Mixing
Vector space objects in schemes - confusion
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Let $R$ be the ring $mathbfCtimesmathbfC$, and consider the affine line $mathbfA^1_R$.
$mathbfA^1_R$ can be given the structure of additive group scheme over $R$, denoted $(mathbfG_a)_R$.
$(mathbfG_a)_R$ carries a functorial multiplicative action of $R$ making it into an $R$-module object in schemes: the free $R$-module scheme of rank $1$.
On the other hand, $R$ is a $mathbfC$-vector space of dimension $2$.
Is $(mathbfG_a)_R$ a $mathbfC$-vector space object in schemes, of dimension $2$?
I am having difficulties to gain intuition about this, because $(mathbfG_a)_R$ is, as a scheme, a disjoint union of two copies of $mathbfA^1_mathbfC$. How does one endow it with a $mathbfC$-vector space object structure?
(likely a very simple question. I’m just a little too confused about something trivial)
ag.algebraic-geometry ac.commutative-algebra algebraic-groups arithmetic-geometry etale-cohomology
asked Sep 3 at 15:44
user128448
854
add a comment |Â
up vote
7
down vote
favorite
Let $R$ be the ring $mathbfCtimesmathbfC$, and consider the affine line $mathbfA^1_R$.
$mathbfA^1_R$ can be given the structure of additive group scheme over $R$, denoted $(mathbfG_a)_R$.
$(mathbfG_a)_R$ carries a functorial multiplicative action of $R$ making it into an $R$-module object in schemes: the free $R$-module scheme of rank $1$.
On the other hand, $R$ is a $mathbfC$-vector space of dimension $2$.
Is $(mathbfG_a)_R$ a $mathbfC$-vector space object in schemes, of dimension $2$?
I am having difficulties to gain intuition about this, because $(mathbfG_a)_R$ is, as a scheme, a disjoint union of two copies of $mathbfA^1_mathbfC$. How does one endow it with a $mathbfC$-vector space object structure?
(likely a very simple question. I’m just a little too confused about something trivial)
ag.algebraic-geometry ac.commutative-algebra algebraic-groups arithmetic-geometry etale-cohomology
asked Sep 3 at 15:44
user128448
854
May be related: mathoverflow.net/questions/268836/…
– Qfwfq
Sep 4 at 7:09
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Let $R$ be the ring $mathbfCtimesmathbfC$, and consider the affine line $mathbfA^1_R$.
$mathbfA^1_R$ can be given the structure of additive group scheme over $R$, denoted $(mathbfG_a)_R$.
$(mathbfG_a)_R$ carries a functorial multiplicative action of $R$ making it into an $R$-module object in schemes: the free $R$-module scheme of rank $1$.
On the other hand, $R$ is a $mathbfC$-vector space of dimension $2$.
Is $(mathbfG_a)_R$ a $mathbfC$-vector space object in schemes, of dimension $2$?
I am having difficulties to gain intuition about this, because $(mathbfG_a)_R$ is, as a scheme, a disjoint union of two copies of $mathbfA^1_mathbfC$. How does one endow it with a $mathbfC$-vector space object structure?
(likely a very simple question. I’m just a little too confused about something trivial)
ag.algebraic-geometry ac.commutative-algebra algebraic-groups arithmetic-geometry etale-cohomology
asked Sep 3 at 15:44
user128448
854
Let $R$ be the ring $mathbfCtimesmathbfC$, and consider the affine line $mathbfA^1_R$.
$mathbfA^1_R$ can be given the structure of additive group scheme over $R$, denoted $(mathbfG_a)_R$.
$(mathbfG_a)_R$ carries a functorial multiplicative action of $R$ making it into an $R$-module object in schemes: the free $R$-module scheme of rank $1$.
On the other hand, $R$ is a $mathbfC$-vector space of dimension $2$.
Is $(mathbfG_a)_R$ a $mathbfC$-vector space object in schemes, of dimension $2$?
