Coordinate ring of a scheme in functorial algebraic geometry
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I will preface this by saying that I am new to algebraic geometry, but I am somewhat experienced with category theory.
I'm just reading the introduction to Milne's notes "Basic Theory of Affine Group Schemes". He uses the functorial point of view here, so I am viewing an affine scheme over $K$ as a representable functor $X: KmathsfAlg to mathsfSets$, and a scheme is likewise defined as a functor satisfying appropriate gluing properties. We can think of some general functor as a generalized scheme.
In section I.3 he has a subsection titled "The canonical coordinate ring of an affine group" but I noticed that his construction seems to define a canonical "coordinate ring" for every sort of "generalized scheme", not just affine group schemes. Indeed, if $X: KmathsfAlg to mathsfSets$ is a functor then $mathrmNat(X, mathbbA^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbbA^1_K: KmathsfAlg to mathsfSets$.
So we have a functor $mathsfSets^KmathsfAlg to KmathsfAlg$ defined by $X mapsto mathrmNat(X, mathbbA^1_K)$.
Moreover, we have an obvious natural transformation $alpha: X to mathrmSpec_K(mathrmNat(X, mathbbA^1_K))$,
where $mathrmSpec_K$ here is just the contravariant Yoneda embedding (since I am thinking of affine schemes as functors rather than ringed spaces). This natural transformation has components $alpha_A: X(A) to mathrmHom(mathrmNat(X, mathbbA^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.
My question is:
Is it reasonable to call $mathrmNat(X, mathbbA^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: KmathsfAlg to mathsfSets$? If not, what should we call this?
Is the functor $mathsfSets^KmathsfAlg to KmathsfAlg$ mapping $X$ to $mathrmNat(X, mathbbA^1_K)$ adjoint (on the left or right) to $mathrmSpec_K: KmathsfAlg^mathrmopp to mathsfSets^KmathsfAlg$? My guess is that it is the left adjoint to $mathrmSpec_K$.
Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrmHom(mathrmNat(X, mathbbA^1_K), A)$? I can see that $X$ is an affine scheme over $K$ if and only if this is an isomorphism. But what if $X$ is not affine? How do we interpret this?
algebraic-geometry ring-theory category-theory schemes algebraic-groups
asked Dec 13 at 2:43
ಠ_ಠ
5,39721242
add a comment |
I will preface this by saying that I am new to algebraic geometry, but I am somewhat experienced with category theory.
I'm just reading the introduction to Milne's notes "Basic Theory of Affine Group Schemes". He uses the functorial point of view here, so I am viewing an affine scheme over $K$ as a representable functor $X: KmathsfAlg to mathsfSets$, and a scheme is likewise defined as a functor satisfying appropriate gluing properties. We can think of some general functor as a generalized scheme.
In section I.3 he has a subsection titled "The canonical coordinate ring of an affine group" but I noticed that his construction seems to define a canonical "coordinate ring" for every sort of "generalized scheme", not just affine group schemes. Indeed, if $X: KmathsfAlg to mathsfSets$ is a functor then $mathrmNat(X, mathbbA^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbbA^1_K: KmathsfAlg to mathsfSets$.
So we have a functor $mathsfSets^KmathsfAlg to KmathsfAlg$ defined by $X mapsto mathrmNat(X, mathbbA^1_K)$.
Moreover, we have an obvious natural transformation $alpha: X to mathrmSpec_K(mathrmNat(X, mathbbA^1_K))$,
where $mathrmSpec_K$ here is just the contravariant Yoneda embedding (since I am thinking of affine schemes as functors rather than ringed spaces). This natural transformation has components $alpha_A: X(A) to mathrmHom(mathrmNat(X, mathbbA^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.
My question is:
Is it reasonable to call $mathrmNat(X, mathbbA^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: KmathsfAlg to mathsfSets$? If not, what should we call this?
Is the functor $mathsfSets^KmathsfAlg to KmathsfAlg$ mapping $X$ to $mathrmNat(X, mathbbA^1_K)$ adjoint (on the left or right) to $mathrmSpec_K: KmathsfAlg^mathrmopp to mathsfSets^KmathsfAlg$? My guess is that it is the left adjoint to $mathrmSpec_K$.
Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrmHom(mathrmNat(X, mathbbA^1_K), A)$? I can see that $X$ is an affine scheme over $K$ if and only if this is an isomorphism. But what if $X$ is not affine? How do we interpret this?
algebraic-geometry ring-theory category-theory schemes algebraic-groups
asked Dec 13 at 2:43
ಠ_ಠ
5,39721242
add a comment |
I will preface this by saying that I am new to algebraic geometry, but I am somewhat experienced with category theory.
