the following conjecture:

The Alexander polynomial of a line arrangement ${\overline{{\mathcal{A}}}}$ has the form

$$\Delta_{{\overline{{\mathcal{A}}}}}(t)=(t-1)^{|{\overline{{\mathcal{A}}}}|-1}\Phi_3(t)^{e_3}[\Phi_2(t)\Phi_4(t)]^{e_4}$$

where $e_3=\beta_3({\overline{{\mathcal{A}}}})$ and $e_2=e_4=\beta_2({\overline{{\mathcal{A}}}})$.

The $\beta_i({\overline{{\mathcal{A}}}})$ are the modular Aomoto-Betti numbers of ${\overline{{\mathcal{A}}}}$, and they depend only on $L({\overline{{\mathcal{A}}}})$ and $i$ (see [@PS Section 3]).

As for question 2. many examples in the literature suggest that in order for $q(t)$ to be different from $1$ it is necessary that ${\overline{{\mathcal{A}}}}$ admits a multinet; the latter is a purely combinatorial notion that may be thought of as a formalisation of the idea of ‘highly symmetric arrangement’ (see \[defn:multinets\]).

We prove the following:

\[thm:AP\] Let ${\overline{{\mathcal{A}}}}$ be a line arrangement s.t. any line passes through at least one of two given points $P_1$ and $P_2$. Then the Alexander polynomial of ${\overline{{\mathcal{A}}}}$ is trivial, i.e. $\Delta_{{\overline{{\mathcal{A}}}}}(t)=(t-1)^{|{\overline{{\mathcal{A}}}}|-1}$.

Since arrangements of this type do not admit multinets, this reinforces the idea that the existence of multinets is necessary in order to have $q(t)\neq 1$; moreover, our result is consistent with the conjecture by Papadima and Suciu, as $\beta_2({\overline{{\mathcal{A}}}})=\beta_3({\overline{{\mathcal{A}}}})=0$.

The key steps of the proof of Theorem \[thm:AP\] are the following:

1.  We associate to ${\overline{{\mathcal{A}}}}$ a threefold $X\subset{\mathbb{P}}^4$ and a fibred threefold $\psi:X'\rightarrow {\mathbb{P}}^1$ such that any fibre of $\psi$ 