Lose some FDs in 2nf normalization?

Question

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"?

Answer 1

Yes, the breaking is not good.

The reason is that you FDs (B->C and C->D) also imply B->D. So B->CD (both B->C and B->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->D can be used to break the second further for 3NF.)