The stable ring of a ring and the Mac Lane resolution

Written by Claude Opus 5.5 (Anthropic) on 21 October 2026. Self-checked by the writing AI. Public domain (CC0).

The stabilization of the free abelian group functor, applied to a ring $A$ of a topos, is a simplicial abelian sheaf $\mathbb Z^{\mathrm{st}}(A)$ with homology $A$ in degree $0$ and none in degree $1$ (Stable derived functors, Example 2.1). This lesson shows that it is a simplicial ring, flat over $\mathbb Z$, and that its augmentation to $A$ is a ring homomorphism. The ring structure comes from the spectrum $A(1)^\times$ of Graded rings and the spectra of modules: the free abelian group functor turns its multiplication into the multiplication of a ring of type $C_\bullet(S_1)$, and the double diagonal of Diagonal rings and adjunction formulas turns that into $\mathbb Z^{\mathrm{st}}(A)$. An $A$-module $M$ gives a $\mathbb Z^{\mathrm{st}}(A)$-module $\mathbb Z^{\mathrm{st}}(M)$.

Two applications follow. Truncating the normalized ring $N\mathbb Z^{\mathrm{st}}(A)$ gives a flat presentation of $A$, which computes the derived tensor product $A\otimes^{\mathbf L}_{\mathbb Z}B$ of two rings. And Mac Lane's theorem says that $A\otimes^{\mathbf L}_{N\mathbb Z^{\mathrm{st}}(A)}N\mathbb Z^{\mathrm{st}}(M)$ is $M$. Combined with the bar construction, it gives a resolution of $M$, functorial in $M$, whose terms are sums of direct summands of the free modules $A(A^p\times M^q)$ on products of copies of $A$ and $M$.

Prerequisites. Diagonal rings and adjunction formulas (Construction 2.1, Lemma 2.2, Proposition 2.3); Graded rings and the spectra of modules (Definitions 1.2 and 3.2, Propositions 2.1 and 2.2); Stabilization of non-additive functors (Definition 3.2, Lemma 3.4, Corollary 3.7); Stable derived functors (Proposition 1.4 and Example 2.1); Differential graded modules in a topos (Propositions 1.3, 2.4–2.6, Lemmas 2.1, 2.2, 3.1 and 3.2); Simplicial rings and modules in a topos, Proposition 2.1; Derived tensor products of simplicial modules (Lemma 1.1, Lemma 5.1, Proposition 5.2); Non-additive functors of complexes in a ringed topos (Section 3 and Construction 4.3); local inductive limits, Section 5 of Homotopy sheaves and the free abelian Whitehead theorem; Theorems 1.1–1.3 of The cotangent complex. From the AI-integrated Stacks project: bounded above complexes of flat modules are K-flat [Stacks, Tag 06YQ]; the standard simplicial resolution of an adjunction [Tag 08NC].

Conventions

$T$ is the topos of sheaves on a site. Rings are associative and unital, not necessarily commutative, and modules are left modules. For a sheaf of sets $X$, $\mathbb Z(X)$ is the free abelian sheaf on $X$, the sheafification of $U\mapsto\mathbb Z[X(U)]$, and $R(X)=R\otimes_{\mathbb Z}\mathbb Z(X)$ the free $R$-module on $X$ for a ring $R$. Since sheafification commutes with finite limits and with tensor products, $\mathbb Z(X\times Y)=\mathbb Z(X)\otimes\mathbb Z(Y)$, and $\mathbb Z(X)\otimes F$ is the sheafification of $U\mapsto\bigoplus_{X(U)}F(U)$, which is exact in $F$. So $\mathbb Z(X)$ is flat. A morphism $X\times Y\to Z$ of sheaves of sets therefore induces a map $\mathbb Z(X)\otimes\mathbb Z(Y)\to\mathbb Z(Z)$, and a monoid $G$ of $T$ gives the ring $\mathbb Z(G)$, with $[g][h]=[gh]$.

Differential graded rings, their modules, derived categories, the derived tensor products $\otimes^{\mathbf L}$ and the category $D(\mathbf{Andg}(T))$ of differential graded rings in nonpositive degrees localized at the quasi-isomorphisms are those of the lesson on differential graded modules; left modules over a ring are modules over its opposite. For a simplicial ring $R$, $NR$ is the normalized differential graded ring, placed in degrees $\leq0$, and $NE$ the normalized module of an $R$-module $E$ (Proposition 2.1 of the lesson on simplicial rings, whose proof uses commutativity only for graded commutativity). For a differential graded ring $P$ in nonpositive degrees, $t_1(P)=(P^{-1}/dP^{-2}\to P^0)$ is its truncated ring (Proposition 1.3 of the lesson on differential graded modules).