I'm just reading the introduction to Milne's notes "Basic Theory of Affine Group Schemes". He uses the functorial point of view here, so I am viewing an affine scheme over $K$ as a representable functor $X: KmathsfAlg to mathsfSets$, and a scheme is likewise defined as a functor satisfying appropriate gluing properties. We can think of some general functor as a generalized scheme.
In section I.3 he has a subsection titled "The canonical coordinate ring of an affine group" but I noticed that his construction seems to define a canonical "coordinate ring" for every sort of "generalized scheme", not just affine group schemes. Indeed, if $X: KmathsfAlg to mathsfSets$ is a functor then $mathrmNat(X, mathbbA^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbbA^1_K: KmathsfAlg to mathsfSets$.
So we have a functor $mathsfSets^KmathsfAlg to KmathsfAlg$ defined by $X mapsto mathrmNat(X, mathbbA^1_K)$.
Moreover, we have an obvious natural transformation $alpha: X to mathrmSpec_K(mathrmNat(X, mathbbA^1_K))$,
where $mathrmSpec_K$ here is just the contravariant Yoneda embedding (since I am thinking of affine schemes as functors rather than ringed spaces). This natural transformation has components $alpha_A: X(A) to mathrmHom(mathrmNat(X, mathbbA^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.
My question is:
Is it reasonable to call $mathrmNat(X, mathbbA^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: KmathsfAlg to mathsfSets$? If not, what should we call this?
Is the functor $mathsfSets^KmathsfAlg to KmathsfAlg$ mapping $X$ to $mathrmNat(X, mathbbA^1_K)$ adjoint (on the left or right) to $mathrmSpec_K: KmathsfAlg^mathrmopp to mathsfSets^KmathsfAlg$? My guess is that it is the left adjoint to $mathrmSpec_K$.
Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrmHom(mathrmNat(X, mathbbA^1_K), A)$? I can see that $X$ is an affine scheme over $K$ if and only if this is an isomorphism. But what if $X$ is not affine? How do we interpret this?
algebraic-geometry ring-theory category-theory schemes algebraic-groups
asked Dec 13 at 2:43
ಠ_ಠ
5,39721242
I will preface this by saying that I am new to algebraic geometry, but I am somewhat experienced with category theory.
I'm just reading the introduction to Milne's notes "Basic Theory of Affine Group Schemes". He uses the functorial point of view here, so I am viewing an affine scheme over $K$ as a representable functor $X: KmathsfAlg to mathsfSets$, and a scheme is likewise defined as a functor satisfying appropriate gluing properties. We can think of some general functor as a generalized scheme.
In section I.3 he has a subsection titled "The canonical coordinate ring of an affine group" but I noticed that his construction seems to define a canonical "coordinate ring" for every sort of "generalized scheme", not just affine group schemes. Indeed, if $X: KmathsfAlg to mathsfSets$ is a functor then $mathrmNat(X, mathbbA^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbbA^1_K: KmathsfAlg to mathsfSets$.
So we have a functor $mathsfSets^KmathsfAlg to KmathsfAlg$ defined by $X mapsto mathrmNat(X, mathbbA^1_K)$.
Moreover, we have an obvious natural transformation $alpha: X to mathrmSpec_K(mathrmNat(X, mathbbA^1_K))$,
where $mathrmSpec_K$ here is just the contravariant Yoneda embedding (since I am thinking of affine schemes as functors rather than ringed spaces). This natural transformation has components $alpha_A: X(A) to mathrmHom(mathrmNat(X, mathbbA^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.
My question is:
Is it reasonable to call $mathrmNat(X, mathbbA^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: KmathsfAlg to mathsfSets$? If not, what should we call this?
Is the functor $mathsfSets^KmathsfAlg to KmathsfAlg$ mapping $X$ to $mathrmNat(X, mathbbA^1_K)$ adjoint (on the left or right) to $mathrmSpec_K: KmathsfAlg^mathrmopp to mathsfSets^KmathsfAlg$? My guess is that it is the left adjoint to $mathrmSpec_K$.
Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrmHom(mathrmNat(X, mathbbA^1_K), A)$? I can see that $X$ is an affine scheme over $K$ if and only if this is an isomorphism. But what if $X$ is not affine? How do we interpret this?
algebraic-geometry ring-theory category-theory schemes algebraic-groups
algebraic-geometry ring-theory category-theory schemes algebraic-groups
asked Dec 13 at 2:43
ಠ_ಠ
5,39721242
asked Dec 13 at 2:43
ಠ_ಠ
5,39721242
edited Dec 13 at 4:33
asked Dec 13 at 2:43
ಠ_ಠ
5,39721242
asked Dec 13 at 2:43
ಠ_ಠ
5,39721242
asked Dec 13 at 2:43
ಠ_ಠ
5,39721242
5,39721242
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Yes.
The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.
$alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.
answered Dec 13 at 3:36
Qiaochu Yuan
276k32580918
Thank you very much for your answer!
– ಠ_ಠ
Dec 13 at 3:37
I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
– ಠ_ಠ
Dec 13 at 4:33
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbbA^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).
answered Dec 13 at 4:00
anon
812
Hmm...good point. Maybe somehow it follows from the affine line $mathbbA^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
– ಠ_ಠ
Dec 13 at 4:27
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
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2 Answers
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2 Answers
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votes
active
oldest
votes
Yes.
The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.
$alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.
answered Dec 13 at 3:36
Qiaochu Yuan
276k32580918
Thank you very much for your answer!
– ಠ_ಠ
Dec 13 at 3:37
I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
– ಠ_ಠ
Dec 13 at 4:33
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
Yes.
The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.
$alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.
answered Dec 13 at 3:36
Qiaochu Yuan
276k32580918
Thank you very much for your answer!
– ಠ_ಠ
Dec 13 at 3:37
I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
– ಠ_ಠ
Dec 13 at 4:33
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
Yes.
The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.
$alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.
answered Dec 13 at 3:36
Qiaochu Yuan
276k32580918
Yes.
The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.
$alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.
answered Dec 13 at 3:36
Qiaochu Yuan
276k32580918
answered Dec 13 at 3:36
Qiaochu Yuan
276k32580918
answered Dec 13 at 3:36
Qiaochu Yuan
276k32580918
answered Dec 13 at 3:36
Qiaochu Yuan
276k32580918
276k32580918
Thank you very much for your answer!
– ಠ_ಠ
Dec 13 at 3:37
I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
– ಠ_ಠ
Dec 13 at 4:33
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
Thank you very much for your answer!
– ಠ_ಠ
Dec 13 at 3:37
I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
– ಠ_ಠ
Dec 13 at 4:33
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
Dec 13 at 4:45
Thank you very much for your answer!
– ಠ_ಠ
Dec 13 at 3:37
Thank you very much for your answer!
– ಠ_ಠ
Dec 13 at 3:37
I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
– ಠ_ಠ
Dec 13 at 4:33
I am also curious: it seems like this construction works if we replace $K$-algebras by any sort of algebraic category, like say groups or non-commutative algebras. Is this a reasonable approach to non-commutative geometry?
– ಠ_ಠ
Dec 13 at 4:33
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
Dec 13 at 4:45
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbbA^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).
answered Dec 13 at 4:00
anon
812
Hmm...good point. Maybe somehow it follows from the affine line $mathbbA^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
– ಠ_ಠ
Dec 13 at 4:27
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbbA^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).
answered Dec 13 at 4:00
anon
812
Hmm...good point. Maybe somehow it follows from the affine line $mathbbA^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
– ಠ_ಠ
Dec 13 at 4:27
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbbA^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).
answered Dec 13 at 4:00
anon
812
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbbA^1_K)$ is a $K$-algebra (with operations defined pointwise),'' Not quite --- it may be a proper class (i.e., the underlying "set" may not be a set).
answered Dec 13 at 4:00
anon
812
answered Dec 13 at 4:00
anon
812
answered Dec 13 at 4:00
anon
812
answered Dec 13 at 4:00
anon
812
812
Hmm...good point. Maybe somehow it follows from the affine line $mathbbA^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
– ಠ_ಠ
Dec 13 at 4:27
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
Hmm...good point. Maybe somehow it follows from the affine line $mathbbA^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
– ಠ_ಠ
Dec 13 at 4:27
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
Dec 13 at 4:45
Hmm...good point. Maybe somehow it follows from the affine line $mathbbA^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
– ಠ_ಠ
Dec 13 at 4:27
Hmm...good point. Maybe somehow it follows from the affine line $mathbbA^1_K$ being representable? Otherwise maybe we can use a Grothendieck Universe to fix it (I don't know much about this to be honest).
– ಠ_ಠ
Dec 13 at 4:27
1
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
Dec 13 at 4:45
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
Dec 13 at 4:45
add a comment |
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