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Why don't these morphisms extend?
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This is part of exercise 16.5.A in Vakil's notes (http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf). To give an example of why the assumptions in the curve-to-projective extension theorem (16.5.1) are necessary, he asks you to show the following morphisms $C backslash p to Y$ do not extend to morphisms $C to Y$:
$C = Y = mathbbA^1_k$, $p=0$, and the map is $t mapsto t^-1$.
$C = mathbbA^2_k$, $Y = mathbbP^1_k$, $p=(0,0)$, and the map is $(x,y) mapsto [x : y]$.
- Same as 2 but $C = mathrmSpec(k[x,y]/(y^2-x^3))$.
I completely understand the first one, and intuitively why the second and third should fail, but don't know how to prove it. Any hints are appreciated.
algebraic-geometry curves projective-schemes
asked Jul 26 at 20:49
ggg
435313
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up vote
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This is part of exercise 16.5.A in Vakil's notes (http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf). To give an example of why the assumptions in the curve-to-projective extension theorem (16.5.1) are necessary, he asks you to show the following morphisms $C backslash p to Y$ do not extend to morphisms $C to Y$:
$C = Y = mathbbA^1_k$, $p=0$, and the map is $t mapsto t^-1$.
$C = mathbbA^2_k$, $Y = mathbbP^1_k$, $p=(0,0)$, and the map is $(x,y) mapsto [x : y]$.
- Same as 2 but $C = mathrmSpec(k[x,y]/(y^2-x^3))$.
I completely understand the first one, and intuitively why the second and third should fail, but don't know how to prove it. Any hints are appreciated.
algebraic-geometry curves projective-schemes
asked Jul 26 at 20:49
ggg
435313
1
(2) Approach to $(0,0)$ along $(0,y)$. The map is constant along that line, equal to $[0:y]=[0:1]$ and that is the limit. Approach $(0,0)$ along $(x,0)$. The maps output is $[x:0]=[1:0]$.
– user577471
Jul 26 at 21:15
Is there a way to rephrase this in the language of schemes? I'd prefer not to take limits. For example, if the morphism were to extend, say to $pi : mathbbA^2_k to mathbbP^1_k$, then it would correspond to two global sections of the line bundle $pi^* mathscrO(1)$ on $mathbbA^1_k$ without common zeros, namely $pi^* X, pi^* Y$ where $X, Y$ are projective coordinates. Since it's an extension, we must have $pi^*X = x$ and $pi^*Y = y$ (this is sketchy), but this means that $V(x)$ maps to $V(X)$ and $V(y)$ maps to $V(Y)$, which can't happen since $V(X) cap V(Y) = emptyset$.
– ggg
Jul 27 at 18:15
Yes, you can Frenchify it. Without any gain in understanding, though. Composing $pi$ with $f_y:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(y)$, you get a map $picirc f_y$ sending $0neq pinmathbbA_k^1$ sent to $[0:1]$. Therefore, $picirc f_y$ should send $0$ to $[0:1]$. Doing the same with $f_x:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(x)$ you get that $0$ should be sent to $[1:0]$.
– user577471
Jul 27 at 20:09
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is part of exercise 16.5.A in Vakil's notes (http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf). To give an example of why the assumptions in the curve-to-projective extension theorem (16.5.1) are necessary, he asks you to show the following morphisms $C backslash p to Y$ do not extend to morphisms $C to Y$:
$C = Y = mathbbA^1_k$, $p=0$, and the map is $t mapsto t^-1$.
$C = mathbbA^2_k$, $Y = mathbbP^1_k$, $p=(0,0)$, and the map is $(x,y) mapsto [x : y]$.
- Same as 2 but $C = mathrmSpec(k[x,y]/(y^2-x^3))$.
I completely understand the first one, and intuitively why the second and third should fail, but don't know how to prove it. Any hints are appreciated.
algebraic-geometry curves projective-schemes
asked Jul 26 at 20:49
ggg
435313
This is part of exercise 16.5.A in Vakil's notes (http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf). To give an example of why the assumptions in the curve-to-projective extension theorem (16.5.1) are necessary, he asks you to show the following morphisms $C backslash p to Y$ do not extend to morphisms $C to Y$:
$C = Y = mathbbA^1_k$, $p=0$, and the map is $t mapsto t^-1$.
$C = mathbbA^2_k$, $Y = mathbbP^1_k$, $p=(0,0)$, and the map is $(x,y) mapsto [x : y]$.
- Same as 2 but $C = mathrmSpec(k[x,y]/(y^2-x^3))$.
I completely understand the first one, and intuitively why the second and third should fail, but don't know how to prove it. Any hints are appreciated.
algebraic-geometry curves projective-schemes
asked Jul 26 at 20:49
ggg
435313
asked Jul 26 at 20:49
ggg
435313
asked Jul 26 at 20:49
ggg
435313
asked Jul 26 at 20:49
ggg
435313
435313
1
(2) Approach to $(0,0)$ along $(0,y)$. The map is constant along that line, equal to $[0:y]=[0:1]$ and that is the limit. Approach $(0,0)$ along $(x,0)$. The maps output is $[x:0]=[1:0]$.
