Existence of $k$-valued Points [closed]
$begingroup$
Let $X$ be a $S$-scheme. (For sake of simplicity assume $S = Spec(mathbb{Z})$)
I have a quite general question about a proper about a property of schemes and advantages of it. I read often that for a field $k$ it is praised as very usefull property if $X$ have a $k$-valued point (i. e. there exist a morphism $Spec(k) to X$).
Could anybody explain the core of the importance(s) of this property for $X$. Maybe with some interesting examples /motivations.
For example I heard that sometimes (for example for curves) it suffice to have only one $k$-valued point to give a rise for an embedding into nice spaces. Could somebody explain how such constructions of embeddings depend on this $k$-valued point? But this is only one of (seemingly) a lot other properties arising from the existence of of such $k$-valued point.
algebraic-geometry schemes
asked Jan 18 at 19:13
KarlPeterKarlPeter
6101315
$endgroup$
closed as too broad by KReiser, Leucippus, Cesareo, Adrian Keister, mrtaurho Jan 19 at 21:59
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $X$ be a $S$-scheme. (For sake of simplicity assume $S = Spec(mathbb{Z})$)
I have a quite general question about a proper about a property of schemes and advantages of it. I read often that for a field $k$ it is praised as very usefull property if $X$ have a $k$-valued point (i. e. there exist a morphism $Spec(k) to X$).
Could anybody explain the core of the importance(s) of this property for $X$. Maybe with some interesting examples /motivations.
For example I heard that sometimes (for example for curves) it suffice to have only one $k$-valued point to give a rise for an embedding into nice spaces. Could somebody explain how such constructions of embeddings depend on this $k$-valued point? But this is only one of (seemingly) a lot other properties arising from the existence of of such $k$-valued point.
algebraic-geometry schemes
asked Jan 18 at 19:13
KarlPeterKarlPeter
6101315
$endgroup$
closed as too broad by KReiser, Leucippus, Cesareo, Adrian Keister, mrtaurho Jan 19 at 21:59
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Do you know that in case of X being a polynomial ring modulo a polynomial, the k-valued points really correspond to solutions of the polynomial?
$endgroup$
– Paul K
Jan 18 at 19:41
$begingroup$
@PaulK: Yes, this is clear. The main point of my question is that the existence of such point allows to show the existence of some quite suprising morphisms. The example that I had in mind was that if $C$ is a curve with cohomology dimensions $h^0(C)=1, h^1(C)=0$ and it contains a $k$.rational point with is also regular then $C cong mathbb{P}^1$. Especially in the proof of it the existence of only one such point allows the concretely construct a morphism $C to mathbb{P}^1$. Motivated by this example I was curious about deeper meaning of this existence of such points.
$endgroup$
– KarlPeter
Jan 18 at 19:51
add a comment |
$begingroup$
Let $X$ be a $S$-scheme. (For sake of simplicity assume $S = Spec(mathbb{Z})$)
I have a quite general question about a proper about a property of schemes and advantages of it. I read often that for a field $k$ it is praised as very usefull property if $X$ have a $k$-valued point (i. e. there exist a morphism $Spec(k) to X$).
Could anybody explain the core of the importance(s) of this property for $X$. Maybe with some interesting examples /motivations.
For example I heard that sometimes (for example for curves) it suffice to have only one $k$-valued point to give a rise for an embedding into nice spaces. Could somebody explain how such constructions of embeddings depend on this $k$-valued point? But this is only one of (seemingly) a lot other properties arising from the existence of of such $k$-valued point.
algebraic-geometry schemes
asked Jan 18 at 19:13
KarlPeterKarlPeter
6101315
$endgroup$
Let $X$ be a $S$-scheme. (For sake of simplicity assume $S = Spec(mathbb{Z})$)
I have a quite general question about a proper about a property of schemes and advantages of it. I read often that for a field $k$ it is praised as very usefull property if $X$ have a $k$-valued point (i. e. there exist a morphism $Spec(k) to X$).
Could anybody explain the core of the importance(s) of this property for $X$. Maybe with some interesting examples /motivations.
