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The Stacks project

10.78 Finite projective modules

Definition 10.78.1. Let R be a ring and M an R-module.

  1. We say that M is locally free if we can cover \mathop{\mathrm{Spec}}(R) by standard opens D(f_ i), i \in I such that M_{f_ i} is a free R_{f_ i}-module for all i \in I.

  2. We say that M is finite locally free if we can choose the covering such that each M_{f_ i} is finite free.

  3. We say that M is finite locally free of rank r if we can choose the covering such that each M_{f_ i} is isomorphic to R_{f_ i}^{\oplus r}.

Note that a finite locally free R-module is automatically finitely presented by Lemma 10.23.2. Moreover, if M is a finite locally free module of rank r over a ring R and if R is nonzero, then r is uniquely determined by Lemma 10.15.8 (because at least one of the localizations R_{f_ i} is a nonzero ring).

Lemma 10.78.2. Let R be a ring and let M be an R-module. The following are equivalent

  1. M is finitely presented and R-flat,

  2. M is finite projective,

  3. M is a direct summand of a finite free R-module,

  4. M is finitely presented and for all \mathfrak p \in \mathop{\mathrm{Spec}}(R) the localization M_{\mathfrak p} is free,

  5. M is finitely presented and for all maximal ideals \mathfrak m \subset R the localization M_{\mathfrak m} is free,

  6. M is finite and locally free,

  7. M is finite locally free, and

  8. M is finite, for every prime \mathfrak p the module M_{\mathfrak p} is free, and the function

    \rho _ M : \mathop{\mathrm{Spec}}(R) \to \mathbf{Z}, \quad \mathfrak p \longmapsto \dim _{\kappa (\mathfrak p)} M \otimes _ R \kappa (\mathfrak p)

    is locally constant in the Zariski topology.

Proof. First suppose M is finite projective, i.e., (2) holds. Take a surjection R^ n \to M and let K be the kernel. Since M is projective, 0 \to K \to R^ n \to M \to 0 splits. Hence (2) \Rightarrow (3). The implication (3) \Rightarrow (2) follows from the fact that a direct summand of a projective is projective, see Lemma 10.77.2.

Assume (3), so we can write K \oplus M \cong R^{\oplus n}. So K is a direct summand of R^ n and thus finitely generated. This shows M = R^{\oplus n}/K is finitely presented. In other words, (3) \Rightarrow (1).

Assume M is finitely presented and flat, i.e., (1) holds. We will prove that (7) holds. Pick any prime \mathfrak p and x_1, \ldots , x_ r \in M which map to a basis of M \otimes _ R \kappa (\mathfrak p). By Nakayama's lemma (in the form of Lemma 10.20.2) these elements generate M_ g for some g \in R, g \not\in \mathfrak p. The corresponding surjection \varphi : R_ g^{\oplus r} \to M_ g has the following two properties: (a) \mathop{\mathrm{Ker}}(\varphi ) is a finite R_ g-module (see Lemma 10.5.3) and (b) \mathop{\mathrm{Ker}}(\varphi ) \otimes \kappa (\mathfrak p) = 0 by flatness of M_ g over R_ g (see Lemma 10.39.12). Hence by Nakayama's lemma again there exists a g' \in R_ g such that \mathop{\mathrm{Ker}}(\varphi )_{g'} = 0. In other words, M_{gg'} is free.

A finite locally free module is a finite module, see Lemma 10.23.2, hence (7) \Rightarrow (6). It is clear that (6) \Rightarrow (7) and that (7) \Rightarrow (8).

