TITLE: Rational slice knot that is not slice
QUESTION [7 upvotes]: Does there exists a knot $K\subset \mathbb{S}^3$ such that

$K$ is not slice
$\exists W^4$, $\partial W = \mathbb{S}^3$ rational homology ball
$\exists $ properly embedded smooth disk $(D,\partial D)\to (W,K)$. ?

In other words $K$ is not slice in $B^4$ but is slice in some rational homology ball.

REPLY [3 votes]: Rationally slice knots in $S^3$ are in abundance due to the collection of Kawauchi's theorems and the recent result of Kim and Wu:
Theorem ([1] + [2]): Any hyperbolic amphicheiral knot $K$ in $S^3$ is rationally slice.
Theorem ([3]): Any fibered, negative amphicheiral knot $K$ in $S^3$ with irreducible Alexander polynomial (called Miyazaki knot) is rationally slice.
Using these theorems, you may find plenty of rationally slice knots that are not smoothly slice.
[1]: Kawauchi, Akio. "The invertibility problem on amphicheiral excellent knots." Proceedings of the Japan Academy, Series A, Mathematical Sciences 55.10 (1979): 399-402.
[2]: Kawauchi, Akio. "Rational-slice knots via strongly negative-amphicheiral knots." Commun. Math. Res. 25.2 (2009): 177-192.
[3]: Kim, Min Hoon, and Zhongtao Wu. "On rational sliceness of Miyazaki's fibered,− amphicheiral knots." Bulletin of the London Mathematical Society 50.3 (2018): 462-476.