I am having difficulties to gain intuition about this, because $(mathbfG_a)_R$ is, as a scheme, a disjoint union of two copies of $mathbfA^1_mathbfC$. How does one endow it with a $mathbfC$-vector space object structure?
(likely a very simple question. I’m just a little too confused about something trivial)
ag.algebraic-geometry ac.commutative-algebra algebraic-groups arithmetic-geometry etale-cohomology
asked Sep 3 at 15:44
user128448
854
edited Sep 3 at 15:54
asked Sep 3 at 15:44
user128448
854
asked Sep 3 at 15:44
user128448
854
asked Sep 3 at 15:44
user128448
854
854
May be related: mathoverflow.net/questions/268836/…
– Qfwfq
Sep 4 at 7:09
add a comment |Â
May be related: mathoverflow.net/questions/268836/…
– Qfwfq
Sep 4 at 7:09
May be related: mathoverflow.net/questions/268836/…
– Qfwfq
Sep 4 at 7:09
May be related: mathoverflow.net/questions/268836/…
– Qfwfq
Sep 4 at 7:09
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
10
down vote
accepted
Briefly, $mathbbA^1_R$ is not a vector space over $mathbbA^1_mathbbC$ in a natural way.
Strictly speaking, saying that $mathbbA^1_R$ is a ring object in schemes in not precise. What is correct is that the morhpism $mathbbA^1_Rto mathrmSpec R$ (adjoint to $Rto R[T]$) is a ring object in the category of schemes over $R$ (i.e. schemes equipped with a morphism into $mathrmSpec R$). [This means that there are addition and product morphisms $+,cdot:mathbbA^1_Rtimes_RmathbbA^1_Rto mathbbA^1_R$ and $R$-sections $0,1:mathrmSpec Rto mathbbA^1_R$ such that the ring axioms (phrased as diagrams) hold.]
In order to speak about $mathbbA^1_R$ as a "module over $mathbbA^1_mathbbC$" in some sense, you need to work in the category of schemes over $mathbbC$, where $mathbbA^1_mathbbCtomathrmSpec mathbbC$ is a ring object.
With this notation, the $2$-dimensional ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space object in schemes over $mathbbC$ would be
$$
mathbbA^1_mathbbCtimes_mathrmSpecmathbbCmathbbA^1_mathbbC=mathbbA^2_mathbbC
$$
equipped with the evident morphism into $mathrmSpecmathbbC$.
In particular, it is not isomorphic to $mathbbA^1_Rto mathrmSpecmathbbC$ (adjoint to $xmapsto (x,x):mathbbCto R[T]$) in the category of schemes over $mathbbC$. Therefore, $mathbbA^1_Rto mathrmSpec mathbbC$ is not a 2-dimensional vector space over $mathbbA^1_mathbbCto mathrmSpecmathbbC$
(or a vector space of any dimension in the said category).
A more advanced way to see what is happening is by looking on:
- the sheaf on (the large Zariski site of) $mathrmSpec R$ represented by $mathbbA^1_Rto mathrmSpec R$,
compared to:
- the sheaf on $mathrmSpec mathbbC$ represented by $mathbbA^1_Rto mathrmSpec mathbbC$.
The first is easily seen to be naturally isomorphic to the sheaf of rings $mathcalO_mathrmSpec R$ (it assigns $Yto mathrmSpec R$ to $Gamma(Y,mathcalO_Y)$), which is essentially
the same as saying that $mathbbA^1_Rto mathrmSpec R$ is a ring object in the category of $R$-schemes.
The second one is just a sheaf of sets and does not posses any natural structure of an $mathcalO_mathbbA^1_mathbbC$-module (which amounts to an ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space structure on $mathbbA^1_Rto mathrmSpecmathbbC$). Rather, it is the set sheaf $mathcalO_mathbbA^1_mathbbCcoprod mathcalO_mathbbA^1_mathbbC$.
