Definition 10.81.1. Let $f: M \to N$ be a map of $R$-modules. Then $f$ is called *universally injective* if for every $R$-module $Q$, the map $f \otimes _ R \text{id}_ Q: M \otimes _ R Q \to N \otimes _ R Q$ is injective. A sequence $0 \to M_1 \to M_2 \to M_3 \to 0$ of $R$-modules is called *universally exact* if it is exact and $M_1 \to M_2$ is universally injective.

## 10.81 Universally injective module maps

Next we discuss universally injective module maps, which are in a sense complementary to flat modules (see Lemma 10.81.5). We follow Lazard's thesis [Autour]; also see [Lam].

Example 10.81.2. Examples of universally exact sequences.

A split short exact sequence is universally exact since tensoring commutes with taking direct sums.

The colimit of a directed system of universally exact sequences is universally exact. This follows from the fact that taking directed colimits is exact and that tensoring commutes with taking colimits. In particular the colimit of a directed system of split exact sequences is universally exact. We will see below that, conversely, any universally exact sequence arises in this way.

Next we give a list of criteria for a short exact sequence to be universally exact. They are analogues of criteria for flatness given above. Parts (3)-(6) below correspond, respectively, to the criteria for flatness given in Lemmas 10.38.11, 10.80.1, 10.80.3, and Theorem 10.80.4.

Theorem 10.81.3. Let

be an exact sequence of $R$-modules. The following are equivalent:

The sequence $0 \to M_1 \to M_2 \to M_3 \to 0$ is universally exact.

For every finitely presented $R$-module $Q$, the sequence

\[ 0 \to M_1 \otimes _ R Q \to M_2 \otimes _ R Q \to M_3 \otimes _ R Q \to 0 \]is exact.

Given elements $x_ i \in M_1$ $(i = 1, \ldots , n)$, $y_ j \in M_2$ $(j = 1, \ldots , m)$, and $a_{ij} \in R$ $(i = 1, \ldots , n, j = 1, \ldots , m)$ such that for all $i$

\[ f_1(x_ i) = \sum \nolimits _ j a_{ij} y_ j, \]there exists $z_ j \in M_1$ $(j =1, \ldots , m)$ such that for all $i$,

\[ x_ i = \sum \nolimits _ j a_{ij} z_ j . \]Given a commutative diagram of $R$-module maps

\[ \xymatrix{ R^ n \ar[r] \ar[d] & R^ m \ar[d] \\ M_1 \ar[r]^{f_1} & M_2 } \]where $m$ and $n$ are integers, there exists a map $R^ m \to M_1$ making the top triangle commute.

For every finitely presented $R$-module $P$, the $R$-module map $\mathop{\mathrm{Hom}}\nolimits _ R(P, M_2) \to \mathop{\mathrm{Hom}}\nolimits _ R(P, M_3)$ is surjective.

The sequence $0 \to M_1 \to M_2 \to M_3 \to 0$ is the colimit of a directed system of split exact sequences of the form

\[ 0 \to M_{1} \to M_{2, i} \to M_{3, i} \to 0 \]where the $M_{3, i}$ are finitely presented.

**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 10.8.12. 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 10.81.2 (2). $\square$

The previous theorem shows that a universally exact sequence is always a colimit of split short exact sequences. If the cokernel of a universally injective map is finitely presented, then in fact the map itself splits:

Lemma 10.81.4. Let

be an exact sequence of $R$-modules. Suppose $M_3$ is of finite presentation. Then

is universally exact if and only if it is split.

**Proof.**
A split short exact sequence is always universally exact, see Example 10.81.2. Conversely, if the sequence is universally exact, then by Theorem 10.81.3 (5) applied to $P = M_3$, the map $M_2 \to M_3$ admits a section.
$\square$

The following lemma shows how universally injective maps are complementary to flat modules.

Lemma 10.81.5. Let $M$ be an $R$-module. Then $M$ is flat if and only if any exact sequence of $R$-modules

is universally exact.

**Proof.**
This follows from Lemma 10.80.3 and Theorem 10.81.3 (5).
$\square$

Example 10.81.6. Non-split and non-flat universally exact sequences.