1. Derived tensor products of rings

Lemma 1.1 (presentations). Let $S$ be a ring, $q\colon Q\to S$ a surjective ring homomorphism with $Q$ flat over $\mathbb Z$, and $J=\ker q$. Then $J$ is flat over $\mathbb Z$, and the complex $Q_J=(J\to Q)$, with $J$ in degree $-1$, $Q$ in degree $0$ and the inclusion as differential, is a differential graded ring in nonpositive degrees for the multiplication of $Q$ on $Q\times Q$, $Q\times J$ and $J\times Q$ and zero on $J\times J$. The projection $Q_J\to S$ is a quasi-isomorphism of differential graded rings.

Proof. An abelian sheaf whose groups of sections have no torsion is flat: its tensor product with a sheaf $G$ is the sheafification of $U\mapsto F(U)\otimes G(U)$, which is exact in $G$ because torsion-free abelian groups are flat, and sheafification is exact. Conversely, the sections of a flat sheaf have no torsion, since multiplication by $n$ on $\mathbb Z$ is injective. So $Q$ has torsion-free sections, hence so has its subsheaf $J$, and $J$ is flat. The products lie in the stated terms because $J$ is a two-sided ideal, and associativity holds because it holds in $Q$. The Leibniz rule needs checking only on $J\times J$, $J\times Q$ and $Q\times J$: for $x,y\in J$, $d(xy)=0$ and $(dx)y-x(dy)=xy-xy=0$; for $x\in J$ and $z\in Q$, $d(xz)=xz=(dx)z$ and $d(zx)=zx=z(dx)$. The cohomology of $Q_J$ is $Q/J=S$ in degree $0$ and $0$ in degree $-1$. $\square$

Proposition 1.2. Let $R$ and $S$ be rings, and $Q\to S$ a presentation as in Lemma 1.1. Put $R\otimes^{\mathbf L}_{\mathbb Z}S=R\otimes_{\mathbb Z}Q_J,$ a differential graded ring in degrees $-1$ and $0$, with the homomorphism $R\to R\otimes_{\mathbb Z}Q_J$, $r\mapsto r\otimes1$.

Up to isomorphism in $D(\mathbf{Andg}(T))$ under $R$, it does not depend on the presentation. In $D(\mathbb Z)$ it is the derived tensor product of $R$ and $S$.
If $R$ and $S$ are commutative, it is isomorphic under $R$, in $D(\mathbf{Andg}(T))$, to the normalization of the derived tensor product $R\otimes^\ell_{\mathbb Z}S$ of Proposition 5.2 of the lesson on derived tensor products.

The same construction applies with the presented ring on the left; by Proposition 2.6 of the lesson on differential graded modules, its derived category is defined up to equivalence.

Proof. (1) Let $Q'\to S$ be a second presentation. The fibre product $Q''=Q\times_SQ'$ is a ring with a surjection onto $S$, and it is flat over $\mathbb Z$, being a subsheaf of $Q\oplus Q'$. The projections to $Q$ and $Q'$ induce homomorphisms of differential graded rings $Q''_{J''}\to Q_J$ and $Q''_{J''}\to Q'_{J'}$, which are quasi-isomorphisms because all three resolve $S$. The three complexes are bounded complexes of flat sheaves, hence K-flat [Tag 06YQ], so $R\otimes-$ preserves these quasi-isomorphisms (Lemma 2.2 of the lesson on differential graded modules), under $R$. For the same reason $R\otimes Q_J$ computes $R\otimes^{\mathbf L}_{\mathbb Z}S$ in $D(\mathbb Z)$.

