Let k be an algebraically closed field and A a finite-dimensional k-algebra. Let M, N be finite-dimensional A-modules.

Theorem. The module N is a degeneration of M if and only if there exists a finite-dimensional A-module C and an exact sequence 0 -> C -> M \oplus C -> N -> 0.

Proof: According to Riedtmann, the existence of an exact sequence 0 -> C -> M \oplus C -> N -> 0 implies that N is a degeneration of M. Thus, only the converse has to be shown. Zwara shows the following stronger assertion:

Theorem. If N is a degeneration of M, then there exists a module Z and a "strict" endomorphism f of Z, such that Z/Zf is isomorphic to N and such that Z is isomorphic to Zf \oplus M.

Remark: Here, an endomorphism f of Z is called strict provided there exists a natural number n such that the image of f is equal to the kernel of f^n. Strict endomorphism are nilpotent (with order of nilpotency n+1), and if f is a strict endomorphism of Z with order of nilpotency n+1 and with kernel N, then the factors Zf^i/Zf^(i+1) for i < n+1 are isomorphic to N, thus Z has a filtration by modules of the form N.

Any endomorphism f of Z gives rise to an exact sequence 0 -> Zf -> Z -> Z/Zf -> 0, thus if Z is isomorphic to Zf \oplus M, then this sequence is as required. We also obtain the sequence 0 -> Ker f -> Z -> Zf -> 0, thus under the same assumption, 0 -> Ker f -> Zf \oplus M -> Zf -> 0. Of course, for a strict endomorphism f, the modules Z/Zf and Ker f are isomorphic.