Let $S$ be a $\mathbb Z^{\ge 0}$-graded ring over $A$. If $S_+$ is finitely generated, then $S$ is a finitely generated graded $A$-algebra?

Question

Let $S$ be a $\mathbb Z^{\ge 0}$-graded ring over $A$.

Suppose $S_+= (a_1, \dots, a_n)$ where $a_i$ are homogeneous of positive degree.

I want to show that $S$ is a finitely generated graded $A$-algebra.

I believe we want to show that $S=A[a_1, \dots, a_n]$.

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If we let $x=s_0 + s_1 + \dots + s_m \in S$ where the $s_i$ are homogeneous of degree $i$, then for $i \ge 1$, we can write $s_i = f_1a_1 + \cdots + f_na_n$ where $f_i \in S$. Writing each $f_i$ in terms of its homogeneous components, we can then write the $f_i$ as linear combinations of the $a_i$ with coefficients in $S$.

Now we just have this loop of writing the homogeneous elements of positive degree as linear combinations of the $a_i$ with cofficients in $S$, then rewriting those coefficients in terms of homogeneous components, then rewriting those homogeneous components of positive degree as linear combinations of the $a_i$ with cofficients in $S$...

How do we go about this?

Answer 1

You already got the idea, so this is just a matter of writing the things down and maybe making the "$\dots$" precise:

Note that it suffices to prove that $S_m\subset A[a_1,\dots,a_n]$ for all $m$. By induction we may assume that $m>0$ and $S_k\subset A[a_1,\dots,a_n]$ for $k<m$. Let $s\in S_m$, by assumption we may write $s=f_1a_1+\dots+f_na_n$ where $f_i\in S$. Note that we can assume that the $f_i$ are either $0$ or homogeneous of degree $m-\deg a_i$ (see below). If $\deg a_i=m$ we have $f_i\in S_0=A$ and thus $f_ia_i\in A[a_1,\dots,a_n]$. If $\deg a_i<m$ we have $f_i\in S_k$ for $k=m-\deg a_i<m$, so by induction $f_i\in A[a_1,\dots,a_n]$ and hence $f_ia_i\in A[a_1,\dots,a_n]$. If $\deg a_i>m$ we necessarily have $f_i=0$.
Thus $s=f_1a_1+\dots+f_na_n\in A[a_1,\dots,a_n]$.

Edit: If some $f_i$ has a non-zero homogeneous component in some other degree than $m-\deg a_i$, also $f_ia_i$ has some non-zero component in degree $\ne m$. In the sum $f_1a_1+\dots+f_na_n$ these will have to cancel out, so we might have ignored these components (in all $f_i$) from the beginning.
A bit more precise: write $f_i = \sum_{l=1}^M f_{il}$ where $f_{il}\in S_l$. Then $$S_m\ni s=\sum_{i=1}^n f_ia_i=\sum_{i=1}^n\sum_{l=1}^Mf_{il}a_i=\sum_{i=1}^nf_{i(m-\deg a_i)}a_i+(\cdots)$$ (If $m-\deg a_i<0$ we simply ignore that term or set $S_{k}=0$ for $k<0$)
The first sum is homogeneous of degree $m$. The second sum is $0$ in its $m$-th homogeneous component. Hence the second sum is $0$ and we can thus replace the $f_i$ by $f_{i(m-\deg a_i)}$ and therefore assume that the $f_i$ are either $0$ or homogenous of degree $m-\deg a_i$.