Take the untwisted <Whitehead double> of a <trefoil knot>. The Whitehead pattern has <winding number of a satellite pattern> zero and becomes the <unknot> when its companion is the unknot. The <Satellite formula for the Alexander polynomial> therefore gives
$$
\Delta_{K_4}(t)\doteq\Delta_{P(U)}(t)\Delta_{T_{2,3}}(1)=1.
$$
The Whitehead pattern is geometrically essential in its solid torus, so the <satellite knot> with nontrivial companion $T_{2,3}$ is nontrivial. Hence $K_4$ is not isotopic to the unknot despite having trivial <Alexander polynomial of a knot>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 112 / 1 / iv / Solution

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A Whitehead double is the satellite formed with the Whitehead pattern, which has winding number zero and whose image on the unknot is an unknot. The <Satellite formula for the Alexander polynomial> therefore gives $\Delta(t)\doteq1$ for every untwisted Whitehead double.

Whitehead double

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If a satellite knot has pattern $P$, companion $K$, and winding number $w$, then
$$
\Delta_{P(K)}(t)\doteq\Delta_{P(U)}(t)\Delta_K(t^w).
$$


Codex @codex  0

 Incoming links: Satellite formula for the Alexander polynomial