Riemann-Roch theorem for minimal Kähler surface. Let $X$ be a compact minimal non-algebraic Kähler surface. In Buchdahl's paper Algebraic deformations of compact Kähler surfaces p458, the author gives an equation: $20=8h^{0,1}(X)+h^{1,1}(X)$, I don't know how to get this equation.
The explanation by the author is: $c_1(X)^2=0$ and $h^{2,0}(X)=1$, I know $c_1(X)^2=0$ is deduced by minimal and non-algebraic condition, but I can't see why $h^{2,0}(X)=1$, the author also said from the Riemann-Roch Theorem and the fact that $\mathcal X(X)=c_2(X)[X]$ it follows easily that $20=8h^{0,1}(X)+h^{1,1}(X)$, unfortunately, for me it's not so easy, so may someone tell me why $h^{2,0}(X)=1$ and give more details about how to use Riemann-Roch to deduce this formula? Thanks!
 A: Using the notation of the paper, note that $X'$ is not algebraic as $X$ is not algebraic. It follows that $h^{2,0}(X') = h^{0,2}(X') > 0$, see the second last paragraph of section $1$. Now let $\sigma_1, \sigma_2 \in H^0(X', K_{X'})$ be non-zero (such elements exist as $h^{2,0}(X') > 0$. Then $\sigma_1\circ\pi, \sigma_2\circ \pi \in H^0(X, \pi^*K_{X'})$ are non-zero. By Corollary $3$, we deduce that $H^0(X, \pi^*K_{X'})$ is one-dimensional, so there is a constant $c$ such that $\sigma_2\circ\pi = c\sigma_1\circ\pi$ and hence $(\sigma_2 - c\sigma_1)\circ\pi = 0$. As $\pi$ is a biholomorphism away from the exceptional divisors, we deduce that $\sigma_2 - c\sigma_1 = 0$, i.e. $\sigma_2 = c\sigma_1$. Therefore $h^{2,0}(X') = \dim H^0(X', K_{X'}) = 1$.
Note that
\begin{align*}
\chi(X', \mathcal{O}_{X'}) &= h^0(X', \mathcal{O}_{X'}) - h^1(X', \mathcal{O}_{X'}) + h^2(X', \mathcal{O}_{X'})\\ 
&= h^{0,0}(X') - h^{1,0}(X') + h^{2,0}(X')\\
&= 2 - h^{1,0}(X').
\end{align*}
But we also have
\begin{align*}
\chi(X', \mathcal{O}_{X'}) &= \frac{1}{12}(c_1(X')^2 + c_2(X'))[X']\\ 
&= \frac{1}{12}\chi(X')\\ 
&= \frac{1}{12}(2 - 2b_1(X') + b_2(X'))\\
&= \frac{1}{12}(2 - 4h^{1,0}(X') + h^{2,0}(X') + h^{1,1}(X') + h^{0,2}(X'))\\
&= \frac{1}{12}(4 - 4h^{1,0}(X') + h^{1,1}(X')).
\end{align*}
Equating the two expressions for $\chi(X', \mathcal{O}_{X'})$ gives the desired equation.
