Definition of affine space as a quasiprojective variety, Shafarevich
$begingroup$
I am quite confused the definition given by Shafaverich:
Definition: A regular map $f:X rightarrow mathbb P^n$ of an irreducible quasiproj. variety $X$ to projective space $Bbb P^n$ is given by an $(m+1)$ truple of form
$$(F_0 : cdots : F_m)$$
of same degree in hom. coord. of $x in Bbb P^n$. We require that for every $x in X$ there exists such an expression for $f$ such that $F_i(x)not=0$ for at least one $i$.
The definition of regular maps between quasiprojective varieties allows us to define isomorphism.
Definition: We say a quasiproj. variety is isomoprhic to a closed subspace of an affine space , an affine variety
In this case, to apply the first definition, we must regard a closed subspace of an affine space as one in $Bbb P^n$? What are we doing here?
I suppose we for "embed" $Bbb A^n$ into $Bbb P^n$, by one of the choice $Bbb A_i^n$ (where $i$th coordinate is nonzero), and show it is independent? Also, are there not more ways to regard $Bbb A^n$ as a subset in $Bbb P^n$?
algebraic-geometry
asked Nov 26 '18 at 8:18
CL.CL.
2,1972825
$endgroup$
add a comment |
$begingroup$
I am quite confused the definition given by Shafaverich:
Definition: A regular map $f:X rightarrow mathbb P^n$ of an irreducible quasiproj. variety $X$ to projective space $Bbb P^n$ is given by an $(m+1)$ truple of form
$$(F_0 : cdots : F_m)$$
of same degree in hom. coord. of $x in Bbb P^n$. We require that for every $x in X$ there exists such an expression for $f$ such that $F_i(x)not=0$ for at least one $i$.
The definition of regular maps between quasiprojective varieties allows us to define isomorphism.
Definition: We say a quasiproj. variety is isomoprhic to a closed subspace of an affine space , an affine variety
In this case, to apply the first definition, we must regard a closed subspace of an affine space as one in $Bbb P^n$? What are we doing here?
I suppose we for "embed" $Bbb A^n$ into $Bbb P^n$, by one of the choice $Bbb A_i^n$ (where $i$th coordinate is nonzero), and show it is independent? Also, are there not more ways to regard $Bbb A^n$ as a subset in $Bbb P^n$?
algebraic-geometry
asked Nov 26 '18 at 8:18
CL.CL.
2,1972825
$endgroup$
$begingroup$
The first definition is the definition of a regular map. How is this connected to the second definition? You would do well to check the precise wording for the second definition, as it looks to have been altered in transcription (the first also has typos).
$endgroup$
– Mark Bennet
Nov 26 '18 at 8:26
$begingroup$
Ok, my first definition is incomplete, I will edit it now. The second definition is word by word. Which I do not understand because we have not defined the notion of maps between quas proj and affine space.
$endgroup$
– CL.
Nov 26 '18 at 8:36
add a comment |
$begingroup$
I am quite confused the definition given by Shafaverich:
Definition: A regular map $f:X rightarrow mathbb P^n$ of an irreducible quasiproj. variety $X$ to projective space $Bbb P^n$ is given by an $(m+1)$ truple of form
$$(F_0 : cdots : F_m)$$
of same degree in hom. coord. of $x in Bbb P^n$. We require that for every $x in X$ there exists such an expression for $f$ such that $F_i(x)not=0$ for at least one $i$.
The definition of regular maps between quasiprojective varieties allows us to define isomorphism.
Definition: We say a quasiproj. variety is isomoprhic to a closed subspace of an affine space , an affine variety
In this case, to apply the first definition, we must regard a closed subspace of an affine space as one in $Bbb P^n$? What are we doing here?
I suppose we for "embed" $Bbb A^n$ into $Bbb P^n$, by one of the choice $Bbb A_i^n$ (where $i$th coordinate is nonzero), and show it is independent? Also, are there not more ways to regard $Bbb A^n$ as a subset in $Bbb P^n$?
algebraic-geometry
asked Nov 26 '18 at 8:18
CL.CL.
