Invertible ideal sheaf.
$begingroup$
I am a little bit confused with the idea of an invertible ideal sheaf. I cannot convince myself that there are invertible ideal sheaves on a scheme $X$ non isomorphic to the structure sheaf of $X$. I would like to know where my reasoning fails.
Given an invertible ideal sheaf $mathcal{I} subset mathcal{O}_X$ it comes equipped with the inclusion morphism into the structure sheaf. If the localization $mathcal{I}_x cong mathcal{O}_{X,x}$ how is possible that the inclusion morphism does not induce an isomorphism between stalks?
algebraic-geometry sheaf-theory
asked Dec 19 '16 at 10:26
AbellanAbellan
1,796721
$endgroup$
add a comment |
$begingroup$
I am a little bit confused with the idea of an invertible ideal sheaf. I cannot convince myself that there are invertible ideal sheaves on a scheme $X$ non isomorphic to the structure sheaf of $X$. I would like to know where my reasoning fails.
Given an invertible ideal sheaf $mathcal{I} subset mathcal{O}_X$ it comes equipped with the inclusion morphism into the structure sheaf. If the localization $mathcal{I}_x cong mathcal{O}_{X,x}$ how is possible that the inclusion morphism does not induce an isomorphism between stalks?
algebraic-geometry sheaf-theory
asked Dec 19 '16 at 10:26
AbellanAbellan
1,796721
$endgroup$
1
$begingroup$
The point is that an isomorphism $I_x cong O_{X,x}$ is abstract, it is not induced by the embedding $I to O_X$. In fact, if $I$ is an invertible ideal, the morphism $I_x to O_{X,x}$ induced by the embedding is zero for some points $x in X$.
$endgroup$
– Sasha
Dec 19 '16 at 10:41
$begingroup$
Hi @Sasha, I knew the reason must be that but I couldn't find myself a nice example. Could you add an example as an answer so I can accept it? Thanks for your time :)
$endgroup$
– Abellan
Dec 19 '16 at 10:44
$begingroup$
One moment in the affine case if $I subset A$ is an invertible ideal if we consider some maximal ideal $I subset mathfrak{m}$ then no element of $I_{mathfrak{m}}$ will be mapped to $1 in A_{mathfrak{m}}$ right?
$endgroup$
– Abellan
Dec 19 '16 at 11:06
1
$begingroup$
The simplest example is $I = O(-1)$ on $P^1$, with the embedding $O(-1) to O$ corresponding to a point $x_0 in P^1$. In homogeneous coordinates $(u:v)$ on $P^1$ if $x_0 = (u_0:v_0)$ then this map is given by $v_0 u - u_0 v in Gamma(P^1,O(1)) = Hom(O(-1),O)$. Then the map $I_x to O_{X,x}$ is an isomorphism for $x ne x_0$ and is zero for $x = x_0$.
$endgroup$
– Sasha
Dec 19 '16 at 11:16
add a comment |
$begingroup$
I am a little bit confused with the idea of an invertible ideal sheaf. I cannot convince myself that there are invertible ideal sheaves on a scheme $X$ non isomorphic to the structure sheaf of $X$. I would like to know where my reasoning fails.
Given an invertible ideal sheaf $mathcal{I} subset mathcal{O}_X$ it comes equipped with the inclusion morphism into the structure sheaf. If the localization $mathcal{I}_x cong mathcal{O}_{X,x}$ how is possible that the inclusion morphism does not induce an isomorphism between stalks?
algebraic-geometry sheaf-theory
asked Dec 19 '16 at 10:26
AbellanAbellan
1,796721
$endgroup$
I am a little bit confused with the idea of an invertible ideal sheaf. I cannot convince myself that there are invertible ideal sheaves on a scheme $X$ non isomorphic to the structure sheaf of $X$. I would like to know where my reasoning fails.