[Edit: You can obtain from $mathbbA^1_R$ a two-dimensional
"$mathrmA^1_mathbbC$-vector space in the category of $mathbbC$-schemes", i.e. $mathbbA^2_mathbbC$, by applying Weil restriction relative
to $mathrmSpec Rto mathrmSpec mathbbC$.]
answered Sep 3 at 18:06
Uriya First
645318
Just to fully understand: in any event, $mathbfA^1_R$ carries an action of $mathbfCtimesmathbfC$. Am I right?
– user128448
Sep 3 at 21:20
1
I would say no, but some data missing in your question. To properly discuss such a question, you need to view $(mathrmSpec?),mathbbCtimesmathbbC$ as a ring object in some category which also contains $mathbbA^1_R$. Then you can ask about the existence of a meaningful action (in the categorical sense). What is true, though, is that $mathbbA^1_Rto mathrmSpec R$ (which is $mathbbA^1_mathbbCtimes mathbbCto mathrmSpec R$) acts on itself in the category of $R$-schemes.
– Uriya First
Sep 4 at 6:04
1
I would also add that the fact that $R$ is a commutative $mathbbC$-algebra, does not mean that $mathrmSpec Rto mathrmSpecmathbbC$ is a ring object in the category of $mathbbC$-schemes.
– Uriya First
Sep 4 at 6:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
Briefly, $mathbbA^1_R$ is not a vector space over $mathbbA^1_mathbbC$ in a natural way.
Strictly speaking, saying that $mathbbA^1_R$ is a ring object in schemes in not precise. What is correct is that the morhpism $mathbbA^1_Rto mathrmSpec R$ (adjoint to $Rto R[T]$) is a ring object in the category of schemes over $R$ (i.e. schemes equipped with a morphism into $mathrmSpec R$). [This means that there are addition and product morphisms $+,cdot:mathbbA^1_Rtimes_RmathbbA^1_Rto mathbbA^1_R$ and $R$-sections $0,1:mathrmSpec Rto mathbbA^1_R$ such that the ring axioms (phrased as diagrams) hold.]
In order to speak about $mathbbA^1_R$ as a "module over $mathbbA^1_mathbbC$" in some sense, you need to work in the category of schemes over $mathbbC$, where $mathbbA^1_mathbbCtomathrmSpec mathbbC$ is a ring object.
With this notation, the $2$-dimensional ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space object in schemes over $mathbbC$ would be
$$
mathbbA^1_mathbbCtimes_mathrmSpecmathbbCmathbbA^1_mathbbC=mathbbA^2_mathbbC
$$
equipped with the evident morphism into $mathrmSpecmathbbC$.
In particular, it is not isomorphic to $mathbbA^1_Rto mathrmSpecmathbbC$ (adjoint to $xmapsto (x,x):mathbbCto R[T]$) in the category of schemes over $mathbbC$. Therefore, $mathbbA^1_Rto mathrmSpec mathbbC$ is not a 2-dimensional vector space over $mathbbA^1_mathbbCto mathrmSpecmathbbC$
(or a vector space of any dimension in the said category).
A more advanced way to see what is happening is by looking on:
- the sheaf on (the large Zariski site of) $mathrmSpec R$ represented by $mathbbA^1_Rto mathrmSpec R$,
compared to:
- the sheaf on $mathrmSpec mathbbC$ represented by $mathbbA^1_Rto mathrmSpec mathbbC$.
The first is easily seen to be naturally isomorphic to the sheaf of rings $mathcalO_mathrmSpec R$ (it assigns $Yto mathrmSpec R$ to $Gamma(Y,mathcalO_Y)$), which is essentially
the same as saying that $mathbbA^1_Rto mathrmSpec R$ is a ring object in the category of $R$-schemes.
The second one is just a sheaf of sets and does not posses any natural structure of an $mathcalO_mathbbA^1_mathbbC$-module (which amounts to an ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space structure on $mathbbA^1_Rto mathrmSpecmathbbC$). Rather, it is the set sheaf $mathcalO_mathbbA^1_mathbbCcoprod mathcalO_mathbbA^1_mathbbC$.
