The Stacks project

Proof. Choose a presentation

of $M$ as an $S$-module. Let $\mathfrak q$ be a prime ideal of $S$ lying over a prime $\mathfrak p$ of $R$. By assumption there exist finitely many elements $k_1, \ldots , k_ t \in K$ such that if we set $K' = \sum Sk_ j \subset K$ then $K'_{\mathfrak p} = K_{\mathfrak p}$ and $K'_{\mathfrak p_ j} = K_{\mathfrak p_ j}$ for $j = 1, \ldots , m$. Setting $M' = S^{\oplus r}/K'$ we deduce that in particular $M'_{\mathfrak q} = M_{\mathfrak q}$. By openness of flatness, see Algebra, Theorem 10.129.4 we conclude that there exists a $g \in S$, $g \not\in \mathfrak q$ such that $M'_ g$ is flat over $R$. Thus $M'_ g \to M_ g$ is a surjective map of flat $R$-modules. Consider the commutative diagram

The bottom arrow is an isomorphism by choice of $k_1, \ldots , k_ t$. The left vertical arrow is an injective map as $R \to \prod R_{\mathfrak p_ j}$ is injective and $M'_ g$ is flat over $R$. Hence the top horizontal arrow is injective, hence an isomorphism. This proves that $M_ g$ is of finite presentation over $S_ g$. We conclude by applying Algebra, Lemma 10.23.2. $\square$

Proof. By assumption $S$ is a quotient of a polynomial ring over $R$. Thus the result follows directly from Lemma 15.25.1. $\square$

Proof. Let $M = \bigoplus M_ d$ be the grading on $M$. Pick homogeneous generators $m_1, \ldots , m_ r \in M$ of $M$. Say $\deg (m_ i) = d_ i \in \mathbf{Z}$. This gives us a presentation

which in each degree $d$ leads to the short exact sequence

By assumption each $M_ d$ is a finite flat $R$-module. By Algebra, Lemma 10.78.5 this implies each $M_ d$ is a finite free $R$-module. Hence we see each $K_ d$ is a finite $R$-module. Also each $K_ d$ is flat over $R$ by Algebra, Lemma 10.39.13. Hence we conclude that each $K_ d$ is finite free by Algebra, Lemma 10.78.5 again.

Let $\mathfrak m$ be the maximal ideal of $R$. By the flatness of $M$ over $R$ the short exact sequences above remain short exact after tensoring with $\kappa = \kappa (\mathfrak m)$. As the ring $S \otimes _ R \kappa $ is Noetherian we see that there exist homogeneous elements $k_1, \ldots , k_ t \in K$ such that the images $\overline{k}_ j$ generate $K \otimes _ R \kappa $ over $S \otimes _ R \kappa $. Say $\deg (k_ j) = e_ j$. Thus for any $d$ the map

becomes surjective after tensoring with $\kappa $. By Nakayama's lemma (Algebra, Lemma 10.20.1) this implies the map is surjective over $R$. Hence $K$ is generated by $k_1, \ldots , k_ t$ over $S$ and we win. $\square$

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. $\square$

Proof. We are going to use that an $A$-module is flat if and only if it is torsion free, see Lemma 15.22.10. By Algebra, Lemma 10.57.10 we can find a graded $A$-algebra $S$ with $S_0 = A$ and generated by finitely many elements in degree $1$, an element $f \in S_1$ and a finite graded $S$-module $N$ such that $B \cong S_{(f)}$ and $M \cong N_{(f)}$. If $M$ is torsion free, then we can take $N$ torsion free by replacing it by $N/N_{tors}$, see Lemma 15.22.2. Similarly, if $B$ is torsion free, then we can take $S$ torsion free by replacing it by $S/S_{tors}$. Hence in case (1), we may apply Lemma 15.25.4 to see that $S$ is a finitely presented $A$-algebra, which implies that $B = S_{(f)}$ is a finitely presented $A$-algebra. To see (2) we may first replace $S$ by a graded polynomial ring, and then we may apply Lemma 15.25.3 to conclude. $\square$

Proof. By assumption we can write $B$ as a quotient of the localization of a polynomial algebra $P = A[x_1, \ldots , x_ n]$ at a prime ideal $\mathfrak q$. In case (1) we consider $M = B$ as a finite module over $P_\mathfrak q$ and in case (2) we consider $M$ as a finite module over $P_\mathfrak q$. In both cases, we have to show that this is a finitely presented $P_\mathfrak q$-module, see Algebra, Lemma 10.6.4 for case (2).

Choose a presentation $0 \to K \to P_\mathfrak q^{\oplus r} \to M \to 0$ which is possible because $M$ is finite over $P_\mathfrak q$. Let $L = P^{\oplus r} \cap K$. Then $K = L_\mathfrak q$, see Algebra, Lemma 10.9.15. Then $N = P^{\oplus r}/L$ is a submodule of $M$ and hence flat by Lemma 15.22.10. Since also $N$ is a finite $P$-module, we see that $N$ is finitely presented as a $P$-module by Lemma 15.25.6. Since localization is exact (Algebra, Proposition 10.9.12) we see that $N_\mathfrak q = M$ and we conclude. $\square$