A finite locally free module is a finitely presented module, see Lemma 10.23.2, hence (7) \Rightarrow (4). Of course (4) implies (5). Since we may check flatness locally (see Lemma 10.39.18) we conclude that (5) implies (1). At this point we have

\xymatrix{ (2) \ar@{<=>}[r] & (3) \ar@{=>}[r] & (1) \ar@{=>}[r] & (7) \ar@{<=>}[r] \ar@{=>}[rd] \ar@{=>}[d] & (6) \\ & & (5) \ar@{=>}[u] & (4) \ar@{=>}[l] & (8) }

Suppose that M satisfies (1), (4), (5), (6), and (7). We will prove that (3) holds. It suffices to show that M is projective. We have to show that \mathop{\mathrm{Hom}}\nolimits _ R(M, -) is exact. Let 0 \to N'' \to N \to N'\to 0 be a short exact sequence of R-module. We have to show that 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N'') \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N') \to 0 is exact. As M is finite locally free there exist a covering \mathop{\mathrm{Spec}}(R) = \bigcup D(f_ i) such that M_{f_ i} is finite free. By Lemma 10.10.2 we see that

0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N'')_{f_ i} \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N)_{f_ i} \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N')_{f_ i} \to 0

is equal to 0 \to \mathop{\mathrm{Hom}}\nolimits _{R_{f_ i}}(M_{f_ i}, N''_{f_ i}) \to \mathop{\mathrm{Hom}}\nolimits _{R_{f_ i}}(M_{f_ i}, N_{f_ i}) \to \mathop{\mathrm{Hom}}\nolimits _{R_{f_ i}}(M_{f_ i}, N'_{f_ i}) \to 0 which is exact as M_{f_ i} is free and as the localization 0 \to N''_{f_ i} \to N_{f_ i} \to N'_{f_ i} \to 0 is exact (as localization is exact). Whence we see that 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N'') \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N') \to 0 is exact by Lemma 10.23.2.

Finally, assume that (8) holds. Pick a maximal ideal \mathfrak m \subset R. Pick x_1, \ldots , x_ r \in M which map to a \kappa (\mathfrak m)-basis of M \otimes _ R \kappa (\mathfrak m) = M/\mathfrak mM. In particular \rho _ M(\mathfrak m) = r. By Nakayama's Lemma 10.20.1 there exists an f \in R, f \not\in \mathfrak m such that x_1, \ldots , x_ r generate M_ f over R_ f. By the assumption that \rho _ M is locally constant there exists a g \in R, g \not\in \mathfrak m such that \rho _ M is constant equal to r on D(g). We claim that

\Psi : R_{fg}^{\oplus r} \longrightarrow M_{fg}, \quad (a_1, \ldots , a_ r) \longmapsto \sum a_ i x_ i

is an isomorphism. This claim will show that M is finite locally free, i.e., that (7) holds. To see the claim it suffices to show that the induced map on localizations \Psi _{\mathfrak p} : R_{\mathfrak p}^{\oplus r} \to M_{\mathfrak p} is an isomorphism for all \mathfrak p \in D(fg), see Lemma 10.23.1. By our choice of f the map \Psi _{\mathfrak p} is surjective. By assumption (8) we have M_{\mathfrak p} \cong R_{\mathfrak p}^{\oplus \rho _ M(\mathfrak p)} and by our choice of g we have \rho _ M(\mathfrak p) = r. Hence \Psi _{\mathfrak p} determines a surjection R_{\mathfrak p}^{\oplus r} \to M_{\mathfrak p} \cong R_{\mathfrak p}^{\oplus r} whence is an isomorphism by Lemma 10.16.4. (Of course this last fact follows from a simple matrix argument also.) \square

Lemma 10.78.3. Let R be a reduced ring and let M be an R-module. Then the equivalent conditions of Lemma 10.78.2 are also equivalent to

  1. M is finite and the function \rho _ M : \mathop{\mathrm{Spec}}(R) \to \mathbf{Z}, \mathfrak p \mapsto \dim _{\kappa (\mathfrak p)} M \otimes _ R \kappa (\mathfrak p) is locally constant in the Zariski topology.

