Is the cylinder an algebraic variety?
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up vote
2
down vote
favorite
$$
cases
x^2 + y^2 = 1 \
z = z
$$
where the 2nd equation can be regarded as $varnothing$.
Is this an algebraic variety?
If yes, is the surface defined by the intersection of 2 (perpendicular) cylinders also an algebraic variety?
Thanks.
algebraic-geometry
 |Â
show 2 more comments
up vote
2
down vote
favorite
$$
cases
x^2 + y^2 = 1 \
z = z
$$
where the 2nd equation can be regarded as $varnothing$.
Is this an algebraic variety?
If yes, is the surface defined by the intersection of 2 (perpendicular) cylinders also an algebraic variety?
Thanks.
algebraic-geometry
1
$V(x^2 + y^2 - r^2) subset mathbb A^3$ is the infinite cylinder. Are you wanting a cylinder with finite height?
â Benjamin Tighe
Sep 10 at 21:01
No, I just thought that a cylinder is kinda weird as an algebraic variety.
â Yan King Yin
Sep 10 at 21:05
3
The infinite cylinder is a real algebraic variety. Real algebraic geometry has quite a different flavor from algebraic geometry over algebraically closed fields such as $mathbbC$.
â Qiaochu Yuan
Sep 10 at 21:10
@QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
â Yan King Yin
Sep 10 at 21:11
It seems that $x^2 + y^2 = 1$ is not expressible as a complex polynomial equation in terms of $z = x + yi$.
â Yan King Yin
Sep 11 at 4:34
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$$
cases
x^2 + y^2 = 1 \
z = z
$$
where the 2nd equation can be regarded as $varnothing$.
Is this an algebraic variety?
If yes, is the surface defined by the intersection of 2 (perpendicular) cylinders also an algebraic variety?
Thanks.
algebraic-geometry
$$
cases
x^2 + y^2 = 1 \
z = z
$$
where the 2nd equation can be regarded as $varnothing$.
Is this an algebraic variety?
If yes, is the surface defined by the intersection of 2 (perpendicular) cylinders also an algebraic variety?
Thanks.
algebraic-geometry
algebraic-geometry
edited Sep 11 at 4:39
asked Sep 10 at 20:33
Yan King Yin
451211
451211
1
$V(x^2 + y^2 - r^2) subset mathbb A^3$ is the infinite cylinder. Are you wanting a cylinder with finite height?
â Benjamin Tighe
Sep 10 at 21:01
No, I just thought that a cylinder is kinda weird as an algebraic variety.
â Yan King Yin
Sep 10 at 21:05
3
The infinite cylinder is a real algebraic variety. Real algebraic geometry has quite a different flavor from algebraic geometry over algebraically closed fields such as $mathbbC$.
â Qiaochu Yuan
Sep 10 at 21:10
@QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
â Yan King Yin
Sep 10 at 21:11
It seems that $x^2 + y^2 = 1$ is not expressible as a complex polynomial equation in terms of $z = x + yi$.
â Yan King Yin
Sep 11 at 4:34
 |Â
show 2 more comments
1
$V(x^2 + y^2 - r^2) subset mathbb A^3$ is the infinite cylinder. Are you wanting a cylinder with finite height?
â Benjamin Tighe
Sep 10 at 21:01
No, I just thought that a cylinder is kinda weird as an algebraic variety.
â Yan King Yin
Sep 10 at 21:05
3
The infinite cylinder is a real algebraic variety. Real algebraic geometry has quite a different flavor from algebraic geometry over algebraically closed fields such as $mathbbC$.
â Qiaochu Yuan
Sep 10 at 21:10
@QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
â Yan King Yin
Sep 10 at 21:11
It seems that $x^2 + y^2 = 1$ is not expressible as a complex polynomial equation in terms of $z = x + yi$.
â Yan King Yin
Sep 11 at 4:34
1
1
$V(x^2 + y^2 - r^2) subset mathbb A^3$ is the infinite cylinder. Are you wanting a cylinder with finite height?
