Mixing
Intermediate step solving Hartshorne Ex II-3.12 a)
$begingroup$
In this exercise we have a surjective graded ring homomorphism $varphi:Sto T$. This induces a morphism $f:$Proj$(T)to $Proj$(S)$ by contraction of ideals. I'm asked to show that $f$ is a closed immersion.
In orther to show that $f^sharp:mathcal{O}_{Proj(S)}to f_*mathcal{O}_{Proj(T)}$ is surjective, I restric it to an open subset of the form $D_+(g)={pin$ Proj$(S)mid gnotin p}$. Surjectivity in these sets would imply surjectivity on the fibers so I'd be done.
I would like to show that $f_*mathcal{O}_{Proj(T)}(D_+(g))=mathcal{O}_{Proj(T)}(D_+(varphi(g)))$ to apply proposition 2.5 of Hartshorne and get a natural homomorphism $S_{(g)}to T_{(varphi(g))}$, which is clearly surjective.
To show this, it is enough to show $f^{-1}(D_+(g))=D_+(varphi(g))$. Explicitily, the firs set, using that $f^{-1}$ is given by ideal extension and that $varphi$ is surjective, is ${varphi(p)in$Proj$(T)mid varphi(g)notinvarphi(p)}$. So what I need to show is that if $gnotin p$, then $varphi(g)notin varphi(p)$. This is obviously false for a general surjective ring homomorphism, but I guess this is true for the graded case.
How can I show this final implication (if it is true)? If it is not true, how can I solve the exercise?
abstract-algebra algebraic-geometry ring-theory graded-rings
asked Jan 26 at 13:26
JaviJavi
3,0212832
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add a comment |
$begingroup$
In this exercise we have a surjective graded ring homomorphism $varphi:Sto T$. This induces a morphism $f:$Proj$(T)to $Proj$(S)$ by contraction of ideals. I'm asked to show that $f$ is a closed immersion.
In orther to show that $f^sharp:mathcal{O}_{Proj(S)}to f_*mathcal{O}_{Proj(T)}$ is surjective, I restric it to an open subset of the form $D_+(g)={pin$ Proj$(S)mid gnotin p}$. Surjectivity in these sets would imply surjectivity on the fibers so I'd be done.
I would like to show that $f_*mathcal{O}_{Proj(T)}(D_+(g))=mathcal{O}_{Proj(T)}(D_+(varphi(g)))$ to apply proposition 2.5 of Hartshorne and get a natural homomorphism $S_{(g)}to T_{(varphi(g))}$, which is clearly surjective.
To show this, it is enough to show $f^{-1}(D_+(g))=D_+(varphi(g))$. Explicitily, the firs set, using that $f^{-1}$ is given by ideal extension and that $varphi$ is surjective, is ${varphi(p)in$Proj$(T)mid varphi(g)notinvarphi(p)}$. So what I need to show is that if $gnotin p$, then $varphi(g)notin varphi(p)$. This is obviously false for a general surjective ring homomorphism, but I guess this is true for the graded case.
How can I show this final implication (if it is true)? If it is not true, how can I solve the exercise?
abstract-algebra algebraic-geometry ring-theory graded-rings
asked Jan 26 at 13:26
JaviJavi
3,0212832
$endgroup$
add a comment |
$begingroup$
In this exercise we have a surjective graded ring homomorphism $varphi:Sto T$. This induces a morphism $f:$Proj$(T)to $Proj$(S)$ by contraction of ideals. I'm asked to show that $f$ is a closed immersion.
In orther to show that $f^sharp:mathcal{O}_{Proj(S)}to f_*mathcal{O}_{Proj(T)}$ is surjective, I restric it to an open subset of the form $D_+(g)={pin$ Proj$(S)mid gnotin p}$. Surjectivity in these sets would imply surjectivity on the fibers so I'd be done.
I would like to show that $f_*mathcal{O}_{Proj(T)}(D_+(g))=mathcal{O}_{Proj(T)}(D_+(varphi(g)))$ to apply proposition 2.5 of Hartshorne and get a natural homomorphism $S_{(g)}to T_{(varphi(g))}$, which is clearly surjective.
To show this, it is enough to show $f^{-1}(D_+(g))=D_+(varphi(g))$. Explicitily, the firs set, using that $f^{-1}$ is given by ideal extension and that $varphi$ is surjective, is ${varphi(p)in$Proj$(T)mid varphi(g)notinvarphi(p)}$. So what I need to show is that if $gnotin p$, then $varphi(g)notin varphi(p)$. This is obviously false for a general surjective ring homomorphism, but I guess this is true for the graded case.
How can I show this final implication (if it is true)? If it is not true, how can I solve the exercise?
abstract-algebra algebraic-geometry ring-theory graded-rings
asked Jan 26 at 13:26
JaviJavi
3,0212832
$endgroup$
In this exercise we have a surjective graded ring homomorphism $varphi:Sto T$. This induces a morphism $f:$Proj$(T)to $Proj$(S)$ by contraction of ideals. I'm asked to show that $f$ is a closed immersion.
