Lose some FDs in 2nf normalization? Suppose we have R=(A,B,C,D) and FDs are {B->C, C->D}. The only candidate key is AB, then A and B are prime; C and D are not prime.
B->C breaks the 2NF requirements because is a partial dependency. So we break the relation into R1(A,B,D) and R2(B,C). But now C->D is lost! Am I wrong? So this breaking is not "good"?
Yes, the breaking is not good. 

The reason is that you FDs (B-&gt;C and C-&gt;D) also imply B-&gt;D. So B-&gt;CD (both B-&gt;C and B-&gt;D) break 2NF and should be used for the table breaking. 

So, the relations, in order to be 2NF should be: (A,B) and (B,C,D). 

(and then the C-&gt;D can be used to break the second further for 3NF.)