[Edit: You can obtain from $mathbbA^1_R$ a two-dimensional
"$mathrmA^1_mathbbC$-vector space in the category of $mathbbC$-schemes", i.e. $mathbbA^2_mathbbC$, by applying Weil restriction relative
to $mathrmSpec Rto mathrmSpec mathbbC$.]
answered Sep 3 at 18:06
Uriya First
645318
Just to fully understand: in any event, $mathbfA^1_R$ carries an action of $mathbfCtimesmathbfC$. Am I right?
– user128448
Sep 3 at 21:20
1
I would say no, but some data missing in your question. To properly discuss such a question, you need to view $(mathrmSpec?),mathbbCtimesmathbbC$ as a ring object in some category which also contains $mathbbA^1_R$. Then you can ask about the existence of a meaningful action (in the categorical sense). What is true, though, is that $mathbbA^1_Rto mathrmSpec R$ (which is $mathbbA^1_mathbbCtimes mathbbCto mathrmSpec R$) acts on itself in the category of $R$-schemes.
– Uriya First
Sep 4 at 6:04
1
I would also add that the fact that $R$ is a commutative $mathbbC$-algebra, does not mean that $mathrmSpec Rto mathrmSpecmathbbC$ is a ring object in the category of $mathbbC$-schemes.
– Uriya First
Sep 4 at 6:07
add a comment |Â
up vote
10
down vote
accepted
Briefly, $mathbbA^1_R$ is not a vector space over $mathbbA^1_mathbbC$ in a natural way.
Strictly speaking, saying that $mathbbA^1_R$ is a ring object in schemes in not precise. What is correct is that the morhpism $mathbbA^1_Rto mathrmSpec R$ (adjoint to $Rto R[T]$) is a ring object in the category of schemes over $R$ (i.e. schemes equipped with a morphism into $mathrmSpec R$). [This means that there are addition and product morphisms $+,cdot:mathbbA^1_Rtimes_RmathbbA^1_Rto mathbbA^1_R$ and $R$-sections $0,1:mathrmSpec Rto mathbbA^1_R$ such that the ring axioms (phrased as diagrams) hold.]
In order to speak about $mathbbA^1_R$ as a "module over $mathbbA^1_mathbbC$" in some sense, you need to work in the category of schemes over $mathbbC$, where $mathbbA^1_mathbbCtomathrmSpec mathbbC$ is a ring object.
With this notation, the $2$-dimensional ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space object in schemes over $mathbbC$ would be
$$
mathbbA^1_mathbbCtimes_mathrmSpecmathbbCmathbbA^1_mathbbC=mathbbA^2_mathbbC
$$
equipped with the evident morphism into $mathrmSpecmathbbC$.
In particular, it is not isomorphic to $mathbbA^1_Rto mathrmSpecmathbbC$ (adjoint to $xmapsto (x,x):mathbbCto R[T]$) in the category of schemes over $mathbbC$. Therefore, $mathbbA^1_Rto mathrmSpec mathbbC$ is not a 2-dimensional vector space over $mathbbA^1_mathbbCto mathrmSpecmathbbC$
(or a vector space of any dimension in the said category).
A more advanced way to see what is happening is by looking on:
- the sheaf on (the large Zariski site of) $mathrmSpec R$ represented by $mathbbA^1_Rto mathrmSpec R$,
compared to:
- the sheaf on $mathrmSpec mathbbC$ represented by $mathbbA^1_Rto mathrmSpec mathbbC$.
The first is easily seen to be naturally isomorphic to the sheaf of rings $mathcalO_mathrmSpec R$ (it assigns $Yto mathrmSpec R$ to $Gamma(Y,mathcalO_Y)$), which is essentially
the same as saying that $mathbbA^1_Rto mathrmSpec R$ is a ring object in the category of $R$-schemes.
The second one is just a sheaf of sets and does not posses any natural structure of an $mathcalO_mathbbA^1_mathbbC$-module (which amounts to an ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space structure on $mathbbA^1_Rto mathrmSpecmathbbC$). Rather, it is the set sheaf $mathcalO_mathbbA^1_mathbbCcoprod mathcalO_mathbbA^1_mathbbC$.
