The Stacks project

Proof. Obviously (1) implies (2).

Next we show (2) implies (3). Let $f_1(x_ i) = \sum _ j a_{ij} y_ j$ be relations as in (3). Let $(d_ j)$ be a basis for $R^ m$, $(e_ i)$ a basis for $R^ n$, and $R^ m \to R^ n$ the map given by $d_ j \mapsto \sum _ i a_{ij} e_ i$. Let $Q$ be the cokernel of $R^ m \to R^ n$. Then tensoring $R^ m \to R^ n \to Q \to 0$ by the map $f_1: M_1 \to M_2$, we get a commutative diagram

where $M_1^{\oplus m} \to M_1^{\oplus n}$ is given by

and $M_2^{\oplus m} \to M_2^{\oplus n}$ is given similarly. We want to show $x = (x_1, \ldots , x_ n) \in M_1^{\oplus n}$ is in the image of $M_1^{\oplus m} \to M_1^{\oplus n}$. By (2) the map $M_1 \otimes Q \to M_2 \otimes Q$ is injective, hence by exactness of the top row it is enough to show $x$ maps to $0$ in $M_2 \otimes Q$, and so by exactness of the bottom row it is enough to show the image of $x$ in $M_2^{\oplus n}$ is in the image of $M_2^{\oplus m} \to M_2^{\oplus n}$. This is true by assumption.

Condition (4) is just a translation of (3) into diagram form.

Next we show (4) implies (5). Let $\varphi : P \to M_3$ be a map from a finitely presented $R$-module $P$. We must show that $\varphi $ lifts to a map $P \to M_2$. Choose a presentation of $P$,

Using freeness of $R^ n$ and $R^ m$, we can construct $h_2: R^ m \to M_2$ and then $h_1: R^ n \to M_1$ such that the following diagram commutes

By (4) there is a map $k_1: R^ m \to M_1$ such that $k_1 \circ g_1 = h_1$. Now define $h'_2: R^ m \to M_2$ by $h_2' = h_2 - f_1 \circ k_1$. Then

Hence by passing to the quotient $h'_2$ defines a map $\varphi ': P \to M_2$ such that $\varphi ' \circ g_2 = h_2'$. In a diagram, we have

where the top triangle commutes. We claim that $\varphi '$ is the desired lift, i.e. that $f_2 \circ \varphi ' = \varphi $. From the definitions we have

Since $g_2$ is surjective, this finishes the proof.

Now we show (5) implies (6). Write $M_{3}$ as the colimit of a directed system of finitely presented modules $M_{3, i}$, see Lemma 10.11.3. Let $M_{2, i}$ be the fiber product of $M_{3, i}$ and $M_{2}$ over $M_{3}$—by definition this is the submodule of $M_2 \times M_{3, i}$ consisting of elements whose two projections onto $M_3$ are equal. Let $M_{1, i}$ be the kernel of the projection $M_{2, i} \to M_{3, i}$. Then we have a directed system of exact sequences

and for each $i$ a map of exact sequences

compatible with the directed system. From the definition of the fiber product $M_{2, i}$, it follows that the map $M_{1, i} \to M_1$ is an isomorphism. By (5) there is a map $M_{3, i} \to M_{2}$ lifting $M_{3, i} \to M_3$, and by the universal property of the fiber product this gives rise to a section of $M_{2, i} \to M_{3, i}$. Hence the sequences

split. Passing to the colimit, we have a commutative diagram

with exact rows and outer vertical maps isomorphisms. Hence $\mathop{\mathrm{colim}}\nolimits M_{2, i} \to M_2$ is also an isomorphism and (6) holds.

Condition (6) implies (1) by Example 10.82.2 (2). $\square$

Proof. A split short exact sequence is always universally exact, see Example 10.82.2. Conversely, if the sequence is universally exact, then by Theorem 10.82.3 (5) applied to $P = M_3$, the map $M_2 \to M_3$ admits a section. $\square$

Proof. This follows from Lemma 10.81.3 and Theorem 10.82.3 (5). $\square$

Proof. Let $0 \to N \to N' \to N'' \to 0$ be a short exact sequence of $R$-modules. Consider the commutative diagram

(we have dropped the $0$'s on the boundary). By assumption the rows give short exact sequences and the arrow $M_2 \otimes N \to M_2 \otimes N'$ is injective. Clearly this implies that $M_1 \otimes N \to M_1 \otimes N'$ is injective and we see that $M_1$ is flat. In particular the left and middle columns give rise to short exact sequences. It follows from a diagram chase that the arrow $M_3 \otimes N \to M_3 \otimes N'$ is injective. Hence $M_3$ is flat. $\square$

