Given a 3-manifold $M$, one can define the Kauffman bracket skein module $K_t(M)$ as the $C$-vector space with basis "links (including the empty link) in $M$ up to ambient isotopy," modulo the skein relations, which can be found in the second paragraph of section two of http://arxiv.org/abs/math/0402102. (Side question - how can I draw these relations in Latex?)

If $S$ is a surface, then $K_t(S\times [0,1])$ has an algebra structure given by stacking one link on top of another. If $S$ is a boundary component of $M$, then $K_t(M)$ is a (left) $K_t(S\times [0,1])$ module, where the left module structure is given by gluing $S\times \{1\}$ to the copy of $S$ in the boundary of $M$. In this situation, we can define a left module map $K_t(A\times [0,1]) \to K_t(M)$ which is uniquely defined by "(empty link in $S\times [0,1]$) maps to (empty link in $M$)." The "peripheral ideal" is the kernel of this module map, and is a left ideal of $K_t(S\times [0,1])$.

The motivation for these definitions comes from knot theory - if $K$ is a knot in $S_3$, then the complement of a small tubular neighborhood of $K$ is a manifold with a torus boundary, and the algebra $K_t(T^2\times [0,1])$ and module $K_t(S^3 \setminus K)$ give information about the knot $K$.

Now I can ask my question: Is there a manifold $M$ with a torus boundary such that the peripheral ideal is trivial?

I've just recently started learning about knot theory, and I'm having a hard time trying to figure this out. One thing that I do know is that $M$ cannot be of the form $S^3 \setminus K$, because of propositions 7 and 8 in http://arxiv.org/abs/math/9812048. I also suspect that $M$ will actually have **two** boundary components which are a torus, but I don't really have a good reason for this.

Also, I suspect this might be a hard question, so any hints about one might approach it would be helpful.