The Stacks project

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

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

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

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.) $\square$

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

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. $\square$

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. $\square$

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. $\square$

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 $M$. $\square$

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. $\square$

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. $\square$