Basic example of a formal affine scheme, functorial point of viewIs there functorial point of view to differential operator?Functorial point of view of spectrum (Looking for reference)Functorial point of view for formal schemesNon-algebraizable Formal Scheme?coherent sheaves on affine formal schemesglobal section of affine $C^infty$-schemeBasic questions about formal schemesFunctorial description of a certain subgroup schemeReference Request: Categorical/Functorial approach to Formal Schemes and Formal GroupsCoproduct in the category of affine schemes, functorial point of view
Basic example of a formal affine scheme, functorial point of view
Is there functorial point of view to differential operator?Functorial point of view of spectrum (Looking for reference)Functorial point of view for formal schemesNon-algebraizable Formal Scheme?coherent sheaves on affine formal schemesglobal section of affine $C^infty$-schemeBasic questions about formal schemesFunctorial description of a certain subgroup schemeReference Request: Categorical/Functorial approach to Formal Schemes and Formal GroupsCoproduct in the category of affine schemes, functorial point of view
$begingroup$
$letopn=operatorname$For my BA thesis I have to describe formal groups from the functorial point of view. I am hence reading Strickland - Formal Schemes and Formal Groups, which is apparently the only article that deals with this topic in that way.
He defines (4.1) an formal scheme as a functor $X: opnCRingsto opnSet$ that is a small filtered colimit of affine schemes i.e., $X(R)=limlimits_rightarrow iX_i(R)$.
The first example (4.2) is given by the functor $widehatmathbb A^1 $ defined as $widehatmathbb A^1(R)mathrel:=opnNil(R)$.
I don't understand why this functor is the colimit over $N$ of the functors $opnspec(mathbbZ[x]/x^N+1)mathrel:=opnHom_opnCRing(mathbbZ[x]/x^N+1,_)$.
I would appreciate it if someone could explain it in general and kindly give an illustrating example. Other simple examples of formal schemes are also highly welcome. Many thanks!
ag.algebraic-geometry ct.category-theory limits-and-colimits formal-schemes
edited 6 hours ago
LSpice
3,1282 gold badges26 silver badges31 bronze badges
asked 10 hours ago
sagirotsagirot
475 bronze badges
$endgroup$
add a comment |
$begingroup$
$letopn=operatorname$For my BA thesis I have to describe formal groups from the functorial point of view. I am hence reading Strickland - Formal Schemes and Formal Groups, which is apparently the only article that deals with this topic in that way.
He defines (4.1) an formal scheme as a functor $X: opnCRingsto opnSet$ that is a small filtered colimit of affine schemes i.e., $X(R)=limlimits_rightarrow iX_i(R)$.
The first example (4.2) is given by the functor $widehatmathbb A^1 $ defined as $widehatmathbb A^1(R)mathrel:=opnNil(R)$.
I don't understand why this functor is the colimit over $N$ of the functors $opnspec(mathbbZ[x]/x^N+1)mathrel:=opnHom_opnCRing(mathbbZ[x]/x^N+1,_)$.
I would appreciate it if someone could explain it in general and kindly give an illustrating example. Other simple examples of formal schemes are also highly welcome. Many thanks!
ag.algebraic-geometry ct.category-theory limits-and-colimits formal-schemes
edited 6 hours ago
LSpice
3,1282 gold badges26 silver badges31 bronze badges
asked 10 hours ago
sagirotsagirot
475 bronze badges
$endgroup$
4
$begingroup$
You seem to be asking a lot of questions about basic examples and definitions from this paper. In order that your BA thesis represent your own work, it is probably better to work these out yourself, or at least to seek help from your advisor or classmates.
$endgroup$
– LSpice
6 hours ago
add a comment |
$begingroup$
$letopn=operatorname$For my BA thesis I have to describe formal groups from the functorial point of view. I am hence reading Strickland - Formal Schemes and Formal Groups, which is apparently the only article that deals with this topic in that way.
He defines (4.1) an formal scheme as a functor $X: opnCRingsto opnSet$ that is a small filtered colimit of affine schemes i.e., $X(R)=limlimits_rightarrow iX_i(R)$.
The first example (4.2) is given by the functor $widehatmathbb A^1 $ defined as $widehatmathbb A^1(R)mathrel:=opnNil(R)$.
I don't understand why this functor is the colimit over $N$ of the functors $opnspec(mathbbZ[x]/x^N+1)mathrel:=opnHom_opnCRing(mathbbZ[x]/x^N+1,_)$.
