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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?

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    $\begingroup$ 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. $\endgroup$ Jun 3, 2015 at 21:14
  • $\begingroup$ Oh yes. I'm asking also that the $\Delta_{K_i}$'s are non invertible as Laurent polynomials. $\endgroup$ Jun 3, 2015 at 23:08

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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.

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Another approach is to note that the Alexander polynomial is unchanged by "doubled delta moves." One way to visualize a doubled delta move is to tie a Borromean rings into three pairs of antiparallel strands. Given any knot, it is reasonable to expect that you can do doubled delta moves to make it prime. So any realizable Alexander Polynomial can be realized by a prime knot. In fact, it seems to be proven in the literature that every Alexander polynomial can be realized by a hyperbolic knot, which is even stronger than being prime.

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  • $\begingroup$ Hi @CheerfulParsnip. Please do you have a reference for your last sentence? Thanks! $\endgroup$
    – GaryMak
    Apr 13, 2020 at 9:53
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    $\begingroup$ @GaryMak I wrote this comment 3 years ago, so I don't remember what I was thinking, but a quick google search yields: "Realizing Alexander polynomials by hyperbolic links" by A. Stoimenow. doi.org/10.1016/j.exmath.2009.06.003 $\endgroup$ Apr 13, 2020 at 16:26

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