– user577471
Jul 26 at 21:15
Is there a way to rephrase this in the language of schemes? I'd prefer not to take limits. For example, if the morphism were to extend, say to $pi : mathbbA^2_k to mathbbP^1_k$, then it would correspond to two global sections of the line bundle $pi^* mathscrO(1)$ on $mathbbA^1_k$ without common zeros, namely $pi^* X, pi^* Y$ where $X, Y$ are projective coordinates. Since it's an extension, we must have $pi^*X = x$ and $pi^*Y = y$ (this is sketchy), but this means that $V(x)$ maps to $V(X)$ and $V(y)$ maps to $V(Y)$, which can't happen since $V(X) cap V(Y) = emptyset$.
– ggg
Jul 27 at 18:15
Yes, you can Frenchify it. Without any gain in understanding, though. Composing $pi$ with $f_y:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(y)$, you get a map $picirc f_y$ sending $0neq pinmathbbA_k^1$ sent to $[0:1]$. Therefore, $picirc f_y$ should send $0$ to $[0:1]$. Doing the same with $f_x:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(x)$ you get that $0$ should be sent to $[1:0]$.
– user577471
Jul 27 at 20:09
add a comment |Â
1
(2) Approach to $(0,0)$ along $(0,y)$. The map is constant along that line, equal to $[0:y]=[0:1]$ and that is the limit. Approach $(0,0)$ along $(x,0)$. The maps output is $[x:0]=[1:0]$.
– user577471
Jul 26 at 21:15
Is there a way to rephrase this in the language of schemes? I'd prefer not to take limits. For example, if the morphism were to extend, say to $pi : mathbbA^2_k to mathbbP^1_k$, then it would correspond to two global sections of the line bundle $pi^* mathscrO(1)$ on $mathbbA^1_k$ without common zeros, namely $pi^* X, pi^* Y$ where $X, Y$ are projective coordinates. Since it's an extension, we must have $pi^*X = x$ and $pi^*Y = y$ (this is sketchy), but this means that $V(x)$ maps to $V(X)$ and $V(y)$ maps to $V(Y)$, which can't happen since $V(X) cap V(Y) = emptyset$.
– ggg
Jul 27 at 18:15
Yes, you can Frenchify it. Without any gain in understanding, though. Composing $pi$ with $f_y:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(y)$, you get a map $picirc f_y$ sending $0neq pinmathbbA_k^1$ sent to $[0:1]$. Therefore, $picirc f_y$ should send $0$ to $[0:1]$. Doing the same with $f_x:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(x)$ you get that $0$ should be sent to $[1:0]$.
– user577471
Jul 27 at 20:09
1
1
(2) Approach to $(0,0)$ along $(0,y)$. The map is constant along that line, equal to $[0:y]=[0:1]$ and that is the limit. Approach $(0,0)$ along $(x,0)$. The maps output is $[x:0]=[1:0]$.
– user577471
Jul 26 at 21:15
(2) Approach to $(0,0)$ along $(0,y)$. The map is constant along that line, equal to $[0:y]=[0:1]$ and that is the limit. Approach $(0,0)$ along $(x,0)$. The maps output is $[x:0]=[1:0]$.
– user577471
Jul 26 at 21:15
Is there a way to rephrase this in the language of schemes? I'd prefer not to take limits. For example, if the morphism were to extend, say to $pi : mathbbA^2_k to mathbbP^1_k$, then it would correspond to two global sections of the line bundle $pi^* mathscrO(1)$ on $mathbbA^1_k$ without common zeros, namely $pi^* X, pi^* Y$ where $X, Y$ are projective coordinates. Since it's an extension, we must have $pi^*X = x$ and $pi^*Y = y$ (this is sketchy), but this means that $V(x)$ maps to $V(X)$ and $V(y)$ maps to $V(Y)$, which can't happen since $V(X) cap V(Y) = emptyset$.
– ggg
Jul 27 at 18:15
Is there a way to rephrase this in the language of schemes? I'd prefer not to take limits. For example, if the morphism were to extend, say to $pi : mathbbA^2_k to mathbbP^1_k$, then it would correspond to two global sections of the line bundle $pi^* mathscrO(1)$ on $mathbbA^1_k$ without common zeros, namely $pi^* X, pi^* Y$ where $X, Y$ are projective coordinates. Since it's an extension, we must have $pi^*X = x$ and $pi^*Y = y$ (this is sketchy), but this means that $V(x)$ maps to $V(X)$ and $V(y)$ maps to $V(Y)$, which can't happen since $V(X) cap V(Y) = emptyset$.
