## 10.77 Projective modules

Some lemmas on projective modules.

Definition 10.77.1. Let $R$ be a ring. An $R$-module $P$ is projective if and only if the functor $\mathop{\mathrm{Hom}}\nolimits _ R(P, -) : \text{Mod}_ R \to \text{Mod}_ R$ is an exact functor.

The functor $\mathop{\mathrm{Hom}}\nolimits _ R(M, - )$ is left exact for any $R$-module $M$, see Lemma 10.10.1. Hence the condition for $P$ to be projective really signifies that given a surjection of $R$-modules $N \to N'$ the map $\mathop{\mathrm{Hom}}\nolimits _ R(P, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(P, N')$ is surjective.

Lemma 10.77.2. Let $R$ be a ring. Let $P$ be an $R$-module. The following are equivalent

1. $P$ is projective,

2. $P$ is a direct summand of a free $R$-module, and

3. $\mathop{\mathrm{Ext}}\nolimits ^1_ R(P, M) = 0$ for every $R$-module $M$.

Proof. Assume $P$ is projective. Choose a surjection $\pi : F \to P$ where $F$ is a free $R$-module. As $P$ is projective there exists a $i \in \mathop{\mathrm{Hom}}\nolimits _ R(P, F)$ such that $\pi \circ i = \text{id}_ P$. In other words $F \cong \mathop{\mathrm{Ker}}(\pi ) \oplus i(P)$ and we see that $P$ is a direct summand of $F$.

Conversely, assume that $P \oplus Q = F$ is a free $R$-module. Note that the free module $F = \bigoplus _{i \in I} R$ is projective as $\mathop{\mathrm{Hom}}\nolimits _ R(F, M) = \prod _{i \in I} M$ and the functor $M \mapsto \prod _{i \in I} M$ is exact. Then $\mathop{\mathrm{Hom}}\nolimits _ R(F, -) = \mathop{\mathrm{Hom}}\nolimits _ R(P, -) \times \mathop{\mathrm{Hom}}\nolimits _ R(Q, -)$ as functors, hence both $P$ and $Q$ are projective.

Assume $P \oplus Q = F$ is a free $R$-module. Then we have a free resolution $F_\bullet$ of the form

$\ldots F \xrightarrow {a} F \xrightarrow {b} F \to P \to 0$

where the maps $a, b$ alternate and are equal to the projector onto $P$ and $Q$. Hence the complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , M)$ is split exact in degrees $\geq 1$, whence we see the vanishing in (3).

Assume $\mathop{\mathrm{Ext}}\nolimits ^1_ R(P, M) = 0$ for every $R$-module $M$. Pick a free resolution $F_\bullet \to P$. Set $M = \mathop{\mathrm{Im}}(F_1 \to F_0) = \mathop{\mathrm{Ker}}(F_0 \to P)$. Consider the element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ R(P, M)$ given by the class of the quotient map $\pi : F_1 \to M$. Since $\xi$ is zero there exists a map $s : F_0 \to M$ such that $\pi = s \circ (F_1 \to F_0)$. Clearly, this means that

$F_0 = \mathop{\mathrm{Ker}}(s) \oplus \mathop{\mathrm{Ker}}(F_0 \to P) = P \oplus \mathop{\mathrm{Ker}}(F_0 \to P)$

and we win. $\square$

Lemma 10.77.3. Let $R$ be a Noetherian ring. Let $P$ be a finite $R$-module. If $\mathop{\mathrm{Ext}}\nolimits ^1_ R(P, M) = 0$ for every finite $R$-module $M$, then $P$ is projective.

This lemma can be strengthened: There is a version for finitely presented $R$-modules if $R$ is not assumed Noetherian. There is a version with $M$ running through all finite length modules in the Noetherian case.

Proof. Choose a surjection $R^{\oplus n} \to P$ with kernel $M$. Since $\mathop{\mathrm{Ext}}\nolimits ^1_ R(P, M) = 0$ this surjection is split and we conclude by Lemma 10.77.2. $\square$

Proof. This is true by the characterization of projectives as direct summands of free modules in Lemma 10.77.2. $\square$

Lemma 10.77.5. Let $R$ be a ring. Let $I \subset R$ be a nilpotent ideal. Let $\overline{P}$ be a projective $R/I$-module. Then there exists a projective $R$-module $P$ such that $P/IP \cong \overline{P}$.

