Mixing
Relative Spectrum respects Pullback
Clash Royale CLAN TAG#URR8PPP
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$requireAMScd$
Let $(X,mathcal O_X)$ be a scheme and $mathcal A$ a quasi-coherent $mathcal O_X$-algebra, i.e. a sheaf of ring $mathcal A$ together with a morphism $mathcal O_X to mathcal A$ making it a quasi-coherent $mathcal O_X$-module.
Let consider the induced scheme $Spec(mathcal A)$ the relative spectrum defined locally via $$Gamma(U, mathcal O_Spec(mathcal A)):= mathcal A(U)$$
gluing provides indeed a scheme.
Now let consider a morphism $g: Y to X$ of schemes.
My question is why the construction above respects pullbacks; therefore why
$$Spec(g^*mathcal A) cong Spec(mathcal A) times_X Y$$
My ideas: I heard that $Spec(mathcal A)$ represents the functor
$$begineqnarray labelconstructions-equation-spec F : mathitSch^opp & longrightarrow & textitSets \ T & longmapsto & F(T) = textpairs (f, varphi ) text with f:T to X, varphi: f^*mathcalA to mathcalO_ T text surjective nonumber endeqnarray$$
so using Yoneda lemma it suffice to show it on level of pairs as above. But how?
By the way: can it also be shown directly on the level of stalks using the definition of pullback functor of sheaves via $$f^*mathcal G:=f^-1mathcal Gotimes _f^-1mathcal O_Ymathcal O_X$$?
Is the induced map $Spec(mathcal A) to X$ set theoretically surjective?
algebraic-geometry
asked Jul 16 at 1:14
KarlPeter
509313
add a comment |Â
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1
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$requireAMScd$
Let $(X,mathcal O_X)$ be a scheme and $mathcal A$ a quasi-coherent $mathcal O_X$-algebra, i.e. a sheaf of ring $mathcal A$ together with a morphism $mathcal O_X to mathcal A$ making it a quasi-coherent $mathcal O_X$-module.
Let consider the induced scheme $Spec(mathcal A)$ the relative spectrum defined locally via $$Gamma(U, mathcal O_Spec(mathcal A)):= mathcal A(U)$$
gluing provides indeed a scheme.
Now let consider a morphism $g: Y to X$ of schemes.
My question is why the construction above respects pullbacks; therefore why
$$Spec(g^*mathcal A) cong Spec(mathcal A) times_X Y$$
My ideas: I heard that $Spec(mathcal A)$ represents the functor
$$begineqnarray labelconstructions-equation-spec F : mathitSch^opp & longrightarrow & textitSets \ T & longmapsto & F(T) = textpairs (f, varphi ) text with f:T to X, varphi: f^*mathcalA to mathcalO_ T text surjective nonumber endeqnarray$$
so using Yoneda lemma it suffice to show it on level of pairs as above. But how?
By the way: can it also be shown directly on the level of stalks using the definition of pullback functor of sheaves via $$f^*mathcal G:=f^-1mathcal Gotimes _f^-1mathcal O_Ymathcal O_X$$?
Is the induced map $Spec(mathcal A) to X$ set theoretically surjective?
algebraic-geometry
asked Jul 16 at 1:14
KarlPeter
509313
1
Hint: check affine-locally and reduce your problem to the morphism of schemes associated to the structure map $Ato B$ which makes $B$ into an $A$-algebra.
– KReiser
Jul 16 at 3:43
@KReiser: Ok. If I consider the local case that $B$ is a $A$-algebra and $f: Spec(R) to Spec(A)$ is the induced morphism by the ring map $A to R$. My problem here is to "translate" $f^*B$ locally to the level of rings, especially how to cope with the functor $f^-1 $ on the level of rings. I guess your hint surgests to compare locally $B otimes_A R$ with $f^*B cong f^-1B otimes_f^-1A R$. Now my problem is how translate $f^-1B$ (resp $f^-1A$) locally to rings?
