## 46.8 Pure extensions

We want to characterize extensions of quasi-coherent sheaves on the big site of an affine schemes in terms of algebra. To do this we introduce the following notion.

Definition 46.8.1. Let $A$ be a ring.

An $A$-module $P$ is said to be *pure projective* if for every universally exact sequence $0 \to K \to M \to N \to 0$ of $A$-module the sequence $0 \to \mathop{\mathrm{Hom}}\nolimits _ A(P, K) \to \mathop{\mathrm{Hom}}\nolimits _ A(P, M) \to \mathop{\mathrm{Hom}}\nolimits _ A(P, N) \to 0$ is exact.

An $A$-module $I$ is said to be *pure injective* if for every universally exact sequence $0 \to K \to M \to N \to 0$ of $A$-module the sequence $0 \to \mathop{\mathrm{Hom}}\nolimits _ A(N, I) \to \mathop{\mathrm{Hom}}\nolimits _ A(M, I) \to \mathop{\mathrm{Hom}}\nolimits _ A(K, I) \to 0$ is exact.

Let's characterize pure projectives.

Lemma 46.8.2. Let $A$ be a ring.

A module is pure projective if and only if it is a direct summand of a direct sum of finitely presented $A$-modules.

For any module $M$ there exists a universally exact sequence $0 \to N \to P \to M \to 0$ with $P$ pure projective.

**Proof.**
First note that a finitely presented $A$-module is pure projective by Algebra, Theorem 10.82.3. Hence a direct summand of a direct sum of finitely presented $A$-modules is indeed pure projective. Let $M$ be any $A$-module. Write $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} P_ i$ as a filtered colimit of finitely presented $A$-modules. Consider the sequence

\[ 0 \to N \to \bigoplus P_ i \to M \to 0. \]

For any finitely presented $A$-module $P$ the map $\mathop{\mathrm{Hom}}\nolimits _ A(P, \bigoplus P_ i) \to \mathop{\mathrm{Hom}}\nolimits _ A(P, M)$ is surjective, as any map $P \to M$ factors through some $P_ i$. Hence by Algebra, Theorem 10.82.3 this sequence is universally exact. This proves (2). If now $M$ is pure projective, then the sequence is split and we see that $M$ is a direct summand of $\bigoplus P_ i$.
$\square$

Let's characterize pure injectives.

Lemma 46.8.3. Let $A$ be a ring. For any $A$-module $M$ set $M^\vee = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(M, \mathbf{Q}/\mathbf{Z})$.

For any $A$-module $M$ the $A$-module $M^\vee $ is pure injective.

An $A$-module $I$ is pure injective if and only if the map $I \to (I^\vee )^\vee $ splits.

For any module $M$ there exists a universally exact sequence $0 \to M \to I \to N \to 0$ with $I$ pure injective.

**Proof.**
We will use the properties of the functor $M \mapsto M^\vee $ found in More on Algebra, Section 15.55 without further mention. Part (1) holds because $\mathop{\mathrm{Hom}}\nolimits _ A(N, M^\vee ) = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(N \otimes _ A M, \mathbf{Q}/\mathbf{Z})$ and because $\mathbf{Q}/\mathbf{Z}$ is injective in the category of abelian groups. Hence if $I \to (I^\vee )^\vee $ is split, then $I$ is pure injective. We claim that for any $A$-module $M$ the evaluation map $ev : M \to (M^\vee )^\vee $ is universally injective. To see this note that $ev^\vee : ((M^\vee )^\vee )^\vee \to M^\vee $ has a right inverse, namely $ev' : M^\vee \to ((M^\vee )^\vee )^\vee $. Then for any $A$-module $N$ applying the exact faithful functor ${}^\vee $ to the map $N \otimes _ A M \to N \otimes _ A (M^\vee )^\vee $ gives

\[ \mathop{\mathrm{Hom}}\nolimits _ A(N, ((M^\vee )^\vee )^\vee ) = \Big(N \otimes _ A (M^\vee )^\vee \Big)^\vee \to \Big(N \otimes _ A M\Big)^\vee = \mathop{\mathrm{Hom}}\nolimits _ A(N, M^\vee ) \]

which is surjective by the existence of the right inverse. The claim follows. The claim implies (3) and the necessity of the condition in (2).
$\square$

Before we continue we make the following observation which we will use frequently in the rest of this section.

Lemma 46.8.4. Let $A$ be a ring.

Let $L \to M \to N$ be a universally exact sequence of $A$-modules. Let $K = \mathop{\mathrm{Im}}(M \to N)$. Then $K \to N$ is universally injective.

Any universally exact complex can be split into universally exact short exact sequences.

**Proof.**
Proof of (1). For any $A$-module $T$ the sequence $L \otimes _ A T \to M \otimes _ A T \to K \otimes _ A T \to 0$ is exact by right exactness of $\otimes $. By assumption the sequence $L \otimes _ A T \to M \otimes _ A T \to N \otimes _ A T$ is exact. Combined this shows that $K \otimes _ A T \to N \otimes _ A T$ is injective.

Part (2) means the following: Suppose that $M^\bullet $ is a universally exact complex of $A$-modules. Set $K^ i = \mathop{\mathrm{Ker}}(d^ i) \subset M^ i$. Then the short exact sequences $0 \to K^ i \to M^ i \to K^{i + 1} \to 0$ are universally exact. This follows immediately from part (1).
$\square$

Definition 46.8.5. Let $A$ be a ring. Let $M$ be an $A$-module.

A *pure projective resolution* $P_\bullet \to M$ is a universally exact sequence

\[ \ldots \to P_1 \to P_0 \to M \to 0 \]

with each $P_ i$ pure projective.