[Edit: You can obtain from $mathbbA^1_R$ a two-dimensional
"$mathrmA^1_mathbbC$-vector space in the category of $mathbbC$-schemes", i.e. $mathbbA^2_mathbbC$, by applying Weil restriction relative
to $mathrmSpec Rto mathrmSpec mathbbC$.]
answered Sep 3 at 18:06
Uriya First
645318
Just to fully understand: in any event, $mathbfA^1_R$ carries an action of $mathbfCtimesmathbfC$. Am I right?
– user128448
Sep 3 at 21:20
1
I would say no, but some data missing in your question. To properly discuss such a question, you need to view $(mathrmSpec?),mathbbCtimesmathbbC$ as a ring object in some category which also contains $mathbbA^1_R$. Then you can ask about the existence of a meaningful action (in the categorical sense). What is true, though, is that $mathbbA^1_Rto mathrmSpec R$ (which is $mathbbA^1_mathbbCtimes mathbbCto mathrmSpec R$) acts on itself in the category of $R$-schemes.
– Uriya First
Sep 4 at 6:04
1
I would also add that the fact that $R$ is a commutative $mathbbC$-algebra, does not mean that $mathrmSpec Rto mathrmSpecmathbbC$ is a ring object in the category of $mathbbC$-schemes.
– Uriya First
Sep 4 at 6:07
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
Briefly, $mathbbA^1_R$ is not a vector space over $mathbbA^1_mathbbC$ in a natural way.
Strictly speaking, saying that $mathbbA^1_R$ is a ring object in schemes in not precise. What is correct is that the morhpism $mathbbA^1_Rto mathrmSpec R$ (adjoint to $Rto R[T]$) is a ring object in the category of schemes over $R$ (i.e. schemes equipped with a morphism into $mathrmSpec R$). [This means that there are addition and product morphisms $+,cdot:mathbbA^1_Rtimes_RmathbbA^1_Rto mathbbA^1_R$ and $R$-sections $0,1:mathrmSpec Rto mathbbA^1_R$ such that the ring axioms (phrased as diagrams) hold.]
In order to speak about $mathbbA^1_R$ as a "module over $mathbbA^1_mathbbC$" in some sense, you need to work in the category of schemes over $mathbbC$, where $mathbbA^1_mathbbCtomathrmSpec mathbbC$ is a ring object.
With this notation, the $2$-dimensional ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space object in schemes over $mathbbC$ would be
$$
mathbbA^1_mathbbCtimes_mathrmSpecmathbbCmathbbA^1_mathbbC=mathbbA^2_mathbbC
$$
equipped with the evident morphism into $mathrmSpecmathbbC$.
In particular, it is not isomorphic to $mathbbA^1_Rto mathrmSpecmathbbC$ (adjoint to $xmapsto (x,x):mathbbCto R[T]$) in the category of schemes over $mathbbC$. Therefore, $mathbbA^1_Rto mathrmSpec mathbbC$ is not a 2-dimensional vector space over $mathbbA^1_mathbbCto mathrmSpecmathbbC$
(or a vector space of any dimension in the said category).
A more advanced way to see what is happening is by looking on:
- the sheaf on (the large Zariski site of) $mathrmSpec R$ represented by $mathbbA^1_Rto mathrmSpec R$,
compared to:
- the sheaf on $mathrmSpec mathbbC$ represented by $mathbbA^1_Rto mathrmSpec mathbbC$.
The first is easily seen to be naturally isomorphic to the sheaf of rings $mathcalO_mathrmSpec R$ (it assigns $Yto mathrmSpec R$ to $Gamma(Y,mathcalO_Y)$), which is essentially
the same as saying that $mathbbA^1_Rto mathrmSpec R$ is a ring object in the category of $R$-schemes.
