdot b_p)epsilon$. The kernel of the "multiplication by $b_0epsilon$" map is not finitely generated.
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– Jason Starr
1 hour ago
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Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.
Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.
Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.
Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$
For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.
In the special case that $Y$ equals $text{Spec} k[t]_{langle trangle}$ and $Z$ equals the singleton of the closed point, then also $mathcal{O}_X(Y)$ equals $k[t]_{langle t rangle} oplus k[t]_{langle t rangle}cdot epsilon$, which is Noetherian. Also $mathcal{O}_X(Ysetminus Z)$ equals $k(t)$, which is Noetherian. Thus, also the ring of sections is Noetherian for every open subset.
There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.
If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.
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add a comment |
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Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.
Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.
Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.
Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$
For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.
In the special case that $Y$ equals $text{Spec} k[t]_{langle trangle}$ and $Z$ equals the singleton of the closed point, then also $mathcal{O}_X(Y)$ equals $k[t]_{langle t rangle} oplus k[t]_{langle t rangle}cdot epsilon$, which is Noetherian. Also $mathcal{O}_X(Ysetminus Z)$ equals $k(t)$, which is Noetherian. Thus, also the ring of sections is Noetherian for every open subset.
There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.
If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.
$endgroup$
add a comment |
$begingroup$
Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.
Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.
Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.
Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$
For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.
In the special case that $Y$ equals $text{Spec} k[t]_{langle trangle}$ and $Z$ equals the singleton of the closed point, then also $mathcal{O}_X(Y)$ equals $k[t]_{langle t rangle} oplus k[t]_{langle t rangle}cdot epsilon$, which is Noetherian. Also $mathcal{O}_X(Ysetminus Z)$ equals $k(t)$, which is Noetherian. Thus, also the ring of sections is Noetherian for every open subset.
There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.
If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.
$endgroup$
Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.
Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.
Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.
Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$
For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.
In the special case that $Y$ equals $text{Spec} k[t]_{langle trangle}$ and $Z$ equals the singleton of the closed point, then also $mathcal{O}_X(Y)$ equals $k[t]_{langle t rangle} oplus k[t]_{langle t rangle}cdot epsilon$, which is Noetherian. Also $mathcal{O}_X(Ysetminus Z)$ equals $k(t)$, which is Noetherian. Thus, also the ring of sections is Noetherian for every open subset.
There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.
If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.
edited 45 mins ago
community wiki
4 revs
Jason Starr
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add a comment |
$begingroup$
A non-coherent ring whose stalks are noetherian, constructed by Harris and Nagata, is described on p. 51 in S. Glaz, Commutative coherent rings, Lecture Notes in Math. 1371, Springer, Berlin, 1989.
(Noetherianness of the stalks of the structure sheaf of a scheme does not imply local noetherianness of the scheme, not even in the affine case.)
$endgroup$
1
$begingroup$
Those example are even simpler. Let $R$ be $mathbb{Z}oplus Iepsilon$ where $I$ is the $mathbb{Z}$-module equal to the direct sum of $mathbb{Z}cdot b_0$ and $mathbb{Z}/pmathbb{Z}cdot b_p$ for every prime integer $p$, where the elements $b_0$ and $b_p$ are placeholders. Every local ring of $R$ is Noetherian, either $mathbb{Q}oplus mathbb{Q}cdot b_0epsilon$ or $mathbb{Z}/pmathbb{Z}oplus (mathbb{Z}/pmathbb{Z}cdot b_0 oplus mathbb{Z}/pmathbb{Z}cdot b_p)epsilon$. The kernel of the "multiplication by $b_0epsilon$" map is not finitely generated.
$endgroup$
– Jason Starr
1 hour ago
add a comment |
$begingroup$
A non-coherent ring whose stalks are noetherian, constructed by Harris and Nagata, is described on p. 51 in S. Glaz, Commutative coherent rings, Lecture Notes in Math. 1371, Springer, Berlin, 1989.
(Noetherianness of the stalks of the structure sheaf of a scheme does not imply local noetherianness of the scheme, not even in the affine case.)
$endgroup$
1
$begingroup$
Those example are even simpler. Let $R$ be $mathbb{Z}oplus Iepsilon$ where $I$ is the $mathbb{Z}$-module equal to the direct sum of $mathbb{Z}cdot b_0$ and $mathbb{Z}/pmathbb{Z}cdot b_p$ for every prime integer $p$, where the elements $b_0$ and $b_p$ are placeholders. Every local ring of $R$ is Noetherian, either $mathbb{Q}oplus mathbb{Q}cdot b_0epsilon$ or $mathbb{Z}/pmathbb{Z}oplus (mathbb{Z}/pmathbb{Z}cdot b_0 oplus mathbb{Z}/pmathbb{Z}cdot b_p)epsilon$. The kernel of the "multiplication by $b_0epsilon$" map is not finitely generated.
