Pushforward of constant sheaf on the generic point.











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Let $X$ be an integral scheme, and $g: eta to X$ the inclusion of its generic point. If $M$ is an abelian group and $M_eta$ is a constant sheaf on $eta$ taking value $M$, then is it true that $g_* M_eta = M_X$?



My thought is yes, as if $U subset X$ is open, then $g_* M_eta (U) = M_eta( g^{-1}(U)) = M$, as every nonempty open set will contain the generic point. However it seems that one may need that $X$ is normal (or some other hypothesis) for this to be true. For reference, this is Exercise 3.7 (Chap II) in Milne's Etale Cohomology.










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  • The argument in your third line is perfectly correct and normality (or any other property $X$ may or may not have) plays absolutely no role in this question.
    – Georges Elencwajg
    Nov 13 at 22:47










  • @GeorgesElencwajg Glad to know I'm not losing my mind, I also thought this was odd. Milne claims that normality is required for this (he goes on to claim that if $X$ is a curve with a node $z$, then there is an exact sequence $0to M_X to g_*M_eta to i_*M_eta to 0$, where $i:Z to X$ is the inclusion of the nodal point). Is it possible he's referring to this scenario in the Etale topology (implicitly, I'm also asking if you know if this is true for the Etale topology)?
    – DKS
    Nov 14 at 0:30






  • 1




    Dear DKS, since I'm not 100% sure of my answer for the étale topology case, I'd rather say nothing than take the risk of misleading you.
    – Georges Elencwajg
    Nov 14 at 20:01















up vote
3
down vote

favorite












Let $X$ be an integral scheme, and $g: eta to X$ the inclusion of its generic point. If $M$ is an abelian group and $M_eta$ is a constant sheaf on $eta$ taking value $M$, then is it true that $g_* M_eta = M_X$?



My thought is yes, as if $U subset X$ is open, then $g_* M_eta (U) = M_eta( g^{-1}(U)) = M$, as every nonempty open set will contain the generic point. However it seems that one may need that $X$ is normal (or some other hypothesis) for this to be true. For reference, this is Exercise 3.7 (Chap II) in Milne's Etale Cohomology.










share|cite|improve this question
























  • The argument in your third line is perfectly correct and normality (or any other property $X$ may or may not have) plays absolutely no role in this question.
    – Georges Elencwajg
    Nov 13 at 22:47










  • @GeorgesElencwajg Glad to know I'm not losing my mind, I also thought this was odd. Milne claims that normality is required for this (he goes on to claim that if $X$ is a curve with a node $z$, then there is an exact sequence $0to M_X to g_*M_eta to i_*M_eta to 0$, where $i:Z to X$ is the inclusion of the nodal point). Is it possible he's referring to this scenario in the Etale topology (implicitly, I'm also asking if you know if this is true for the Etale topology)?
    – DKS
    Nov 14 at 0:30






  • 1




    Dear DKS, since I'm not 100% sure of my answer for the étale topology case, I'd rather say nothing than take the risk of misleading you.
    – Georges Elencwajg
    Nov 14 at 20:01













up vote
3
down vote

favorite









up vote
3
down vote

favorite











Let $X$ be an integral scheme, and $g: eta to X$ the inclusion of its generic point. If $M$ is an abelian group and $M_eta$ is a constant sheaf on $eta$ taking value $M$, then is it true that $g_* M_eta = M_X$?



My thought is yes, as if $U subset X$ is open, then $g_* M_eta (U) = M_eta( g^{-1}(U)) = M$, as every nonempty open set will contain the generic point. However it seems that one may need that $X$ is normal (or some other hypothesis) for this to be true. For reference, this is Exercise 3.7 (Chap II) in Milne's Etale Cohomology.










share|cite|improve this question















Let $X$ be an integral scheme, and $g: eta to X$ the inclusion of its generic point. If $M$ is an abelian group and $M_eta$ is a constant sheaf on $eta$ taking value $M$, then is it true that $g_* M_eta = M_X$?



