Projective Cover

Definition

An epimorphism of modules $f:U\twoheadrightarrow V$ is called essential if no proper submodule of $U$ is mapped surjectively onto $V$ by $f$. Equivalently, whenever $g:W\to U$ is a map such that $fg$ is an epimorphism, then $g$ is an epimorphism.

Definition

A projective cover of a module $U$ is an essential epimorphism $P\stackrel{\tau }{\twoheadrightarrow }U$ where $P$ is projective.

Remark that a projective cover does not always exist. However, if it exists, then it is unique up to isomorphism. In addition, ^mjip80 gives a case where the projective cover exists.

Uniqueness

Proposition

Suppose that $f:P\to U$ is a projective cover of a module $U$ and $g:Q\to U$ is an epimorphism where $Q$ is projective. Then $Q=Q_1\oplus Q_2$ so that $g$ has the component $g=(g_1,0)$ with respect to the direct sum decomposition and $g_1:Q_1\to U$ satisfies that the diagram commutes

where $\gamma$ is an isomorphism.

If any exists, the projective covers of a module $U$ are all isomorphic, by isomorphisms that commute with the essential epimorphism.

\begin{proof} Since $Q$ is projective, there exists $\beta$ such that $g=f\circ \beta$. Since $P$ is projective, there exists $\alpha$ such that $f=g\circ \alpha$.

Then $f=f(\beta \alpha )$ yields that $\beta$ is an epimorphism. As $P$ is projective, $\beta$ splits and so $Q=Q_1\oplus Q_2$ with $Q_2=\ker \beta$. Thus, $\gamma =\beta {|}_{Q_1}$ is what we desire.

Furthermore, assume that $f:P\twoheadrightarrow U$ and $g:Q\twoheadrightarrow U$ are distinct projective covers, then by the argument above, one can check $Q=Q_1\oplus Q_2$ and $Q_1\simeq P$. Since $g$ is essential, $Q_2=0$ and so $P\simeq Q$. \end{proof}

Existence

For an Artinian ring $A$, recall that a simple $A$-module always has the projective cover (see here), $U/\mathrm{Rad}(U)$ is semisimple for all $A$-module $U$ (see here), and $U\to U/\mathrm{Rad}(U)$ is essential by Nakayama lemma. Then we can prove the following theorem.

Theorem

Let $A$ be an Artinian ring, and let $U$ be $A$-module. Then $U$ has a projective cover.

\begin{proof} We only prove for $U$ finitely generated.

Note that $U/\mathrm{Rad}U\simeq S_1\oplus \cdots \oplus S_n$ with $S_i$ simple, and then by ^koarlo $h:P\to U/\mathrm{Rad}U$ is a projective cover of $U/\mathrm{Rad}U$, where $P:=P_{S_1}\oplus \cdots \oplus P_{S_n}$. Now consider the natural surjection

$$g:U\twoheadrightarrow U/\mathrm{Rad}U.$$

Since $P$ is projective, there exists a lift $f:P\to U$ such that the following diagram commutes by ^9f21f9.

Since $h$ is a projective cover and $g$ is essential by Nakayama lemma, we conclude that $f:P\to U$ is a projective cover of $U$ by ^q7sonb. Therefore, $U$ admits a projective cover. \end{proof}

Remark. In particular, $U$ has the same projective cover as $U/\mathrm{Rad}U$.