[proofplan] We construct the resolution inductively by repeatedly presenting each kernel as a quotient of a free module. First choose a free module $F_0$ surjecting onto $M$; then choose a free module $F_1$ surjecting onto $\ker \varepsilon$; then repeat this construction on the successive kernels. Exactness follows because each differential is defined by composing a surjection onto the previous kernel with the kernel inclusion. Finally, [free modules are projective](/theorems/4201) by the basis-lifting property, so the same exact sequence is a projective resolution. [/proofplan] [step:Present $M$ as a quotient of a free left $R$-module] Let $S_0$ be the underlying set of $M$. Define $F_0$ to be the free left $R$-module on $S_0$, written \begin{align*} F_0 := R^{(S_0)}. \end{align*} Thus every element of $F_0$ is a finite formal sum $\sum_{s \in S_0} r_s e_s$, where $r_s \in R$, all but finitely many $r_s$ are zero, and $(e_s)_{s \in S_0}$ is the standard basis indexed by $S_0$. Define the $R$-[linear map](/page/Linear%20Map) \begin{align*} \varepsilon: F_0 &\to M \\ \sum_{s \in S_0} r_s e_s &\mapsto \sum_{s \in S_0} r_s s. \end{align*} The sum on the right is finite, so the map is well-defined. For every $m \in M$, the basis element $e_m \in F_0$ satisfies $\varepsilon(e_m) = m$, so $\varepsilon$ is surjective. Define \begin{align*} K_1 := \ker \varepsilon \subset F_0. \end{align*} [/step] [step:Construct the successive free modules and differentials] Assume that, for some integer $n \geq 1$, the left $R$-module $K_n$ has been defined as a submodule of $F_{n-1}$. Let $S_n$ be the underlying set of $K_n$, and define \begin{align*} F_n := R^{(S_n)} \end{align*} to be the free left $R$-module on $S_n$. Define the $R$-linear map \begin{align*} \pi_n: F_n &\to K_n \\ \sum_{s \in S_n} r_s e_s &\mapsto \sum_{s \in S_n} r_s s. \end{align*} As before, $\pi_n$ is surjective because $\pi_n(e_k) = k$ for every $k \in K_n$. Let \begin{align*} \iota_n: K_n &\to F_{n-1} \end{align*} be the inclusion map. Define the differential \begin{align*} d_n: F_n &\to F_{n-1} \\ x &\mapsto \iota_n(\pi_n(x)). \end{align*} Finally define \begin{align*} K_{n+1} := \ker d_n \subset F_n. \end{align*} By induction, this constructs free left $R$-modules $F_n$ for all integers $n \geq 0$ and $R$-linear maps $d_n: F_n \to F_{n-1}$ for all integers $n \geq 1$. [/step] [step:Verify that consecutive differentials compose to zero] We first check the composition at $F_1$. Since $d_1(F_1) = \iota_1(\pi_1(F_1)) = K_1 = \ker \varepsilon$, we have \begin{align*} \varepsilon \circ d_1 = 0. \end{align*} Now let $n \geq 1$. By construction, \begin{align*} K_{n+1} = \ker d_n. \end{align*} Since $d_{n+1}$ is defined as the composite of the surjection $\pi_{n+1}:F_{n+1}\to K_{n+1}$ with the inclusion $K_{n+1}\subset F_n$, its image lies in $K_{n+1}$. Therefore \begin{align*} \operatorname{im} d_{n+1} \subseteq K_{n+1} = \ker d_n, \end{align*} and hence \begin{align*} d_n \circ d_{n+1} = 0. \end{align*} [/step] [step:Verify exactness at every term of the augmented sequence] Exactness at $M$ means that $\varepsilon:F_0\to M$ is surjective, which was proved in the first step. Exactness at $F_0$ means \begin{align*} \operatorname{im} d_1 = \ker \varepsilon. \end{align*} By construction, $d_1=\iota_1\circ \pi_1$, where $\pi_1:F_1\to K_1$ is surjective and $\iota_1:K_1\to F_0$ is the inclusion. Hence \begin{align*} \operatorname{im} d_1 = K_1 = \ker \varepsilon. \end{align*} Now fix an integer $n \geq 1$. Exactness at $F_n$ means \begin{align*} \operatorname{im} d_{n+1} = \ker d_n. \end{align*} By definition, $K_{n+1}=\ker d_n$. Also $d_{n+1}=\iota_{n+1}\circ \pi_{n+1}$, where $\pi_{n+1}:F_{n+1}\to K_{n+1}$ is surjective and $\iota_{n+1}:K_{n+1}\to F_n$ is the inclusion. Therefore \begin{align*} \operatorname{im} d_{n+1} = K_{n+1} = \ker d_n. \end{align*} Thus the augmented sequence is exact at every term. [/step] [step:Show that the free resolution is also a projective resolution] It remains to justify the final assertion. Let $F$ be a free left $R$-module with basis $(e_s)_{s \in S}$ for some set $S$. To prove that $F$ is projective, let $A$ and $B$ be left $R$-modules, let $q:A\to B$ be a surjective $R$-linear map, and let $f:F\to B$ be an $R$-linear map. For each $s\in S$, choose an element $a_s\in A$ such that \begin{align*} q(a_s)=f(e_s), \end{align*} which is possible because $q$ is surjective. By the universal property of the free module $F$, there is a unique $R$-linear map \begin{align*} \widetilde f:F&\to A \end{align*} such that $\widetilde f(e_s)=a_s$ for every $s\in S$. For every basis element $e_s$, we have \begin{align*} (q\circ \widetilde f)(e_s)=q(a_s)=f(e_s). \end{align*} Since two $R$-linear maps out of a free module agree when they agree on a basis, it follows that \begin{align*} q\circ \widetilde f=f. \end{align*} Thus $F$ is projective. Each $F_n$ in the exact augmented sequence constructed above is free, hence projective. Therefore the same augmented exact sequence is a projective resolution of $M$. [/step]