A *pure injective resolution* $M \to I^\bullet $ is a universally exact sequence

\[ 0 \to M \to I^0 \to I^1 \to \ldots \]

with each $I^ i$ pure injective.

These resolutions satisfy the usual uniqueness properties among the class of all universally exact left or right resolutions.

Lemma 46.8.6. Let $A$ be a ring.

Any $A$-module has a pure projective resolution.

Let $M \to N$ be a map of $A$-modules. Let $P_\bullet \to M$ be a pure projective resolution and let $N_\bullet \to N$ be a universally exact resolution.

There exists a map of complexes $P_\bullet \to N_\bullet $ inducing the given map

\[ M = \mathop{\mathrm{Coker}}(P_1 \to P_0) \to \mathop{\mathrm{Coker}}(N_1 \to N_0) = N \]

two maps $\alpha , \beta : P_\bullet \to N_\bullet $ inducing the same map $M \to N$ are homotopic.

**Proof.**
Part (1) follows immediately from Lemma 46.8.2. Before we prove (2) and (3) note that by Lemma 46.8.4 we can split the universally exact complex $N_\bullet \to N \to 0$ into universally exact short exact sequences $0 \to K_0 \to N_0 \to N \to 0$ and $0 \to K_ i \to N_ i \to K_{i - 1} \to 0$.

Proof of (2). Because $P_0$ is pure projective we can find a map $P_0 \to N_0$ lifting the map $P_0 \to M \to N$. We obtain an induced map $P_1 \to F_0 \to N_0$ wich ends up in $K_0$. Since $P_1$ is pure projective we may lift this to a map $P_1 \to N_1$. This in turn induces a map $P_2 \to P_1 \to N_1$ which maps to zero into $N_0$, i.e., into $K_1$. Hence we may lift to get a map $P_2 \to N_2$. Repeat.

Proof of (3). To show that $\alpha , \beta $ are homotopic it suffices to show the difference $\gamma = \alpha - \beta $ is homotopic to zero. Note that the image of $\gamma _0 : P_0 \to N_0$ is contained in $K_0$. Hence we may lift $\gamma _0$ to a map $h_0 : P_0 \to N_1$. Consider the map $\gamma _1' = \gamma _1 - h_0 \circ d_{P, 1} : P_1 \to N_1$. By our choice of $h_0$ we see that the image of $\gamma _1'$ is contained in $K_1$. Since $P_1$ is pure projective may lift $\gamma _1'$ to a map $h_1 : P_1 \to N_2$. At this point we have $\gamma _1 = h_0 \circ d_{F, 1} + d_{N, 2} \circ h_1$. Repeat.
$\square$

Lemma 46.8.7. Let $A$ be a ring.

Any $A$-module has a pure injective resolution.

Let $M \to N$ be a map of $A$-modules. Let $M \to M^\bullet $ be a universally exact resolution and let $N \to I^\bullet $ be a pure injective resolution.

There exists a map of complexes $M^\bullet \to I^\bullet $ inducing the given map

\[ M = \mathop{\mathrm{Ker}}(M^0 \to M^1) \to \mathop{\mathrm{Ker}}(I^0 \to I^1) = N \]

two maps $\alpha , \beta : M^\bullet \to I^\bullet $ inducing the same map $M \to N$ are homotopic.

**Proof.**
This lemma is dual to Lemma 46.8.6. The proof is identical, except one has to reverse all the arrows.
$\square$

Using the material above we can define pure extension groups as follows. Let $A$ be a ring and let $M$, $N$ be $A$-modules. Choose a pure injective resolution $N \to I^\bullet $. By Lemma 46.8.7 the complex

\[ \mathop{\mathrm{Hom}}\nolimits _ A(M, I^\bullet ) \]

is well defined up to homotopy. Hence its $i$th cohomology module is a well defined invariant of $M$ and $N$.

Definition 46.8.8. Let $A$ be a ring and let $M$, $N$ be $A$-modules. The $i$th *pure extension module* $\text{Pext}^ i_ A(M, N)$ is the $i$th cohomology module of the complex $\mathop{\mathrm{Hom}}\nolimits _ A(M, I^\bullet )$ where $I^\bullet $ is a pure injective resolution of $N$.

Warning: It is not true that an exact sequence of $A$-modules gives rise to a long exact sequence of pure extensions groups. (You need a universally exact sequence for this.) We collect some facts which are obvious from the material above.

Lemma 46.8.9. Let $A$ be a ring.

$\text{Pext}^ i_ A(M, N) = 0$ for $i > 0$ whenever $N$ is pure injective,

$\text{Pext}^ i_ A(M, N) = 0$ for $i > 0$ whenever $M$ is pure projective, in particular if $M$ is an $A$-module of finite presentation,

$\text{Pext}^ i_ A(M, N)$ is also the $i$th cohomology module of the complex $\mathop{\mathrm{Hom}}\nolimits _ A(P_\bullet , N)$ where $P_\bullet $ is a pure projective resolution of $M$.

**Proof.**
To see (3) consider the double complex

\[ A^{\bullet , \bullet } = \mathop{\mathrm{Hom}}\nolimits _ A(P_\bullet , I^\bullet ) \]

Each of its rows is exact except in degree $0$ where its cohomology is $\mathop{\mathrm{Hom}}\nolimits _ A(M, I^ q)$. Each of its columns is exact except in degree $0$ where its cohomology is $\mathop{\mathrm{Hom}}\nolimits _ A(P_ p, N)$. Hence the two spectral sequences associated to this complex in Homology, Section 12.25 degenerate, giving the equality.
$\square$

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