– KarlPeter
Jul 17 at 0:04
Suppose $g:YtooperatornameSpec A$ is our morphism, and $widetildeB$ is our sheaf of algebras. WLOG we may assume $Y=operatornameSpec R$ because isomorphisms may be checked affine-locally. Then $g^*(widetildeB)=widetildeBotimes_A R$ as a sheaf on $Y$, so $operatornameSpec(g^*widetildeB)=operatornameSpec Botimes_A R$ which is exactly the fiber product on the RHS. On the level of rings, $f^-1$ just changes the ring we consider the module over from $R$ to $A$ by letting $ain A$ act by $f(a)in R$ while not disturbing the module.
– KReiser
Jul 17 at 0:29
@KReiser:Let $varphi: A to R$ a rind morphism inducing the morphism $f: Spec(R) to Spec(A)$ and the $A$-algebra $B$ becomes via $widetildeB$ a $mathcalO_A$-module. For an open set $U subset Spec(R)$ (wlog consider $U= D(r):= Spec(R) - V(r)$ for $r in R$ ) the local sections of $f^-1widetildeB$ on $U = D(r)$ are defined via $$f^-1(widetildeB)(U):= varinjlim_f(U) subset V (widetildeB)(V) $$ I don't see from here how to see your argument that $f^-1$ on the level of ring just changes it. The problem is what is $f(U)= f(D(r)$.
– KarlPeter
Jul 17 at 10:29
We're over an affine space and therefore the only thing we need to do is consider global sections by the equivalence between modules over $mathcalO_X(X)$ and modules over $mathcalO_X$.
– KReiser
Jul 17 at 17:02
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$requireAMScd$
Let $(X,mathcal O_X)$ be a scheme and $mathcal A$ a quasi-coherent $mathcal O_X$-algebra, i.e. a sheaf of ring $mathcal A$ together with a morphism $mathcal O_X to mathcal A$ making it a quasi-coherent $mathcal O_X$-module.
Let consider the induced scheme $Spec(mathcal A)$ the relative spectrum defined locally via $$Gamma(U, mathcal O_Spec(mathcal A)):= mathcal A(U)$$
gluing provides indeed a scheme.
Now let consider a morphism $g: Y to X$ of schemes.
My question is why the construction above respects pullbacks; therefore why
$$Spec(g^*mathcal A) cong Spec(mathcal A) times_X Y$$
My ideas: I heard that $Spec(mathcal A)$ represents the functor
$$begineqnarray labelconstructions-equation-spec F : mathitSch^opp & longrightarrow & textitSets \ T & longmapsto & F(T) = textpairs (f, varphi ) text with f:T to X, varphi: f^*mathcalA to mathcalO_ T text surjective nonumber endeqnarray$$
so using Yoneda lemma it suffice to show it on level of pairs as above. But how?
By the way: can it also be shown directly on the level of stalks using the definition of pullback functor of sheaves via $$f^*mathcal G:=f^-1mathcal Gotimes _f^-1mathcal O_Ymathcal O_X$$?
Is the induced map $Spec(mathcal A) to X$ set theoretically surjective?
algebraic-geometry
asked Jul 16 at 1:14
KarlPeter
509313
$requireAMScd$
Let $(X,mathcal O_X)$ be a scheme and $mathcal A$ a quasi-coherent $mathcal O_X$-algebra, i.e. a sheaf of ring $mathcal A$ together with a morphism $mathcal O_X to mathcal A$ making it a quasi-coherent $mathcal O_X$-module.
Let consider the induced scheme $Spec(mathcal A)$ the relative spectrum defined locally via $$Gamma(U, mathcal O_Spec(mathcal A)):= mathcal A(U)$$
gluing provides indeed a scheme.
Now let consider a morphism $g: Y to X$ of schemes.
My question is why the construction above respects pullbacks; therefore why
$$Spec(g^*mathcal A) cong Spec(mathcal A) times_X Y$$
My ideas: I heard that $Spec(mathcal A)$ represents the functor
$$begineqnarray labelconstructions-equation-spec F : mathitSch^opp & longrightarrow & textitSets \ T & longmapsto & F(T) = textpairs (f, varphi ) text with f:T to X, varphi: f^*mathcalA to mathcalO_ T text surjective nonumber endeqnarray$$
so using Yoneda lemma it suffice to show it on level of pairs as above. But how?