I would appreciate it if someone could explain it in general and kindly give an illustrating example. Other simple examples of formal schemes are also highly welcome. Many thanks!
ag.algebraic-geometry ct.category-theory limits-and-colimits formal-schemes
edited 6 hours ago
LSpice
3,1282 gold badges26 silver badges31 bronze badges
asked 10 hours ago
sagirotsagirot
475 bronze badges
$endgroup$
$letopn=operatorname$For my BA thesis I have to describe formal groups from the functorial point of view. I am hence reading Strickland - Formal Schemes and Formal Groups, which is apparently the only article that deals with this topic in that way.
He defines (4.1) an formal scheme as a functor $X: opnCRingsto opnSet$ that is a small filtered colimit of affine schemes i.e., $X(R)=limlimits_rightarrow iX_i(R)$.
The first example (4.2) is given by the functor $widehatmathbb A^1 $ defined as $widehatmathbb A^1(R)mathrel:=opnNil(R)$.
I don't understand why this functor is the colimit over $N$ of the functors $opnspec(mathbbZ[x]/x^N+1)mathrel:=opnHom_opnCRing(mathbbZ[x]/x^N+1,_)$.
I would appreciate it if someone could explain it in general and kindly give an illustrating example. Other simple examples of formal schemes are also highly welcome. Many thanks!
ag.algebraic-geometry ct.category-theory limits-and-colimits formal-schemes
ag.algebraic-geometry ct.category-theory limits-and-colimits formal-schemes
edited 6 hours ago
LSpice
3,1282 gold badges26 silver badges31 bronze badges
asked 10 hours ago
sagirotsagirot
475 bronze badges
edited 6 hours ago
LSpice
3,1282 gold badges26 silver badges31 bronze badges
asked 10 hours ago
sagirotsagirot
475 bronze badges
edited 6 hours ago
LSpice
3,1282 gold badges26 silver badges31 bronze badges
edited 6 hours ago
LSpice
3,1282 gold badges26 silver badges31 bronze badges
edited 6 hours ago
LSpice
3,1282 gold badges26 silver badges31 bronze badges
3,1282 gold badges26 silver badges31 bronze badges
asked 10 hours ago
sagirotsagirot
475 bronze badges
asked 10 hours ago
sagirotsagirot
475 bronze badges
asked 10 hours ago
sagirotsagirot
475 bronze badges
475 bronze badges
4
$begingroup$
You seem to be asking a lot of questions about basic examples and definitions from this paper. In order that your BA thesis represent your own work, it is probably better to work these out yourself, or at least to seek help from your advisor or classmates.
$endgroup$
– LSpice
6 hours ago
add a comment |
4
$begingroup$
You seem to be asking a lot of questions about basic examples and definitions from this paper. In order that your BA thesis represent your own work, it is probably better to work these out yourself, or at least to seek help from your advisor or classmates.
$endgroup$
– LSpice
6 hours ago
4
4
$begingroup$
You seem to be asking a lot of questions about basic examples and definitions from this paper. In order that your BA thesis represent your own work, it is probably better to work these out yourself, or at least to seek help from your advisor or classmates.
$endgroup$
– LSpice
6 hours ago
$begingroup$
You seem to be asking a lot of questions about basic examples and definitions from this paper. In order that your BA thesis represent your own work, it is probably better to work these out yourself, or at least to seek help from your advisor or classmates.
$endgroup$
– LSpice
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It might be illuminating to first work the example of (ordinary) affine space $mathbbA^1_mathbbZ$ over the integers.
As a functor, $mathbbA^1_mathbbZ$ is the forgetful functor $mathitRings^mathrmoprightarrowmathitSets$, sending a ring $R$ to its underlying set. It is representable by $mathbbZ[t]$, as can be seen by the isomorphism $$mathrmHom_mathitRings(mathbbZ[t],R)cong R.$$
(As a homomorphism $fcolonmathbbZ[t]rightarrow R$ is determined by its value $f(t)$ at $t$, we can define an isomorphism $mathrmHom_mathitRings(mathbbZ[t],R)rightarrow R$ by $fmapsto f(t)$ for all $finmathrmHom_mathitRings(mathbbZ[t],R)$.)
$widehatmathbbA^1_mathbbZ$ is similar: there is an isomorphism
$$mathrmHom_mathitRings(mathbbZ[t]/(t^n),R)congmathrmNil_n(R),$$
where $mathrmNil_n(R)$ denotes the set of nilpotent elements $r$ of $R$ of order $n$ (i.e. $r^n=0$).
