If C ⊆ P 2 is a plane curve, then g e n u s ( C ) = 1 2 ( d − 1 ) ( d − 2 ). Compare with example in the notesIn fact, the first part of the exercise is easy to solve using the Hilbert polynomial of C.The part of comparing with the example 0.1.3 is what I don't understand.What do you think the author wants here as an answer?

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If   C  ⊆            P        2   is a plane curve, then   g  e  n  u  s  (  C  )  =      1    2    (  d  −  1  )  (  d  −  2  ). Compare with example in the notesIn fact, the first part of the exercise is easy to solve using the Hilbert polynomial of C.The part of comparing with the example 0.1.3 is what I don&#039;t understand.What do you think the author wants here as an answer?

Phillip Fletcher · Accepted Answer

Step 1The topological (=geometric) genus and the arithmetic genus indeed coincide for a smooth curve   C  ⊂            P        2    (      C    ), and the common genus is       1    2    (  d  −  1  )  (  d  −  2  ) if the degree of C is d.These genera however no longer coincide in the singular case:For example the curve       y    2    z  =      x    3   has geometric genus   g  (  C  )  =  0, since             P        1    (      C    ) is homeomorphic to C under the map   (  t  :  v  )  ↦  (      t    2    v  :      t    3    :      v    3    ).Step 2However its arithmetic genus is       p    a    (  C  )  =      1    2    (  3  −  1  )  (  3  −  2  )  =  1, exactly as for any plane projective curve of degree   d  =  3.We always have   g  (  C  )  ≤      p    a    (  C  ). The difference between these numbers is a subtle invariant of the singularities of C and there are formulas for the difference       p    a    (  C  )  −  g  (  C  )

If C subseteq P^2 is a plane curve of degree d, then its arithmetic genus g(C) is equal to 1/2 (d-1)(d-2). Compare this to example 0.1.3 (of the notes of course).

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