The second one is just a sheaf of sets and does not posses any natural structure of an $mathcalO_mathbbA^1_mathbbC$-module (which amounts to an ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space structure on $mathbbA^1_Rto mathrmSpecmathbbC$). Rather, it is the set sheaf $mathcalO_mathbbA^1_mathbbCcoprod mathcalO_mathbbA^1_mathbbC$.
[Edit: You can obtain from $mathbbA^1_R$ a two-dimensional
"$mathrmA^1_mathbbC$-vector space in the category of $mathbbC$-schemes", i.e. $mathbbA^2_mathbbC$, by applying Weil restriction relative
to $mathrmSpec Rto mathrmSpec mathbbC$.]
answered Sep 3 at 18:06
Uriya First
645318
Briefly, $mathbbA^1_R$ is not a vector space over $mathbbA^1_mathbbC$ in a natural way.
Strictly speaking, saying that $mathbbA^1_R$ is a ring object in schemes in not precise. What is correct is that the morhpism $mathbbA^1_Rto mathrmSpec R$ (adjoint to $Rto R[T]$) is a ring object in the category of schemes over $R$ (i.e. schemes equipped with a morphism into $mathrmSpec R$). [This means that there are addition and product morphisms $+,cdot:mathbbA^1_Rtimes_RmathbbA^1_Rto mathbbA^1_R$ and $R$-sections $0,1:mathrmSpec Rto mathbbA^1_R$ such that the ring axioms (phrased as diagrams) hold.]
In order to speak about $mathbbA^1_R$ as a "module over $mathbbA^1_mathbbC$" in some sense, you need to work in the category of schemes over $mathbbC$, where $mathbbA^1_mathbbCtomathrmSpec mathbbC$ is a ring object.
With this notation, the $2$-dimensional ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space object in schemes over $mathbbC$ would be
$$
mathbbA^1_mathbbCtimes_mathrmSpecmathbbCmathbbA^1_mathbbC=mathbbA^2_mathbbC
$$
equipped with the evident morphism into $mathrmSpecmathbbC$.
In particular, it is not isomorphic to $mathbbA^1_Rto mathrmSpecmathbbC$ (adjoint to $xmapsto (x,x):mathbbCto R[T]$) in the category of schemes over $mathbbC$. Therefore, $mathbbA^1_Rto mathrmSpec mathbbC$ is not a 2-dimensional vector space over $mathbbA^1_mathbbCto mathrmSpecmathbbC$
(or a vector space of any dimension in the said category).
A more advanced way to see what is happening is by looking on:
- the sheaf on (the large Zariski site of) $mathrmSpec R$ represented by $mathbbA^1_Rto mathrmSpec R$,
compared to:
- the sheaf on $mathrmSpec mathbbC$ represented by $mathbbA^1_Rto mathrmSpec mathbbC$.
The first is easily seen to be naturally isomorphic to the sheaf of rings $mathcalO_mathrmSpec R$ (it assigns $Yto mathrmSpec R$ to $Gamma(Y,mathcalO_Y)$), which is essentially
the same as saying that $mathbbA^1_Rto mathrmSpec R$ is a ring object in the category of $R$-schemes.
The second one is just a sheaf of sets and does not posses any natural structure of an $mathcalO_mathbbA^1_mathbbC$-module (which amounts to an ($mathbbA^1_mathbbCto mathrmSpecmathbbC$)-vector space structure on $mathbbA^1_Rto mathrmSpecmathbbC$). Rather, it is the set sheaf $mathcalO_mathbbA^1_mathbbCcoprod mathcalO_mathbbA^1_mathbbC$.
[Edit: You can obtain from $mathbbA^1_R$ a two-dimensional
"$mathrmA^1_mathbbC$-vector space in the category of $mathbbC$-schemes", i.e. $mathbbA^2_mathbbC$, by applying Weil restriction relative
to $mathrmSpec Rto mathrmSpec mathbbC$.]
answered Sep 3 at 18:06
Uriya First
645318
edited Sep 4 at 6:09
answered Sep 3 at 18:06
Uriya First
645318
answered Sep 3 at 18:06
Uriya First
645318
answered Sep 3 at 18:06
Uriya First
645318
645318
Just to fully understand: in any event, $mathbfA^1_R$ carries an action of $mathbfCtimesmathbfC$. Am I right?
