Is the Alexander Ideal of a Link always Principal? It is known that the Alexander ideal of a knot (i.e., a link of one component) is always a principal ideal since any tame knot in $S^3$ has a square presentation (Rolfsen, D. Knots and Links, pp. 206-207). Does the same hold for a link of more than one component?
 A: The answer depends if you are working in the multivariable case or not. In the single variable case the ideal is principal. In the multivariable case, if the link has $3$ or fewer components, the answer is yes, but in general the answer is no. Here are some details.
Let $L$ be an $n$-component link and let $X_L$ denote its exterior. Let's start with the one-variable case. Consider the covering space $\widehat{X}$ corresponding to the kernel of the map $\pi_1(X_L) \stackrel{ab}{\to} H_1(X_L) \to \mathbb{Z}$ where the latter map sends each meridian to $1$. The one-variable Alexander polynomial $\Delta_L(t)$ is the order of the $\mathbb{Z}[t^{\pm 1}]$-module $H_1(\widehat{X}_L)$ and the Alexander ideal is the $0$-th elementary ideal of $H_1(\widehat{X}_L)$. This module is presented by $tA-A^T$, where $A$ is any Seifert matrix $A$ for $L$. Now recall that if a module $M$ admits a square presentation matrix, then the elementary ideal $E_0(M)$ is principal (generated by the determinant of the matrix). So in this case the Alexander ideal is principal.
Let's move on to the multivariable case. This time, let $\widetilde{X}_L$ denote the covering space corresponding to the kernel of the map $\pi_1(X_L) \stackrel{ab}{\to} H_1(X_L) \cong \mathbb{Z}^n$. This time the Alexander ideal is the $0$-th elementary ideal of the $\mathbb{Z}[t_1^{\pm 1},\ldots,t_n^{\pm 1}]$-module $H_1(\widetilde{X}_L)$. For links with 3 or fewer components, this module has a square presentation matrix so the Alexander ideal is principal (such matrices can be found using "C-complexes"). Let $I_n$ denote the augmentation ideal i.e. the kernel of the augmentation map $\mathbb{Z}[t_1^{\pm 1},\ldots,t_n^{\pm 1}] \to \mathbb{Z}$ that maps each $t_i$ to $1$. As explained in Kawauchi's "a survey of knot theory" Proposition 7.3.9, for $n \geq 4$, we have $E_0(H_1(\widetilde{X}_L))=I_n^s \cdot (\Delta_L(t_1,...,t_n))$, where $s={n-1 \choose 2}$. Thus, in particular (since $n \geq 4$) if $\Delta_L(t_1,\ldots,t_n) \neq 0$, then the ideal is not principal.
