Topology of algebraic variety Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Variety and algebraic curvesHow to tell if algebraic set is a variety?Rational parametrization of algebraic varietyQing Liu's definition of an algebraic variety, a non-separated lineDegree of an algebraic variety given by a polynomial parametrizationHow does a complex algebraic variety know about its analytic topology?$GL(mathbb C)$ as an Algebraic Groupalgebraic variety of dimension 0When is a complex manifold of the form $mathbbC^g/Lambda$ an algebraic variety?Why is the algebraic torus an affine variety?

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Topology of algebraic variety



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Variety and algebraic curvesHow to tell if algebraic set is a variety?Rational parametrization of algebraic varietyQing Liu's definition of an algebraic variety, a non-separated lineDegree of an algebraic variety given by a polynomial parametrizationHow does a complex algebraic variety know about its analytic topology?$GL(mathbb C)$ as an Algebraic Groupalgebraic variety of dimension 0When is a complex manifold of the form $mathbbC^g/Lambda$ an algebraic variety?Why is the algebraic torus an affine variety?










0












$begingroup$


I have an algebraic variety given by a polynomial $1+x(1+y)^2$ over $mathbb C$. Is there any reasonable way to see how topologically the algebraic variety defined by this polynomial looks like?










share|cite|improve this question









$endgroup$



migrated from mathoverflow.net Mar 26 at 16:53


This question came from our site for professional mathematicians.













  • 1




    $begingroup$
    In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
    $endgroup$
    – Nicolas Hemelsoet
    Mar 27 at 10:27






  • 1




    $begingroup$
    Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
    $endgroup$
    – reuns
    Mar 27 at 11:41
















0












$begingroup$


I have an algebraic variety given by a polynomial $1+x(1+y)^2$ over $mathbb C$. Is there any reasonable way to see how topologically the algebraic variety defined by this polynomial looks like?










share|cite|improve this question









$endgroup$



migrated from mathoverflow.net Mar 26 at 16:53


This question came from our site for professional mathematicians.













  • 1




    $begingroup$
    In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
    $endgroup$
    – Nicolas Hemelsoet
    Mar 27 at 10:27






  • 1




    $begingroup$
    Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
    $endgroup$
    – reuns
    Mar 27 at 11:41














0












0








0





$begingroup$


I have an algebraic variety given by a polynomial $1+x(1+y)^2$ over $mathbb C$. Is there any reasonable way to see how topologically the algebraic variety defined by this polynomial looks like?










share|cite|improve this question









$endgroup$




I have an algebraic variety given by a polynomial $1+x(1+y)^2$ over $mathbb C$. Is there any reasonable way to see how topologically the algebraic variety defined by this polynomial looks like?







algebraic-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 26 at 15:22







Adam Klah











migrated from mathoverflow.net Mar 26 at 16:53


This question came from our site for professional mathematicians.









migrated from mathoverflow.net Mar 26 at 16:53


This question came from our site for professional mathematicians.









  • 1




    $begingroup$
    In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
    $endgroup$
    – Nicolas Hemelsoet
    Mar 27 at 10:27






  • 1




    $begingroup$
    Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
    $endgroup$
    – reuns
    Mar 27 at 11:41













  • 1




    $begingroup$
    In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
    $endgroup$
    – Nicolas Hemelsoet
    Mar 27 at 10:27






  • 1




    $begingroup$
    Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
    $endgroup$
    – reuns
    Mar 27 at 11:41








1




1




$begingroup$
In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
$endgroup$
– Nicolas Hemelsoet
Mar 27 at 10:27




$begingroup$
In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
$endgroup$
– Nicolas Hemelsoet
Mar 27 at 10:27




1




1




$begingroup$
Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
$endgroup$
– reuns
Mar 27 at 11:41





$begingroup$
Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
$endgroup$
– reuns
Mar 27 at 11:41











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$begingroup$

The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?






share|cite|improve this answer









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    0












    $begingroup$

    The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?






        share|cite|improve this answer









        $endgroup$



        The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 27 at 9:44









        quantumquantum

        538210




        538210



























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