(We define an isomorphism as before, but now the element $f(t)$ that we send $f$ to must be nilpotent in $R$, for it to preserve the ring structure: $f(t)^n=f(t^n)=f(0)=0$.)
The answer from here own was changed following Dmitri Pavlov's comment:
As (co)limits of presheaves are computed objectwise, we see that $widehatmathbbA^1_mathbbZ$ sends $R$ to the colimit $mathrmcolim(mathrmNil_n(R))=bigcup_ngeq0mathrmNil_n(R)=mathrmNil(R)$.
answered 7 hours ago
Théo de Oliveira SantosThéo de Oliveira Santos
8821 gold badge3 silver badges22 bronze badges
$endgroup$
$begingroup$
(Sorry for the probably very overcomplicated proof.)
$endgroup$
– Théo de Oliveira Santos
7 hours ago
3
$begingroup$
Why do you need EGA here? Once you observed that Nil_n computes nilpotents of order n, you can immediately conclude the proof by observing that colimits of functors CRing→Set are computed objectwise and Nil(R)=⋃_n≥0Nil_n(R) because any nilpotent element of R has finite order.
$endgroup$
– Dmitri Pavlov
6 hours ago
$begingroup$
@DmitriPavlov That's precisely the fact I was missing. Thank you!
$endgroup$
– Théo de Oliveira Santos
6 hours ago
add a comment |
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$begingroup$
It might be illuminating to first work the example of (ordinary) affine space $mathbbA^1_mathbbZ$ over the integers.
As a functor, $mathbbA^1_mathbbZ$ is the forgetful functor $mathitRings^mathrmoprightarrowmathitSets$, sending a ring $R$ to its underlying set. It is representable by $mathbbZ[t]$, as can be seen by the isomorphism $$mathrmHom_mathitRings(mathbbZ[t],R)cong R.$$
(As a homomorphism $fcolonmathbbZ[t]rightarrow R$ is determined by its value $f(t)$ at $t$, we can define an isomorphism $mathrmHom_mathitRings(mathbbZ[t],R)rightarrow R$ by $fmapsto f(t)$ for all $finmathrmHom_mathitRings(mathbbZ[t],R)$.)
$widehatmathbbA^1_mathbbZ$ is similar: there is an isomorphism
$$mathrmHom_mathitRings(mathbbZ[t]/(t^n),R)congmathrmNil_n(R),$$
where $mathrmNil_n(R)$ denotes the set of nilpotent elements $r$ of $R$ of order $n$ (i.e. $r^n=0$).
(We define an isomorphism as before, but now the element $f(t)$ that we send $f$ to must be nilpotent in $R$, for it to preserve the ring structure: $f(t)^n=f(t^n)=f(0)=0$.)
The answer from here own was changed following Dmitri Pavlov's comment:
As (co)limits of presheaves are computed objectwise, we see that $widehatmathbbA^1_mathbbZ$ sends $R$ to the colimit $mathrmcolim(mathrmNil_n(R))=bigcup_ngeq0mathrmNil_n(R)=mathrmNil(R)$.
answered 7 hours ago
Théo de Oliveira SantosThéo de Oliveira Santos
8821 gold badge3 silver badges22 bronze badges
$endgroup$
$begingroup$
(Sorry for the probably very overcomplicated proof.)
$endgroup$
– Théo de Oliveira Santos
7 hours ago
3
$begingroup$
Why do you need EGA here? Once you observed that Nil_n computes nilpotents of order n, you can immediately conclude the proof by observing that colimits of functors CRing→Set are computed objectwise and Nil(R)=⋃_n≥0Nil_n(R) because any nilpotent element of R has finite order.
$endgroup$
– Dmitri Pavlov
6 hours ago
$begingroup$
@DmitriPavlov That's precisely the fact I was missing. Thank you!
$endgroup$
– Théo de Oliveira Santos
6 hours ago
add a comment |
$begingroup$
It might be illuminating to first work the example of (ordinary) affine space $mathbbA^1_mathbbZ$ over the integers.
As a functor, $mathbbA^1_mathbbZ$ is the forgetful functor $mathitRings^mathrmoprightarrowmathitSets$, sending a ring $R$ to its underlying set. It is representable by $mathbbZ[t]$, as can be seen by the isomorphism $$mathrmHom_mathitRings(mathbbZ[t],R)cong R.$$
(As a homomorphism $fcolonmathbbZ[t]rightarrow R$ is determined by its value $f(t)$ at $t$, we can define an isomorphism $mathrmHom_mathitRings(mathbbZ[t],R)rightarrow R$ by $fmapsto f(t)$ for all $finmathrmHom_mathitRings(mathbbZ[t],R)$.)