Proof. Pick a maximal ideal \mathfrak m \subset R. Pick x_1, \ldots , x_ r \in M which map to a \kappa (\mathfrak m)-basis of M \otimes _ R \kappa (\mathfrak m) = M/\mathfrak mM. In particular \rho _ M(\mathfrak m) = r. By Nakayama's Lemma 10.20.1 there exists an f \in R, f \not\in \mathfrak m such that x_1, \ldots , x_ r generate M_ f over R_ f. By the assumption that \rho _ M is locally constant there exists a g \in R, g \not\in \mathfrak m such that \rho _ M is constant equal to r on D(g). We claim that

\Psi : R_{fg}^{\oplus r} \longrightarrow M_{fg}, \quad (a_1, \ldots , a_ r) \longmapsto \sum a_ i x_ i

is an isomorphism. This claim will show that M is finite locally free, i.e., that (7) holds. Since \Psi is surjective, it suffices to show that \Psi is injective. Since R_{fg} is reduced, it suffices to show that \Psi is injective after localization at all minimal primes \mathfrak p of R_{fg}, see Lemma 10.25.2. However, we know that R_\mathfrak p = \kappa (\mathfrak p) by Lemma 10.25.1 and \rho _ M(\mathfrak p) = r hence \Psi _\mathfrak p : R_\mathfrak p^{\oplus r} \to M \otimes _ R \kappa (\mathfrak p) is an isomorphism as a surjective map of finite dimensional vector spaces of the same dimension. \square

Remark 10.78.4. It is not true that a finite R-module which is R-flat is automatically projective. A counter example is where R = \mathcal{C}^\infty (\mathbf{R}) is the ring of infinitely differentiable functions on \mathbf{R}, and M = R_{\mathfrak m} = R/I where \mathfrak m = \{ f \in R \mid f(0) = 0\} and I = \{ f \in R \mid \exists \epsilon , \epsilon > 0 : f(x) = 0\ \forall x, |x| < \epsilon \} .

Lemma 10.78.5. (Warning: see Remark 10.78.4.) Suppose R is a local ring, and M is a finite flat R-module. Then M is finite free.

Proof. Follows from the equational criterion of flatness, see Lemma 10.39.11. Namely, suppose that x_1, \ldots , x_ r \in M map to a basis of M/\mathfrak mM. By Nakayama's Lemma 10.20.1 these elements generate M. We want to show there is no relation among the x_ i. Instead, we will show by induction on n that if x_1, \ldots , x_ n \in M are linearly independent in the vector space M/\mathfrak mM then they are independent over R.

The base case of the induction is where we have x \in M, x \not\in \mathfrak mM and a relation fx = 0. By the equational criterion there exist y_ j \in M and a_ j \in R such that x = \sum a_ j y_ j and fa_ j = 0 for all j. Since x \not\in \mathfrak mM we see that at least one a_ j is a unit and hence f = 0 .

Suppose that \sum f_ i x_ i is a relation among x_1, \ldots , x_ n. By our choice of x_ i we have f_ i \in \mathfrak m. According to the equational criterion of flatness there exist a_{ij} \in R and y_ j \in M such that x_ i = \sum a_{ij} y_ j and \sum f_ i a_{ij} = 0. Since x_ n \not\in \mathfrak mM we see that a_{nj}\not\in \mathfrak m for at least one j. Since \sum f_ i a_{ij} = 0 we get f_ n = \sum _{i = 1}^{n-1} (-a_{ij}/a_{nj}) f_ i. The relation \sum f_ i x_ i = 0 now can be rewritten as \sum _{i = 1}^{n-1} f_ i( x_ i + (-a_{ij}/a_{nj}) x_ n) = 0. Note that the elements x_ i + (-a_{ij}/a_{nj}) x_ n map to n-1 linearly independent elements of M/\mathfrak mM. By induction assumption we get that all the f_ i, i \leq n-1 have to be zero, and also f_ n = \sum _{i = 1}^{n-1} (-a_{ij}/a_{nj}) f_ i. This proves the induction step. \square

Lemma 10.78.6. Let R \to S be a flat local homomorphism of local rings. Let M be a finite R-module. Then M is finite projective over R if and only if M \otimes _ R S is finite projective over S.