My thought is yes, as if $U subset X$ is open, then $g_* M_eta (U) = M_eta( g^{-1}(U)) = M$, as every nonempty open set will contain the generic point. However it seems that one may need that $X$ is normal (or some other hypothesis) for this to be true. For reference, this is Exercise 3.7 (Chap II) in Milne's Etale Cohomology.







algebraic-geometry






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share|cite|improve this question













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share|cite|improve this question








edited Nov 13 at 22:10

























asked Nov 13 at 22:04









DKS

600312




600312












  • The argument in your third line is perfectly correct and normality (or any other property $X$ may or may not have) plays absolutely no role in this question.
    – Georges Elencwajg
    Nov 13 at 22:47










  • @GeorgesElencwajg Glad to know I'm not losing my mind, I also thought this was odd. Milne claims that normality is required for this (he goes on to claim that if $X$ is a curve with a node $z$, then there is an exact sequence $0to M_X to g_*M_eta to i_*M_eta to 0$, where $i:Z to X$ is the inclusion of the nodal point). Is it possible he's referring to this scenario in the Etale topology (implicitly, I'm also asking if you know if this is true for the Etale topology)?
    – DKS
    Nov 14 at 0:30






  • 1




    Dear DKS, since I'm not 100% sure of my answer for the étale topology case, I'd rather say nothing than take the risk of misleading you.
    – Georges Elencwajg
    Nov 14 at 20:01


















  • The argument in your third line is perfectly correct and normality (or any other property $X$ may or may not have) plays absolutely no role in this question.
    – Georges Elencwajg
    Nov 13 at 22:47










  • @GeorgesElencwajg Glad to know I'm not losing my mind, I also thought this was odd. Milne claims that normality is required for this (he goes on to claim that if $X$ is a curve with a node $z$, then there is an exact sequence $0to M_X to g_*M_eta to i_*M_eta to 0$, where $i:Z to X$ is the inclusion of the nodal point). Is it possible he's referring to this scenario in the Etale topology (implicitly, I'm also asking if you know if this is true for the Etale topology)?
    – DKS
    Nov 14 at 0:30






  • 1




    Dear DKS, since I'm not 100% sure of my answer for the étale topology case, I'd rather say nothing than take the risk of misleading you.
    – Georges Elencwajg
    Nov 14 at 20:01
















The argument in your third line is perfectly correct and normality (or any other property $X$ may or may not have) plays absolutely no role in this question.
– Georges Elencwajg
Nov 13 at 22:47




The argument in your third line is perfectly correct and normality (or any other property $X$ may or may not have) plays absolutely no role in this question.
– Georges Elencwajg
Nov 13 at 22:47












@GeorgesElencwajg Glad to know I'm not losing my mind, I also thought this was odd. Milne claims that normality is required for this (he goes on to claim that if $X$ is a curve with a node $z$, then there is an exact sequence $0to M_X to g_*M_eta to i_*M_eta to 0$, where $i:Z to X$ is the inclusion of the nodal point). Is it possible he's referring to this scenario in the Etale topology (implicitly, I'm also asking if you know if this is true for the Etale topology)?
– DKS
Nov 14 at 0:30




@GeorgesElencwajg Glad to know I'm not losing my mind, I also thought this was odd. Milne claims that normality is required for this (he goes on to claim that if $X$ is a curve with a node $z$, then there is an exact sequence $0to M_X to g_*M_eta to i_*M_eta to 0$, where $i:Z to X$ is the inclusion of the nodal point). Is it possible he's referring to this scenario in the Etale topology (implicitly, I'm also asking if you know if this is true for the Etale topology)?
– DKS
Nov 14 at 0:30




1




1




Dear DKS, since I'm not 100% sure of my answer for the étale topology case, I'd rather say nothing than take the risk of misleading you.
– Georges Elencwajg
Nov 14 at 20:01




Dear DKS, since I'm not 100% sure of my answer for the étale topology case, I'd rather say nothing than take the risk of misleading you.
– Georges Elencwajg
Nov 14 at 20:01















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