By the way: can it also be shown directly on the level of stalks using the definition of pullback functor of sheaves via $$f^*mathcal G:=f^-1mathcal Gotimes _f^-1mathcal O_Ymathcal O_X$$?
Is the induced map $Spec(mathcal A) to X$ set theoretically surjective?
algebraic-geometry
asked Jul 16 at 1:14
KarlPeter
509313
edited Jul 16 at 1:20
asked Jul 16 at 1:14
KarlPeter
509313
asked Jul 16 at 1:14
KarlPeter
509313
asked Jul 16 at 1:14
KarlPeter
509313
509313
1
Hint: check affine-locally and reduce your problem to the morphism of schemes associated to the structure map $Ato B$ which makes $B$ into an $A$-algebra.
– KReiser
Jul 16 at 3:43
@KReiser: Ok. If I consider the local case that $B$ is a $A$-algebra and $f: Spec(R) to Spec(A)$ is the induced morphism by the ring map $A to R$. My problem here is to "translate" $f^*B$ locally to the level of rings, especially how to cope with the functor $f^-1 $ on the level of rings. I guess your hint surgests to compare locally $B otimes_A R$ with $f^*B cong f^-1B otimes_f^-1A R$. Now my problem is how translate $f^-1B$ (resp $f^-1A$) locally to rings?
– KarlPeter
Jul 17 at 0:04
Suppose $g:YtooperatornameSpec A$ is our morphism, and $widetildeB$ is our sheaf of algebras. WLOG we may assume $Y=operatornameSpec R$ because isomorphisms may be checked affine-locally. Then $g^*(widetildeB)=widetildeBotimes_A R$ as a sheaf on $Y$, so $operatornameSpec(g^*widetildeB)=operatornameSpec Botimes_A R$ which is exactly the fiber product on the RHS. On the level of rings, $f^-1$ just changes the ring we consider the module over from $R$ to $A$ by letting $ain A$ act by $f(a)in R$ while not disturbing the module.
– KReiser
Jul 17 at 0:29
@KReiser:Let $varphi: A to R$ a rind morphism inducing the morphism $f: Spec(R) to Spec(A)$ and the $A$-algebra $B$ becomes via $widetildeB$ a $mathcalO_A$-module. For an open set $U subset Spec(R)$ (wlog consider $U= D(r):= Spec(R) - V(r)$ for $r in R$ ) the local sections of $f^-1widetildeB$ on $U = D(r)$ are defined via $$f^-1(widetildeB)(U):= varinjlim_f(U) subset V (widetildeB)(V) $$ I don't see from here how to see your argument that $f^-1$ on the level of ring just changes it. The problem is what is $f(U)= f(D(r)$.
– KarlPeter
Jul 17 at 10:29
We're over an affine space and therefore the only thing we need to do is consider global sections by the equivalence between modules over $mathcalO_X(X)$ and modules over $mathcalO_X$.
– KReiser
Jul 17 at 17:02
add a comment |Â
1
Hint: check affine-locally and reduce your problem to the morphism of schemes associated to the structure map $Ato B$ which makes $B$ into an $A$-algebra.
– KReiser
Jul 16 at 3:43
@KReiser: Ok. If I consider the local case that $B$ is a $A$-algebra and $f: Spec(R) to Spec(A)$ is the induced morphism by the ring map $A to R$. My problem here is to "translate" $f^*B$ locally to the level of rings, especially how to cope with the functor $f^-1 $ on the level of rings. I guess your hint surgests to compare locally $B otimes_A R$ with $f^*B cong f^-1B otimes_f^-1A R$. Now my problem is how translate $f^-1B$ (resp $f^-1A$) locally to rings?
– KarlPeter
Jul 17 at 0:04
Suppose $g:YtooperatornameSpec A$ is our morphism, and $widetildeB$ is our sheaf of algebras. WLOG we may assume $Y=operatornameSpec R$ because isomorphisms may be checked affine-locally. Then $g^*(widetildeB)=widetildeBotimes_A R$ as a sheaf on $Y$, so $operatornameSpec(g^*widetildeB)=operatornameSpec Botimes_A R$ which is exactly the fiber product on the RHS. On the level of rings, $f^-1$ just changes the ring we consider the module over from $R$ to $A$ by letting $ain A$ act by $f(a)in R$ while not disturbing the module.