(2) Let $P=P^{\mathrm{alg}}(S)\to S$ be the resolution of Lemma 5.1 of the lesson on derived tensor products, over the constant ring $\mathbb Z$: each $P_n$ is a polynomial ring, flat over $\mathbb Z$, and $H_0P=S$, $H_1P=0$. So $P_0\to S$ is a presentation, and $t_1(NP)=(N_1P/dN_2P\to P_0)$ is $Q_J$ for $Q=P_0$: the differential is injective because $H_1P=0$, with image the kernel of $P_0\to S$. The projection $NP\to t_1(NP)$ is a homomorphism of differential graded rings and a quasi-isomorphism, both complexes resolving $S$. Tensoring with $R$, which commutes with normalization because $R$ is constant, gives quasi-isomorphisms of differential graded rings $N(R\otimes_{\mathbb Z}P)\to R\otimes t_1(NP)=R\otimes Q_J$ under $R$. By Proposition 5.2 of the lesson on derived tensor products, $R\otimes_{\mathbb Z}P$ is $R\otimes^\ell_{\mathbb Z}S$. $\square$

A homomorphism $\varphi\colon P\to P'$ of differential graded rings in nonpositive degrees, or more generally a morphism of $D(\mathbf{Andg}(T))$, gives adjoint functors between $D(P)$ and $D(P')$, equivalences when $\varphi$ is a quasi-isomorphism (Propositions 2.5 and 2.6 of the lesson on differential graded modules). This applies in particular to $R\to R\otimes^{\mathbf L}_{\mathbb Z}S$.

2. The stable ring of a ring

Construction 2.1. Let $A$ be a ring, and $A(1)^\times$ its spectrum, an $S_1$-graded monoid of $T$ (Definition 3.2 of the lesson on graded rings). Applying $\mathbb Z(-)$ componentwise gives the $S_1$-graded ring $\mathbb Z(A(1)^\times)$, with the multiplications $\mathbb Z(A\langle m\rangle(p))\otimes\mathbb Z(A\langle n\rangle(p'))\to$ $\mathbb Z(A\langle m+n\rangle(p,p'))$ induced by those of $A(1)^\times$. Let $Q_A=\mathbb Z\bigl(C_\bullet(A(1)^\times)\bigr)=C_\bullet\bigl(\mathbb Z(A(1)^\times)\bigr),$ a ring of type $C_\bullet(S_1)$. By Propositions 2.1 and 2.2 of that lesson, $Q_A^{\mathrm{red}}(-1)=C_\bullet(\mathbb Z(A(1)^\times)^{\mathrm{red}}(-1))$ is a ring of type $C_\bullet(S_0)$. Put $\mathbb Z^{\mathrm{st}}(A)=\Delta^2Q_A^{\mathrm{red}}(-1),$ a simplicial ring (Construction 2.1 of the lesson on diagonal rings).

By Lemma 2.2 of the lesson on diagonal rings, $\mathbb Z^{\mathrm{st}}(A)=\Delta C_\bullet(\Delta X)$ with $X=\mathbb Z(A(1)^\times)^{\mathrm{red}}(-1)$. The diagram $\Delta X$ of $\bar\Delta'$ is the diagram $D_F(A)$ of Definition 3.2 of the lesson on stabilization for the functor $F=\mathbb Z(-)$ and the constant simplicial abelian sheaf $A$, and $\Delta C_\bullet(D_F(A))$ is its stabilization $F^{\mathrm{st}}(A)$. So the underlying simplicial abelian sheaf of $\mathbb Z^{\mathrm{st}}(A)$ is the stabilization of $\mathbb Z(-)$ at $A$, whose homology is $A$ in degree $0$ and $0$ in degree $1$ (Example 2.1 (3) and (4) of the lesson on stable derived functors). For a simplicial ring $A$, $\mathbb Z^{\mathrm{st}}(A)$ is the diagonal of $q\mapsto\mathbb Z^{\mathrm{st}}(A_q)$, again a simplicial ring.

Proposition 2.2.

The terms of $\mathbb Z^{\mathrm{st}}(A)$ are direct sums of direct summands of abelian sheaves $\mathbb Z(A^k)$; in particular they are flat. In degree $0$, $\begin{aligned}&\mathbb Z^{\mathrm{st}}(A)_0\\&\quad =\mathbb Z(A)\oplus I(A)\oplus I(A)\oplus\cdots,\end{aligned}$ the sum over $k\geq0$ of the values at the ordinals $[k-1]$, where $I(A)$ is the kernel of the augmentation $\mathbb Z(A)\to\mathbb Z$, $[a]\mapsto1$. The summand of $[k-1]$ times that of $[l-1]$ lies in that of $[k+l-1]$, through the multiplication $[a][b]=[ab]$.
Let $s\colon\mathbb Z(A)\to A$, $[a]\mapsto a$, a ring homomorphism. The stabilization of the natural transformation $s\colon\mathbb Z(M)\to M$, followed by the canonical quasi-isomorphism $\operatorname{id}^{\mathrm{st}}(A)\to A$ of Corollary 3.7 of the lesson on stabilization, is a homomorphism of simplicial rings $\varepsilon\colon\mathbb Z^{\mathrm{st}}(A)\to A,$ the augmentation. In degree $0$ it is $s$ on every summand of (1). It induces an isomorphism $H_0\mathbb Z^{\mathrm{st}}(A)\cong A$.