$widehatmathbbA^1_mathbbZ$ is similar: there is an isomorphism
$$mathrmHom_mathitRings(mathbbZ[t]/(t^n),R)congmathrmNil_n(R),$$
where $mathrmNil_n(R)$ denotes the set of nilpotent elements $r$ of $R$ of order $n$ (i.e. $r^n=0$).
(We define an isomorphism as before, but now the element $f(t)$ that we send $f$ to must be nilpotent in $R$, for it to preserve the ring structure: $f(t)^n=f(t^n)=f(0)=0$.)
The answer from here own was changed following Dmitri Pavlov's comment:
As (co)limits of presheaves are computed objectwise, we see that $widehatmathbbA^1_mathbbZ$ sends $R$ to the colimit $mathrmcolim(mathrmNil_n(R))=bigcup_ngeq0mathrmNil_n(R)=mathrmNil(R)$.
answered 7 hours ago
Théo de Oliveira SantosThéo de Oliveira Santos
8821 gold badge3 silver badges22 bronze badges
$endgroup$
$begingroup$
(Sorry for the probably very overcomplicated proof.)
$endgroup$
– Théo de Oliveira Santos
7 hours ago
3
$begingroup$
Why do you need EGA here? Once you observed that Nil_n computes nilpotents of order n, you can immediately conclude the proof by observing that colimits of functors CRing→Set are computed objectwise and Nil(R)=⋃_n≥0Nil_n(R) because any nilpotent element of R has finite order.
$endgroup$
– Dmitri Pavlov
6 hours ago
$begingroup$
@DmitriPavlov That's precisely the fact I was missing. Thank you!
$endgroup$
– Théo de Oliveira Santos
6 hours ago
add a comment |
$begingroup$
It might be illuminating to first work the example of (ordinary) affine space $mathbbA^1_mathbbZ$ over the integers.
As a functor, $mathbbA^1_mathbbZ$ is the forgetful functor $mathitRings^mathrmoprightarrowmathitSets$, sending a ring $R$ to its underlying set. It is representable by $mathbbZ[t]$, as can be seen by the isomorphism $$mathrmHom_mathitRings(mathbbZ[t],R)cong R.$$
(As a homomorphism $fcolonmathbbZ[t]rightarrow R$ is determined by its value $f(t)$ at $t$, we can define an isomorphism $mathrmHom_mathitRings(mathbbZ[t],R)rightarrow R$ by $fmapsto f(t)$ for all $finmathrmHom_mathitRings(mathbbZ[t],R)$.)
$widehatmathbbA^1_mathbbZ$ is similar: there is an isomorphism
$$mathrmHom_mathitRings(mathbbZ[t]/(t^n),R)congmathrmNil_n(R),$$
where $mathrmNil_n(R)$ denotes the set of nilpotent elements $r$ of $R$ of order $n$ (i.e. $r^n=0$).
(We define an isomorphism as before, but now the element $f(t)$ that we send $f$ to must be nilpotent in $R$, for it to preserve the ring structure: $f(t)^n=f(t^n)=f(0)=0$.)
The answer from here own was changed following Dmitri Pavlov's comment:
As (co)limits of presheaves are computed objectwise, we see that $widehatmathbbA^1_mathbbZ$ sends $R$ to the colimit $mathrmcolim(mathrmNil_n(R))=bigcup_ngeq0mathrmNil_n(R)=mathrmNil(R)$.
answered 7 hours ago
Théo de Oliveira SantosThéo de Oliveira Santos
8821 gold badge3 silver badges22 bronze badges
$endgroup$
It might be illuminating to first work the example of (ordinary) affine space $mathbbA^1_mathbbZ$ over the integers.
As a functor, $mathbbA^1_mathbbZ$ is the forgetful functor $mathitRings^mathrmoprightarrowmathitSets$, sending a ring $R$ to its underlying set. It is representable by $mathbbZ[t]$, as can be seen by the isomorphism $$mathrmHom_mathitRings(mathbbZ[t],R)cong R.$$
(As a homomorphism $fcolonmathbbZ[t]rightarrow R$ is determined by its value $f(t)$ at $t$, we can define an isomorphism $mathrmHom_mathitRings(mathbbZ[t],R)rightarrow R$ by $fmapsto f(t)$ for all $finmathrmHom_mathitRings(mathbbZ[t],R)$.)