Proof. Omitted. $\square$

Proof. Omitted. $\square$

Proof. Omitted. $\square$

Proof. Let $N$ be an $R$-module. We have to show that $N \to N \otimes _ R S$ is injective. As $S$ is faithfully flat as an $R$-module, it suffices to prove this after tensoring with $S$. Hence it suffices to show that $N \otimes _ R S \to N \otimes _ R S \otimes _ R S$, $n \otimes s \mapsto n \otimes 1 \otimes s$ is injective. This is true because there is a retraction, namely, $n \otimes s \otimes s' \mapsto n \otimes ss'$. $\square$

Proof. Let $N$ be an $R$-module. Let $\mathfrak q$ be a prime of $S$ lying over the prime $\mathfrak p$ of $R$. Then we have

Moreover, the same thing holds for $M'$ and localization is exact. Also, if $N$ is an $R_{\mathfrak p}$-module, then $N_{\mathfrak p} = N$. Using this the equivalences can be proved in a straightforward manner.

For example, suppose that (5) holds. Let $K = \mathop{\mathrm{Ker}}(M \otimes _ R N \to M' \otimes _ R N)$. By the remarks above we see that $K_{\mathfrak m} = 0$ for each maximal ideal $\mathfrak m$ of $S$. Hence $K = 0$ by Lemma 10.23.1. Thus (1) holds. Conversely, suppose that (1) holds. Take any $\mathfrak q \subset S$ lying over $\mathfrak p \subset R$. Take any module $N$ over $R_{\mathfrak p}$. Then by assumption $\mathop{\mathrm{Ker}}(M \otimes _ R N \to M' \otimes _ R N) = 0$. Hence by the formulae above and the fact that $N = N_{\mathfrak p}$ we see that $\mathop{\mathrm{Ker}}(M_{\mathfrak q} \otimes _{R_{\mathfrak p}} N \to M'_{\mathfrak q} \otimes _{R_{\mathfrak p}} N) = 0$. In other words (4) holds. Of course (4) $\Rightarrow $ (5) is immediate. Hence (1), (4) and (5) are all equivalent. We omit the proof of the other equivalences. $\square$

Proof. You can prove this using Lemma 10.82.12 but you can also prove it directly as follows. Assume $M \to M'$ is $A$-universally injective. Let $Q$ be an $A$-module. Then $Q \otimes _ A M \to Q \otimes _ A M'$ is injective. Since localization is exact we see that $(S')^{-1}(Q \otimes _ A M) \to (S')^{-1}(Q \otimes _ A M')$ is injective. As $(S')^{-1}(Q \otimes _ A M) = Q \otimes _ A (S')^{-1}M$ and similarly for $M'$ we see that $Q \otimes _ A (S')^{-1}M \to Q \otimes _ A (S')^{-1}M'$ is injective, hence $(S')^{-1}M \to (S')^{-1}M'$ is universally injective as a map of $A$-modules. This proves the first part of (1). To see (2) we can use the following two facts: (a) if $Q$ is an $S^{-1}A$-module, then $Q \otimes _ A S^{-1}A = Q$, i.e., tensoring with $Q$ over $A$ is the same thing as tensoring with $Q$ over $S^{-1}A$, (b) if $M$ is any $A$-module on which the elements of $S$ are invertible, then $M \otimes _ A Q = M \otimes _{S^{-1}A} S^{-1}Q$. Part (2) follows from this immediately. $\square$

Proof. It suffices to show that $M \otimes _ R Q \to M' \otimes _ R Q$ is injective for every finite $R$-module $Q$, see Theorem 10.82.3. Then $Q$ has a finite filtration $0 = Q_0 \subset Q_1 \subset \ldots \subset Q_ n = Q$ by submodules whose subquotients are isomorphic to cyclic modules $R/I_ i$, see Lemma 10.5.4. Since $M'$ is flat, we obtain a filtration

of $M' \otimes _ R Q$ by submodules $M' \otimes _ R Q_ i$ whose successive quotients are $M' \otimes _ R R/I_ i = M'/I_ iM'$. A simple induction argument shows that it suffices to check $M/I_ i M \to M'/I_ i M'$ is injective. Note that the collection of finitely generated ideals $I'_ i \subset I_ i$ is a directed set. Thus $M/I_ iM = \mathop{\mathrm{colim}}\nolimits M/I'_ iM$ is a filtered colimit, similarly for $M'$, the maps $M/I'_ iM \to M'/I'_ i M'$ are injective by assumption, and since filtered colimits are exact (Lemma 10.8.8) we conclude. $\square$