Corollary 2

Let $\mathfrak{a}$ be a two-sided ideal of A and M an A-module such that the $(A/\mathfrak{a})$-module $M/\mathfrak{a}M$ is free. Suppose that one of the following conditions is satisfied:

(i) The module M is finitely generated, and $\mathfrak{a}$ is contained in the radical of A.

(ii) The ideal $\mathfrak{a}$ is nilpotent.

Then M has a projective cover.

More precisely, let P be a free A-module, $(e_i)_{i\in I}$ a basis of P, and let $u: P\rightarrow M$ be a homomorphism such that the canonical images of the elements $u(e_i)$ of $M/\mathfrak{a}M$ form a basis of the $(A/\mathfrak{a})$-module $M/\mathfrak{a}M$. Then $(P, u)$ is a projective cover of M.

The $(A/\mathfrak{a}$)-module $P/\mathfrak{a}P$ is free, and the homomorphism $\overline{u}$ from $P/\mathfrak{a}P$ to $M/\mathfrak{a}M$ deduced from $u$ by passing to the quotients transforms a basis of $P/\mathfrak{a}P$ into a basis of $M/\mathfrak{a}M$, hence is bijective.

If $\mathfrak{a}$ is nilpotent, then it suffices to apply Proposition 10. Now suppose that the ring A is nonzero and that the module M is finitely generated; then so is the $(A/\mathfrak{a}$)-module $M/\mathfrak{a}M$, and consequently also the $(A/\mathfrak{a}$)-module $P/\mathfrak{a}P$. Every basis of $P/\mathfrak{a}P$ is then finite. It follows that the set I is finite and that the A-module P is finitely generated. We then apply Proposition 10 again.