Lemma 15.25.4. Let $R$ be a ring. Let $S = \bigoplus _{n \geq 0} S_ n$ be a graded $R$-algebra. Let $M = \bigoplus _{d \in \mathbf{Z}} M_ d$ be a graded $S$-module. Assume $S$ is finitely generated as an $R$-algebra, assume $S_0$ is a finite $R$-algebra, and assume there exist finitely many primes $\mathfrak p_ j$, $i = 1, \ldots , m$ such that $R \to \prod R_{\mathfrak p_ j}$ is injective.

If $S$ is flat over $R$, then $S$ is a finitely presented $R$-algebra.

If $M$ is flat as an $R$-module and finite as an $S$-module, then $M$ is finitely presented as an $S$-module.

**Proof.**
As $S$ is finitely generated as an $R$-algebra, it is finitely generated as an $S_0$ algebra, say by homogeneous elements $t_1, \ldots , t_ n \in S$ of degrees $d_1, \ldots , d_ n > 0$. Set $P = R[x_1, \ldots , x_ n]$ with $\deg (x_ i) = d_ i$. The ring map $P \to S$, $x_ i \to t_ i$ is finite as $S_0$ is a finite $R$-module. To prove (1) it suffices to prove that $S$ is a finitely presented $P$-module. To prove (2) it suffices to prove that $M$ is a finitely presented $P$-module. Thus it suffices to prove that if $S = P$ is a graded polynomial ring and $M$ is a finite $S$-module flat over $R$, then $M$ is finitely presented as an $S$-module. By Lemma 15.25.3 we see $M_{\mathfrak p}$ is a finitely presented $S_{\mathfrak p}$-module for every prime $\mathfrak p$ of $R$. Thus the result follows from Lemma 15.25.1.
$\square$

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