Decomposition that does not preserve functional dependency

Question

When can a BCNF decomposition not preserve functional dependency... I am trying to figure this out for say R=(V,W,X,Y,Z)

Answer 1

Taken from Database Design and Relational Theory:

R = (S, J, T)

{S, J} -> {T}
{T} -> {J}

This is not in BCNF, since T -> J holds and T is not a key.

Decomposing it into R1 = (T, J) and R2 = (T, S) with {T} and {T, S} being keys resp. leads to BCNF.

However, the dependency {S, J} -> {T} is lost.