Alexander-Conway polynomial of an unlinked knot... I had asked this elsewhere earlier in the week but I decided I am more likely to get an answer here:
Is it true that for all unlinks, the Alexander-Conway polynomial is equivalent to 0?
It seems intuitive and I can't come up with a counter example, but I am not exactly knowledgeable enough in knot theory to come up with a proof.
Thanks in advance!
 A: Using the skein relation in which $\nabla(\text{unknot}) = 1$ and 
$$
\nabla(L_+) - \nabla(L_-) = z \nabla(L_0),
$$
where, as usual, $L_+, L_-, L_0$ denote three links that differ only at a specific crossing, consider $L_0$ to be the splittable link.
You really have to sketch these local diagrams to make sense of the following.
The local changes to form $L_+$ and $L_-$ will be made in a disk chosen so that the two strands in $L_0$ come from two unlinked components that run parallel to one another without crossing.  Replacing these parallel strands with positive and negative crossings has the effect of fusing the components together and producing isotopic links $L_+$ and $L_-$.
Therefore, 
$$
z \nabla(L_0) = \nabla(L_+) - \nabla(L_-) = 0.
$$
But these are polynomials, so $\nabla(L_0) = 0$.
A: Yes, it is true that all unlinks have Alexander-Conway polynomial equal to zero.  
This comes from the more general fact that  the Alexander polynomial of a splittable link is always 0. A link is splittable if its components can be seperated by a plane in $\mathbb{R}^3$, which is exactly what you would want it to mean. And the unlink obviously falls into this category. I found this fact on mathworld.
