# 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?

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 $$S=A[a_1, \dots, a_n]$$.

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 $$S$$...

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 $$k. Let $$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 we have $$f_i\in S_k$$ for $$k=m-\deg a_i, so by induction $$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 $$m-\deg a_i$$.

• Why are the $f_i$ either $0$ or homogeneous of degree $m-\deg a_i$? Aug 10, 2021 at 19:37
• It wasn't quite correct as I wrote it: they don't have to be homogeneous but we can choose them so that they satisfy this condition, I will edit the answer Aug 10, 2021 at 19:41
• I think I worked out how you got that. Let me do the case showing $S_1 \subset A[a_1, \dots, a_n]$. Let $s \in S_1$. By assumption, we may write $s=f_1a_1 + \cdots + f_na_n$ where $f_i \in S$. Take any $f_i$. We may write $f_i = b_0 + \cdots + b_k$ where $b_j \in S_j$. Then $f_i a_i = b_0a_1 + \cdots + b_ka_i$. Since $s\in S_1$ we must have $\deg b_j a_i = \deg b_j +\deg a_i = 1$ and hence $\deg b_j = 1-\deg a_i$. Since $\deg a_i >0$, then $b_j =0$ for $1 \le j \le k$ and so $f_i = b_0 \in S_0=A$. Then $f_ia_i \in A[a_1, \dots, a_n]$ for each $i$ and hence $s \in A[a_1, \dots, a_n]$. Correct? Aug 10, 2021 at 19:53
• The $b_j$ don't have to be zero for $j\geq1$, the point is that in the end they will cancel out with the other $b$'s from the other $f$'s, so we can ignore them. Consider for example $a_1=X,a_2=Y$ in $k[X,Y]$ and $S=X+Y$. Then let $f_1=1+Y,f_2=1-X$, we have $s=f_1a_1+f_2a_2$ but the $f$'s are not homogeneous! But the stuff which is not in degree $0$ will cancel in the sum: $(1+Y)X+(1-X)Y = (X + Y) + (YX-XY)$, so that's why we could have ignored the terms that are not in degree $0$ in the beginning, i.e. just choose $f_1=f_2=1$ Aug 10, 2021 at 19:59