The Stacks project

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 . It should read .

Comment #5583 by Tamás on

In the proof of 10.76.6 for injectivity I think you need that the maps are isomorphisms for all , not just that they are injective. This is what is proven, but then it immediately implies that is an isomorphism by the snake lemma, without using the "Nakayama" argument at the beginning of the proof.

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05CD. Beware of the difference between the letter 'O' and the digit '0'.