Every finite-dimensional module has a projective cover, unique up to isomorphism over the target

Choose a finite $k$-basis $m_1,\dots ,m_r$ of $M$. It also generates $M$ as an $A$-module, so sending the standard generators to the $m_i$ gives a surjection $\epsilon :A^r\to M$. Among the direct summands $P$ of the finite-dimensional module $A^r$ for which $\epsilon {\mathrm{\mid }}_P:P\to M$ is surjective, choose one of minimal $k$-dimension; the family is nonempty because it contains $A^r$. Put $\pi :=\epsilon {\mathrm{\mid }}_P$, and write $j:P\to A^r$ and $s:A^r\to P$ for the inclusion and projection, with $sj=1_P$. For any surjection $q:E\to X$ and map $u:P\to X$, choose $y_i\in E$ with $q(y_i)=u(s(e_i))$ for the finitely many standard basis vectors $e_i$. These finite choices need no AC by [L1]. The formula $v({\sum }_i{a}_i{e}_i)={\sum }_i{a}_i{y}_i$ defines an $A$-linear map $v:A^r\to E$ with $qv=us$, so $q(vj)=usj=u$. Thus $P$ is projective by [F2], including when $r=0$.

Put $K=\ker \pi$, and suppose $N\le P$ satisfies $N+K=P$. Then $\pi {\mathrm{\mid }}_N:N\to M$ is surjective. Projectivity of $P$ supplies a map $h:P\to N$ with $(\pi {\mathrm{\mid }}_N)h=\pi$; write $f:P\to P$ for $h$ followed by the inclusion. Thus $\pi f=\pi$, and hence $\pi f^n=\pi$ for every $n\ge 1$.

Since $P$ is finite-dimensional, the kernels and images of the powers of $f$ stabilize. For a sufficiently large $n$, $P=\ker (f^n)\oplus \mathrm{im}(f^n):$ the intersection is zero because $f^n(x)=0$ for $x=f^n(y)$ implies $f^{2n}(y)=0$ and stabilization gives $f^n(y)=0$, while rank-nullity gives that the two dimensions sum to ${\dim }_{k}P$. The restriction of $\pi$ to $\mathrm{im}(f^n)$ is surjective because $\pi f^n=\pi$. Minimality of $P$ in step 1.1 therefore forces $\ker (f^n)=0$. Hence $f$ is injective and thus bijective, so $P=f(P)\subseteq N$. Therefore $N=P$, $K$ is superfluous, and $\pi$ is a projective cover.

Now let $\pi :P\to M$ and $\rho :Q\to M$ be projective covers. Both sources are finite-dimensional: lift a finite $k$-basis of $M$ to $P$, let $P_0$ be the submodule generated by those lifts, and observe that $P=P_0+\ker \pi$; superfluity gives $P=P_0$, which is finite-dimensional because $A$ is. The same argument applies to $Q$. Projectivity yields maps $f:P\to Q$ and $g:Q\to P$ with $\rho f=\pi$ and $\pi g=\rho$. Then $\pi (1_P-gf)=0$, so $(1_P-gf)(P)\subseteq \ker \pi$. Hence $P=gf(P)+\ker \pi$, and the superfluity of $\ker \pi$ gives $gf(P)=P$. Thus $gf$ is surjective, hence bijective on the finite-dimensional module $P$. The same argument shows that $fg$ is bijective on $Q$. Since $gf$ is bijective, $f$ is injective; since $fg$ is surjective, $f$ is surjective. Therefore $f$ is an isomorphism, and it still satisfies $\rho \circ f=\pi$.

Steps 1.1, 2.1, 3.1, and 4.1 prove existence and uniqueness up to isomorphism over the target.