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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.

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So do you have another approach to prove the claim? –  LinAlgMan Jan 22 '13 at 15:56
In fact, as I reasoned above: $f_M = f_{TM} + f_{M/TM}$ sum of (Hilbert) polynomials. Then $\deg f_{M/TM} < \deg f_M$ if and only if $f_M$ and $f_{TM}$ are of the same degree and have the same leading coefficient. I need to show that the injectiveness of $T$ implies this. –  LinAlgMan Jan 27 '13 at 13:15

1 Answer 1

Consider the exact sequence of filtrations $$ 0 \longrightarrow F_n(L) \longrightarrow F_n(M) \longrightarrow F_n(M/L) \longrightarrow 0 $$ By Artin-Rees Lemma $F_n(L) = L \cap F_n(M)$ is a good filtration. Let $L = TM$ where $T : M \to M$ is injective endomorphism. Then there is an induces linear transformation of vector spaces $$ T_n : F_n(M) \to F_n(M) \ . $$ Then $T_n$ is injective. But $T_n$ is a linear endomorphism of finite dimensional vector spaces and hence also surjective. Therefore, $$ \dim F_n(M) = \dim F_n(L) $$ and hence they have the same Hilbert polynomial.

As I reasoned above, $$ f_M(n) = f_{TM}(n) + f_{M/TM}(n)$$ immplies that $f_M$ and $f_{TM}$ have the same degree and leading coefficient and thus $f_{M/TM}$ is strictly of lesser degree.

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