In this section we discuss some implications of the type “flat $+$ finite type $\Rightarrow $ finite presentation”. We will revisit this result in the chapter on flatness, see More on Flatness, Section 38.1. A first result of this type was proved in Algebra, Lemma 10.108.6.
Lemma 15.25.1. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]$ be a polynomial ring over $R$. Let $M$ be an $S$-module. Assume
there exist finitely many primes $\mathfrak p_1, \ldots , \mathfrak p_ m$ of $R$ such that the map $R \to \prod R_{\mathfrak p_ j}$ is injective,
$M$ is a finite $S$-module,
$M$ flat over $R$, and
for every prime $\mathfrak p$ of $R$ the module $M_{\mathfrak p}$ is of finite presentation over $S_{\mathfrak p}$.
Then $M$ is of finite presentation over $S$.
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. $\square$
Lemma 15.25.2. Let $R \to S$ be a ring homomorphism. Assume
there exist finitely many primes $\mathfrak p_1, \ldots , \mathfrak p_ m$ of $R$ such that the map $R \to \prod R_{\mathfrak p_ j}$ is injective,
$R \to S$ is of finite type,
$S$ flat over $R$, and
for every prime $\mathfrak p$ of $R$ the ring $S_{\mathfrak p}$ is of finite presentation over $R_{\mathfrak p}$.
Then $S$ is of finite presentation over $R$.
Proof. By assumption $S$ is a quotient of a polynomial ring over $R$. Thus the result follows directly from Lemma 15.25.1. $\square$
Lemma 15.25.3. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]$ be a graded polynomial algebra over $R$, i.e., $\deg (x_ i) > 0$ but not necessarily equal to $1$. Let $M$ be a graded $S$-module. Assume
$R$ is a local ring,
$M$ is a finite $S$-module, and
$M$ is flat over $R$.
Then $M$ is finitely presented as an $S$-module.
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. $\square$
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. $\square$
Remark 15.25.5. Let $R$ be a ring. When does $R$ satisfy the condition mentioned in Lemmas 15.25.1, 15.25.2, and 15.25.4? This holds if
$R$ is local,
$R$ is Noetherian,
$R$ is a domain,
$R$ is a reduced ring with finitely many minimal primes, or
$R$ has finitely many weakly associated primes, see Algebra, Lemma 10.66.17.
Thus these lemmas hold in all cases listed above.
The following lemma will be improved on in More on Flatness, Proposition 38.13.10.
Lemma 15.25.6.reference Let $A$ be a valuation ring. Let $A \to B$ be a ring map of finite type. Let $M$ be a finite $B$-module.
If $B$ is flat over $A$, then $B$ is a finitely presented $A$-algebra.
If $M$ is flat as an $A$-module, then $M$ is finitely presented as a $B$-module.
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. $\square$
Lemma 15.25.7. Let $A$ be a valuation ring. Let $A \to B$ be a local homomorphism which is essentially of finite type. Let $M$ be a finite $B$-module.
If $B$ is flat over $A$, then $B$ is essentially of finite presentation over $A$.
If $M$ is flat as an $A$-module, then $M$ is finitely presented as a $B$-module.
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. $\square$
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