$widehatmathbbA^1_mathbbZ$ is similar: there is an isomorphism
$$mathrmHom_mathitRings(mathbbZ[t]/(t^n),R)congmathrmNil_n(R),$$
where $mathrmNil_n(R)$ denotes the set of nilpotent elements $r$ of $R$ of order $n$ (i.e. $r^n=0$).
(We define an isomorphism as before, but now the element $f(t)$ that we send $f$ to must be nilpotent in $R$, for it to preserve the ring structure: $f(t)^n=f(t^n)=f(0)=0$.)
The answer from here own was changed following Dmitri Pavlov's comment:
As (co)limits of presheaves are computed objectwise, we see that $widehatmathbbA^1_mathbbZ$ sends $R$ to the colimit $mathrmcolim(mathrmNil_n(R))=bigcup_ngeq0mathrmNil_n(R)=mathrmNil(R)$.
answered 7 hours ago
Théo de Oliveira SantosThéo de Oliveira Santos
8821 gold badge3 silver badges22 bronze badges
edited 6 hours ago
answered 7 hours ago
Théo de Oliveira SantosThéo de Oliveira Santos
8821 gold badge3 silver badges22 bronze badges
answered 7 hours ago
Théo de Oliveira SantosThéo de Oliveira Santos
8821 gold badge3 silver badges22 bronze badges
answered 7 hours ago
Théo de Oliveira SantosThéo de Oliveira Santos
8821 gold badge3 silver badges22 bronze badges
8821 gold badge3 silver badges22 bronze badges
$begingroup$
(Sorry for the probably very overcomplicated proof.)
$endgroup$
– Théo de Oliveira Santos
7 hours ago
3
$begingroup$
Why do you need EGA here? Once you observed that Nil_n computes nilpotents of order n, you can immediately conclude the proof by observing that colimits of functors CRing→Set are computed objectwise and Nil(R)=⋃_n≥0Nil_n(R) because any nilpotent element of R has finite order.
$endgroup$
– Dmitri Pavlov
6 hours ago
$begingroup$
@DmitriPavlov That's precisely the fact I was missing. Thank you!
$endgroup$
– Théo de Oliveira Santos
6 hours ago
add a comment |
$begingroup$
(Sorry for the probably very overcomplicated proof.)
$endgroup$
– Théo de Oliveira Santos
7 hours ago
3
$begingroup$
Why do you need EGA here? Once you observed that Nil_n computes nilpotents of order n, you can immediately conclude the proof by observing that colimits of functors CRing→Set are computed objectwise and Nil(R)=⋃_n≥0Nil_n(R) because any nilpotent element of R has finite order.
$endgroup$
– Dmitri Pavlov
6 hours ago
$begingroup$
@DmitriPavlov That's precisely the fact I was missing. Thank you!
$endgroup$
– Théo de Oliveira Santos
6 hours ago
$begingroup$
(Sorry for the probably very overcomplicated proof.)
$endgroup$
– Théo de Oliveira Santos
7 hours ago
$begingroup$
(Sorry for the probably very overcomplicated proof.)
$endgroup$
– Théo de Oliveira Santos
7 hours ago
3
3
$begingroup$
Why do you need EGA here? Once you observed that Nil_n computes nilpotents of order n, you can immediately conclude the proof by observing that colimits of functors CRing→Set are computed objectwise and Nil(R)=⋃_n≥0Nil_n(R) because any nilpotent element of R has finite order.
$endgroup$
– Dmitri Pavlov
6 hours ago
$begingroup$
Why do you need EGA here? Once you observed that Nil_n computes nilpotents of order n, you can immediately conclude the proof by observing that colimits of functors CRing→Set are computed objectwise and Nil(R)=⋃_n≥0Nil_n(R) because any nilpotent element of R has finite order.
$endgroup$
– Dmitri Pavlov
6 hours ago
$begingroup$
@DmitriPavlov That's precisely the fact I was missing. Thank you!
$endgroup$
– Théo de Oliveira Santos
6 hours ago
$begingroup$
@DmitriPavlov That's precisely the fact I was missing. Thank you!
$endgroup$
– Théo de Oliveira Santos
6 hours ago
add a comment |
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$begingroup$
You seem to be asking a lot of questions about basic examples and definitions from this paper. In order that your BA thesis represent your own work, it is probably better to work these out yourself, or at least to seek help from your advisor or classmates.
$endgroup$
– LSpice
6 hours ago