Proof. By Lemma 10.78.2 being finite projective over a local ring is the same thing as being finite free. Suppose that M \otimes _ R S is a finite free S-module. Pick x_1, \ldots , x_ r \in M whose images in M/\mathfrak m_ RM form a basis over \kappa (\mathfrak m). Then we see that x_1 \otimes 1, \ldots , x_ r \otimes 1 are a basis for M \otimes _ R S. This implies that the map R^{\oplus r} \to M, (a_ i) \mapsto \sum a_ i x_ i becomes an isomorphism after tensoring with S. By faithful flatness of R \to S, see Lemma 10.39.17 we see that it is an isomorphism. \square

Lemma 10.78.7. Let R be a semi-local ring. Let M be a finite locally free module. If M has constant rank, then M is free. In particular, if R has connected spectrum, then M is free.

Proof. Omitted. Hints: First show that M/\mathfrak m_ iM has the same dimension d for all maximal ideal \mathfrak m_1, \ldots , \mathfrak m_ n of R using the rank is constant. Next, show that there exist elements x_1, \ldots , x_ d \in M which form a basis for each M/\mathfrak m_ iM by the Chinese remainder theorem. Finally show that x_1, \ldots , x_ d is a basis for M. \square

Here is a technical lemma that is used in the chapter on groupoids.

Lemma 10.78.8. Let R be a local ring with maximal ideal \mathfrak m and infinite residue field. Let R \to S be a ring map. Let M be an S-module and let N \subset M be an R-submodule. Assume

  1. S is semi-local and \mathfrak mS is contained in the Jacobson radical of S,

  2. M is a finite free S-module, and

  3. N generates M as an S-module.

Then N contains an S-basis of M.

Proof. Assume M is free of rank n. Let I \subset S be the Jacobson radical. By Nakayama's Lemma 10.20.1 a sequence of elements m_1, \ldots , m_ n is a basis for M if and only if \overline{m}_ i \in M/IM generate M/IM. Hence we may replace M by M/IM, N by N/(N \cap IM), R by R/\mathfrak m, and S by S/IS. In this case we see that S is a finite product of fields S = k_1 \times \ldots \times k_ r and M = k_1^{\oplus n} \times \ldots \times k_ r^{\oplus n}. The fact that N \subset M generates M as an S-module means that there exist x_ j \in N such that a linear combination \sum a_ j x_ j with a_ j \in S has a nonzero component in each factor k_ i^{\oplus n}. Because R = k is an infinite field, this means that also some linear combination y = \sum c_ j x_ j with c_ j \in k has a nonzero component in each factor. Hence y \in N generates a free direct summand Sy \subset M. By induction on n the result holds for M/Sy and the submodule \overline{N} = N/(N \cap Sy). In other words there exist \overline{y}_2, \ldots , \overline{y}_ n in \overline{N} which (freely) generate M/Sy. Then y, y_2, \ldots , y_ n (freely) generate M and we win. \square

Lemma 10.78.9. Let R be ring. Let L, M, N be R-modules. The canonical map

\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \otimes _ R L \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N \otimes _ R L)

is an isomorphism if M is finite projective.

Proof. By Lemma 10.78.2 we see that M is finitely presented as well as finite locally free. By Lemmas 10.10.2 and 10.12.16 formation of the left and right hand side of the arrow commutes with localization. We may check that our map is an isomorphism after localization, see Lemma 10.23.2. Thus we may assume M is finite free. In this case the lemma is immediate. \square


Comments (4)

Comment #2296 by Dario Weißmann on

In Lemma 10.77.5 we apply Lemma 10.77.2 to get that the module is finite flat. Then we should apply Lemma 10.77.4 to see that it is finite free (this reference is missing). And I think the should be chosen such that their residue classes form a basis of and not just generate it.

In the hints of Lemma 10.77.6 one should use the rank is constant instead of the spectrum is connected.

Comment #6825 by David Liu on

I don't understand one part in Lemma 10.78.6. Why is a basis of ?

Comment #6967 by on

This is because if is a finite free module over a local ring , then a set of elements , of is a basis for over if and only if their images in are a -basis for this vector space. OK?


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