Minimum required crossings in a link diagram for a $k$-component Brunnian link What is the minimum number, $s$, of crossings in a link diagram for $k$ (component) links fully knotted together such that cutting any single link frees all individual component links--becomes an unlink?  (We say such links are Brunnian.)
Clearly if $k=2$ (Hopf link), then $s=2$.
If $k=3$, the Borromean rings show that $s \leq 6$.
Might there be an approach based on roots of some knot polynomial of order $k$?
 A: I believe you are asking how many crossings a Brunnian link must have. It is possible for a Brunnian link with $k$ components to have $8k-8$ crossings

but $6k$ is possible using a rubber band chain that closes up.

I don't know if this is asymptotically the best possible.
A: There are Brunnian links with $2, 6, 14, 26$ crossings with $2,3,4, 5$ components respectively: 




On the other hand, Doug Zare shows that there are $k$-component Brunnian links with crossing number $6k$. Here is another picture for $k=3$:

This seems plausibly optimal (ie linear growth of
crossings, with coefficient $6$). 
There is a trivial lower bound on the crossing number
of $2k$ for a $k$-component Brunnian link, $k>2$.
Define the number of crossings
of a component as the number of crossings that it
goes through, counted with multiplicity (so self-crossings
count twice). 
 If the number of crossings
were $< 2k$, when we add up the number of crossings
of all of the components, we get $< 4k$. So one component must have at most 2
crossings. Since each component must cross any other
component an even number of times, we see that this
component must cross a single other component twice. Then
the two components cannot be linked, so we can slide
the two crossing one off of the other (possibly over other components
in the diagram) to see that it has a trivial component,
and hence the link isn't Brunnian. 
