# Alexander polynomial of sums

A standard argument of knot theory reveals that the Alexander polynomial of the sum of two knots $K= K_1 \# K_2$ is equal to the product of the Alexander polynomial of the two summands $K_1$ and $K_2$.

Suppose that the Alexander polynomial $\Delta_K(t)$ of a knot $K \subset S^3$ can be factorized as $\Delta_K(t)= \Delta_{K_1}(t) \cdot \Delta_{K_2}(t)$, where $K_1$ and $K_2$ are two knots in $S^3$ with non trivial Alexander polynomial (not a unit). Is it true that $K$ is a non prime knot?

• Note that there are knots with trivial Alexander polynomials and not every knot is prime. So you have to assume at least that the Alexander polynomial of the knot admits a factorization into two non-trivial Alexander polynomials to get a more interesting answer. Jun 3, 2015 at 21:14
• Oh yes. I'm asking also that the $\Delta_{K_i}$'s are non invertible as Laurent polynomials. Jun 3, 2015 at 23:08

No: let $K$ be $8_{15}$, let $K_1$ be the trefoil, and let $K_2$ be $7_2$. The Alexander polynomial for $K$ is $(t-1+t^{-1})(3t-5+3t^{-1})$, yet $K$ is prime.