In spite of Lemma 10.81.4, it is possible to have a short exact sequence of $R$-modules

\[ 0 \to M_1 \to M_2 \to M_3 \to 0 \]that is universally exact but non-split. For instance, take $R = \mathbf{Z}$, let $M_1 = \bigoplus _{n=1}^{\infty } \mathbf{Z}$, let $M_{2} = \prod _{n = 1}^{\infty } \mathbf{Z}$, and let $M_{3}$ be the cokernel of the inclusion $M_1 \to M_2$. Then $M_1, M_2, M_3$ are all flat since they are torsion-free (More on Algebra, Lemma 15.22.11), so by Lemma 10.81.5,

\[ 0 \to M_1 \to M_2 \to M_3 \to 0 \]is universally exact. However there can be no section $s: M_3 \to M_2$. In fact, if $x$ is the image of $(2, 2^2, 2^3, \ldots ) \in M_2$ in $M_3$, then any module map $s: M_3 \to M_2$ must kill $x$. This is because $x \in 2^ n M_3$ for any $n \geq 1$, hence $s(x)$ is divisible by $2^ n$ for all $n \geq 1$ and so must be $0$.

In spite of Lemma 10.81.5, it is possible to have a short exact sequence of $R$-modules

\[ 0 \to M_1 \to M_2 \to M_3 \to 0 \]that is universally exact but with $M_1, M_2, M_3$ all non-flat. In fact if $M$ is any non-flat module, just take the split exact sequence

\[ 0 \to M \to M \oplus M \to M \to 0. \]For instance over $R = \mathbf{Z}$, take $M$ to be any torsion module.

Taking the direct sum of an exact sequence as in (1) with one as in (2), we get a short exact sequence of $R$-modules

\[ 0 \to M_1 \to M_2 \to M_3 \to 0 \]that is universally exact, non-split, and such that $M_1, M_2, M_3$ are all non-flat.

Lemma 10.81.7. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules, and suppose $M_2$ is flat. Then $M_1$ and $M_3$ are flat.

**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. $\square$

Lemma 10.81.8. Let $R$ be a ring. Let $M \to M'$ be a universally injective $R$-module map. Then for any $R$-module $N$ the map $M \otimes _ R N \to M' \otimes _ R N$ is universally injective.

**Proof.**
Omitted.
$\square$

Lemma 10.81.9. Let $R$ be a ring. A composition of universally injective $R$-module maps is universally injective.

**Proof.**
Omitted.
$\square$

Lemma 10.81.10. Let $R$ be a ring. Let $M \to M'$ and $M' \to M''$ be $R$-module maps. If their composition $M \to M''$ is universally injective, then $M \to M'$ is universally injective.

**Proof.**
Omitted.
$\square$

Lemma 10.81.11. Let $R \to S$ be a faithfully flat ring map. Then $R \to S$ is universally injective as a map of $R$-modules. In particular $R \cap IS = I$ for any ideal $I \subset R$.

**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, $n \otimes s \otimes s' \mapsto n \otimes ss'$.
$\square$

Lemma 10.81.12. Let $R \to S$ be a ring map. Let $M \to M'$ be a map of $S$-modules. The following are equivalent

$M \to M'$ is universally injective as a map of $R$-modules,

for each prime $\mathfrak q$ of $S$ the map $M_{\mathfrak q} \to M'_{\mathfrak q}$ is universally injective as a map of $R$-modules,

for each maximal ideal $\mathfrak m$ of $S$ the map $M_{\mathfrak m} \to M'_{\mathfrak m}$ is universally injective as a map of $R$-modules,

for each prime $\mathfrak q$ of $S$ the map $M_{\mathfrak q} \to M'_{\mathfrak q}$ is universally injective as a map of $R_{\mathfrak p}$-modules, where $\mathfrak p$ is the inverse image of $\mathfrak q$ in $R$, and

for each maximal ideal $\mathfrak m$ of $S$ the map $M_{\mathfrak m} \to M'_{\mathfrak m}$ is universally injective as a map of $R_{\mathfrak p}$-modules, where $\mathfrak p$ is the inverse image of $\mathfrak m$ in $R$.

**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 10.22.1. 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. $\square$

Lemma 10.81.13. Let $\varphi : A \to B$ be a ring map. Let $S \subset A$ and $S' \subset B$ be multiplicative subsets such that $\varphi (S) \subset S'$. Let $M \to M'$ be a map of $B$-modules.

If $M \to M'$ is universally injective as a map of $A$-modules, then $(S')^{-1}M \to (S')^{-1}M'$ is universally injective as a map of $A$-modules and as a map of $S^{-1}A$-modules.

If $M$ and $M'$ are $(S')^{-1}B$-modules, then $M \to M'$ is universally injective as a map of $A$-modules if and only if it is universally injective as a map of $S^{-1}A$-modules.

**Proof.**
You can prove this using Lemma 10.81.12 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.
$\square$

Lemma 10.81.14. Let $R$ be a ring and let $M \to M'$ be a map of $R$-modules. If $M'$ is flat, then $M \to M'$ is universally injective if and only if $M/IM \to M'/IM'$ is injective for every finitely generated ideal $I$ of $R$.

**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 10.81.3. 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. $\square$

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