– KReiser
Jul 17 at 0:29
@KReiser:Let $varphi: A to R$ a rind morphism inducing the morphism $f: Spec(R) to Spec(A)$ and the $A$-algebra $B$ becomes via $widetildeB$ a $mathcalO_A$-module. For an open set $U subset Spec(R)$ (wlog consider $U= D(r):= Spec(R) - V(r)$ for $r in R$ ) the local sections of $f^-1widetildeB$ on $U = D(r)$ are defined via $$f^-1(widetildeB)(U):= varinjlim_f(U) subset V (widetildeB)(V) $$ I don't see from here how to see your argument that $f^-1$ on the level of ring just changes it. The problem is what is $f(U)= f(D(r)$.
– KarlPeter
Jul 17 at 10:29
We're over an affine space and therefore the only thing we need to do is consider global sections by the equivalence between modules over $mathcalO_X(X)$ and modules over $mathcalO_X$.
– KReiser
Jul 17 at 17:02
1
1
Hint: check affine-locally and reduce your problem to the morphism of schemes associated to the structure map $Ato B$ which makes $B$ into an $A$-algebra.
– KReiser
Jul 16 at 3:43
Hint: check affine-locally and reduce your problem to the morphism of schemes associated to the structure map $Ato B$ which makes $B$ into an $A$-algebra.
– KReiser
Jul 16 at 3:43
@KReiser: Ok. If I consider the local case that $B$ is a $A$-algebra and $f: Spec(R) to Spec(A)$ is the induced morphism by the ring map $A to R$. My problem here is to "translate" $f^*B$ locally to the level of rings, especially how to cope with the functor $f^-1 $ on the level of rings. I guess your hint surgests to compare locally $B otimes_A R$ with $f^*B cong f^-1B otimes_f^-1A R$. Now my problem is how translate $f^-1B$ (resp $f^-1A$) locally to rings?
– KarlPeter
Jul 17 at 0:04
@KReiser: Ok. If I consider the local case that $B$ is a $A$-algebra and $f: Spec(R) to Spec(A)$ is the induced morphism by the ring map $A to R$. My problem here is to "translate" $f^*B$ locally to the level of rings, especially how to cope with the functor $f^-1 $ on the level of rings. I guess your hint surgests to compare locally $B otimes_A R$ with $f^*B cong f^-1B otimes_f^-1A R$. Now my problem is how translate $f^-1B$ (resp $f^-1A$) locally to rings?
– KarlPeter
Jul 17 at 0:04
Suppose $g:YtooperatornameSpec A$ is our morphism, and $widetildeB$ is our sheaf of algebras. WLOG we may assume $Y=operatornameSpec R$ because isomorphisms may be checked affine-locally. Then $g^*(widetildeB)=widetildeBotimes_A R$ as a sheaf on $Y$, so $operatornameSpec(g^*widetildeB)=operatornameSpec Botimes_A R$ which is exactly the fiber product on the RHS. On the level of rings, $f^-1$ just changes the ring we consider the module over from $R$ to $A$ by letting $ain A$ act by $f(a)in R$ while not disturbing the module.
– KReiser
Jul 17 at 0:29
Suppose $g:YtooperatornameSpec A$ is our morphism, and $widetildeB$ is our sheaf of algebras. WLOG we may assume $Y=operatornameSpec R$ because isomorphisms may be checked affine-locally. Then $g^*(widetildeB)=widetildeBotimes_A R$ as a sheaf on $Y$, so $operatornameSpec(g^*widetildeB)=operatornameSpec Botimes_A R$ which is exactly the fiber product on the RHS. On the level of rings, $f^-1$ just changes the ring we consider the module over from $R$ to $A$ by letting $ain A$ act by $f(a)in R$ while not disturbing the module.
