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Tag 08XX

45.4. Projective covers

In this section we briefly discuss projective covers.

Definition 45.4.1. 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.

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 45.4.2. 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 $P = \mathop{\rm Ker}(\alpha) \oplus P'$. Since $P' \to M$ is surjective and since $P \to M$ is essential we conclude that $\mathop{\rm Ker}(\alpha)$ is zero as desired. $\square$

Here is an example where projective covers exist.

Lemma 45.4.3. 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 10.19.1). 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. $\square$

    The code snippet corresponding to this tag is a part of the file dualizing.tex and is located in lines 425–488 (see updates for more information).

    \section{Projective covers}
    \label{section-projective-cover}
    
    \noindent
    In this section we briefly discuss projective covers.
    
    \begin{definition}
    \label{definition-projective-cover}
    Let $R$ be a ring. A surjection $P \to M$ of $R$-modules is said
    to be a {\it projective cover}, or sometimes a {\it projective envelope},
    if $P$ is a projective $R$-module and $P \to M$ is an essential
    surjection.
    \end{definition}
    
    \noindent
    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.
    
    \begin{lemma}
    \label{lemma-projective-cover-unique}
    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.
    \end{lemma}
    
    \begin{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
    $P = \Ker(\alpha) \oplus P'$. Since $P' \to M$ is surjective
    and since $P \to M$ is essential we conclude that $\Ker(\alpha)$
    is zero as desired.
    \end{proof}
    
    \noindent
    Here is an example where projective covers exist.
    
    \begin{lemma}
    \label{lemma-projective-covers-local}
    Let $(R, \mathfrak m, \kappa)$ be a local ring. Any finite $R$-module has
    a projective cover.
    \end{lemma}
    
    \begin{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 \ref{algebra-lemma-NAK}). 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.
    \end{proof}

    Comments (2)

    Comment #2607 by Dario WeiƟmann on June 23, 2017 a 6:02 pm UTC

    Typo in lemma 45.4.3: $N\subset R^{\oplus r}$ instead of $N \subset R^{\oplus R}$.

    Comment #2630 by Johan (site) on July 7, 2017 a 12:45 pm UTC

    Thanks, fixed here.

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