Is $D(f)$ the smallest open set of $operatornameSpecB$ such that $D_+(f)subset D(f)$?

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Let $B$ be a graded ring and $rho:operatornameProjBto operatornameSpecB$ the canonical injection, that is, $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$. For any non-nilpotent homogeneous $fin B_+$, is $D(f)$ the smallest open set of $operatornameSpecB$ such that $D_+(f)subset D(f)$?




algebraic-geometry affine-schemes projective-schemes




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    I don't understand what you mean by the canonical injection $mathrmProj Bto mathrmSpec B$; for example, if $B=k[x_0,...,x_n]$ this is saying that $mathbb P^n$ embeds in $mathrm A^n$, which is not true.
    – Devlin Mallory
    2 days ago










  • $rho$ is not a morphism of schemes. $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$.
    – Born to be proud
    2 days ago














up vote
0
down vote

favorite
1












Let $B$ be a graded ring and $rho:operatornameProjBto operatornameSpecB$ the canonical injection, that is, $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$. For any non-nilpotent homogeneous $fin B_+$, is $D(f)$ the smallest open set of $operatornameSpecB$ such that $D_+(f)subset D(f)$?




algebraic-geometry affine-schemes projective-schemes




share|cite|improve this question

















  • 1




    I don't understand what you mean by the canonical injection $mathrmProj Bto mathrmSpec B$; for example, if $B=k[x_0,...,x_n]$ this is saying that $mathbb P^n$ embeds in $mathrm A^n$, which is not true.
    – Devlin Mallory
    2 days ago










  • $rho$ is not a morphism of schemes. $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$.
    – Born to be proud
    2 days ago












up vote
0
down vote

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Let $B$ be a graded ring and $rho:operatornameProjBto operatornameSpecB$ the canonical injection, that is, $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$. For any non-nilpotent homogeneous $fin B_+$, is $D(f)$ the smallest open set of $operatornameSpecB$ such that $D_+(f)subset D(f)$?




algebraic-geometry affine-schemes projective-schemes




share|cite|improve this question













Let $B$ be a graded ring and $rho:operatornameProjBto operatornameSpecB$ the canonical injection, that is, $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$. For any non-nilpotent homogeneous $fin B_+$, is $D(f)$ the smallest open set of $operatornameSpecB$ such that $D_+(f)subset D(f)$?





algebraic-geometry affine-schemes projective-schemes





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago
























asked 2 days ago









Born to be proud

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  • 1




    I don't understand what you mean by the canonical injection $mathrmProj Bto mathrmSpec B$; for example, if $B=k[x_0,...,x_n]$ this is saying that $mathbb P^n$ embeds in $mathrm A^n$, which is not true.
    – Devlin Mallory
    2 days ago










  • $rho$ is not a morphism of schemes. $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$.
    – Born to be proud
    2 days ago












  • 1




    I don't understand what you mean by the canonical injection $mathrmProj Bto mathrmSpec B$; for example, if $B=k[x_0,...,x_n]$ this is saying that $mathbb P^n$ embeds in $mathrm A^n$, which is not true.
    – Devlin Mallory
    2 days ago










  • $rho$ is not a morphism of schemes. $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$.
    – Born to be proud
    2 days ago







1




1




I don't understand what you mean by the canonical injection $mathrmProj Bto mathrmSpec B$; for example, if $B=k[x_0,...,x_n]$ this is saying that $mathbb P^n$ embeds in $mathrm A^n$, which is not true.
– Devlin Mallory
2 days ago




I don't understand what you mean by the canonical injection $mathrmProj Bto mathrmSpec B$; for example, if $B=k[x_0,...,x_n]$ this is saying that $mathbb P^n$ embeds in $mathrm A^n$, which is not true.
– Devlin Mallory
2 days ago












$rho$ is not a morphism of schemes. $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$.
– Born to be proud
2 days ago




$rho$ is not a morphism of schemes. $forall mathfrak pin operatornameProjB$, $rho(mathfrak p)=mathfrak p$.
– Born to be proud
2 days ago










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Let $B_0=B$, then $B_+=0$, so $operatornameProjB=varnothing$.
However, it doesn't make any sense.



Can anybody give a counterexample such that $D_+(f)neq varnothing$






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    up vote
    0
    down vote













    Let $B_0=B$, then $B_+=0$, so $operatornameProjB=varnothing$.
    However, it doesn't make any sense.



    Can anybody give a counterexample such that $D_+(f)neq varnothing$






    share|cite|improve this answer



























      up vote
      0
      down vote













      Let $B_0=B$, then $B_+=0$, so $operatornameProjB=varnothing$.
      However, it doesn't make any sense.



      Can anybody give a counterexample such that $D_+(f)neq varnothing$






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        Let $B_0=B$, then $B_+=0$, so $operatornameProjB=varnothing$.
        However, it doesn't make any sense.



        Can anybody give a counterexample such that $D_+(f)neq varnothing$






        share|cite|improve this answer















        Let $B_0=B$, then $B_+=0$, so $operatornameProjB=varnothing$.
        However, it doesn't make any sense.



        Can anybody give a counterexample such that $D_+(f)neq varnothing$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago


























        answered 2 days ago









        Born to be proud

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