Betti Numbers and the Néron-Severi group. In the Paper "On the Mordell-Weil lattices" it is proved that the rank $\rho$ of the Néron-Severi group of a rational elliptic surface equals 10. Without any further explanation, it is stated that "for a rational surface, $\rho = b_2$ holds". I don't quite understand this. If there is an easy answer, it would be great if someone could give me a hint. If not, a reference to the literature would be much appreciated (I am not vey familiar with algebraic topology).
The statement can be found in the proof of Lemma 10.1, page 28:
http://www.rkmath.rikkyo.ac.jp/math/shioda/papers/mwl.pdf
 A: About why $\rho=b_2$ for a rational surface: this has a few ingredients. Maybe I'm giving more details than you need, but others might find them useful.


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*The Hodge Decomposition Theorem, which says that (for any smooth $X$) the cohomology group $H^2(X,\mathbf{C})$ decomposes as a direct sum of subspaces $H^{2,0} \oplus H^{1,1} \oplus H^{0,2}$. Each of these is a vector space of differential forms of a certain type: $H^{2,0}$ is the vector space of holomorphic 2-forms on $X$, and $H^{0,2}$ is its complex conjugate, so it has the same dimension as $H^{2,0}$.

*The fact that rational varieties have no holomorphic differential forms, so in particular here $H^{2,0}=0$. Together with 1, this tells you that $H^2(X,\mathbf{C})=H^{1,1}$, the space of so-called $(1,1)$-forms. 

*Finally, we need the Leftschetz $(1,1)$-Theorem, which says (again in general) that an element of $H^{1,1} \cap H^2(X,\mathbf{Z})$ (an integral $(1,1)$-form) is algebraic, that is, it is the class of a divisor on $X$. 
Now by 2, we know that the vector space $H^{1,1}$ is spanned by the subgroup $H^2(X,\mathbf{Z})$, so 3 tells us that $H^{1,1}$ is spanned by algebraic classes. The span of the algebraic classes is exactly the Neron-Severi group tensored with $\mathbf{C}$, so we have $NS(X) \otimes \mathbf{C} = H^{1,1}=H^2(X,\mathbf{C})$, hence $\rho=b_2$.
