# graded homomorphism of finitely-generated, free modules with minimality condition

The present question is a follow up of this question: finite-free-graded modules and the grading of their duals.

Let $k$ be a field, $S=k[x_1,\dots,x_n]$ and $\phi: F \rightarrow G$ a graded homomorphism of graded, finitely-generated, free $S$-modules $F,G$. Denote by $S_+$ the irrelevant ideal of $S$ and suppose that $\phi(F) \subset S_+ G$. Suppose additionally that the maximal degree that appears in $G$ is $m$ and that there are no degrees greater than $m$ in $F$. Now taking the dual of $\phi$ gives $\phi^*: Hom(G,S) \rightarrow Hom(F,S)$. Then the smallest degree that appears in $Hom(G,S)$ is $-m$, let's denote the corresponding component by $S(-m)$.

How can we show that $\phi^*( S(-m)) = 0$?

PS: I can see roughly why this must be the case, but i am looking for a rigorous proof, which is going to further help me understand how the gradings work.

-

Let $S = k[x]$ and $\mathfrak{m} = (x)$. Consider the following exact sequence $$0 \to S(-1) \stackrel{\cdot x}{\to} S \to S/ (x) \to 0.$$ Let $F = S(-1), G = S$, and $\phi$ the multiplication by $x$. Taking $Hom_S(-,S)$ of the exact sequence implies $$0 \to Hom_S(S/(x),S) \to Hom_S(S,S) \stackrel{\cdot x}{\to} Hom_S(S(-1),S)$$ is exact. Since $Hom_S(S/(x),S) = 0$, $\phi^*$ is injective. Then $Hom_S(S,S) \cong S \neq 0$ implies that $\phi^* \neq 0$.
This condition implies that $F$ is Artinian since $\mathfrak{m}^l G = 0$ for $l \gg 0$ implies $\mathfrak{m}$ is associated to $G$. Then $G$ can not be a free $S$-module.