– user128448
Sep 3 at 21:20
1
I would say no, but some data missing in your question. To properly discuss such a question, you need to view $(mathrmSpec?),mathbbCtimesmathbbC$ as a ring object in some category which also contains $mathbbA^1_R$. Then you can ask about the existence of a meaningful action (in the categorical sense). What is true, though, is that $mathbbA^1_Rto mathrmSpec R$ (which is $mathbbA^1_mathbbCtimes mathbbCto mathrmSpec R$) acts on itself in the category of $R$-schemes.
– Uriya First
Sep 4 at 6:04
1
I would also add that the fact that $R$ is a commutative $mathbbC$-algebra, does not mean that $mathrmSpec Rto mathrmSpecmathbbC$ is a ring object in the category of $mathbbC$-schemes.
– Uriya First
Sep 4 at 6:07
add a comment |Â
Just to fully understand: in any event, $mathbfA^1_R$ carries an action of $mathbfCtimesmathbfC$. Am I right?
– user128448
Sep 3 at 21:20
1
I would say no, but some data missing in your question. To properly discuss such a question, you need to view $(mathrmSpec?),mathbbCtimesmathbbC$ as a ring object in some category which also contains $mathbbA^1_R$. Then you can ask about the existence of a meaningful action (in the categorical sense). What is true, though, is that $mathbbA^1_Rto mathrmSpec R$ (which is $mathbbA^1_mathbbCtimes mathbbCto mathrmSpec R$) acts on itself in the category of $R$-schemes.
– Uriya First
Sep 4 at 6:04
1
I would also add that the fact that $R$ is a commutative $mathbbC$-algebra, does not mean that $mathrmSpec Rto mathrmSpecmathbbC$ is a ring object in the category of $mathbbC$-schemes.
– Uriya First
Sep 4 at 6:07
Just to fully understand: in any event, $mathbfA^1_R$ carries an action of $mathbfCtimesmathbfC$. Am I right?
– user128448
Sep 3 at 21:20
Just to fully understand: in any event, $mathbfA^1_R$ carries an action of $mathbfCtimesmathbfC$. Am I right?
– user128448
Sep 3 at 21:20
1
1
I would say no, but some data missing in your question. To properly discuss such a question, you need to view $(mathrmSpec?),mathbbCtimesmathbbC$ as a ring object in some category which also contains $mathbbA^1_R$. Then you can ask about the existence of a meaningful action (in the categorical sense). What is true, though, is that $mathbbA^1_Rto mathrmSpec R$ (which is $mathbbA^1_mathbbCtimes mathbbCto mathrmSpec R$) acts on itself in the category of $R$-schemes.
– Uriya First
Sep 4 at 6:04
I would say no, but some data missing in your question. To properly discuss such a question, you need to view $(mathrmSpec?),mathbbCtimesmathbbC$ as a ring object in some category which also contains $mathbbA^1_R$. Then you can ask about the existence of a meaningful action (in the categorical sense). What is true, though, is that $mathbbA^1_Rto mathrmSpec R$ (which is $mathbbA^1_mathbbCtimes mathbbCto mathrmSpec R$) acts on itself in the category of $R$-schemes.
– Uriya First
Sep 4 at 6:04
1
1
I would also add that the fact that $R$ is a commutative $mathbbC$-algebra, does not mean that $mathrmSpec Rto mathrmSpecmathbbC$ is a ring object in the category of $mathbbC$-schemes.
– Uriya First
Sep 4 at 6:07
I would also add that the fact that $R$ is a commutative $mathbbC$-algebra, does not mean that $mathrmSpec Rto mathrmSpecmathbbC$ is a ring object in the category of $mathbbC$-schemes.
– Uriya First
Sep 4 at 6:07
add a comment |Â
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May be related: mathoverflow.net/questions/268836/…
– Qfwfq
Sep 4 at 7:09