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 $X: Kmathsf{Alg} to mathsf{Sets}$, 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 $X: Kmathsf{Alg} to mathsf{Sets}$ is a functor then $mathrm{Nat}(X, mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbb{A}^1_K: Kmathsf{Alg} to mathsf{Sets}$.
So we have a functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ defined by $X mapsto mathrm{Nat}(X, mathbb{A}^1_K)$.
Moreover, we have an obvious natural transformation $alpha: X to mathrm{Spec_K}(mathrm{Nat}(X, mathbb{A}^1_K))$,
where $mathrm{Spec}_K$ 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 $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.
My question is:
Is it reasonable to call $mathrm{Nat}(X, mathbb{A}^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: Kmathsf{Alg} to mathsf{Sets}$? If not, what should we call this?
Is the functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ mapping $X$ to $mathrm{Nat}(X, mathbb{A}^1_K)$ adjoint (on the left or right) to $mathrm{Spec}_K: Kmathsf{Alg}^{mathrm{opp}} to mathsf{Sets}^{Kmathsf{Alg}}$? My guess is that it is the left adjoint to $mathrm{Spec}_K$.
Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$? 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
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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 $X: Kmathsf{Alg} to mathsf{Sets}$, 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 $X: Kmathsf{Alg} to mathsf{Sets}$ is a functor then $mathrm{Nat}(X, mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbb{A}^1_K: Kmathsf{Alg} to mathsf{Sets}$.
So we have a functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ defined by $X mapsto mathrm{Nat}(X, mathbb{A}^1_K)$.
Moreover, we have an obvious natural transformation $alpha: X to mathrm{Spec_K}(mathrm{Nat}(X, mathbb{A}^1_K))$,
where $mathrm{Spec}_K$ 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 $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.
My question is:
Is it reasonable to call $mathrm{Nat}(X, mathbb{A}^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: Kmathsf{Alg} to mathsf{Sets}$? If not, what should we call this?
Is the functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ mapping $X$ to $mathrm{Nat}(X, mathbb{A}^1_K)$ adjoint (on the left or right) to $mathrm{Spec}_K: Kmathsf{Alg}^{mathrm{opp}} to mathsf{Sets}^{Kmathsf{Alg}}$? My guess is that it is the left adjoint to $mathrm{Spec}_K$.
Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$? 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
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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 $X: Kmathsf{Alg} to mathsf{Sets}$, 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 $X: Kmathsf{Alg} to mathsf{Sets}$ is a functor then $mathrm{Nat}(X, mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbb{A}^1_K: Kmathsf{Alg} to mathsf{Sets}$.
So we have a functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ defined by $X mapsto mathrm{Nat}(X, mathbb{A}^1_K)$.
Moreover, we have an obvious natural transformation $alpha: X to mathrm{Spec_K}(mathrm{Nat}(X, mathbb{A}^1_K))$,
where $mathrm{Spec}_K$ 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 $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.
My question is:
Is it reasonable to call $mathrm{Nat}(X, mathbb{A}^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: Kmathsf{Alg} to mathsf{Sets}$? If not, what should we call this?
Is the functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ mapping $X$ to $mathrm{Nat}(X, mathbb{A}^1_K)$ adjoint (on the left or right) to $mathrm{Spec}_K: Kmathsf{Alg}^{mathrm{opp}} to mathsf{Sets}^{Kmathsf{Alg}}$? My guess is that it is the left adjoint to $mathrm{Spec}_K$.
Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$? 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
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 $X: Kmathsf{Alg} to mathsf{Sets}$, 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 $X: Kmathsf{Alg} to mathsf{Sets}$ is a functor then $mathrm{Nat}(X, mathbb{A}^1_K)$ is a $K$-algebra (with operations defined pointwise), since the affine line over $K$ is the forgetful functor $mathbb{A}^1_K: Kmathsf{Alg} to mathsf{Sets}$.
So we have a functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ defined by $X mapsto mathrm{Nat}(X, mathbb{A}^1_K)$.
Moreover, we have an obvious natural transformation $alpha: X to mathrm{Spec_K}(mathrm{Nat}(X, mathbb{A}^1_K))$,
where $mathrm{Spec}_K$ 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 $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$ given by $x mapsto (f mapsto f_A(x))$.
My question is:
Is it reasonable to call $mathrm{Nat}(X, mathbb{A}^1_K), A)$ the coordinate ring for any "generalize scheme" given by a functor $X: Kmathsf{Alg} to mathsf{Sets}$? If not, what should we call this?
Is the functor $mathsf{Sets}^{Kmathsf{Alg}} to Kmathsf{Alg}$ mapping $X$ to $mathrm{Nat}(X, mathbb{A}^1_K)$ adjoint (on the left or right) to $mathrm{Spec}_K: Kmathsf{Alg}^{mathrm{opp}} to mathsf{Sets}^{Kmathsf{Alg}}$? My guess is that it is the left adjoint to $mathrm{Spec}_K$.
Is there a name and interpretation for this natural transformation $alpha_A: X(A) to mathrm{Hom}(mathrm{Nat}(X, mathbb{A}^1_K), A)$? 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
edited 1 hour ago
asked 3 hours ago
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2 Answers
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oldest
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up vote
3
down vote
accepted
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.
Thank you very much for your answer!
– ಠ_ಠ
2 hours ago
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
1 hour ago
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
1 hour ago
add a comment |
up vote
1
down vote
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ 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).
Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ 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
1 hour ago
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
1 hour ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
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.
Thank you very much for your answer!
– ಠ_ಠ
2 hours ago
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
1 hour ago
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
1 hour ago
add a comment |
up vote
3
down vote
accepted
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.
Thank you very much for your answer!
– ಠ_ಠ
2 hours ago
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
1 hour ago
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
1 hour ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
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.
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 2 hours ago
Qiaochu Yuan
276k32579916
276k32579916
Thank you very much for your answer!
– ಠ_ಠ
2 hours ago
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
1 hour ago
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
1 hour ago
add a comment |
Thank you very much for your answer!
– ಠ_ಠ
2 hours ago
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
1 hour ago
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
1 hour ago
Thank you very much for your answer!
– ಠ_ಠ
2 hours ago
Thank you very much for your answer!
– ಠ_ಠ
2 hours ago
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
1 hour ago
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
1 hour ago
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
1 hour ago
ಠ_ಠ: there's not really anything here to do geometry with.
– Qiaochu Yuan
1 hour ago
add a comment |
up vote
1
down vote
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ 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).
Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ 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
1 hour ago
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
1 hour ago
add a comment |
up vote
1
down vote
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ 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).
Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ 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
1 hour ago
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
1 hour ago
add a comment |
up vote
1
down vote
up vote
1
down vote
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ 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).
"Indeed, if $X:KAlg→Sets$ is a functor then $Nat(X,mathbb{A}^1_K)$ 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 2 hours ago
anon
462
462
Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ 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
1 hour ago
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
1 hour ago
add a comment |
Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ 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
1 hour ago
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
1 hour ago
Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ 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
1 hour ago
Hmm...good point. Maybe somehow it follows from the affine line $mathbb{A}^1_K$ 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
1 hour ago
1
1
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
1 hour ago
We can require $X$ to be small (a small colimit of representables); I think that should fix it.
– Qiaochu Yuan
1 hour ago
add a comment |
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