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The Alexander module is the first integral homology of the infinite cyclic cover of a knot exterior. Its deck transformation makes it a module over $\mathbb{Z}[t^{\pm 1}]$, and its first elementary ideal determines the Alexander polynomial of a knot.

Replace the open knot complement by the compact knot exterior $E_K$; they have the same homotopy type. The abelianizationsends an oriented meridian of a knot to $1$. The connected covering space corresponding to its kernel is the infinite cyclic cover of a knot exterior ${\tilde{E}}_K$. Choose a generator $t$ of its deck transformation group. Thenis the Alexander module of a knot over $\Lambda =\mathbb{Z}[t,t^{-1}]$. It is a finitely presented torsion module, and its order, equivalently the greatest common divisor of the appropriate maximal matrix minors of an Alexander matrix, is the Alexander polynomial of a knot, up to a unit $\pm t^m$.

For a fibered knot, the infinite cyclic cover of a knot exterior is $\Sigma \times \mathbb{R}$. Write the mapping torus with the gluing convention in the question. The deck transformationhas exactly that quotient: $(x,1)$ and $(\varphi (x),0)$ are identified. Choose $t=D$. Its action on $H_1(\Sigma ;\mathbb{Z})$ is ${\varphi }_{\ast }$. If one instead chooses the generator projecting in the positive circle direction, its action is ${\varphi }_{\ast }^{-1}$; this replaces $t$ by $t^{-1}$ and produces an associate for a knot.

Choose an integral basis for $H_1(\Sigma )\cong {\mathbb{Z}}^{2g}$ and let $\Phi$ be the matrix of ${\varphi }_{\ast }$. The Alexander module of a knot has the finite free module presentationThere are no further relations: using $tv=\Phi v$ and the invertibility of $\Phi$, every Laurent combination reduces uniquely to an integral combination of the chosen basis elements. Taking the order of this square presentation gives the monodromy formulaFor a disk fiber both sides are $1$, using the empty determinant convention.