Suppose $M$ is a finitely generated graded $k[x_1,\dots, x_n]$-module where $k$ is a field. Then the Hilbert function is defined to be $$H_M(s)=\dim_k M_s$$
It is said that $M_s$ must be finite-dimensional as a $k$-vector space, or else the submodule $\bigoplus_s^\infty M_i$ would not be finitely generated.
I don't find this very obvious, although it seems to be a trivial fact as Eisenbud writes it like a comment. Is it because if $\bigoplus_s^\infty M_i$ is finitely generated by $m_1,\dots, m_r$ where $m_i\in M_{k(i)}$ for some $k(i)\geqslant s$, then we can say that each element $m\in M_s$ is linear combination of a subset $\{m_j\}$ over $k$ where the $m_j$ are in $M_s$ for degree reasons? But if this is correct, why do we need to consider the larger submodule $\bigoplus_s^\infty M_i$ instead of just $M_s$?
Any help is greatly appreciated. Thanks in advance.
Let $M$ be a finitely-generated graded $k[x_1,\dots,x_n]$-module and let the minimal degree of $M$ be $s$. If $a_1,a_2 \dots$ is a $k$-basis of $M_s$, then $a_1,a_2,\dots$ are part of a minimal generating set of $M$. Indeed, no $a_i$ can be written as a $k[x_1,\dots,x_n]$-linear combination of $\{a_j\}_{j \ne i}$ and elements of higher degree, for degree reasons.
We thus consider the submodule $\bigoplus^{\infty}_{s} M_i$ of $M$ in your context that we may argue from the minimal degree. If we don't do this, there may be elements in degree $s$ that can be produced as a linear combination involving also generators in smaller degree; they are not minimal generators of $M$ but are of $\bigoplus^{\infty}_{s} M_i$.