Proof. By Lemma 10.77.2 we can choose a set $A$ and a direct sum decomposition $\bigoplus _{\alpha \in A} R/I = \overline{P} \oplus \overline{K}$ for some $R/I$-module $\overline{K}$. Write $F = \bigoplus _{\alpha \in A} R$ for the free $R$-module on $A$. Choose a lift $p : F \to F$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$ of $\bigoplus _{\alpha \in A} R/I$. Note that $p^2 - p \in \text{End}_ R(F)$ is a nilpotent endomorphism of $F$ (as $I$ is nilpotent and the matrix entries of $p^2 - p$ are in $I$; more precisely, if $I^ n = 0$, then $(p^2 - p)^ n = 0$). Hence by Lemma 10.32.7 we can modify our choice of $p$ and assume that $p$ is a projector. Set $P = \mathop{\mathrm{Im}}(p)$. $\square$

Lemma 10.77.6. Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Let $\overline{P}$ be a finite projective $R/I$-module. Then there exists a finite projective $R$-module $P$ such that $P/IP \cong \overline{P}$.

Proof. Recall that $\overline{P}$ is a direct summand of a free $R/I$-module $\bigoplus _{\alpha \in A} R/I$ by Lemma 10.77.2. As $\overline{P}$ is finite, it follows that $\overline{P}$ is contained in $\bigoplus _{\alpha \in A'} R/I$ for some $A' \subset A$ finite. Hence we may assume we have a direct sum decomposition $(R/I)^{\oplus n} = \overline{P} \oplus \overline{K}$ for some $n$ and some $R/I$-module $\overline{K}$. Choose a lift $p \in \text{Mat}(n \times n, R)$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$ of $(R/I)^{\oplus n}$. Note that $p^2 - p \in \text{Mat}(n \times n, R)$ is nilpotent: as $I$ is locally nilpotent and the matrix entries $c_{ij}$ of $p^2 - p$ are in $I$ we have $c_{ij}^ t = 0$ for some $t > 0$ and then $(p^2 - p)^{tn^2} = 0$ (by looking at the matrix coefficients). Hence by Lemma 10.32.7 we can modify our choice of $p$ and assume that $p$ is a projector. Set $P = \mathop{\mathrm{Im}}(p)$. $\square$

Lemma 10.77.7. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

1. $I$ is nilpotent,

2. $M/IM$ is a projective $R/I$-module,

3. $M$ is a flat $R$-module.

Then $M$ is a projective $R$-module.

Proof. By Lemma 10.77.5 we can find a projective $R$-module $P$ and an isomorphism $P/IP \to M/IM$. We are going to show that $M$ is isomorphic to $P$ which will finish the proof. Because $P$ is projective we can lift the map $P \to P/IP \to M/IM$ to an $R$-module map $P \to M$ which is an isomorphism modulo $I$. Since $I^ n = 0$ for some $n$, we can use the filtrations

\begin{align*} 0 = I^ nM \subset I^{n - 1}M \subset \ldots \subset IM \subset M \\ 0 = I^ nP \subset I^{n - 1}P \subset \ldots \subset IP \subset P \end{align*}

to see that it suffices to show that the induced maps $I^ aP/I^{a + 1}P \to I^ aM/I^{a + 1}M$ are bijective. Since both $P$ and $M$ are flat $R$-modules we can identify this with the map

$I^ a/I^{a + 1} \otimes _{R/I} P/IP \longrightarrow I^ a/I^{a + 1} \otimes _{R/I} M/IM$

induced by $P \to M$. Since we chose $P \to M$ such that the induced map $P/IP \to M/IM$ is an isomorphism, we win. $\square$

## Comments (4)

Comment #545 by Paxton on

There is a small typo in the very last paragraph of Lemma 10.72.2 in the definition of $M$. It should read $M = im(F^1 \to F^0) = ker(F^0 \to P)$.

Comment #5583 by Tamás on

In the proof of 10.76.6 for injectivity I think you need that the maps $I^aP/I^{a+1}P\to I^aM/I^{a+1}M$ are isomorphisms for all $a$, not just that they are injective. This is what is proven, but then it immediately implies that $P\to M$ is an isomorphism by the snake lemma, without using the "Nakayama" argument at the beginning of the proof.

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