2,1972825
$endgroup$
I am quite confused the definition given by Shafaverich:
Definition: A regular map $f:X rightarrow mathbb P^n$ of an irreducible quasiproj. variety $X$ to projective space $Bbb P^n$ is given by an $(m+1)$ truple of form
$$(F_0 : cdots : F_m)$$
of same degree in hom. coord. of $x in Bbb P^n$. We require that for every $x in X$ there exists such an expression for $f$ such that $F_i(x)not=0$ for at least one $i$.
The definition of regular maps between quasiprojective varieties allows us to define isomorphism.
Definition: We say a quasiproj. variety is isomoprhic to a closed subspace of an affine space , an affine variety
In this case, to apply the first definition, we must regard a closed subspace of an affine space as one in $Bbb P^n$? What are we doing here?
I suppose we for "embed" $Bbb A^n$ into $Bbb P^n$, by one of the choice $Bbb A_i^n$ (where $i$th coordinate is nonzero), and show it is independent? Also, are there not more ways to regard $Bbb A^n$ as a subset in $Bbb P^n$?
algebraic-geometry
algebraic-geometry
asked Nov 26 '18 at 8:18
CL.CL.
2,1972825
asked Nov 26 '18 at 8:18
CL.CL.
2,1972825
edited Nov 26 '18 at 8:37
CL.
asked Nov 26 '18 at 8:18
CL.CL.
2,1972825
asked Nov 26 '18 at 8:18
CL.CL.
2,1972825
asked Nov 26 '18 at 8:18
CL.CL.
2,1972825
2,1972825
$begingroup$
The first definition is the definition of a regular map. How is this connected to the second definition? You would do well to check the precise wording for the second definition, as it looks to have been altered in transcription (the first also has typos).
$endgroup$
– Mark Bennet
Nov 26 '18 at 8:26
$begingroup$
Ok, my first definition is incomplete, I will edit it now. The second definition is word by word. Which I do not understand because we have not defined the notion of maps between quas proj and affine space.
$endgroup$
– CL.
Nov 26 '18 at 8:36
add a comment |
$begingroup$
The first definition is the definition of a regular map. How is this connected to the second definition? You would do well to check the precise wording for the second definition, as it looks to have been altered in transcription (the first also has typos).
$endgroup$
– Mark Bennet
Nov 26 '18 at 8:26
$begingroup$
Ok, my first definition is incomplete, I will edit it now. The second definition is word by word. Which I do not understand because we have not defined the notion of maps between quas proj and affine space.
$endgroup$
– CL.
Nov 26 '18 at 8:36
$begingroup$
The first definition is the definition of a regular map. How is this connected to the second definition? You would do well to check the precise wording for the second definition, as it looks to have been altered in transcription (the first also has typos).
$endgroup$
– Mark Bennet
Nov 26 '18 at 8:26
$begingroup$
The first definition is the definition of a regular map. How is this connected to the second definition? You would do well to check the precise wording for the second definition, as it looks to have been altered in transcription (the first also has typos).
$endgroup$
– Mark Bennet
Nov 26 '18 at 8:26
$begingroup$
Ok, my first definition is incomplete, I will edit it now. The second definition is word by word. Which I do not understand because we have not defined the notion of maps between quas proj and affine space.
$endgroup$
– CL.
Nov 26 '18 at 8:36
$begingroup$
Ok, my first definition is incomplete, I will edit it now. The second definition is word by word. Which I do not understand because we have not defined the notion of maps between quas proj and affine space.
$endgroup$
– CL.
Nov 26 '18 at 8:36
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
There's not much deep going on here, it's just a definition. In spirit, we want closed subsets of affine space $Bbb A^n$ to be our "affine varieties", but sometimes you have a situation where $X$ is a quasiprojective variety that just happens to be also isomorphic to a closed subset of $Bbb A^n$. Since we believe that "up to isomorphism" is the proper equivalence to put on varieties, we will call these things affine varieties as well.
