Smooth curve of genus 1 in P1C×P1C.Show that { ( ( x 0 : x 1 ) , ( y 0 : y 1 ) ) : ( x 0 2 + x 1 2 ) ( y 0 2 + y 1 2 ) = x 0 x 1 y 0 y 1 } ⊂ P C 1 × P C 1 is a smooth curve of genus 1.The author briefly talked about the concept genus in this section without assuming much knowledge in topology. One proposition said that the genus of smooth curve of degree d in P C 2 is ( d − 1 2 ) .Then in one exercise I showed that the arithmetic genus ( − 1 ) dim ⁡ X ⋅ ( χ X ( 0 ) − 1 ) of a smooth curve in P C 2 agrees with the geometric genus, where χ X is the Hilbert polynomial of the projective variety X. The author then stated that this can be generalized to any smooth projective curve over C.So I think I should use that formula for this exercise. The smoothness can be shown by the Segre embedding. Now the formula says that the genus is 1 − χ X ( 0 ) .And χ X ( d ) = ( 3 + d 3 ) − ( 3 + d − 4 3 ) . But I cannot get 1. I must have made some mistake when calculating χ X ( d ). But I couldn't figure out where.

Question

Smooth curve of genus 1 in P1C×P1C.Show that  {  (  (      x    0    :      x    1    )  ,  (      y    0    :      y    1    )  )  :  (      x    0    2    +      x    1    2    )  (      y    0    2    +      y    1    2    )  =      x    0        x    1        y    0        y    1    }  ⊂            P                      C              1    ×            P                      C              1  is a smooth curve of genus 1.The author briefly talked about the concept genus in this section without assuming much knowledge in topology. One proposition said that the genus of smooth curve of degree d in             P                      C              2   is                     (                            d        −        1            2                      )            .Then in one exercise I showed that the arithmetic genus   (  −  1      )          dim      ⁡      X        ⋅  (      χ    X    (  0  )  −  1  ) of a smooth curve in             P                      C              2   agrees with the geometric genus, where       χ    X   is the Hilbert polynomial of the projective variety X. The author then stated that this can be generalized to any smooth projective curve over C.So I think I should use that formula for this exercise. The smoothness can be shown by the Segre embedding. Now the formula says that the genus is  1  −      χ    X    (  0  )  .And       χ    X    (  d  )  =                    (                            3        +        d            3                      )              −                    (                            3        +        d        −        4            3                      )            . But I cannot get 1. I must have made some mistake when calculating       χ    X    (  d  ). But I couldn&#039;t figure out where.

Camden Stanton · Accepted Answer

Step 1Using the Segre embedding             P        1    ×            P        1    ↪            P        3   given by   (      x    0    :      x    1    )  ×  (      y    0    :      y    1    )  ↦  (      x    0        y    0    ,      x    0        y    1    ,      x    1        y    0    ,      x    1        y    1    ), we can rewrite the curve as an intersection of two hypersurfaces in             P        3  . Explicitly, these hypersurfaces will be   x  w  =  y  z and       x    2    +      y    2    +      z    2    +      w    2    =  x  w. These are both quadrics, so that we have a free resolution of             O        X   (X is the curve):  0  →            O                                P                3              (  −  4  )  →            O                                P                3              (  −  2      )    2    →            O                                P                3              →            O        X    →  0Twisting by   d  &amp;gt;&amp;gt;  0, we get  0  →            O                                P                3              (  −  4  +  d  )  →            O                                P                3              (  −  2  +  d      )    2    →            O                                P                3              (  d  )  →            O        X    (  d  )  →  0By additivity of Euler characteristics, the Hilbert polynomial of X is given as the alternating sum                    (                            3        +        d            3                      )              −  2                    (                            3        +        d        −        2            3                      )              +                    (                            3        +        d        −        4            3                      )              =  4  d  .Putting   d  =  0 gives   g  =  1 by your formula.--Step 2Here is another way to compute the genus, using Hurwitz&#039; formula: Let X be the curve again, thought of as sitting inside             P        1    ×            P        1  . Then consider the projection to             P        1   This gives a   2  :  1 map to             P        1  , and by Hurwitz&#039; formula we have  2  g  −  2  =  2  (  0  −  2  )  +  deg  ⁡  R  ,where degR is the degree of the ramification divisor, that is, degree of the defining equation of the points of             P        1    ×            P        1   where the map fails to be   2  :  1. One can compute that this degree is 4 (by the abc-formula). Hence   2  g  −  2  =  −  4  +  4, which implies that   g  =  1.

Smooth curve of genus 1 in P_C^1 times P_C^1.

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