Alexander polynomial of any knot evaluated at 1 is $\pm$ 1

Question

I'm supposed to prove for a knot theory homework assignment that the Alexander polynomial of any knot (as opposed to link) is $\pm1$. From examples, I'm pretty convinced that this is true, but I have no idea where to start with proving it. I was thinking I could use the matrix associated with the knots somehow (I've noticed that, at least in those that I've computed, there is at least one column which does not sum to 0, unlike with links), but I get stuck trying to show in a general case that the determinant has to be $\pm1$. Any ideas about where to start or what kind of proof would work would be greatly appreciated.

Answer 1

Against advice, I kept thinking about this in terms of matrices and found a pretty beautiful solution. Basically, it all hinges on how you label the knot to set up a system of equations. When you're evaluating at 1, overcrossings all get assigned zero coefficients, so we only really want to worry about undercrossings. So given a knot, pick a starting strand and orientation, label that strand's starting crossing as crossing 1, and continue following that strand along, adding labels to crossings only when the strand reaches an undercrossing. The resulting system of equations corresponds to a lower triangular matrix, with an additional $\pm1$ entry in the top right corner. Since we delete a row and column when computing the Alexander polynomial, just delete the bottom row and rightmost column, and the resulting matrix is lower triangular with diagonal entries of $\pm1$, which implies that the determinant is $\pm1$. Kind of a hand-wavey way to prove this, but I thought it was pretty cool how it worked out.