Nothing about "quasiprojective" was important here; if we define more general types of varieties (I don't know in how much generality Shafarevich does this), then we will always call such varieties affine varieties when they are isomorphic to a closed subset of $Bbb A^n$.
So a good example is the one given right here after the line you mention. If you take $X:=Bbb A^1smallsetminus{0}$, then $X$ is quasiprojective because it is open in $Bbb A^1$, which is open in $Bbb P^1$ under the identification $$Bbb A^1simeq{[x:y]mid xneq0}subsetBbb P^1,$$ but on the other hand it is isomorphic to the closed subset ${(x,y)mid xy=1}subsetBbb A^2$. Since it "looks like" a closed subset of $Bbb A^n$, we will call $X$ an affine variety.
I want to emphasize further that Shafarevich is not saying that every quasiprojective variety can be embedded as a closed subset of affine space. For instance, $Bbb P^1$ is a quasiprojective variety, which is not an affine variety. If you know stuff about regular functions then you know that the only globally defined regular functions on $Bbb P^1$ are constants (this is true for any $Bbb P^m$), but this is never true for closed subsets $ZsubsetBbb A^n$ unless $Z$ is a single point.
answered Nov 26 '18 at 18:07
Alex MathersAlex Mathers
10.8k21344
$endgroup$
1
$begingroup$
My problem is perhaps even more elementary, (i) what does it mean to have a quasi projective variety isomorphic to an affine variety? We have only defined isomoprhisms for between quasiprojective varieties and projective spaces. This leads to an even more fundamental question (ii) why is a closed affine set a quasiprojective vareity - which is defined as an open set of a closed projective set. We can regard affine space as, say A_0^n in P^n, and a closed set of affine space, is also a closed set of P^n with A_0^n. These definitions are really confusing me. I will update post.
$endgroup$
– CL.
Nov 26 '18 at 19:33
1
$begingroup$
@André3000 yes, my mistake. corrected now
$endgroup$
– Alex Mathers
Nov 26 '18 at 21:55
$begingroup$
@CL. I want to double check what I was about to write, let me get back to this in a few hours
$endgroup$
– Alex Mathers
Nov 26 '18 at 22:03
$begingroup$
@CL. Okay, so using the definitions of subspace topology, you can show that if $Y$ is a closed subset of an open subspace of $X$ if and only if $Y$ is an open subset of a closed subspace of $X$. Using this, by choosing an identification of $Bbb A^n$ with an open subset of $Bbb P^n$, we see that every closed subset of affine space $Bbb A^n$ can be considered as a quasiprojective variety.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
$begingroup$
Now using the fact that you know what an isomorphism of quasiprojective varieties is, we can say that a quasiprojective variety $X$ is affine if and only if if it isomorphic (as quasiprojective varieties) to a closed subset of some affine space $Bbb A^n$.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
|
show 2 more comments
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$begingroup$
There's not much deep going on here, it's just a definition. In spirit, we want closed subsets of affine space $Bbb A^n$ to be our "affine varieties", but sometimes you have a situation where $X$ is a quasiprojective variety that just happens to be also isomorphic to a closed subset of $Bbb A^n$. Since we believe that "up to isomorphism" is the proper equivalence to put on varieties, we will call these things affine varieties as well.
Nothing about "quasiprojective" was important here; if we define more general types of varieties (I don't know in how much generality Shafarevich does this), then we will always call such varieties affine varieties when they are isomorphic to a closed subset of $Bbb A^n$.
So a good example is the one given right here after the line you mention. If you take $X:=Bbb A^1smallsetminus{0}$, then $X$ is quasiprojective because it is open in $Bbb A^1$, which is open in $Bbb P^1$ under the identification $$Bbb A^1simeq{[x:y]mid xneq0}subsetBbb P^1,$$ but on the other hand it is isomorphic to the closed subset ${(x,y)mid xy=1}subsetBbb A^2$. Since it "looks like" a closed subset of $Bbb A^n$, we will call $X$ an affine variety.
