## 15.70 Modules which are close to being projective

There seem to be many different of definitions in the literature of “almost projective modules”. In this section we discuss just one of the many possibilities.

Lemma 15.70.1. Let $R$ be a ring. Let $M$, $N$ be $R$-modules.

1. Given an $R$-module map $\varphi : M \to N$ the following are equivalent: (a) $\varphi$ factors through a projective $R$-module, and (b) $\varphi$ factors through a free $R$-module.

2. The set of $\varphi : M \to N$ satisfying the equivalent conditions of (1) is an $R$-submodule of $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$.

3. Given maps $\psi : M' \to M$ and $\xi : N \to N'$, if $\varphi : M \to N$ satisfies the equivalent conditions of (1), then $\xi \circ \varphi \circ \psi : M' \to N'$ does too.

Proof. The equivalence of (1)(a) and (1)(b) follows from Algebra, Lemma 10.77.2. If $\varphi : M \to N$ and $\varphi ' : M \to N$ factor through the modules $P$ and $P'$ then $\varphi + \varphi '$ factors through $P \oplus P'$ and $\lambda \varphi$ factors through $P$ for all $\lambda \in R$. This proves (2). If $\varphi : M \to N$ factors through the module $P$ and $\psi$ and $\xi$ are as in (3), then $\xi \circ \varphi \circ \psi$ factors through $P$. This proves (3). $\square$

Lemma 15.70.2. Let $R$ be a ring. Let $\varphi : M \to N$ be an $R$-module map. If $\varphi$ factors through a projective module and $M$ is a finite $R$-module, then $\varphi$ factors through a finite projective module.

Proof. By Lemma 15.70.1 we can factor $\varphi = \tau \circ \sigma$ where the target of $\sigma$ is $\bigoplus _{i \in I} R$ for some set $I$. Choose generators $x_1, \ldots , x_ n$ for $M$. Write $\sigma (x_ j) = (a_{ji})_{i \in I}$. For each $j$ only a finite number of $a_{ij}$ are nonzero. Hence $\sigma$ has image contained in a finite free $R$-module and we conclude. $\square$

Let $R$ be a ring. Observe that an $R$-module is projective if and only if the identity on $R$ factors through a projective module.

Lemma 15.70.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. The following conditions are equivalent

1. for every $a \in I$ the map $a : M \to M$ factors through a projective $R$-module,

2. for every $a \in I$ the map $a : M \to M$ factors through a free $R$-module, and

3. $\mathop{\mathrm{Ext}}\nolimits ^1_ R(M, N)$ is annihilated by $I$ for every $R$-module $N$.

Proof. The equivalence of (1) and (2) follows from Lemma 15.70.1. If (1) holds, then (3) holds because $\mathop{\mathrm{Ext}}\nolimits ^1_ R(P, N)$ for any $N$ and any projective module $P$. Conversely, assume (3) holds. Choose a short exact sequence $0 \to N \to P \to M \to 0$ with $P$ projective (or even free). By assumption the corresponding element of $\mathop{\mathrm{Ext}}\nolimits ^1_ R(M, N)$ is annihilated by $I$. Hence for every $a \in I$ the map $a : M \to M$ can be factored through the surjection $P \to M$ and we conclude (1) holds. $\square$

In order to comfortably talk about modules satisfying the equivalent conditions of Lemma 15.70.3 we give the property a name.

Definition 15.70.4. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. We say $M$ is $I$-projective1 if the equivalent conditions of Lemma 15.70.3 hold.

Modules annihilated by $I$ are $I$-projective.

Lemma 15.70.5. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. If $M$ is annihilated by $I$, then $M$ is $I$-projective.

Proof. Immediate from the definition and the fact that the zero module is projective. $\square$

Lemma 15.70.6. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let

$0 \to K \to P \to M \to 0$

be a short exact sequence of $R$-modules. If $M$ is $I$-projective and $P$ is projective, then $K$ is $I$-projective.

Proof. The element $\text{id}_ K \in \mathop{\mathrm{Hom}}\nolimits _ R(K, K)$ maps to the class of the given extension in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(M, K)$. Since by assumption this class is annihilated by any $a \in I$ we see that $a : K \to K$ factors through $K \to P$ and we conclude. $\square$

Lemma 15.70.7. Let $R$ be a ring. Let $I \subset R$ be an ideal. If $M$ is a finite, $I$-projective $R$-module, then $M^\vee = \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ is $I$-projective.

Proof. Assume $M$ is finite and $I$-projective. Choose a short exact sequence $0 \to K \to R^{\oplus r} \to M \to 0$. This produces an injection $M^\vee \to R^{\oplus r} = (R^{\oplus r})^\vee$. Since the extension class in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(M, K)$ corresponding to the short exact sequence is annihilated by $I$, we see that for any $a \in I$ we can find a map $M \to R^{\oplus r}$ such that the composition with the given map $R^{\oplus r} \to M$ is equal to $a : M \to M$. Taking duals we find that $a : M^\vee \to M^\vee$ factors through the map $M^\vee \to R^{\oplus r}$ given above and we conclude. $\square$

[1] This is nonstandard notation.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).