# $(S_f)_0$ is a finitely generated algebra if $S$ is. [duplicate]

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)

## marked as duplicate by user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 18 at 21:34

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