Jordan form distinction from Alexander polynomial difference for g_j = z1^2 f_j + h_j singularities
Prove that for the surface singularity germs g_0(z_1,z_2,z_3) = z_1^2 f_0(z_1,z_2,z_3) + h_0(z_2,z_3) and g_1(z_1,z_2,z_3) = z_1^2 f_1(z_1,z_2,z_3) + h_1(z_2,z_3) in C^3, where f_0 and f_1 are reduced, irreducible, homogeneous polynomials of the same degree d whose projective plane curves C_0 = V_{P^2}(f_0) and C_1 = V_{P^2}(f_1) form a Zariski pair, and h_0 and h_1 are convenient homogeneous polynomials in z_2 and z_3 of degree d+3 chosen so that z_1^2 f_j(0,z_2,z_3)+h_j(z_2,z_3) is Newton non-degenerate and Sing(V_{P^2}(f_j)) ∩ V_{P^2}(h_j) = ∅, if the Alexander polynomials of C_0 and C_1 are different, then the Jordan normal form of the monodromy of g_0 at the origin is different from that of g_1, and therefore the surface germs (V(g_0),0) and (V(g_1),0) have distinct embedded topologies in C^3. Do not assume that the singularities of C_0 and C_1 are Newton non-degenerate.
References
Conjecture If the Alexander polynomials of $C_0$ and $C_1$ are different, then the Jordan form of the monodromy of $g_0$ is different from that of $g_1$, and therefore the surface germs $(V(g_0),\mathbf{0})$ and $(V(g_1),\mathbf{0})$ have distinct embedded topologies in $\mathbb{C}3$.