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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 $(*)$.

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    $\begingroup$ In response to your first bolded question, if you have a lower bound on the degree of an arbitrary nonzero element, then you know that any filtered level below that degree must be trivial. Otherwise, you'd be able to find an element of strictly lower degree, a contradiction. $\endgroup$
    – jben2021
    Commented Aug 23, 2020 at 18:28
  • $\begingroup$ @jben2021: Thanks for the answer. That makes sense. $\endgroup$
    – J. Doe
    Commented Aug 23, 2020 at 18:32
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    $\begingroup$ In response to your $d_j>0$ question, if all the $\alpha_{s_j}$, $d_j$, and $\text{deg}(m_s)$ are positive and must satisfy that constraint, you have only finitely many ways of meeting this constraint. If the $d_j$ were allowed to be negative, however, you could use this to introduce some slack into your values for the $\alpha_{s_j}$ and $\text{deg}(m_s)$. I hope that's clear. Let me know if my wording is confusing. $\endgroup$
    – jben2021
    Commented Aug 23, 2020 at 18:34
  • $\begingroup$ @jben2021: Thanks again. That makes sense as well. $\endgroup$
    – J. Doe
    Commented Aug 23, 2020 at 18:36

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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.

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  • $\begingroup$ Thanks so much, your answer + comments answer all my questions. $\endgroup$
    – J. Doe
    Commented Aug 23, 2020 at 18:49

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