I want to emphasize further that Shafarevich is not saying that every quasiprojective variety can be embedded as a closed subset of affine space. For instance, $Bbb P^1$ is a quasiprojective variety, which is not an affine variety. If you know stuff about regular functions then you know that the only globally defined regular functions on $Bbb P^1$ are constants (this is true for any $Bbb P^m$), but this is never true for closed subsets $ZsubsetBbb A^n$ unless $Z$ is a single point.
answered Nov 26 '18 at 18:07
Alex MathersAlex Mathers
10.8k21344
$endgroup$
1
$begingroup$
My problem is perhaps even more elementary, (i) what does it mean to have a quasi projective variety isomorphic to an affine variety? We have only defined isomoprhisms for between quasiprojective varieties and projective spaces. This leads to an even more fundamental question (ii) why is a closed affine set a quasiprojective vareity - which is defined as an open set of a closed projective set. We can regard affine space as, say A_0^n in P^n, and a closed set of affine space, is also a closed set of P^n with A_0^n. These definitions are really confusing me. I will update post.
$endgroup$
– CL.
Nov 26 '18 at 19:33
1
$begingroup$
@André3000 yes, my mistake. corrected now
$endgroup$
– Alex Mathers
Nov 26 '18 at 21:55
$begingroup$
@CL. I want to double check what I was about to write, let me get back to this in a few hours
$endgroup$
– Alex Mathers
Nov 26 '18 at 22:03
$begingroup$
@CL. Okay, so using the definitions of subspace topology, you can show that if $Y$ is a closed subset of an open subspace of $X$ if and only if $Y$ is an open subset of a closed subspace of $X$. Using this, by choosing an identification of $Bbb A^n$ with an open subset of $Bbb P^n$, we see that every closed subset of affine space $Bbb A^n$ can be considered as a quasiprojective variety.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
$begingroup$
Now using the fact that you know what an isomorphism of quasiprojective varieties is, we can say that a quasiprojective variety $X$ is affine if and only if if it isomorphic (as quasiprojective varieties) to a closed subset of some affine space $Bbb A^n$.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
|
show 2 more comments
$begingroup$
There's not much deep going on here, it's just a definition. In spirit, we want closed subsets of affine space $Bbb A^n$ to be our "affine varieties", but sometimes you have a situation where $X$ is a quasiprojective variety that just happens to be also isomorphic to a closed subset of $Bbb A^n$. Since we believe that "up to isomorphism" is the proper equivalence to put on varieties, we will call these things affine varieties as well.
Nothing about "quasiprojective" was important here; if we define more general types of varieties (I don't know in how much generality Shafarevich does this), then we will always call such varieties affine varieties when they are isomorphic to a closed subset of $Bbb A^n$.
So a good example is the one given right here after the line you mention. If you take $X:=Bbb A^1smallsetminus{0}$, then $X$ is quasiprojective because it is open in $Bbb A^1$, which is open in $Bbb P^1$ under the identification $$Bbb A^1simeq{[x:y]mid xneq0}subsetBbb P^1,$$ but on the other hand it is isomorphic to the closed subset ${(x,y)mid xy=1}subsetBbb A^2$. Since it "looks like" a closed subset of $Bbb A^n$, we will call $X$ an affine variety.
I want to emphasize further that Shafarevich is not saying that every quasiprojective variety can be embedded as a closed subset of affine space. For instance, $Bbb P^1$ is a quasiprojective variety, which is not an affine variety. If you know stuff about regular functions then you know that the only globally defined regular functions on $Bbb P^1$ are constants (this is true for any $Bbb P^m$), but this is never true for closed subsets $ZsubsetBbb A^n$ unless $Z$ is a single point.
answered Nov 26 '18 at 18:07
Alex MathersAlex Mathers
10.8k21344
$endgroup$
1
$begingroup$
My problem is perhaps even more elementary, (i) what does it mean to have a quasi projective variety isomorphic to an affine variety? We have only defined isomoprhisms for between quasiprojective varieties and projective spaces. This leads to an even more fundamental question (ii) why is a closed affine set a quasiprojective vareity - which is defined as an open set of a closed projective set. We can regard affine space as, say A_0^n in P^n, and a closed set of affine space, is also a closed set of P^n with A_0^n. These definitions are really confusing me. I will update post.