Proof. (1) The term of $\Delta^2Q_A^{\mathrm{red}}(-1)$ in degree $r$ is the sum, over the $r$-simplices $\sigma$ of the nerve of $\bar\Delta'$, of the values at $(r,\ldots,r)$ of $\mathbb Z(A\langle k\rangle)^{\mathrm{red}}[-\mathbf 1]=$ $\Gamma\bigl((N\mathbb Z(A\langle k\rangle))_{\geq\mathbf 1}[-\mathbf 1]\bigr)$, where $k$ is the size of the first vertex of $\sigma$. The terms of the normalization are direct summands of the values $\mathbb Z(A\langle k\rangle(p))=\mathbb Z(A^{p_0\cdots p_{k-1}})$ (Theorem 1.1 of The cotangent complex), and $\Gamma$ takes finite sums of them. In degree $0$, the value for $k=0$ is $\mathbb Z(A)$, and for $k\geq1$ it is the normalized part of $\mathbb Z(A\langle k\rangle)$ in multidegree $\mathbf 1$: the intersection of the kernels of the faces $d_1$ in each coordinate, which go to $\mathbb Z(A\langle k\rangle\text{ at a vertex})=\mathbb Z(\mathrm{pt})=\mathbb Z$. This is $I(A)$. The multiplication is that of $Q_A$, on the summands of $[k-1]$ and $[l-1]$ of $N_0(\bar\Delta')$.

(2) The map $s\colon\mathbb Z(A\langle n\rangle(p))\to A\langle n\rangle(p)$ is additive and, by the formula for the multiplication of spectra, multiplicative from the ring $\mathbb Z(A(1)^\times)$ to the ring $A(1)$ (Remark 3.3 of the lesson on graded rings). So it is a homomorphism of $S_1$-graded rings, and $\Delta^2C_\bullet((-)^{\mathrm{red}}(-1))$ of it is a homomorphism of simplicial rings $\mathbb Z^{\mathrm{st}}(A)\to\Delta^2C_\bullet(A(1)^{\mathrm{red}}(-1))$. Here $A(1)^{\mathrm{red}}(-1)=cA$, the trivial $S_0$-graded ring, and $\Delta^2C_\bullet(cA)$ is the simplicial ring $A\otimes\mathbb Z(N(\bar\Delta'))$, the ring of the simplicial monoid $N(\bar\Delta')$ under the sum of strings, with coefficients in $A$ (Proposition 3.4 (2) of the lesson on diagonal rings). Its augmentation to $A$, induced by $\sigma\mapsto1$, is a homomorphism of simplicial rings. On underlying simplicial abelian sheaves this composite is the stabilization of $s$ followed by the augmentation of the simplicial replacement of the constant diagram $A$, which is the map of Corollary 3.7 of the lesson on stabilization. In degree $0$ the composite is $s$ on the summand of each ordinal. On $H_0$ it is the isomorphism of Example 2.1 (3) of the lesson on stable derived functors. $\square$

Remark 2.3. The projection of $\mathbb Z^{\mathrm{st}}(A)_0$ onto the summand $\mathbb Z(A)$ of the empty ordinal, followed by $s$, is a ring homomorphism, but it is not compatible with the faces. Let $\delta\colon[-1]\to[0]$ be the $1$-simplex from the empty ordinal to $[0]$, and $x=[1]$ in the summand of $\delta$, with value $\mathbb Z(A)$ constant in the simplicial direction. Then $d_1x=[1]$ in the summand of $[-1]$, and $d_0x$ is the image of $[1]$ under the structure map $\mathbb Z(A)\to I(A)$, $[a]\mapsto[a]-[0]$, in the summand of $[0]$ (Lemma 3.4 of the lesson on stabilization). The projection followed by $s$ gives $1$ and $0$, while $\varepsilon$ gives $1$ and $1$. So the augmentation in degree $0$ is $s$ on all summands, not only on the first.

Proposition 2.4. Let $A$ and $B$ be rings.

