I'm working through McMullen's paper "The Alexander polynomial of a 3-manifold and the Thurston norm on cohomology" and have a question concerning the following setup:
Given a link complement $(X, p)$ with $G = \pi_1(X)$, the Alexander polynomial $\Delta$ is the first order of the Alexander module, which is the $\mathbb{Z}[G/G']$-module $H_1(X_{\inf}, \pi^{-1}(p); \mathbb{Z})$ where $\pi: X_{\inf} \to X$ is the maximal free abelian cover of $X$.
Given a homomorphism $\phi: \pi_1(X) \to \mathbb{Z}$ we have another covering space $\pi_{\phi}: X_{\phi} \to X$ characterized by $(\pi_{\phi})_*(\pi_1(X_{\phi})) = \ker \phi$. The first homology of this cover $H_1(X_{\phi}, \pi_{\phi}^{-1}(p))$ is a $\mathbb{Z}[\mathbb{Z}]$-module under the action of deck transformations, as in the case of the Alexander module.
Now what I don't understand is that McMullen claims that the first elementary ideal of $H_1(X_{\phi}, \pi_{\phi}^{-1}(p))$ is just the image under "$\phi$" of the first elementary ideal of the Alexander module. (The map $\phi$ is in quotes because I'm referring to the induced map from $\mathbb{Z}[G/G'] \to \mathbb{Z}[G/\ker \phi]$.) As a consequence, if $\Delta_{\phi}$ is the first order of $H_1(X_{\phi}, \pi_{\phi}^{-1}(p))$, we have $\phi(\Delta) = \Delta_{\phi}$.
Why is this? McMullen only uses this relation when $\phi$ is primitive, i.e. surjective, but it seems it could be true for any map onto a free abelian group. Perhaps it can be proved using the reduction to algebra by the equation $$ H_1(X_{\phi}, \pi_{\phi}^{-1}(p)) \cong m(G)/m(\ker \phi)m(G) $$ where $m(H)$ is the ideal of $\mathbb{Z}[H]$ generated by $(h - 1: h \in H)$. For example, from here we can fit the Alexander module and $H_1(X_{\phi}, \pi_{\phi}^{-1}(p))$ into an exact sequence of $\mathbb{Z}[G]$-modules, but I still can't come to the conclusion.
