# Finite Presentations on Graded $A$-modules over Noetherian rings.

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.