Infinite family of different prime knots with trivial Alexander polynomial I am looking for infinite families of prime knots that have all Alexander polynomial equals to 1. I wrote "families" (and not "family") since perhaps there are different constructions out there.
Moreover, is there such a family for which all knots are of genus one (just as the Whitehead doubles)?
 A: The P(p,q,r) pretzel knots with p, q, r odd integers have a genus 1 Seifert surface that just consists of two disks connected with three twisted ribbons, with p, q, r twists, respectively. So they have a Seifert matrix $A = \left(\begin{matrix}p+q & q+1 \\ q-1 & q+r\end{matrix}\right)/2$.
From this one computes that the determinant of P(p,q,r) is $|\det(A + A^{\top})| =pq+qr+rp$. For genus 1 knots, the Alexander polynomial is 1 if and only if the determinant is 1. This gives you plenty of examples of genus 1 knots with Alexander polynomial 1, e.g. P(-3,5,7).
More generally, given any knot $K$ with Alexander polynomial 1 and a Seifert surface $\Sigma$, you can construct further knots of the same genus as $K$ simply by tying a knot into a band of $\Sigma$ without changing the framing of the core curve of the band (this is often called "infection").
A: Whitehead doubles are one such family of genus one knots, see this MO answer
A: Another family of examples is given by the "generalised Kinoshita-Terasaka" knots, here is a picture from Lickorish' "An introduction to Knot Theory".

Here $d$ is assumed to be even. Of course, this is not a family of genus 1 knots.
