Let $R$ be a Noetherian ring and $A$ be a finitely generated graded $R$-algebra. Then we know $A$ is Noetherian by the Hilbert Basis theorem. Let $M$ be a finitely generated $A$-module. Then $M$ is a Noetherian $A$-module.

The text I am reading says that there exists free graded $A$-modules $L,L’$ and degree preserving maps $\varphi, \psi’$ such that $$L’\xrightarrow{\varphi} L\xrightarrow{\psi}M \rightarrow 0 $$ Is exact.

My question is: What exactly are the free-graded $A$-modules in this case? To say that a graded $A$-module is free is to say that it is of the form $A(n_1)\oplus A(n_2)\oplus \cdots A(n_l)$ for some $n_i \in \mathbb{Z}$ where $A(l)$ represents the twisting of the graded ring $A$ by $l$, i.e. $A(l)_m = A_{m+l}$ for all $m\in \mathbb{Z}$.

My attempt: We can construct $L$ easily: Since $M$ is finitely generated, choose homogeneous generators $x_1 \in M_{n_1} ,\cdots, x_r \in M_{n_r}$ and define the map $$\psi: A(-n_1) \oplus \cdots \oplus A(-n_r) \rightarrow M \text{ defined by} \psi(a_1,\cdots, a_r) = a_1 x_1 + \cdots + a_rx_r $$

Then clearly $\psi$ is a surjective, degree preserving map. So $L\xrightarrow{\psi} M \rightarrow 0$ is done.

It seems intuitive that $L’$ should be $\ker \psi$. It is evident that $L’$ is finitely generated but can we be sure that $L’$ is free as well?


Notice that $M'=\ker\psi $ is a finitely generated graded module. Hence you can repeat your argument replacing $M $ by $M'$. Then you get a surjective map from a free module $L'$ to $M'$. Composing this with the inclusion map $M'\subseteq L $ gives you the required exact sequence.


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