### An incompressible surface

Have a look at the this torus, embedded in a standard way in $S^3$.  The shaded disk $D$ is inside the torus with its boundary on the torus itself.  Consider $\partial D$, the boundary of $D$.  It represents a nontrivial loop in the torus, but since it bounds a disk in $S^3$, the loop is trivial in $S^3$.  Such a disk is called a compressing disk for the torus, and we can compress the torus by cutting along $\partial D$ and gluing in two copies of $D$ as shown in the second picture.  This is also called surgering along $\partial D$.  In this case, we obtain a sphere by compressing along $D$.

Notice that when we compress along $D$ we decrease the genus of our surface by one.  With that in mind, consider the Seifert surface $F$ for the knot $K$ we looked at previously.  Are there any compressing disks for $F$?  Suppose there was a compressing disk $D$ and $\partial D$ didn’t separate $F$ into two components .  Then we could compress $F$ along $D$, obtaining a new surface $F'$.  Since we didn’t touch anything near the boundary of $F$, $F'$ is still a Seifert surface for $K$.  But $F$ is a punctured torus, so $F'$ must be a disk, and the only knot bounding a disk is the unknot.

On the other hand, if there was a compressing disk $D$ such that $\partial D$ did separate $F$ into two components, then one of these components would have to be an annulus.   As above, we would get a disk bounding the knot $K$.  (Why?  Working out these details is a fun exercise using Euler characteristic arguments.)

So if you believe that the trefoil $K$ is knotted, then the surface $F$ must be incompressible!