Inverse image functors

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Let $I_Y$ be the ideal sheaf of $Y subset X$. How to compute $j^*I_Y$ where $j$ is the inclusion ? I'm especially interested when $j$ is the diagonal embedding of $Y$ in $Y times Y$.



What I tried : if $Y = Bbb A^1$ we have the sequence $0 to k[u] to k[x,y] to k[t] to 0$ where the maps are $u mapsto x-y$ and $ x mapsto t, y mapsto t$, corresponding to $0 to I_Y to mathcal O_Y times Y to mathcal j_*O_Y to 0$.



Pulling back gives $j^*I_Y to mathcal O_Y to mathcal O_Y to 0$ and the first morphism becomes zero so this description doesn't seem useful to compute it.



Morally $j^*I_Y$ shoud be "$I_Y_$" but it seems to be zero by definition. Thanks for clarification !







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  • The sequence you write in the second paragraph is not correct, it is supposed to be a sequence of $k[x,y]$ modules and the inclusion of $k[u]to k[x,y]$ is not. It should be $0to (x-y)k[x,y]to k[x,y]to k[t]to 0$. For any (coherent) sheaf $F$ on the product, $j^*F=Fotimes mathcalO_Y$. So, $I_Y|Y=I_Y/I_Y^2$.
    – Mohan
    Jul 25 at 12:16










  • @Mohan : thank you. Could you explain why $I_Y otimes mathcal O_Y = I_Y/I_Y^2$ ?
    – student
    Jul 25 at 20:13






  • 1




    Think affine. If $Isubset A$ is an ideal, and $M$ any module, use the exact sequence $0to Ito Ato A/Ito 0$ and tensor it with $M$ to see that $Motimes_A A/I=M/IM$.
    – Mohan
    Jul 25 at 21:18










  • @Mohan : Ah of course, thanks for the explanations !
    – student
    Jul 25 at 22:29










  • $X$ is a scheme and $Y$ is a closed subset, that is a closed subscheme, of $X$: am I right?
    – Armando j18eos
    Jul 26 at 15:55















up vote
1
down vote

favorite
1












Let $I_Y$ be the ideal sheaf of $Y subset X$. How to compute $j^*I_Y$ where $j$ is the inclusion ? I'm especially interested when $j$ is the diagonal embedding of $Y$ in $Y times Y$.



What I tried : if $Y = Bbb A^1$ we have the sequence $0 to k[u] to k[x,y] to k[t] to 0$ where the maps are $u mapsto x-y$ and $ x mapsto t, y mapsto t$, corresponding to $0 to I_Y to mathcal O_Y times Y to mathcal j_*O_Y to 0$.



Pulling back gives $j^*I_Y to mathcal O_Y to mathcal O_Y to 0$ and the first morphism becomes zero so this description doesn't seem useful to compute it.



Morally $j^*I_Y$ shoud be "$I_Y_$" but it seems to be zero by definition. Thanks for clarification !







share|cite|improve this question



















  • The sequence you write in the second paragraph is not correct, it is supposed to be a sequence of $k[x,y]$ modules and the inclusion of $k[u]to k[x,y]$ is not. It should be $0to (x-y)k[x,y]to k[x,y]to k[t]to 0$. For any (coherent) sheaf $F$ on the product, $j^*F=Fotimes mathcalO_Y$. So, $I_Y|Y=I_Y/I_Y^2$.
    – Mohan
    Jul 25 at 12:16










  • @Mohan : thank you. Could you explain why $I_Y otimes mathcal O_Y = I_Y/I_Y^2$ ?
    – student
    Jul 25 at 20:13






  • 1




    Think affine. If $Isubset A$ is an ideal, and $M$ any module, use the exact sequence $0to Ito Ato A/Ito 0$ and tensor it with $M$ to see that $Motimes_A A/I=M/IM$.
    – Mohan
    Jul 25 at 21:18










  • @Mohan : Ah of course, thanks for the explanations !
    – student
    Jul 25 at 22:29










  • $X$ is a scheme and $Y$ is a closed subset, that is a closed subscheme, of $X$: am I right?
    – Armando j18eos
    Jul 26 at 15:55













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
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1





Let $I_Y$ be the ideal sheaf of $Y subset X$. How to compute $j^*I_Y$ where $j$ is the inclusion ? I'm especially interested when $j$ is the diagonal embedding of $Y$ in $Y times Y$.



What I tried : if $Y = Bbb A^1$ we have the sequence $0 to k[u] to k[x,y] to k[t] to 0$ where the maps are $u mapsto x-y$ and $ x mapsto t, y mapsto t$, corresponding to $0 to I_Y to mathcal O_Y times Y to mathcal j_*O_Y to 0$.



Pulling back gives $j^*I_Y to mathcal O_Y to mathcal O_Y to 0$ and the first morphism becomes zero so this description doesn't seem useful to compute it.



Morally $j^*I_Y$ shoud be "$I_Y_$" but it seems to be zero by definition. Thanks for clarification !







share|cite|improve this question











Let $I_Y$ be the ideal sheaf of $Y subset X$. How to compute $j^*I_Y$ where $j$ is the inclusion ? I'm especially interested when $j$ is the diagonal embedding of $Y$ in $Y times Y$.



