15.25 Flatness and finiteness conditions
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
0 \to K \to S^{\oplus r} \to M \to 0
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
\xymatrix{ M'_ g \ar[r] \ar[d] & M_ g \ar[d] \\ \prod (M'_ g)_{\mathfrak p_ j} \ar[r] & \prod (M_ g)_{\mathfrak p_ j} }
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
0 \to K \to \bigoplus \nolimits _{i = 1, \ldots , r} S(-d_ i) \to M \to 0
which in each degree d leads to the short exact sequence
0 \to K_ d \to \bigoplus \nolimits _{i = 1, \ldots , r} S_{d - d_ i} \to M_ d \to 0.
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
\bigoplus \nolimits _{j = 1, \ldots , t} S_{d - e_ j} \longrightarrow K_ d
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
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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