$A$-module is free if and only if equation involving Hilbert-Poincaré series holds.

Question

Let $A = \oplus_{i \ge 0} A_i$ be a nonnegatively graded commutative algebra and $M$ a nonnegatively graded $A$-module. Assume in addition that $A_0 = k$ and all vector spaces $A_i$ and $M_j$ are finite dimensional. How do I see that the $A$-module $M$ is free if and only if the following equation holds:$$\sum_{i \ge 0} (\text{dim}_k M_i) \cdot t^i = \left( \sum_{i \ge 0} (\text{dim}_k A_i) \cdot t^i\right)\left( \sum_{i \ge 0} (\text{dim}_k (M/A_{>0}M)_i) \cdot t^i\right)?$$

Answer 1

Suppose that equation holds. Choose a homogeneous basis for the graded vector space $M/A_{>0}M$ and lift it to a set $S$ of homogeneous elements of $M$. There is then a canonical map $\varphi:F(S)\to M$ of graded $A$-modules, where $F(S)$ is the free module on $S$ (with the obvious grading). That equation says exactly that $F(S)_i$ and $M_i$ have the same dimension over $k$ for each $i$. Since this dimension is finite, in order to show $\varphi$ is an isomorphism, it suffices to show that it is surjective.

Let us show that $\varphi_i:F(S)_i\to M_i$ is surjective by induction on $i$. Suppose that $\varphi_j:F(S)_j\to M_j$ is surjective for all $j<i$. Then the image of $\varphi_i$ contains all of $(A_{>0}M)_i$ (since every such element is in the submodule of $M$ generated by $\bigoplus_{j=0}^{i-1} M_j$). By our choice of $S$, the composition of $\varphi_i$ with the quotient map $M_i\to (M/A_{>0}M)_i$ is surjective (in fact, an isomorphism). It follows that $\varphi_i$ is surjective.

For the converse, if $M$ is not only free as a module but free as a graded module (i.e., it has a homogeneous basis), then the equation obviously holds. To show $M$ is free as a graded module, as in the argument above, you can construct a surjection $\varphi:F(S)\to M$ from a graded-free module which induces an isomorphism $F(S)/A_{>0}F(S)\to M/A_{>0}M$. By the argument in this answer, such a $\varphi$ is an isomorphism.