$endgroup$
– CL.
Nov 26 '18 at 19:33
1
$begingroup$
@André3000 yes, my mistake. corrected now
$endgroup$
– Alex Mathers
Nov 26 '18 at 21:55
$begingroup$
@CL. I want to double check what I was about to write, let me get back to this in a few hours
$endgroup$
– Alex Mathers
Nov 26 '18 at 22:03
$begingroup$
@CL. Okay, so using the definitions of subspace topology, you can show that if $Y$ is a closed subset of an open subspace of $X$ if and only if $Y$ is an open subset of a closed subspace of $X$. Using this, by choosing an identification of $Bbb A^n$ with an open subset of $Bbb P^n$, we see that every closed subset of affine space $Bbb A^n$ can be considered as a quasiprojective variety.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
$begingroup$
Now using the fact that you know what an isomorphism of quasiprojective varieties is, we can say that a quasiprojective variety $X$ is affine if and only if if it isomorphic (as quasiprojective varieties) to a closed subset of some affine space $Bbb A^n$.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
|
show 2 more comments
$begingroup$
There's not much deep going on here, it's just a definition. In spirit, we want closed subsets of affine space $Bbb A^n$ to be our "affine varieties", but sometimes you have a situation where $X$ is a quasiprojective variety that just happens to be also isomorphic to a closed subset of $Bbb A^n$. Since we believe that "up to isomorphism" is the proper equivalence to put on varieties, we will call these things affine varieties as well.
Nothing about "quasiprojective" was important here; if we define more general types of varieties (I don't know in how much generality Shafarevich does this), then we will always call such varieties affine varieties when they are isomorphic to a closed subset of $Bbb A^n$.
So a good example is the one given right here after the line you mention. If you take $X:=Bbb A^1smallsetminus{0}$, then $X$ is quasiprojective because it is open in $Bbb A^1$, which is open in $Bbb P^1$ under the identification $$Bbb A^1simeq{[x:y]mid xneq0}subsetBbb P^1,$$ but on the other hand it is isomorphic to the closed subset ${(x,y)mid xy=1}subsetBbb A^2$. Since it "looks like" a closed subset of $Bbb A^n$, we will call $X$ an affine variety.
I want to emphasize further that Shafarevich is not saying that every quasiprojective variety can be embedded as a closed subset of affine space. For instance, $Bbb P^1$ is a quasiprojective variety, which is not an affine variety. If you know stuff about regular functions then you know that the only globally defined regular functions on $Bbb P^1$ are constants (this is true for any $Bbb P^m$), but this is never true for closed subsets $ZsubsetBbb A^n$ unless $Z$ is a single point.
answered Nov 26 '18 at 18:07
Alex MathersAlex Mathers
10.8k21344
$endgroup$
There's not much deep going on here, it's just a definition. In spirit, we want closed subsets of affine space $Bbb A^n$ to be our "affine varieties", but sometimes you have a situation where $X$ is a quasiprojective variety that just happens to be also isomorphic to a closed subset of $Bbb A^n$. Since we believe that "up to isomorphism" is the proper equivalence to put on varieties, we will call these things affine varieties as well.
Nothing about "quasiprojective" was important here; if we define more general types of varieties (I don't know in how much generality Shafarevich does this), then we will always call such varieties affine varieties when they are isomorphic to a closed subset of $Bbb A^n$.
