Lemma 10.76.2. 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 $i \circ \pi = \text{id}_ P$. 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

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

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. $\square$

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