Let $\mathcal{O}$ be a finitely-generated $K$-algebra where $K$ is a field and let $M$ be a finitely-generated $\mathcal{O}$-module.
For every good filtration $0 = M_0 \subset M_1 \subset M_2 \subset ...$ we define the Hilbert function $f_M(n) = \dim_K M_n$. One can show that for large $n \gg 0$ the function $f_M(n)$ is a polynomial $\bar{f}_M(n).$ Define $d(M) = \deg \bar{f}_M$. In the following I will omit the bar notation.
Now let $L \subset M$ be a submodule of $M$. Then using the exact sequence $$ 0 \to L \to M \to M/L \to 0$$ one can show that $d(M) = \max \left\{ d(L) , d(M/L) \right\}$ by noting that $$ f_M(n) = f_L(n) + f_{M/L}(n) \ . $$
Assume $M \ne 0$ and let $T : M \to M$ be an injective endomorphism. Then I want to prove that $d(M/TM) < d(M)$.
If $T$ is also surjective or $TM = M$ then we won since this implies $f_M$ and $f_{TM}$ have the same degree and leading coefficient and thus $f_{M/TM}$ must be a polynomial of lesser degree.
The problem is what to do if $T$ is not surjective. An example in mind is to take $M = \mathbb{Z}$ and $T : x \mapsto 2x$, but this example is excluded here since $M$ is defined over a $K$-algebra when $K$ is a field, so it must share some properties of vector spaces. In finite-dimensional vector spaces $T : V \to V$ injective implies $T$ is surjective. If this is true also in our case, then we won.
I would appreciate help here.