I want to prove the following:
If $M=\bigoplus_{i \in \mathbb{Z}} M_i$ is a finitely generated graded
$R$-module ($R=\bigoplus_{i \in \mathbb{N}_0} R_i$ is also a graded
ring and finitely generated as an $R_0$-algebra), then each $M_i$ is
finitely generated and $M_i=\{0\}$ for all $i\ll 0$.
There is this proof (Proposition 1.17, page 7) that I am having trouble understanding and wish if someone can help me out a little.
My understanding so far has been:
Let $m_1, …, m_r$ be a set of generators of $M$ as an $R$-module. We can further assume that each $m_i \in M_i$ because even if we didn't start off with a homogenous system of generators, we can still take their homogenous components and generate $M$ from those. Then any $m \in M\setminus\{0\}$ can be uniquely written as a finite sum $$m=\sum_{i=1}^r a_im_{i}, \quad m_{i} \in M_{i}, a_i \in R.$$
Let $d=\min\{\deg(m_i)\}$ so that $\deg(m)\ge d$. Why does it follow from this that $M_i=\{0\}$ for all $i\ll 0$?
Next, suppose that $r_1, …, r_p$ are the generators of $R$ as an $R_0$-algebra. We might assume also that each $r_i$ is homogenous of positive degree $d_i$.
The $R_0$-module $M_i$ in the decomposition of $M$ is then generated by the elements $$ r_{1}^{\alpha_{s_1}}\cdots r_{p}^{\alpha_{s_p}}m_{s}, \quad \text{for}\; s=1, …, r \qquad (*)$$
where for each s, $\sum_{j=1}^p \alpha_{s_j}d_j + \deg(m_s) = i$. It is then written that because each $d_j>0$ and the elements $m_s$ are finite, it must be that $M_i$ is finitely generated. I don't understand what $d_j>0$ has to do with anything. I also need help convincing myself that $M_i$ really is generated by elements of the form $(*)$.
Best Answer
In response to your question about the $M_i$ being generated by those elements ($\ast$), we already have by assumption that $M$ as a whole is generated as an $R$-module by the $m_1,\dots,m_s$. This gives the first decomposition you wrote down (before "Let $d = \min\dots$"). Now, since $R$ is finitely generated as an $R_0$-algebra, say by the $r_1,\dots,r_p$, which can be taken to be homogeneous as you mentioned, we know we can write any such $r$ in that first decomposition as $r_1^{\alpha_1}\dots r_p^{\alpha_p}$. Then substituting this into the first decomposition should give you what you want.