$t_1(N\mathbb Z^{\mathrm{st}}(A))\to A$, induced by $\varepsilon$, is $Q_J$ for the presentation $\varepsilon_0\colon\mathbb Z^{\mathrm{st}}(A)_0\to A$ of Lemma 1.1.
Let $B^{\mathrm{st}}(A)=\mathbb Z^{\mathrm{st}}(A)\otimes_{\mathbb Z}B$, a simplicial ring. There is an isomorphism, in $D(\mathbf{Andg}(T))$ and under $B$, $A\otimes^{\mathbf L}_{\mathbb Z}B\cong t_1\bigl(NB^{\mathrm{st}}(A)\bigr),$ and so a homomorphism $NB^{\mathrm{st}}(A)\to A\otimes^{\mathbf L}_{\mathbb Z}B$, the projection onto $t_1$.
For every $n$, restriction of scalars along this homomorphism induces an equivalence between the full subcategories of $D(A\otimes^{\mathbf L}_{\mathbb Z}B)$ and of $D(NB^{\mathrm{st}}(A))$ formed by the objects with cohomology in $[n,n+1]$.

Proof. (1) The term $\mathbb Z^{\mathrm{st}}(A)_0=N_0\mathbb Z^{\mathrm{st}}(A)$ is flat, and $\varepsilon_0$ is surjective because $s\colon\mathbb Z(A)\to A$ is. Since $H_1\mathbb Z^{\mathrm{st}}(A)=0$, the differential $N_1/dN_2\to N_0$ is injective; its image is the kernel of $\varepsilon_0$, because $H_0(\varepsilon)$ is an isomorphism. The multiplication of $t_1$ is that of $Q_J$, since products of two terms of degree $-1$ vanish in $t_1$.

(2) By (1) and Proposition 1.2, $A\otimes^{\mathbf L}_{\mathbb Z}B=t_1(N\mathbb Z^{\mathrm{st}}(A))\otimes_{\mathbb Z}B$. Tensoring with $B$ is right exact, so it commutes with the quotient defining $t_1$, and it commutes with normalization, which is the image of an idempotent made of faces and degeneracies. Hence this is $t_1(N(\mathbb Z^{\mathrm{st}}(A)\otimes B))=t_1(NB^{\mathrm{st}}(A))$, as differential graded rings.

(3) This is Proposition 1.3 of the lesson on differential graded modules for $n=1$, together with Proposition 2.6 there for the isomorphism of (2). $\square$

3. Modules and Mac Lane's theorem

Construction 3.1. Let $M$ be an $A$-module. Its spectrum $M(1)^\times$ is an $S_1$-graded $A(1)^\times$-object, so $X_M=\mathbb Z(C_\bullet(M(1)^\times))=C_\bullet(\mathbb Z(M(1)^\times))$ is a $Q_A$-module of type $C_\bullet(S_1)$, and $\mathbb Z^{\mathrm{st}}(M)=\Delta^2X_M^{\mathrm{red}}(-1)$ is a $\mathbb Z^{\mathrm{st}}(A)$-module, whose underlying simplicial abelian sheaf is the stabilization of $\mathbb Z(-)$ at $M$, as in Construction 2.1. Its terms are direct sums of direct summands of sheaves $\mathbb Z(M^k)$, hence flat. The stabilization of $s\colon\mathbb Z(M)\to M$, followed by the quasi-isomorphism of Corollary 3.7 of the lesson on stabilization, is a map of modules $\varepsilon_M\colon\mathbb Z^{\mathrm{st}}(M)\to M$ over $\varepsilon$, by the argument of Proposition 2.2 (2). Everything is functorial in $M$. For a simplicial ring $A$ and a simplicial $A$-module $M$, the same construction, applied in each degree and followed by the diagonal, makes $\mathbb Z^{\mathrm{st}}(M)$ a $\mathbb Z^{\mathrm{st}}(A)$-module with an augmentation over $\varepsilon$.

Let $B_A=N\mathbb Z^{\mathrm{st}}(A)$, with the homomorphism $\varepsilon\colon B_A\to A$ of differential graded rings, $A$ being concentrated in degree $0$. The map $N\varepsilon_M\colon N\mathbb Z^{\mathrm{st}}(M)\to M$ is $B_A$-linear, $M$ being a $B_A$-module through $\varepsilon$. By the adjunction between extension and restriction of scalars along $\varepsilon$ (Proposition 2.5 of the lesson on differential graded modules), it corresponds to a morphism $c_M\colon A\otimes^{\mathbf L}_{B_A}N\mathbb Z^{\mathrm{st}}(M)\to M$ of $D(A)$, the derived category of $A$-modules, natural in $M$.