What I tried : if $Y = Bbb A^1$ we have the sequence $0 to k[u] to k[x,y] to k[t] to 0$ where the maps are $u mapsto x-y$ and $ x mapsto t, y mapsto t$, corresponding to $0 to I_Y to mathcal O_Y times Y to mathcal j_*O_Y to 0$.



Pulling back gives $j^*I_Y to mathcal O_Y to mathcal O_Y to 0$ and the first morphism becomes zero so this description doesn't seem useful to compute it.



Morally $j^*I_Y$ shoud be "$I_Y_$" but it seems to be zero by definition. Thanks for clarification !









share|cite|improve this question










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share|cite|improve this question









asked Jul 25 at 10:37









student

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  • The sequence you write in the second paragraph is not correct, it is supposed to be a sequence of $k[x,y]$ modules and the inclusion of $k[u]to k[x,y]$ is not. It should be $0to (x-y)k[x,y]to k[x,y]to k[t]to 0$. For any (coherent) sheaf $F$ on the product, $j^*F=Fotimes mathcalO_Y$. So, $I_Y|Y=I_Y/I_Y^2$.
    – Mohan
    Jul 25 at 12:16










  • @Mohan : thank you. Could you explain why $I_Y otimes mathcal O_Y = I_Y/I_Y^2$ ?
    – student
    Jul 25 at 20:13






  • 1




    Think affine. If $Isubset A$ is an ideal, and $M$ any module, use the exact sequence $0to Ito Ato A/Ito 0$ and tensor it with $M$ to see that $Motimes_A A/I=M/IM$.
    – Mohan
    Jul 25 at 21:18










  • @Mohan : Ah of course, thanks for the explanations !
    – student
    Jul 25 at 22:29










  • $X$ is a scheme and $Y$ is a closed subset, that is a closed subscheme, of $X$: am I right?
    – Armando j18eos
    Jul 26 at 15:55

















  • The sequence you write in the second paragraph is not correct, it is supposed to be a sequence of $k[x,y]$ modules and the inclusion of $k[u]to k[x,y]$ is not. It should be $0to (x-y)k[x,y]to k[x,y]to k[t]to 0$. For any (coherent) sheaf $F$ on the product, $j^*F=Fotimes mathcalO_Y$. So, $I_Y|Y=I_Y/I_Y^2$.
    – Mohan
    Jul 25 at 12:16










  • @Mohan : thank you. Could you explain why $I_Y otimes mathcal O_Y = I_Y/I_Y^2$ ?
    – student
    Jul 25 at 20:13






  • 1




    Think affine. If $Isubset A$ is an ideal, and $M$ any module, use the exact sequence $0to Ito Ato A/Ito 0$ and tensor it with $M$ to see that $Motimes_A A/I=M/IM$.
    – Mohan
    Jul 25 at 21:18










  • @Mohan : Ah of course, thanks for the explanations !
    – student
    Jul 25 at 22:29










  • $X$ is a scheme and $Y$ is a closed subset, that is a closed subscheme, of $X$: am I right?
    – Armando j18eos
    Jul 26 at 15:55
















The sequence you write in the second paragraph is not correct, it is supposed to be a sequence of $k[x,y]$ modules and the inclusion of $k[u]to k[x,y]$ is not. It should be $0to (x-y)k[x,y]to k[x,y]to k[t]to 0$. For any (coherent) sheaf $F$ on the product, $j^*F=Fotimes mathcalO_Y$. So, $I_Y|Y=I_Y/I_Y^2$.
– Mohan
Jul 25 at 12:16




The sequence you write in the second paragraph is not correct, it is supposed to be a sequence of $k[x,y]$ modules and the inclusion of $k[u]to k[x,y]$ is not. It should be $0to (x-y)k[x,y]to k[x,y]to k[t]to 0$. For any (coherent) sheaf $F$ on the product, $j^*F=Fotimes mathcalO_Y$. So, $I_Y|Y=I_Y/I_Y^2$.
– Mohan
Jul 25 at 12:16












@Mohan : thank you. Could you explain why $I_Y otimes mathcal O_Y = I_Y/I_Y^2$ ?
– student
Jul 25 at 20:13




@Mohan : thank you. Could you explain why $I_Y otimes mathcal O_Y = I_Y/I_Y^2$ ?
– student
Jul 25 at 20:13




1




1




Think affine. If $Isubset A$ is an ideal, and $M$ any module, use the exact sequence $0to Ito Ato A/Ito 0$ and tensor it with $M$ to see that $Motimes_A A/I=M/IM$.
– Mohan
Jul 25 at 21:18




Think affine. If $Isubset A$ is an ideal, and $M$ any module, use the exact sequence $0to Ito Ato A/Ito 0$ and tensor it with $M$ to see that $Motimes_A A/I=M/IM$.
– Mohan
Jul 25 at 21:18












@Mohan : Ah of course, thanks for the explanations !
– student
Jul 25 at 22:29




@Mohan : Ah of course, thanks for the explanations !
– student
Jul 25 at 22:29












$X$ is a scheme and $Y$ is a closed subset, that is a closed subscheme, of $X$: am I right?
– Armando j18eos
Jul 26 at 15:55





$X$ is a scheme and $Y$ is a closed subset, that is a closed subscheme, of $X$: am I right?
– Armando j18eos
Jul 26 at 15:55
















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