Definition 47.4.1. Let $R$ be a ring. A surjection $P \to M$ of $R$-modules is said to be a projective cover, or sometimes a projective envelope, if $P$ is a projective $R$-module and $P \to M$ is an essential surjection.
47.4 Projective covers
In this section we briefly discuss projective covers.
Projective covers do not always exist. For example, if $k$ is a field and $R = k[x]$ is the polynomial ring over $k$, then the module $M = R/(x)$ does not have a projective cover. Namely, for any surjection $f : P \to M$ with $P$ projective over $R$, the proper submodule $(x - 1)P$ surjects onto $M$. Hence $f$ is not essential.
Lemma 47.4.2. Let $R$ be a ring and let $M$ be an $R$-module. If a projective cover of $M$ exists, then it is unique up to isomorphism.
Proof. Let $P \to M$ and $P' \to M$ be projective covers. Because $P$ is a projective $R$-module and $P' \to M$ is surjective, we can find an $R$-module map $\alpha : P \to P'$ compatible with the maps to $M$. Since $P' \to M$ is essential, we see that $\alpha $ is surjective. As $P'$ is a projective $R$-module we can choose a direct sum decomposition $P = \mathop{\mathrm{Ker}}(\alpha ) \oplus P'$. Since $P' \to M$ is surjective and since $P \to M$ is essential we conclude that $\mathop{\mathrm{Ker}}(\alpha )$ is zero as desired. $\square$
Here is an example where projective covers exist.
Lemma 47.4.3. Let $(R, \mathfrak m, \kappa )$ be a local ring. Any finite $R$-module has a projective cover.
Proof. Let $M$ be a finite $R$-module. Let $r = \dim _\kappa (M/\mathfrak m M)$. Choose $x_1, \ldots , x_ r \in M$ mapping to a basis of $M/\mathfrak m M$. Consider the map $f : R^{\oplus r} \to M$. By Nakayama's lemma this is a surjection (Algebra, Lemma 10.20.1). If $N \subset R^{\oplus r}$ is a proper submodule, then $N/\mathfrak m N \to \kappa ^{\oplus r}$ is not surjective (by Nakayama's lemma again) hence $N/\mathfrak m N \to M/\mathfrak m M$ is not surjective. Thus $f$ is an essential surjection. $\square$
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