– ggg
Jul 27 at 18:15
Yes, you can Frenchify it. Without any gain in understanding, though. Composing $pi$ with $f_y:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(y)$, you get a map $picirc f_y$ sending $0neq pinmathbbA_k^1$ sent to $[0:1]$. Therefore, $picirc f_y$ should send $0$ to $[0:1]$. Doing the same with $f_x:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(x)$ you get that $0$ should be sent to $[1:0]$.
– user577471
Jul 27 at 20:09
Yes, you can Frenchify it. Without any gain in understanding, though. Composing $pi$ with $f_y:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(y)$, you get a map $picirc f_y$ sending $0neq pinmathbbA_k^1$ sent to $[0:1]$. Therefore, $picirc f_y$ should send $0$ to $[0:1]$. Doing the same with $f_x:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(x)$ you get that $0$ should be sent to $[1:0]$.
– user577471
Jul 27 at 20:09
add a comment |Â
1 Answer
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As 2) has been adressed in the comments, this answer will address 3).
Consider the function on the image of $C$ given by $[x:y]mapsto y/x$. If the morphism extended to all of $C$, $y/x$ would be a function in $k[x,y]/(y^2-x^3)$, which is clearly wrong.
What's going on here is that singular curves are non-normal, and we're picking an element which is integral over the coordinate algebra but not contained in it - this is only possible when the curve is not normal.
answered Jul 27 at 0:45
KReiser
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1 Answer
1
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
As 2) has been adressed in the comments, this answer will address 3).
Consider the function on the image of $C$ given by $[x:y]mapsto y/x$. If the morphism extended to all of $C$, $y/x$ would be a function in $k[x,y]/(y^2-x^3)$, which is clearly wrong.
What's going on here is that singular curves are non-normal, and we're picking an element which is integral over the coordinate algebra but not contained in it - this is only possible when the curve is not normal.
answered Jul 27 at 0:45
KReiser
7,47011230
add a comment |Â
up vote
0
down vote
As 2) has been adressed in the comments, this answer will address 3).
Consider the function on the image of $C$ given by $[x:y]mapsto y/x$. If the morphism extended to all of $C$, $y/x$ would be a function in $k[x,y]/(y^2-x^3)$, which is clearly wrong.
What's going on here is that singular curves are non-normal, and we're picking an element which is integral over the coordinate algebra but not contained in it - this is only possible when the curve is not normal.
answered Jul 27 at 0:45
KReiser
7,47011230
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As 2) has been adressed in the comments, this answer will address 3).
Consider the function on the image of $C$ given by $[x:y]mapsto y/x$. If the morphism extended to all of $C$, $y/x$ would be a function in $k[x,y]/(y^2-x^3)$, which is clearly wrong.
What's going on here is that singular curves are non-normal, and we're picking an element which is integral over the coordinate algebra but not contained in it - this is only possible when the curve is not normal.
answered Jul 27 at 0:45
KReiser
7,47011230
As 2) has been adressed in the comments, this answer will address 3).
Consider the function on the image of $C$ given by $[x:y]mapsto y/x$. If the morphism extended to all of $C$, $y/x$ would be a function in $k[x,y]/(y^2-x^3)$, which is clearly wrong.
What's going on here is that singular curves are non-normal, and we're picking an element which is integral over the coordinate algebra but not contained in it - this is only possible when the curve is not normal.
answered Jul 27 at 0:45
KReiser
7,47011230
answered Jul 27 at 0:45
KReiser
7,47011230
answered Jul 27 at 0:45
KReiser
7,47011230
answered Jul 27 at 0:45
KReiser
7,47011230
7,47011230
add a comment |Â
add a comment |Â
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1
(2) Approach to $(0,0)$ along $(0,y)$. The map is constant along that line, equal to $[0:y]=[0:1]$ and that is the limit. Approach $(0,0)$ along $(x,0)$. The maps output is $[x:0]=[1:0]$.
– user577471
Jul 26 at 21:15
Is there a way to rephrase this in the language of schemes? I'd prefer not to take limits. For example, if the morphism were to extend, say to $pi : mathbbA^2_k to mathbbP^1_k$, then it would correspond to two global sections of the line bundle $pi^* mathscrO(1)$ on $mathbbA^1_k$ without common zeros, namely $pi^* X, pi^* Y$ where $X, Y$ are projective coordinates. Since it's an extension, we must have $pi^*X = x$ and $pi^*Y = y$ (this is sketchy), but this means that $V(x)$ maps to $V(X)$ and $V(y)$ maps to $V(Y)$, which can't happen since $V(X) cap V(Y) = emptyset$.
– ggg
Jul 27 at 18:15
Yes, you can Frenchify it. Without any gain in understanding, though. Composing $pi$ with $f_y:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(y)$, you get a map $picirc f_y$ sending $0neq pinmathbbA_k^1$ sent to $[0:1]$. Therefore, $picirc f_y$ should send $0$ to $[0:1]$. Doing the same with $f_x:mathbbA_k^1tomathbbA_k^2$ given by $k[x,y]to k[x,y]/(x)$ you get that $0$ should be sent to $[1:0]$.
– user577471
Jul 27 at 20:09