â Benjamin Tighe
Sep 10 at 21:01
$V(x^2 + y^2 - r^2) subset mathbb A^3$ is the infinite cylinder. Are you wanting a cylinder with finite height?
â Benjamin Tighe
Sep 10 at 21:01
No, I just thought that a cylinder is kinda weird as an algebraic variety.
â Yan King Yin
Sep 10 at 21:05
No, I just thought that a cylinder is kinda weird as an algebraic variety.
â Yan King Yin
Sep 10 at 21:05
3
3
The infinite cylinder is a real algebraic variety. Real algebraic geometry has quite a different flavor from algebraic geometry over algebraically closed fields such as $mathbbC$.
â Qiaochu Yuan
Sep 10 at 21:10
The infinite cylinder is a real algebraic variety. Real algebraic geometry has quite a different flavor from algebraic geometry over algebraically closed fields such as $mathbbC$.
â Qiaochu Yuan
Sep 10 at 21:10
@QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
â Yan King Yin
Sep 10 at 21:11
@QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
â Yan King Yin
Sep 10 at 21:11
It seems that $x^2 + y^2 = 1$ is not expressible as a complex polynomial equation in terms of $z = x + yi$.
â Yan King Yin
Sep 11 at 4:34
It seems that $x^2 + y^2 = 1$ is not expressible as a complex polynomial equation in terms of $z = x + yi$.
â Yan King Yin
Sep 11 at 4:34
 |Â
show 2 more comments
1 Answer
1
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oldest
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up vote
1
down vote
$V(x^2+y^2-1, x^2 + z^2 - 1) = (x,y,z) in k mid x^2 + y^2 = 1 wedge x^2 + z^2 = 1$ is the set of points defined by the intersection of two perpendicular cylinders. So is $W(x^2+y^2-1, y^2 + z^2 - 1)$.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$V(x^2+y^2-1, x^2 + z^2 - 1) = (x,y,z) in k mid x^2 + y^2 = 1 wedge x^2 + z^2 = 1$ is the set of points defined by the intersection of two perpendicular cylinders. So is $W(x^2+y^2-1, y^2 + z^2 - 1)$.
add a comment |Â
up vote
1
down vote
$V(x^2+y^2-1, x^2 + z^2 - 1) = (x,y,z) in k mid x^2 + y^2 = 1 wedge x^2 + z^2 = 1$ is the set of points defined by the intersection of two perpendicular cylinders. So is $W(x^2+y^2-1, y^2 + z^2 - 1)$.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$V(x^2+y^2-1, x^2 + z^2 - 1) = (x,y,z) in k mid x^2 + y^2 = 1 wedge x^2 + z^2 = 1$ is the set of points defined by the intersection of two perpendicular cylinders. So is $W(x^2+y^2-1, y^2 + z^2 - 1)$.
$V(x^2+y^2-1, x^2 + z^2 - 1) = (x,y,z) in k mid x^2 + y^2 = 1 wedge x^2 + z^2 = 1$ is the set of points defined by the intersection of two perpendicular cylinders. So is $W(x^2+y^2-1, y^2 + z^2 - 1)$.
answered Sep 11 at 7:33
Kelly Brower
214
214
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1
$V(x^2 + y^2 - r^2) subset mathbb A^3$ is the infinite cylinder. Are you wanting a cylinder with finite height?
â Benjamin Tighe
Sep 10 at 21:01
No, I just thought that a cylinder is kinda weird as an algebraic variety.
â Yan King Yin
Sep 10 at 21:05
3
The infinite cylinder is a real algebraic variety. Real algebraic geometry has quite a different flavor from algebraic geometry over algebraically closed fields such as $mathbbC$.
â Qiaochu Yuan
Sep 10 at 21:10
@QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
â Yan King Yin
Sep 10 at 21:11
It seems that $x^2 + y^2 = 1$ is not expressible as a complex polynomial equation in terms of $z = x + yi$.
â Yan King Yin
Sep 11 at 4:34