$endgroup$
– Jason Starr
1 hour ago
add a comment |
$begingroup$
A non-coherent ring whose stalks are noetherian, constructed by Harris and Nagata, is described on p. 51 in S. Glaz, Commutative coherent rings, Lecture Notes in Math. 1371, Springer, Berlin, 1989.
(Noetherianness of the stalks of the structure sheaf of a scheme does not imply local noetherianness of the scheme, not even in the affine case.)
$endgroup$
A non-coherent ring whose stalks are noetherian, constructed by Harris and Nagata, is described on p. 51 in S. Glaz, Commutative coherent rings, Lecture Notes in Math. 1371, Springer, Berlin, 1989.
(Noetherianness of the stalks of the structure sheaf of a scheme does not imply local noetherianness of the scheme, not even in the affine case.)
answered 2 hours ago
Fred RohrerFred Rohrer
4,43111734
4,43111734
1
$begingroup$
Those example are even simpler. Let $R$ be $mathbb{Z}oplus Iepsilon$ where $I$ is the $mathbb{Z}$-module equal to the direct sum of $mathbb{Z}cdot b_0$ and $mathbb{Z}/pmathbb{Z}cdot b_p$ for every prime integer $p$, where the elements $b_0$ and $b_p$ are placeholders. Every local ring of $R$ is Noetherian, either $mathbb{Q}oplus mathbb{Q}cdot b_0epsilon$ or $mathbb{Z}/pmathbb{Z}oplus (mathbb{Z}/pmathbb{Z}cdot b_0 oplus mathbb{Z}/pmathbb{Z}cdot b_p)epsilon$. The kernel of the "multiplication by $b_0epsilon$" map is not finitely generated.
$endgroup$
– Jason Starr
1 hour ago
add a comment |
1
$begingroup$
Those example are even simpler. Let $R$ be $mathbb{Z}oplus Iepsilon$ where $I$ is the $mathbb{Z}$-module equal to the direct sum of $mathbb{Z}cdot b_0$ and $mathbb{Z}/pmathbb{Z}cdot b_p$ for every prime integer $p$, where the elements $b_0$ and $b_p$ are placeholders. Every local ring of $R$ is Noetherian, either $mathbb{Q}oplus mathbb{Q}cdot b_0epsilon$ or $mathbb{Z}/pmathbb{Z}oplus (mathbb{Z}/pmathbb{Z}cdot b_0 oplus mathbb{Z}/pmathbb{Z}cdot b_p)epsilon$. The kernel of the "multiplication by $b_0epsilon$" map is not finitely generated.
$endgroup$
– Jason Starr
1 hour ago
1
1
$begingroup$
Those example are even simpler. Let $R$ be $mathbb{Z}oplus Iepsilon$ where $I$ is the $mathbb{Z}$-module equal to the direct sum of $mathbb{Z}cdot b_0$ and $mathbb{Z}/pmathbb{Z}cdot b_p$ for every prime integer $p$, where the elements $b_0$ and $b_p$ are placeholders. Every local ring of $R$ is Noetherian, either $mathbb{Q}oplus mathbb{Q}cdot b_0epsilon$ or $mathbb{Z}/pmathbb{Z}oplus (mathbb{Z}/pmathbb{Z}cdot b_0 oplus mathbb{Z}/pmathbb{Z}cdot b_p)epsilon$. The kernel of the "multiplication by $b_0epsilon$" map is not finitely generated.
$endgroup$
– Jason Starr
1 hour ago
$begingroup$
Those example are even simpler. Let $R$ be $mathbb{Z}oplus Iepsilon$ where $I$ is the $mathbb{Z}$-module equal to the direct sum of $mathbb{Z}cdot b_0$ and $mathbb{Z}/pmathbb{Z}cdot b_p$ for every prime integer $p$, where the elements $b_0$ and $b_p$ are placeholders. Every local ring of $R$ is Noetherian, either $mathbb{Q}oplus mathbb{Q}cdot b_0epsilon$ or $mathbb{Z}/pmathbb{Z}oplus (mathbb{Z}/pmathbb{Z}cdot b_0 oplus mathbb{Z}/pmathbb{Z}cdot b_p)epsilon$. The kernel of the "multiplication by $b_0epsilon$" map is not finitely generated.
$endgroup$
– Jason Starr
1 hour ago
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Gaussian is a new contributor. Be nice, and check out our Code of Conduct.
Gaussian is a new contributor. Be nice, and check out our Code of Conduct.
Gaussian is a new contributor. Be nice, and check out our Code of Conduct.
Gaussian is a new contributor. Be nice, and check out our Code of Conduct.
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