Given an invertible ideal sheaf $mathcal{I} subset mathcal{O}_X$ it comes equipped with the inclusion morphism into the structure sheaf. If the localization $mathcal{I}_x cong mathcal{O}_{X,x}$ how is possible that the inclusion morphism does not induce an isomorphism between stalks?
algebraic-geometry sheaf-theory
algebraic-geometry sheaf-theory
asked Dec 19 '16 at 10:26
AbellanAbellan
1,796721
asked Dec 19 '16 at 10:26
AbellanAbellan
1,796721
asked Dec 19 '16 at 10:26
AbellanAbellan
1,796721
asked Dec 19 '16 at 10:26
AbellanAbellan
1,796721
asked Dec 19 '16 at 10:26
AbellanAbellan
1,796721
1,796721
1
$begingroup$
The point is that an isomorphism $I_x cong O_{X,x}$ is abstract, it is not induced by the embedding $I to O_X$. In fact, if $I$ is an invertible ideal, the morphism $I_x to O_{X,x}$ induced by the embedding is zero for some points $x in X$.
$endgroup$
– Sasha
Dec 19 '16 at 10:41
$begingroup$
Hi @Sasha, I knew the reason must be that but I couldn't find myself a nice example. Could you add an example as an answer so I can accept it? Thanks for your time :)
$endgroup$
– Abellan
Dec 19 '16 at 10:44
$begingroup$
One moment in the affine case if $I subset A$ is an invertible ideal if we consider some maximal ideal $I subset mathfrak{m}$ then no element of $I_{mathfrak{m}}$ will be mapped to $1 in A_{mathfrak{m}}$ right?
$endgroup$
– Abellan
Dec 19 '16 at 11:06
1
$begingroup$
The simplest example is $I = O(-1)$ on $P^1$, with the embedding $O(-1) to O$ corresponding to a point $x_0 in P^1$. In homogeneous coordinates $(u:v)$ on $P^1$ if $x_0 = (u_0:v_0)$ then this map is given by $v_0 u - u_0 v in Gamma(P^1,O(1)) = Hom(O(-1),O)$. Then the map $I_x to O_{X,x}$ is an isomorphism for $x ne x_0$ and is zero for $x = x_0$.
$endgroup$
– Sasha
Dec 19 '16 at 11:16
add a comment |
1
$begingroup$
The point is that an isomorphism $I_x cong O_{X,x}$ is abstract, it is not induced by the embedding $I to O_X$. In fact, if $I$ is an invertible ideal, the morphism $I_x to O_{X,x}$ induced by the embedding is zero for some points $x in X$.
$endgroup$
– Sasha
Dec 19 '16 at 10:41
$begingroup$
Hi @Sasha, I knew the reason must be that but I couldn't find myself a nice example. Could you add an example as an answer so I can accept it? Thanks for your time :)
$endgroup$
– Abellan
Dec 19 '16 at 10:44
$begingroup$
One moment in the affine case if $I subset A$ is an invertible ideal if we consider some maximal ideal $I subset mathfrak{m}$ then no element of $I_{mathfrak{m}}$ will be mapped to $1 in A_{mathfrak{m}}$ right?
$endgroup$
– Abellan
Dec 19 '16 at 11:06
1
$begingroup$
The simplest example is $I = O(-1)$ on $P^1$, with the embedding $O(-1) to O$ corresponding to a point $x_0 in P^1$. In homogeneous coordinates $(u:v)$ on $P^1$ if $x_0 = (u_0:v_0)$ then this map is given by $v_0 u - u_0 v in Gamma(P^1,O(1)) = Hom(O(-1),O)$. Then the map $I_x to O_{X,x}$ is an isomorphism for $x ne x_0$ and is zero for $x = x_0$.
$endgroup$
– Sasha
Dec 19 '16 at 11:16
1
1
$begingroup$
The point is that an isomorphism $I_x cong O_{X,x}$ is abstract, it is not induced by the embedding $I to O_X$. In fact, if $I$ is an invertible ideal, the morphism $I_x to O_{X,x}$ induced by the embedding is zero for some points $x in X$.
