Is the cylinder an algebraic variety?

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$$
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.










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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?
    – Benjamin Tighe
    Sep 10 at 21:01










  • No, I just thought that a cylinder is kinda weird as an algebraic variety.
    – 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 $mathbbC$.
    – Qiaochu Yuan
    Sep 10 at 21:10










  • @QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
    – 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 $z = x + yi$.
    – Yan King Yin
    Sep 11 at 4:34














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.










share|cite|improve this question



















  • 1




    $V(x^2 + y^2 - r^2) subset mathbb A^3$ is the infinite cylinder. Are you wanting a cylinder with finite height?
    – Benjamin Tighe
    Sep 10 at 21:01










  • No, I just thought that a cylinder is kinda weird as an algebraic variety.
    – 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 $mathbbC$.
    – Qiaochu Yuan
    Sep 10 at 21:10










  • @QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
    – 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 $z = x + yi$.
    – Yan King Yin
    Sep 11 at 4:34












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.










share|cite|improve this question















$$
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






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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?
    – Benjamin Tighe
    Sep 10 at 21:01










  • No, I just thought that a cylinder is kinda weird as an algebraic variety.
    – 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 $mathbbC$.
    – Qiaochu Yuan
    Sep 10 at 21:10










  • @QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
    – 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 $z = x + yi$.
    – Yan King Yin
    Sep 11 at 4:34












  • 1




    $V(x^2 + y^2 - r^2) subset mathbb A^3$ is the infinite cylinder. Are you wanting a cylinder with finite height?
    – Benjamin Tighe
    Sep 10 at 21:01










  • No, I just thought that a cylinder is kinda weird as an algebraic variety.
    – 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 $mathbbC$.
    – Qiaochu Yuan
    Sep 10 at 21:10










  • @QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
    – 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 $z = x + yi$.
    – 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?
– 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?
– Benjamin Tighe
Sep 10 at 21:01












No, I just thought that a cylinder is kinda weird as an algebraic variety.
– Yan King Yin
Sep 10 at 21:05




No, I just thought that a cylinder is kinda weird as an algebraic variety.
– 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 $mathbbC$.
– 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 $mathbbC$.
– Qiaochu Yuan
Sep 10 at 21:10












@QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
– Yan King Yin
Sep 10 at 21:11




@QiaochuYuan: so, if the field is $mathbbC$ then the cylinder won't be a variety?
– 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 $z = x + yi$.
– 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 $z = x + yi$.
– Yan King Yin
Sep 11 at 4:34










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$V(x^2+y^2-1, x^2 + z^2 - 1) = (x,y,z) in k mid x^2 + y^2 = 1 wedge x^2 + z^2 = 1$ is the set of points defined by the intersection of two perpendicular cylinders. So is $W(x^2+y^2-1, y^2 + z^2 - 1)$.






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    $V(x^2+y^2-1, x^2 + z^2 - 1) = (x,y,z) in k mid x^2 + y^2 = 1 wedge x^2 + z^2 = 1$ is the set of points defined by the intersection of two perpendicular cylinders. So is $W(x^2+y^2-1, y^2 + z^2 - 1)$.






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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 $W(x^2+y^2-1, y^2 + z^2 - 1)$.






      share|cite|improve this answer






















        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 $W(x^2+y^2-1, y^2 + z^2 - 1)$.






        share|cite|improve this answer












        $V(x^2+y^2-1, x^2 + z^2 - 1) = (x,y,z) in k mid x^2 + y^2 = 1 wedge x^2 + z^2 = 1$ is the set of points defined by the intersection of two perpendicular cylinders. So is $W(x^2+y^2-1, y^2 + z^2 - 1)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 11 at 7:33









        Kelly Brower

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