Let $K \subset S^3$ be a knot with exterior $X$ and let $X_\infty$ be the infinite cyclic cover of $X$. Then, by the action of the deck translations, $A(K) = H_1(X_\infty, \mathbb{Z})$ is a $\mathbb{Z}[t^{\pm 1}]$-module which is usually called the Alexander module of $K$. Note that $\mathbb{Z}[t^{\pm 1}]$ is not a PID and so, while $A(K)$ is finitely generated as a $\mathbb{Z}[t^{\pm 1}]$-module, we can not apply the usual structure theorem.
In some circumstances, $A(K)$ is isomorphic to the product of cyclic $\mathbb{Z}[t^{\pm 1}]$-modules - for example when $K$ is the trefoil, $A(K) \cong \mathbb{Z}[t^{\pm 1}] / (t^2 -t + 1)$. In Milnor's paper "Infinite cyclic covers" he mentions that the Alexander module for the pretzel knot $P(25, -3, 13)$ is not a direct sum of cyclic modules.
Are there some necessary/sufficient conditions for the Alexander module to be a sum of cyclic modules?