Show that the following map is surjective
$begingroup$
Situation:
Consider the curves $ C_1 = x^2 + y^2 +x^2y^2 = 1 $ and $C_2 = w^2 = 1 - z^4 .$
Define a map $lambda$ from $C_1$ to $C_2$ with $lambda(x,y) = (x,(1+x^2)y)$.
This map should be bijective. injective was ok, but I failed to show that this map is surjective.
number-theory curves
asked Jan 17 at 11:00
MemoriesMemories
10611
$endgroup$
add a comment |
$begingroup$
Situation:
Consider the curves $ C_1 = x^2 + y^2 +x^2y^2 = 1 $ and $C_2 = w^2 = 1 - z^4 .$
Define a map $lambda$ from $C_1$ to $C_2$ with $lambda(x,y) = (x,(1+x^2)y)$.
This map should be bijective. injective was ok, but I failed to show that this map is surjective.
number-theory curves
asked Jan 17 at 11:00
MemoriesMemories
10611
$endgroup$
add a comment |
$begingroup$
Situation:
Consider the curves $ C_1 = x^2 + y^2 +x^2y^2 = 1 $ and $C_2 = w^2 = 1 - z^4 .$
Define a map $lambda$ from $C_1$ to $C_2$ with $lambda(x,y) = (x,(1+x^2)y)$.
This map should be bijective. injective was ok, but I failed to show that this map is surjective.
number-theory curves
asked Jan 17 at 11:00
MemoriesMemories
10611
$endgroup$
Situation:
Consider the curves $ C_1 = x^2 + y^2 +x^2y^2 = 1 $ and $C_2 = w^2 = 1 - z^4 .$
Define a map $lambda$ from $C_1$ to $C_2$ with $lambda(x,y) = (x,(1+x^2)y)$.
This map should be bijective. injective was ok, but I failed to show that this map is surjective.
number-theory curves
number-theory curves
asked Jan 17 at 11:00
MemoriesMemories
10611
asked Jan 17 at 11:00
MemoriesMemories
10611
asked Jan 17 at 11:00
MemoriesMemories
10611
asked Jan 17 at 11:00
MemoriesMemories
10611
asked Jan 17 at 11:00
MemoriesMemories
10611
10611
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Any point $(z,w)$ in $C_2$ is the image of the point $(x,y)=(z,frac w {1+z^{2}})$. All you have to do is to verify the condition $x^{2}+x^{2}+x^{2}y^{2}=1$ using the condition $w^{2}=1-z^{4}$.
answered Jan 17 at 11:41
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any point $(z,w)$ in $C_2$ is the image of the point $(x,y)=(z,frac w {1+z^{2}})$. All you have to do is to verify the condition $x^{2}+x^{2}+x^{2}y^{2}=1$ using the condition $w^{2}=1-z^{4}$.
answered Jan 17 at 11:41
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
Any point $(z,w)$ in $C_2$ is the image of the point $(x,y)=(z,frac w {1+z^{2}})$. All you have to do is to verify the condition $x^{2}+x^{2}+x^{2}y^{2}=1$ using the condition $w^{2}=1-z^{4}$.
answered Jan 17 at 11:41
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
Any point $(z,w)$ in $C_2$ is the image of the point $(x,y)=(z,frac w {1+z^{2}})$. All you have to do is to verify the condition $x^{2}+x^{2}+x^{2}y^{2}=1$ using the condition $w^{2}=1-z^{4}$.
answered Jan 17 at 11:41
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
Any point $(z,w)$ in $C_2$ is the image of the point $(x,y)=(z,frac w {1+z^{2}})$. All you have to do is to verify the condition $x^{2}+x^{2}+x^{2}y^{2}=1$ using the condition $w^{2}=1-z^{4}$.
answered Jan 17 at 11:41
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 11:41
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 11:41
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 11:41
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
59.2k42161
add a comment |
add a comment |
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