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Let $A, S$ be commutative rings with identity, and assume $S$ is a finitely generated $\mathbb{Z}^{\geq 0}$-graded $A$-algebra. If $f\in S$ is a homogeneous element of positive degree, $S_f$ is a $\mathbb{Z}$-graded $A$-algebra. Is $(S_f)_0$ (the ring of degree $0$ elements) finitely generated as an $A$-algebra?

It seems to come down to a simple combinatorics problem with exponents that i'm too dumb to do.

(why?: I'm trying to show quasiprojective $A$-schemes have finite type)

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marked as duplicate by user26857 abstract-algebra Feb 18 at 21:34

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Yea ok it seems to be true:

If $g_1,...,g_k$ generate $S$ as an $A$ algebra, and if $$s:= \frac{g_1^{e_1}\cdot...\cdot g_k^{e_k}}{f^n} \in (S_f)_0$$ then for each $i = 1,...,k$ we may write $$e_i = deg(f)\cdot q_i + p_i$$ where $0\leq p_i< deg(f)$. Then $$s=\bigg(\prod_{i=1}^k\big(\frac{g_i^{deg(f)}}{f^{deg(g_i)}}\big)^{q_i} \bigg)\cdot \frac{g_1^{p_1}\cdot...\cdot g_k^{p_k}}{f^{n'}} $$

So $(S_f)_0$ is generated by $$\frac{g_i^{deg(f)}}{f^{deg(g_i)}}~~~~~~i =1,...,n$$

together with all $$\frac{g_1^{p_1}\cdot...\cdot g_k^{p_k}}{f^{n'}} \in (S_f)_0~~~with~~~~~0\leq p_i<deg(f) ~~for~all~i$$

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