So a good example is the one given right here after the line you mention. If you take $X:=Bbb A^1smallsetminus{0}$, then $X$ is quasiprojective because it is open in $Bbb A^1$, which is open in $Bbb P^1$ under the identification $$Bbb A^1simeq{[x:y]mid xneq0}subsetBbb P^1,$$ but on the other hand it is isomorphic to the closed subset ${(x,y)mid xy=1}subsetBbb A^2$. Since it "looks like" a closed subset of $Bbb A^n$, we will call $X$ an affine variety.
I want to emphasize further that Shafarevich is not saying that every quasiprojective variety can be embedded as a closed subset of affine space. For instance, $Bbb P^1$ is a quasiprojective variety, which is not an affine variety. If you know stuff about regular functions then you know that the only globally defined regular functions on $Bbb P^1$ are constants (this is true for any $Bbb P^m$), but this is never true for closed subsets $ZsubsetBbb A^n$ unless $Z$ is a single point.
answered Nov 26 '18 at 18:07
Alex MathersAlex Mathers
10.8k21344
edited Nov 26 '18 at 21:54
answered Nov 26 '18 at 18:07
Alex MathersAlex Mathers
10.8k21344
answered Nov 26 '18 at 18:07
Alex MathersAlex Mathers
10.8k21344
answered Nov 26 '18 at 18:07
Alex MathersAlex Mathers
10.8k21344
10.8k21344
1
$begingroup$
My problem is perhaps even more elementary, (i) what does it mean to have a quasi projective variety isomorphic to an affine variety? We have only defined isomoprhisms for between quasiprojective varieties and projective spaces. This leads to an even more fundamental question (ii) why is a closed affine set a quasiprojective vareity - which is defined as an open set of a closed projective set. We can regard affine space as, say A_0^n in P^n, and a closed set of affine space, is also a closed set of P^n with A_0^n. These definitions are really confusing me. I will update post.
$endgroup$
– CL.
Nov 26 '18 at 19:33
1
$begingroup$
@André3000 yes, my mistake. corrected now
$endgroup$
– Alex Mathers
Nov 26 '18 at 21:55
$begingroup$
@CL. I want to double check what I was about to write, let me get back to this in a few hours
$endgroup$
– Alex Mathers
Nov 26 '18 at 22:03
$begingroup$
@CL. Okay, so using the definitions of subspace topology, you can show that if $Y$ is a closed subset of an open subspace of $X$ if and only if $Y$ is an open subset of a closed subspace of $X$. Using this, by choosing an identification of $Bbb A^n$ with an open subset of $Bbb P^n$, we see that every closed subset of affine space $Bbb A^n$ can be considered as a quasiprojective variety.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
$begingroup$
Now using the fact that you know what an isomorphism of quasiprojective varieties is, we can say that a quasiprojective variety $X$ is affine if and only if if it isomorphic (as quasiprojective varieties) to a closed subset of some affine space $Bbb A^n$.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
|
show 2 more comments
1
$begingroup$
My problem is perhaps even more elementary, (i) what does it mean to have a quasi projective variety isomorphic to an affine variety? We have only defined isomoprhisms for between quasiprojective varieties and projective spaces. This leads to an even more fundamental question (ii) why is a closed affine set a quasiprojective vareity - which is defined as an open set of a closed projective set. We can regard affine space as, say A_0^n in P^n, and a closed set of affine space, is also a closed set of P^n with A_0^n. These definitions are really confusing me. I will update post.
$endgroup$
– CL.
Nov 26 '18 at 19:33
1
$begingroup$
@André3000 yes, my mistake. corrected now
$endgroup$
– Alex Mathers
Nov 26 '18 at 21:55
$begingroup$
@CL. I want to double check what I was about to write, let me get back to this in a few hours
$endgroup$
– Alex Mathers
Nov 26 '18 at 22:03
$begingroup$
@CL. Okay, so using the definitions of subspace topology, you can show that if $Y$ is a closed subset of an open subspace of $X$ if and only if $Y$ is an open subset of a closed subspace of $X$. Using this, by choosing an identification of $Bbb A^n$ with an open subset of $Bbb P^n$, we see that every closed subset of affine space $Bbb A^n$ can be considered as a quasiprojective variety.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
$begingroup$
Now using the fact that you know what an isomorphism of quasiprojective varieties is, we can say that a quasiprojective variety $X$ is affine if and only if if it isomorphic (as quasiprojective varieties) to a closed subset of some affine space $Bbb A^n$.