$endgroup$
– Sasha
Dec 19 '16 at 10:41
$begingroup$
The point is that an isomorphism $I_x cong O_{X,x}$ is abstract, it is not induced by the embedding $I to O_X$. In fact, if $I$ is an invertible ideal, the morphism $I_x to O_{X,x}$ induced by the embedding is zero for some points $x in X$.
$endgroup$
– Sasha
Dec 19 '16 at 10:41
$begingroup$
Hi @Sasha, I knew the reason must be that but I couldn't find myself a nice example. Could you add an example as an answer so I can accept it? Thanks for your time :)
$endgroup$
– Abellan
Dec 19 '16 at 10:44
$begingroup$
Hi @Sasha, I knew the reason must be that but I couldn't find myself a nice example. Could you add an example as an answer so I can accept it? Thanks for your time :)
$endgroup$
– Abellan
Dec 19 '16 at 10:44
$begingroup$
One moment in the affine case if $I subset A$ is an invertible ideal if we consider some maximal ideal $I subset mathfrak{m}$ then no element of $I_{mathfrak{m}}$ will be mapped to $1 in A_{mathfrak{m}}$ right?
$endgroup$
– Abellan
Dec 19 '16 at 11:06
$begingroup$
One moment in the affine case if $I subset A$ is an invertible ideal if we consider some maximal ideal $I subset mathfrak{m}$ then no element of $I_{mathfrak{m}}$ will be mapped to $1 in A_{mathfrak{m}}$ right?
$endgroup$
– Abellan
Dec 19 '16 at 11:06
1
1
$begingroup$
The simplest example is $I = O(-1)$ on $P^1$, with the embedding $O(-1) to O$ corresponding to a point $x_0 in P^1$. In homogeneous coordinates $(u:v)$ on $P^1$ if $x_0 = (u_0:v_0)$ then this map is given by $v_0 u - u_0 v in Gamma(P^1,O(1)) = Hom(O(-1),O)$. Then the map $I_x to O_{X,x}$ is an isomorphism for $x ne x_0$ and is zero for $x = x_0$.
$endgroup$
– Sasha
Dec 19 '16 at 11:16
$begingroup$
The simplest example is $I = O(-1)$ on $P^1$, with the embedding $O(-1) to O$ corresponding to a point $x_0 in P^1$. In homogeneous coordinates $(u:v)$ on $P^1$ if $x_0 = (u_0:v_0)$ then this map is given by $v_0 u - u_0 v in Gamma(P^1,O(1)) = Hom(O(-1),O)$. Then the map $I_x to O_{X,x}$ is an isomorphism for $x ne x_0$ and is zero for $x = x_0$.
$endgroup$
– Sasha
Dec 19 '16 at 11:16
add a comment |
1 Answer
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Consider another example : Let $X=Spec A$ where $A=k[x,y]/y^2=x^3-x$ and $k$ some algebraically closed field. $X$ is a nonsingular elliptic curve, hence $A$ is a Dedekind domain and every prime ideal is maximal. We want to show that every maximal ideal is invertible i.e. $A_p cong m_p$. Let $m$ be the maximal ideal and $p$ any prime ideal. If $m notsubset p$ then $(A-p) cap m neq emptyset$ so $A_p cong m_p$. In case $msubset p$ we have $m =p$ (since every prime ideal is maximal) and $A_m$ is a DVR hence $mA$ is a principal ideal i.e invertible.
answered Jan 6 at 17:14
GregoryEGregoryE
211518
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$begingroup$
Consider another example : Let $X=Spec A$ where $A=k[x,y]/y^2=x^3-x$ and $k$ some algebraically closed field. $X$ is a nonsingular elliptic curve, hence $A$ is a Dedekind domain and every prime ideal is maximal. We want to show that every maximal ideal is invertible i.e. $A_p cong m_p$. Let $m$ be the maximal ideal and $p$ any prime ideal. If $m notsubset p$ then $(A-p) cap m neq emptyset$ so $A_p cong m_p$. In case $msubset p$ we have $m =p$ (since every prime ideal is maximal) and $A_m$ is a DVR hence $mA$ is a principal ideal i.e invertible.
