## Tag `05CF`

Chapter 10: Commutative Algebra > Section 10.76: Projective modules

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. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 17960–17969 (see updates for more information).

```
\begin{lemma}
\label{lemma-characterize-projective}
Let $R$ be a ring. Let $P$ be an $R$-module.
The following are equivalent
\begin{enumerate}
\item $P$ is projective,
\item $P$ is a direct summand of a free $R$-module, and
\item $\Ext^1_R(P, M) = 0$ for every $R$-module $M$.
\end{enumerate}
\end{lemma}
\begin{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
$i \in \Hom_R(P, F)$ such that $i \circ \pi = \text{id}_P$.
In other words $F \cong \Ker(\pi) \oplus i(P)$ and we see
that $P$ is a direct summand of $F$.
\medskip\noindent
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 $\Hom_R(F, M) = \prod_{i \in I} M$ and the functor
$M \mapsto \prod_{i \in I} M$ is exact.
Then $\Hom_R(F, -) = \Hom_R(P, -) \times \Hom_R(Q, -)$
as functors, hence both $P$ and $Q$ are projective.
\medskip\noindent
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 $\Hom_R(F_\bullet, M)$ is split
exact in degrees $\geq 1$, whence we see the vanishing in (3).
\medskip\noindent
Assume $\Ext^1_R(P, M) = 0$ for every $R$-module $M$.
Pick a free resolution $F_\bullet \to P$. Set
$M = \Im(F_1 \to F_0) = \Ker(F_0 \to P)$.
Consider the element $\xi \in \Ext^1_R(P, M)$ 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
$$
F_0 = \Ker(s) \oplus \Ker(F_0 \to P) =
P \oplus \Ker(F_0 \to P)
$$
and we win.
\end{proof}
```

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