In orther to show that $f^sharp:mathcal{O}_{Proj(S)}to f_*mathcal{O}_{Proj(T)}$ is surjective, I restric it to an open subset of the form $D_+(g)={pin$ Proj$(S)mid gnotin p}$. Surjectivity in these sets would imply surjectivity on the fibers so I'd be done.
I would like to show that $f_*mathcal{O}_{Proj(T)}(D_+(g))=mathcal{O}_{Proj(T)}(D_+(varphi(g)))$ to apply proposition 2.5 of Hartshorne and get a natural homomorphism $S_{(g)}to T_{(varphi(g))}$, which is clearly surjective.
To show this, it is enough to show $f^{-1}(D_+(g))=D_+(varphi(g))$. Explicitily, the firs set, using that $f^{-1}$ is given by ideal extension and that $varphi$ is surjective, is ${varphi(p)in$Proj$(T)mid varphi(g)notinvarphi(p)}$. So what I need to show is that if $gnotin p$, then $varphi(g)notin varphi(p)$. This is obviously false for a general surjective ring homomorphism, but I guess this is true for the graded case.
How can I show this final implication (if it is true)? If it is not true, how can I solve the exercise?
abstract-algebra algebraic-geometry ring-theory graded-rings
abstract-algebra algebraic-geometry ring-theory graded-rings
asked Jan 26 at 13:26
JaviJavi
3,0212832
asked Jan 26 at 13:26
JaviJavi
3,0212832
edited Jan 26 at 14:28
Javi
asked Jan 26 at 13:26
JaviJavi
3,0212832
asked Jan 26 at 13:26
JaviJavi
3,0212832
asked Jan 26 at 13:26
JaviJavi
3,0212832
3,0212832
add a comment |
add a comment |
1 Answer
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I think it is better to use the adjunction relation from exercise 1.8, which tells us that it is equivalent to prove surjectivity of the homomorphism $mathcal{O}_{Proj(S),varphi^{-1}(p)}tomathcal{O}_{Proj(T),p}$ (using that $f^{-1}G_x=G_{f(x)}$ for a sheaf $G$). Then it is possible to apply proposition 2.5a), which gives us the homomorphism $S_{(varphi^{-1}(p))}to T_{(p)}$.
answered Jan 26 at 18:03
JaviJavi
3,0212832
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think it is better to use the adjunction relation from exercise 1.8, which tells us that it is equivalent to prove surjectivity of the homomorphism $mathcal{O}_{Proj(S),varphi^{-1}(p)}tomathcal{O}_{Proj(T),p}$ (using that $f^{-1}G_x=G_{f(x)}$ for a sheaf $G$). Then it is possible to apply proposition 2.5a), which gives us the homomorphism $S_{(varphi^{-1}(p))}to T_{(p)}$.
answered Jan 26 at 18:03
JaviJavi
3,0212832
$endgroup$
add a comment |
$begingroup$
I think it is better to use the adjunction relation from exercise 1.8, which tells us that it is equivalent to prove surjectivity of the homomorphism $mathcal{O}_{Proj(S),varphi^{-1}(p)}tomathcal{O}_{Proj(T),p}$ (using that $f^{-1}G_x=G_{f(x)}$ for a sheaf $G$). Then it is possible to apply proposition 2.5a), which gives us the homomorphism $S_{(varphi^{-1}(p))}to T_{(p)}$.
answered Jan 26 at 18:03
JaviJavi
3,0212832
$endgroup$
add a comment |
$begingroup$
I think it is better to use the adjunction relation from exercise 1.8, which tells us that it is equivalent to prove surjectivity of the homomorphism $mathcal{O}_{Proj(S),varphi^{-1}(p)}tomathcal{O}_{Proj(T),p}$ (using that $f^{-1}G_x=G_{f(x)}$ for a sheaf $G$). Then it is possible to apply proposition 2.5a), which gives us the homomorphism $S_{(varphi^{-1}(p))}to T_{(p)}$.
answered Jan 26 at 18:03
JaviJavi
3,0212832
$endgroup$
I think it is better to use the adjunction relation from exercise 1.8, which tells us that it is equivalent to prove surjectivity of the homomorphism $mathcal{O}_{Proj(S),varphi^{-1}(p)}tomathcal{O}_{Proj(T),p}$ (using that $f^{-1}G_x=G_{f(x)}$ for a sheaf $G$). Then it is possible to apply proposition 2.5a), which gives us the homomorphism $S_{(varphi^{-1}(p))}to T_{(p)}$.
answered Jan 26 at 18:03
JaviJavi
3,0212832
edited Jan 26 at 18:37
answered Jan 26 at 18:03
JaviJavi
3,0212832
answered Jan 26 at 18:03
JaviJavi
3,0212832
answered Jan 26 at 18:03
JaviJavi
3,0212832
3,0212832
add a comment |
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