$endgroup$
– Alex Mathers
Nov 26 '18 at 23:38
1
1
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My problem is perhaps even more elementary, (i) what does it mean to have a quasi projective variety isomorphic to an affine variety? We have only defined isomoprhisms for between quasiprojective varieties and projective spaces. This leads to an even more fundamental question (ii) why is a closed affine set a quasiprojective vareity - which is defined as an open set of a closed projective set. We can regard affine space as, say A_0^n in P^n, and a closed set of affine space, is also a closed set of P^n with A_0^n. These definitions are really confusing me. I will update post.
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– CL.
Nov 26 '18 at 19:33
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My problem is perhaps even more elementary, (i) what does it mean to have a quasi projective variety isomorphic to an affine variety? We have only defined isomoprhisms for between quasiprojective varieties and projective spaces. This leads to an even more fundamental question (ii) why is a closed affine set a quasiprojective vareity - which is defined as an open set of a closed projective set. We can regard affine space as, say A_0^n in P^n, and a closed set of affine space, is also a closed set of P^n with A_0^n. These definitions are really confusing me. I will update post.
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– CL.
Nov 26 '18 at 19:33
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@André3000 yes, my mistake. corrected now
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– Alex Mathers
Nov 26 '18 at 21:55
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@André3000 yes, my mistake. corrected now
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– Alex Mathers
Nov 26 '18 at 21:55
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@CL. I want to double check what I was about to write, let me get back to this in a few hours
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– Alex Mathers
Nov 26 '18 at 22:03
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@CL. I want to double check what I was about to write, let me get back to this in a few hours
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– Alex Mathers
Nov 26 '18 at 22:03
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@CL. Okay, so using the definitions of subspace topology, you can show that if $Y$ is a closed subset of an open subspace of $X$ if and only if $Y$ is an open subset of a closed subspace of $X$. Using this, by choosing an identification of $Bbb A^n$ with an open subset of $Bbb P^n$, we see that every closed subset of affine space $Bbb A^n$ can be considered as a quasiprojective variety.
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– Alex Mathers
Nov 26 '18 at 23:38
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@CL. Okay, so using the definitions of subspace topology, you can show that if $Y$ is a closed subset of an open subspace of $X$ if and only if $Y$ is an open subset of a closed subspace of $X$. Using this, by choosing an identification of $Bbb A^n$ with an open subset of $Bbb P^n$, we see that every closed subset of affine space $Bbb A^n$ can be considered as a quasiprojective variety.
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– Alex Mathers
Nov 26 '18 at 23:38
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Now using the fact that you know what an isomorphism of quasiprojective varieties is, we can say that a quasiprojective variety $X$ is affine if and only if if it isomorphic (as quasiprojective varieties) to a closed subset of some affine space $Bbb A^n$.
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– Alex Mathers
Nov 26 '18 at 23:38
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Now using the fact that you know what an isomorphism of quasiprojective varieties is, we can say that a quasiprojective variety $X$ is affine if and only if if it isomorphic (as quasiprojective varieties) to a closed subset of some affine space $Bbb A^n$.
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– Alex Mathers
Nov 26 '18 at 23:38
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The first definition is the definition of a regular map. How is this connected to the second definition? You would do well to check the precise wording for the second definition, as it looks to have been altered in transcription (the first also has typos).
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– Mark Bennet
Nov 26 '18 at 8:26
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Ok, my first definition is incomplete, I will edit it now. The second definition is word by word. Which I do not understand because we have not defined the notion of maps between quas proj and affine space.
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– CL.
Nov 26 '18 at 8:36