The Stacks project

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 $i \circ \pi = \text{id}_ P$. 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

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

and we win. $\square$

Proof. This is true by the characterization of projectives as direct summands of free modules in Lemma 10.76.2. $\square$

Proof. By Lemma 10.76.2 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 10.31.7 we can modify our choice of $p$ and assume that $p$ is a projector. Set $P = \mathop{\mathrm{Im}}(p)$. $\square$

Proof. Recall that $\overline{P}$ is a direct summand of a free $R/I$-module $\bigoplus _{\alpha \in A} R/I$ by Lemma 10.76.2. 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 10.31.7 we can modify our choice of $p$ and assume that $p$ is a projector. Set $P = \mathop{\mathrm{Im}}(p)$. $\square$

Proof. By Lemma 10.76.4 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 $I$. By Nakayama's Lemma 10.19.1 the map $P \to M$ is surjective. It remains to show that $P \to M$ is injective. Since $I^ n = 0$ for some $n$, we can use the filtrations

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 injective. Since both $P$ and $M$ are flat $R$-modules we can identify this with the map

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. $\square$