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 $S$...

How do we go about this?


1 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 $m-\deg a_i$.

  • $\begingroup$ Why are the $f_i$ either $0$ or homogeneous of degree $m-\deg a_i$? $\endgroup$
    – user5826
    Aug 10, 2021 at 19:37
  • $\begingroup$ 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 $\endgroup$
    – leoli1
    Aug 10, 2021 at 19:41
  • $\begingroup$ 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? $\endgroup$
    – user5826
    Aug 10, 2021 at 19:53
  • $\begingroup$ 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 $f_1=f_2=1$ $\endgroup$
    – leoli1
    Aug 10, 2021 at 19:59

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