Demystifying homomorphic multiplication in RNS FV variant I am trying to understand a full RNS FV SWHE variant proposed in this paper.

My problem is with the homomorphic multiplication. Original ciphertexts are represented in RNS representation using base (say $q$ with $k$ co-prime moduli). In order to accommodate for coefficient growth, an auxiliary RNS base (say $\mathcal{B}_{sk}$ with $\ell$ co-prime moduli) is introduced. Note that $\{ q \cup \mathcal{B}_{sk} \}$ moduli are co-primes too and $k$ needs not to be equal to $\ell$. What I don't understand is the step (S3) in Algorithm 3 in page 13:


The function "FastBconv" takes as input an extended ciphertext$^*$ in base $\{ q \cup \mathcal{B}_{sk} \}$. How to convert it to obtain a ciphertext in base $ \mathcal{B}_{sk}$ only?
There is also a strange notation of dots ($\dots$) used in the same line in the algorithm, what is that supposed to mean?


*: an extended ciphertext can be viewed as a matrix of $k+\ell$ rows and $n$ columns where $n$ is the underlying ring's polynomial modulus degree.

I appreciate any help in explaining the algorithm.
Yes, this can be confusing at first. Maybe starting with a more recent paper could help: https://eprint.iacr.org/2018/117

In the paper I cited above they have improved the base conversion and scaling in RNS, but also made the procedures simpler as a side effect. 

First, they expend the base of the ciphertext from $Q$ to $Q*P$, then they do the tensoring in $QP$, then they scale it down by $t/Q$ getting the result mod $P$, and then expend again the ciphertext (which is now in base $P$) to $Q*P$ and only keep the $Q$ part.

As for your question, if you have the ciphertext in base $QP$, and you want it  only in base $P$, you can simply throw away the base $Q$ (delete the rows of the polynomial mod each $q_{i}$). Of course this is only valide If the value of the coefficients are within $0,P-1$. But this will always be the case since the auxiliary base $P$ is at least as big as $Q$. After multiplication this is not anymore the case because the coefficients will be of size $Q^2$ (unless $P &gt; Q^2$). However after having them scaled down, they are back within the range of $Q$.