I am studying the Riemann-Roch theorem for surfaces and would like to apply it to an easy surface, for example to $\mathbb{P}^2$.I assume to be working over an algebraically closed field. The version of the Riemann-Roch theorem I am using is the following. Let $D$ be an arbitrary divisor on a surface $X$, and denote the canonical divisor by $K$. Then \begin{equation*} l(D)-s(D)+l(K-D) = \frac{1}{2} D. (D-K) + 1 + p_a(X). \end{equation*}
I am stuck with the left hand side of the Riemann Roch equation for surfaces in this case. I have already found that the right hand side of the Riemann-Roch theorem, which equals $\frac12 (d+1)(d+2)$ for a divisor $D$ of degree $d$. The superabundance $s(D)$ of any divisor of degree $d$ on $\mathbb{P}^2$ is equal to zero. As the canonical divisor $K$ has degree $-3$, we have that $l(D) = \binom{2+d}{2}$, and $l(D) = \binom{-1-d}{2} = \binom{2+d}{2}$. But then I would get that the left hand side equals $\frac12 (d+1)(d+2)$, which equals TWO times the right hand side. Does anyone see what is wrong here?