Relation between the Alexander module of a link and intermediate free abelian covers

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.

• you have to be careful: you are right that Fox derivatives give you a presentation matrix, and you can get the multivariable and the onevariable Alexander polynomial out of it. But the problem is that as a presentation matrix for the Alexander module it is not a square matrix, so you have to take the minors and then the gcd. Taking minors is functorial, but taking gcd's isn't. For example, if the presentation matrix is $(x-1,y-1)$ for two different meridians, then the order is $gcd(x-1,y-1)=1$, but if you abelianize the presentation matrix is $(t-1,t-1)$, so the gcd of the minors is $t-1$. – Stefan Friedl May 8 '14 at 6:43
• Thanks for that insight. As you point out, the statement in my question that says "if the Alexander ideal is functorial, then the gcd is also" is incorrect. In fact, McMullen only uses $(\Delta_{\phi}) = I_{\phi} = \phi_*(I)$ (these are the ideals generated by the minors). – Daniel Copeland May 8 '14 at 15:39