Lemma 10.76.2. Let $R$ be a ring. Let $P$ be an $R$-module. The following are equivalent

$P$ is projective,

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

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

Lemma 10.76.2. Let $R$ be a ring. Let $P$ be an $R$-module. The following are equivalent

$P$ is projective,

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

$\mathop{\mathrm{Ext}}\nolimits ^1_ R(P, M) = 0$ 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. $\square$

Your email address will not be published.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.

## Comments (0)

There are also: