What is the Inverse of $n$ in this algorithm for FFT over rings?

Question

On this algorithm, the last step involves $n^{-1}$, but I'm in doubt about the inverse of $n$ in what ring? It surely isn't talking about $1/n$ as these are operations on a ring of integers.

Answer 1

It surely isn't talking about 1/ as these are operations on a ring of integers.

but with some rounding? When it does does it do some rounding? Because it can't be just plain 1/

It looks to me like it is operations in $\mathbb Z/q\mathbb Z$, which is a field as $q$ has been assumed to be prime. Since $n$ is a power of $2$ it is certainly nonzero in $\mathbb Z/q\mathbb Z$, and therefore it has an inverse in that ring.

There is no rounding or anything involved. The inverse is perfectly well defined $\pmod q$.