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Help me please! How can I find any canonical divisor of $B=\left \{ w_{0}^{3}+w_{1}^{3}+w_{2}^{2}w_0=0 \right \}\subseteq \mathbb{P}^{2}$?

Thanks in advance.

  • 1
    That appears to be a smooth cubic, so you can use the adjunction formula to see that $\omega_B\simeq \mathcal{O}_B(-3+3)\simeq \mathcal{O}_B$. Thus the canonical divisor is trivial.2012-08-27

2 Answers 2

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That is a smooth curve of degree $d = 3$ and genus $g = \frac{1}{2}(d - 1)(d - 2) = 1$. Using Riemann-Roch Theorem we can prove that a divisor $D$ is canonical iff $\deg(D) = 2g - 2$ and $\ell(D) = g$. Since $g = 1$, the zero divisor $D = 0$ is canonical. That is the argument if you are working with the classical language of algebraic geometry.

If you are working with schemes, Matt's comment works well.

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The answer of Matt in the comment is right. In the cases, where you cannot apply the adjunction formula, you can maybe use for example Macaulay2 or Singular to compute the canonical divisor. Here is the code for your problem with output in Macaulay2:

i5 : R=QQ[w_0..w_2]

o5 = R

o5 : PolynomialRing

i6 : I=ideal(w_0^3+w_1^3+w_0 * w_2^2);

i7 : B=Proj(R/I)

o7 = B

o7 : ProjectiveVariety

i8 : cotangentSheaf B

o8 = OO_B ^1

o8 : coherent sheaf on B, free