Answer to Question #24419 in Algebra for siavosh

Question

Show that : every homomorphic image of a Dedekind module is again Dedekind.

Answer 1

If M is Dedekind module thenfor any submodule N of M we have N*N' = M, where N' =(M:N) in RT^-1, and T={t -
non zero divisor, that tm=0 implies m=0}. If we take any K - submodule of M
then submodules in M/K are only submodules N that contain K. But property
N*N' = M will be preserved since K c M and (M:K) c (M:N).