– KReiser
Jul 17 at 0:29
@KReiser:Let $varphi: A to R$ a rind morphism inducing the morphism $f: Spec(R) to Spec(A)$ and the $A$-algebra $B$ becomes via $widetildeB$ a $mathcalO_A$-module. For an open set $U subset Spec(R)$ (wlog consider $U= D(r):= Spec(R) - V(r)$ for $r in R$ ) the local sections of $f^-1widetildeB$ on $U = D(r)$ are defined via $$f^-1(widetildeB)(U):= varinjlim_f(U) subset V (widetildeB)(V) $$ I don't see from here how to see your argument that $f^-1$ on the level of ring just changes it. The problem is what is $f(U)= f(D(r)$.
– KarlPeter
Jul 17 at 10:29
@KReiser:Let $varphi: A to R$ a rind morphism inducing the morphism $f: Spec(R) to Spec(A)$ and the $A$-algebra $B$ becomes via $widetildeB$ a $mathcalO_A$-module. For an open set $U subset Spec(R)$ (wlog consider $U= D(r):= Spec(R) - V(r)$ for $r in R$ ) the local sections of $f^-1widetildeB$ on $U = D(r)$ are defined via $$f^-1(widetildeB)(U):= varinjlim_f(U) subset V (widetildeB)(V) $$ I don't see from here how to see your argument that $f^-1$ on the level of ring just changes it. The problem is what is $f(U)= f(D(r)$.
– KarlPeter
Jul 17 at 10:29
We're over an affine space and therefore the only thing we need to do is consider global sections by the equivalence between modules over $mathcalO_X(X)$ and modules over $mathcalO_X$.
– KReiser
Jul 17 at 17:02
We're over an affine space and therefore the only thing we need to do is consider global sections by the equivalence between modules over $mathcalO_X(X)$ and modules over $mathcalO_X$.
– KReiser
Jul 17 at 17:02
add a comment |Â
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1
Hint: check affine-locally and reduce your problem to the morphism of schemes associated to the structure map $Ato B$ which makes $B$ into an $A$-algebra.
– KReiser
Jul 16 at 3:43
@KReiser: Ok. If I consider the local case that $B$ is a $A$-algebra and $f: Spec(R) to Spec(A)$ is the induced morphism by the ring map $A to R$. My problem here is to "translate" $f^*B$ locally to the level of rings, especially how to cope with the functor $f^-1 $ on the level of rings. I guess your hint surgests to compare locally $B otimes_A R$ with $f^*B cong f^-1B otimes_f^-1A R$. Now my problem is how translate $f^-1B$ (resp $f^-1A$) locally to rings?
– KarlPeter
Jul 17 at 0:04
Suppose $g:YtooperatornameSpec A$ is our morphism, and $widetildeB$ is our sheaf of algebras. WLOG we may assume $Y=operatornameSpec R$ because isomorphisms may be checked affine-locally. Then $g^*(widetildeB)=widetildeBotimes_A R$ as a sheaf on $Y$, so $operatornameSpec(g^*widetildeB)=operatornameSpec Botimes_A R$ which is exactly the fiber product on the RHS. On the level of rings, $f^-1$ just changes the ring we consider the module over from $R$ to $A$ by letting $ain A$ act by $f(a)in R$ while not disturbing the module.
– KReiser
Jul 17 at 0:29
@KReiser:Let $varphi: A to R$ a rind morphism inducing the morphism $f: Spec(R) to Spec(A)$ and the $A$-algebra $B$ becomes via $widetildeB$ a $mathcalO_A$-module. For an open set $U subset Spec(R)$ (wlog consider $U= D(r):= Spec(R) - V(r)$ for $r in R$ ) the local sections of $f^-1widetildeB$ on $U = D(r)$ are defined via $$f^-1(widetildeB)(U):= varinjlim_f(U) subset V (widetildeB)(V) $$ I don't see from here how to see your argument that $f^-1$ on the level of ring just changes it. The problem is what is $f(U)= f(D(r)$.
– KarlPeter
Jul 17 at 10:29
We're over an affine space and therefore the only thing we need to do is consider global sections by the equivalence between modules over $mathcalO_X(X)$ and modules over $mathcalO_X$.
– KReiser
Jul 17 at 17:02