An example of morphism not of finite type
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This is example 3.2.2 of Chapter II of Hartshorne's algebraic geometry. If P is a point of a variety V, with local ring $O_P$, then $Spec O_P$ in general of finite type. I did not see an example right now. Thank you.
algebraic-geometry commutative-algebra
asked Nov 14 at 22:48
Peter Liu
19514
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This is example 3.2.2 of Chapter II of Hartshorne's algebraic geometry. If P is a point of a variety V, with local ring $O_P$, then $Spec O_P$ in general of finite type. I did not see an example right now. Thank you.
algebraic-geometry commutative-algebra
asked Nov 14 at 22:48
Peter Liu
19514
1
Have you tried...any examples? It's arguably harder to find an example where it is of finite type.
– Eric Wofsey
Nov 14 at 22:53
For instance, have you tried $V=mathbb{A}^1$?
– Eric Wofsey
Nov 14 at 22:55
@EricWofsey For affine line, say, if you take the generic point, then you get $k(x)$, but this should be a finitely generated $k-$algebra, as it is generated by $x$ and $x^{-1}$, am I right? or I made some mistakes?
– Peter Liu
Nov 14 at 23:02
$k(x)$ is not generated by $x$ and $x^{-1}$: for instance, it also has elements like $(x+1)^{-1}$.
– Eric Wofsey
Nov 14 at 23:03
@EricWofsey Aha, I see it. I just took a very good point. If I take $(x)$, then $k[x]_{(x)}$ is not finitely generated $k-algebra$ as we need generators like $x-m$.
– Peter Liu
Nov 14 at 23:04
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is example 3.2.2 of Chapter II of Hartshorne's algebraic geometry. If P is a point of a variety V, with local ring $O_P$, then $Spec O_P$ in general of finite type. I did not see an example right now. Thank you.
algebraic-geometry commutative-algebra
asked Nov 14 at 22:48
Peter Liu
19514
This is example 3.2.2 of Chapter II of Hartshorne's algebraic geometry. If P is a point of a variety V, with local ring $O_P$, then $Spec O_P$ in general of finite type. I did not see an example right now. Thank you.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Nov 14 at 22:48
Peter Liu
19514
asked Nov 14 at 22:48
Peter Liu
19514
asked Nov 14 at 22:48
Peter Liu
19514
asked Nov 14 at 22:48
Peter Liu
19514
asked Nov 14 at 22:48
Peter Liu
19514
19514
1
Have you tried...any examples? It's arguably harder to find an example where it is of finite type.
– Eric Wofsey
Nov 14 at 22:53
For instance, have you tried $V=mathbb{A}^1$?
– Eric Wofsey
Nov 14 at 22:55
@EricWofsey For affine line, say, if you take the generic point, then you get $k(x)$, but this should be a finitely generated $k-$algebra, as it is generated by $x$ and $x^{-1}$, am I right? or I made some mistakes?
– Peter Liu
Nov 14 at 23:02
$k(x)$ is not generated by $x$ and $x^{-1}$: for instance, it also has elements like $(x+1)^{-1}$.
– Eric Wofsey
Nov 14 at 23:03
@EricWofsey Aha, I see it. I just took a very good point. If I take $(x)$, then $k[x]_{(x)}$ is not finitely generated $k-algebra$ as we need generators like $x-m$.
– Peter Liu
Nov 14 at 23:04
|
show 1 more comment
1
Have you tried...any examples? It's arguably harder to find an example where it is of finite type.
– Eric Wofsey
Nov 14 at 22:53
For instance, have you tried $V=mathbb{A}^1$?
– Eric Wofsey
Nov 14 at 22:55
@EricWofsey For affine line, say, if you take the generic point, then you get $k(x)$, but this should be a finitely generated $k-$algebra, as it is generated by $x$ and $x^{-1}$, am I right? or I made some mistakes?
– Peter Liu
Nov 14 at 23:02
$k(x)$ is not generated by $x$ and $x^{-1}$: for instance, it also has elements like $(x+1)^{-1}$.
– Eric Wofsey
Nov 14 at 23:03
@EricWofsey Aha, I see it. I just took a very good point. If I take $(x)$, then $k[x]_{(x)}$ is not finitely generated $k-algebra$ as we need generators like $x-m$.
– Peter Liu
Nov 14 at 23:04
1
1
Have you tried...any examples? It's arguably harder to find an example where it is of finite type.
– Eric Wofsey
Nov 14 at 22:53
Have you tried...any examples? It's arguably harder to find an example where it is of finite type.
– Eric Wofsey
Nov 14 at 22:53
For instance, have you tried $V=mathbb{A}^1$?
– Eric Wofsey
Nov 14 at 22:55
For instance, have you tried $V=mathbb{A}^1$?
– Eric Wofsey
Nov 14 at 22:55
@EricWofsey For affine line, say, if you take the generic point, then you get $k(x)$, but this should be a finitely generated $k-$algebra, as it is generated by $x$ and $x^{-1}$, am I right? or I made some mistakes?
– Peter Liu
Nov 14 at 23:02
@EricWofsey For affine line, say, if you take the generic point, then you get $k(x)$, but this should be a finitely generated $k-$algebra, as it is generated by $x$ and $x^{-1}$, am I right? or I made some mistakes?
– Peter Liu
Nov 14 at 23:02
$k(x)$ is not generated by $x$ and $x^{-1}$: for instance, it also has elements like $(x+1)^{-1}$.
– Eric Wofsey
Nov 14 at 23:03
$k(x)$ is not generated by $x$ and $x^{-1}$: for instance, it also has elements like $(x+1)^{-1}$.
– Eric Wofsey
Nov 14 at 23:03
@EricWofsey Aha, I see it. I just took a very good point. If I take $(x)$, then $k[x]_{(x)}$ is not finitely generated $k-algebra$ as we need generators like $x-m$.
– Peter Liu
Nov 14 at 23:04
@EricWofsey Aha, I see it. I just took a very good point. If I take $(x)$, then $k[x]_{(x)}$ is not finitely generated $k-algebra$ as we need generators like $x-m$.
– Peter Liu
Nov 14 at 23:04
|
show 1 more comment
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Have you tried...any examples? It's arguably harder to find an example where it is of finite type.
– Eric Wofsey
Nov 14 at 22:53
For instance, have you tried $V=mathbb{A}^1$?
– Eric Wofsey
Nov 14 at 22:55
@EricWofsey For affine line, say, if you take the generic point, then you get $k(x)$, but this should be a finitely generated $k-$algebra, as it is generated by $x$ and $x^{-1}$, am I right? or I made some mistakes?
– Peter Liu
Nov 14 at 23:02
$k(x)$ is not generated by $x$ and $x^{-1}$: for instance, it also has elements like $(x+1)^{-1}$.
– Eric Wofsey
Nov 14 at 23:03
@EricWofsey Aha, I see it. I just took a very good point. If I take $(x)$, then $k[x]_{(x)}$ is not finitely generated $k-algebra$ as we need generators like $x-m$.
– Peter Liu
Nov 14 at 23:04