For example I heard that sometimes (for example for curves) it suffice to have only one $k$-valued point to give a rise for an embedding into nice spaces. Could somebody explain how such constructions of embeddings depend on this $k$-valued point? But this is only one of (seemingly) a lot other properties arising from the existence of of such $k$-valued point.
algebraic-geometry schemes
algebraic-geometry schemes
asked Jan 18 at 19:13
KarlPeterKarlPeter
6101315
asked Jan 18 at 19:13
KarlPeterKarlPeter
6101315
asked Jan 18 at 19:13
KarlPeterKarlPeter
6101315
asked Jan 18 at 19:13
KarlPeterKarlPeter
6101315
asked Jan 18 at 19:13
KarlPeterKarlPeter
6101315
6101315
closed as too broad by KReiser, Leucippus, Cesareo, Adrian Keister, mrtaurho Jan 19 at 21:59
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by KReiser, Leucippus, Cesareo, Adrian Keister, mrtaurho Jan 19 at 21:59
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Do you know that in case of X being a polynomial ring modulo a polynomial, the k-valued points really correspond to solutions of the polynomial?
$endgroup$
– Paul K
Jan 18 at 19:41
$begingroup$
@PaulK: Yes, this is clear. The main point of my question is that the existence of such point allows to show the existence of some quite suprising morphisms. The example that I had in mind was that if $C$ is a curve with cohomology dimensions $h^0(C)=1, h^1(C)=0$ and it contains a $k$.rational point with is also regular then $C cong mathbb{P}^1$. Especially in the proof of it the existence of only one such point allows the concretely construct a morphism $C to mathbb{P}^1$. Motivated by this example I was curious about deeper meaning of this existence of such points.
$endgroup$
– KarlPeter
Jan 18 at 19:51
add a comment |
$begingroup$
Do you know that in case of X being a polynomial ring modulo a polynomial, the k-valued points really correspond to solutions of the polynomial?
$endgroup$
– Paul K
Jan 18 at 19:41
$begingroup$
@PaulK: Yes, this is clear. The main point of my question is that the existence of such point allows to show the existence of some quite suprising morphisms. The example that I had in mind was that if $C$ is a curve with cohomology dimensions $h^0(C)=1, h^1(C)=0$ and it contains a $k$.rational point with is also regular then $C cong mathbb{P}^1$. Especially in the proof of it the existence of only one such point allows the concretely construct a morphism $C to mathbb{P}^1$. Motivated by this example I was curious about deeper meaning of this existence of such points.
$endgroup$
– KarlPeter
Jan 18 at 19:51
$begingroup$
Do you know that in case of X being a polynomial ring modulo a polynomial, the k-valued points really correspond to solutions of the polynomial?
$endgroup$
– Paul K
Jan 18 at 19:41
$begingroup$
Do you know that in case of X being a polynomial ring modulo a polynomial, the k-valued points really correspond to solutions of the polynomial?
$endgroup$
– Paul K
Jan 18 at 19:41
$begingroup$
@PaulK: Yes, this is clear. The main point of my question is that the existence of such point allows to show the existence of some quite suprising morphisms. The example that I had in mind was that if $C$ is a curve with cohomology dimensions $h^0(C)=1, h^1(C)=0$ and it contains a $k$.rational point with is also regular then $C cong mathbb{P}^1$. Especially in the proof of it the existence of only one such point allows the concretely construct a morphism $C to mathbb{P}^1$. Motivated by this example I was curious about deeper meaning of this existence of such points.
$endgroup$
– KarlPeter
Jan 18 at 19:51
$begingroup$
@PaulK: Yes, this is clear. The main point of my question is that the existence of such point allows to show the existence of some quite suprising morphisms. The example that I had in mind was that if $C$ is a curve with cohomology dimensions $h^0(C)=1, h^1(C)=0$ and it contains a $k$.rational point with is also regular then $C cong mathbb{P}^1$. Especially in the proof of it the existence of only one such point allows the concretely construct a morphism $C to mathbb{P}^1$. Motivated by this example I was curious about deeper meaning of this existence of such points.
$endgroup$
– KarlPeter
Jan 18 at 19:51
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$begingroup$
Do you know that in case of X being a polynomial ring modulo a polynomial, the k-valued points really correspond to solutions of the polynomial?
$endgroup$
– Paul K
Jan 18 at 19:41
$begingroup$
@PaulK: Yes, this is clear. The main point of my question is that the existence of such point allows to show the existence of some quite suprising morphisms. The example that I had in mind was that if $C$ is a curve with cohomology dimensions $h^0(C)=1, h^1(C)=0$ and it contains a $k$.rational point with is also regular then $C cong mathbb{P}^1$. Especially in the proof of it the existence of only one such point allows the concretely construct a morphism $C to mathbb{P}^1$. Motivated by this example I was curious about deeper meaning of this existence of such points.
$endgroup$
– KarlPeter
Jan 18 at 19:51