answered Jan 6 at 17:14
GregoryEGregoryE
211518
$endgroup$
add a comment |
$begingroup$
Consider another example : Let $X=Spec A$ where $A=k[x,y]/y^2=x^3-x$ and $k$ some algebraically closed field. $X$ is a nonsingular elliptic curve, hence $A$ is a Dedekind domain and every prime ideal is maximal. We want to show that every maximal ideal is invertible i.e. $A_p cong m_p$. Let $m$ be the maximal ideal and $p$ any prime ideal. If $m notsubset p$ then $(A-p) cap m neq emptyset$ so $A_p cong m_p$. In case $msubset p$ we have $m =p$ (since every prime ideal is maximal) and $A_m$ is a DVR hence $mA$ is a principal ideal i.e invertible.
answered Jan 6 at 17:14
GregoryEGregoryE
211518
$endgroup$
add a comment |
$begingroup$
Consider another example : Let $X=Spec A$ where $A=k[x,y]/y^2=x^3-x$ and $k$ some algebraically closed field. $X$ is a nonsingular elliptic curve, hence $A$ is a Dedekind domain and every prime ideal is maximal. We want to show that every maximal ideal is invertible i.e. $A_p cong m_p$. Let $m$ be the maximal ideal and $p$ any prime ideal. If $m notsubset p$ then $(A-p) cap m neq emptyset$ so $A_p cong m_p$. In case $msubset p$ we have $m =p$ (since every prime ideal is maximal) and $A_m$ is a DVR hence $mA$ is a principal ideal i.e invertible.
answered Jan 6 at 17:14
GregoryEGregoryE
211518
$endgroup$
Consider another example : Let $X=Spec A$ where $A=k[x,y]/y^2=x^3-x$ and $k$ some algebraically closed field. $X$ is a nonsingular elliptic curve, hence $A$ is a Dedekind domain and every prime ideal is maximal. We want to show that every maximal ideal is invertible i.e. $A_p cong m_p$. Let $m$ be the maximal ideal and $p$ any prime ideal. If $m notsubset p$ then $(A-p) cap m neq emptyset$ so $A_p cong m_p$. In case $msubset p$ we have $m =p$ (since every prime ideal is maximal) and $A_m$ is a DVR hence $mA$ is a principal ideal i.e invertible.
answered Jan 6 at 17:14
GregoryEGregoryE
211518
answered Jan 6 at 17:14
GregoryEGregoryE
211518
answered Jan 6 at 17:14
GregoryEGregoryE
211518
answered Jan 6 at 17:14
GregoryEGregoryE
211518
211518
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1
$begingroup$
The point is that an isomorphism $I_x cong O_{X,x}$ is abstract, it is not induced by the embedding $I to O_X$. In fact, if $I$ is an invertible ideal, the morphism $I_x to O_{X,x}$ induced by the embedding is zero for some points $x in X$.
$endgroup$
– Sasha
Dec 19 '16 at 10:41
$begingroup$
Hi @Sasha, I knew the reason must be that but I couldn't find myself a nice example. Could you add an example as an answer so I can accept it? Thanks for your time :)
$endgroup$
– Abellan
Dec 19 '16 at 10:44
$begingroup$
One moment in the affine case if $I subset A$ is an invertible ideal if we consider some maximal ideal $I subset mathfrak{m}$ then no element of $I_{mathfrak{m}}$ will be mapped to $1 in A_{mathfrak{m}}$ right?
$endgroup$
– Abellan
Dec 19 '16 at 11:06
1
$begingroup$
The simplest example is $I = O(-1)$ on $P^1$, with the embedding $O(-1) to O$ corresponding to a point $x_0 in P^1$. In homogeneous coordinates $(u:v)$ on $P^1$ if $x_0 = (u_0:v_0)$ then this map is given by $v_0 u - u_0 v in Gamma(P^1,O(1)) = Hom(O(-1),O)$. Then the map $I_x to O_{X,x}$ is an isomorphism for $x ne x_0$ and is zero for $x = x_0$.
$endgroup$
– Sasha
Dec 19 '16 at 11:16