Theorem 3.2 (Mac Lane). For every $A$-module $M$, $c_M$ is an isomorphism.

Proof. Restriction of scalars from $D(A)$ to $D(\mathbb Z)$ detects isomorphisms, so it suffices to show that $c_M$ is a quasi-isomorphism. Choose a quasi-isomorphism $P'\to A$ of right $B_A$-modules from a K-flat module (Lemma 2.1 of the lesson on differential graded modules). For every left $B_A$-module $N$, $A\otimes^{\mathbf L}_{B_A}N$ is represented by $P'\otimes_{B_A}N$ (Lemma 2.2 there), and $c_M$ by the composite $P'\otimes_{B_A}N\mathbb Z^{\mathrm{st}}(M)\to$ $A\otimes_{B_A}N\mathbb Z^{\mathrm{st}}(M)\to$ $M$.

Step 1: $M=A$. Here $\mathbb Z^{\mathrm{st}}(M)$ is $\mathbb Z^{\mathrm{st}}(A)$ as a module over itself, $A\otimes_{B_A}B_A=A$, and $c_A$ is the identity of $A$.

Step 2: $M=A^r$. The inclusions of the summands induce a $\mathbb Z^{\mathrm{st}}(A)$-linear map $\mathbb Z^{\mathrm{st}}(A)^{\oplus r}\to\mathbb Z^{\mathrm{st}}(A^r)$, compatible with the augmentations. It is a quasi-isomorphism by the additivity of stable derived functors (Proposition 1.4 of the lesson on stable derived functors), since $\mathbb Z^{\mathrm{st}}$ is its own stable derived functor (Example 2.1 (1) there). So $c_{A^r}$ is a quasi-isomorphism by Step 1.

Step 3: local inductive limits. Let $M=\varinjlim_LM_i$ be the local inductive limit of a local system of modules, $M_i$ being a module over $T/U$ for $i\in L(U)$. All constructions commute with restriction to $T/U$. The functor $\mathbb Z^{\mathrm{st}}$ commutes with local inductive limits (proof of Example 2.1 (4) of the lesson on stable derived functors), and so do normalization and $P'\otimes_{B_A}-$. Local inductive limits are exact and commute with homology (Section 3 of the lesson on non-additive functors). So if every $c_{M_i}$ is a quasi-isomorphism, so is $c_M$.

Step 4: $M=A(X)$ for a sheaf of sets $X$. The abelian sheaf $\mathbb Z(X)$ is flat, so by the theorem of Deligne and Lazard (Theorem 3.2 of the lesson on non-additive functors, for the ring $\mathbb Z$) it is the local inductive limit of a local system of finite free abelian sheaves $\mathbb Z_U^{r}$. Tensoring with $A$ commutes with local inductive limits, so $A(X)=A\otimes\mathbb Z(X)$ is the local inductive limit of the modules $A_U^r$. Steps 2 and 3, over each $T/U$, give the claim.

Step 5: general $M$. Let $P\to M$ be the standard resolution of Construction 4.3 of the lesson on non-additive functors: a simplicial $A$-module with $P_q=A(X_q)$ for sheaves of sets $X_q$, whose augmentation to the constant simplicial module $M$ is a quasi-isomorphism. Since $\mathbb Z^{\mathrm{st}}$ carries quasi-isomorphisms to quasi-isomorphisms (Example 2.1 (1) of the lesson on stable derived functors), $\mathbb Z^{\mathrm{st}}(P)\to\mathbb Z^{\mathrm{st}}(M)$ is a quasi-isomorphism of $\mathbb Z^{\mathrm{st}}(A)$-modules, and by naturality it suffices to show that $c_P\colon P'\otimes_{B_A}N\mathbb Z^{\mathrm{st}}(P)\to NP$ is a quasi-isomorphism. Write $\mathbb Z^{\mathrm{st}}(P)$ as the diagonal of the bisimplicial module $W$ with $W_{q,\bullet}=\mathbb Z^{\mathrm{st}}(P_q)$, on which $\mathbb Z^{\mathrm{st}}(A)$ acts in the second index and is constant in the first. The shuffle map $\operatorname{Tot}NW\to N\operatorname{diag}W$ is a quasi-isomorphism (Lemma 2.6 of the lesson on multisimplicial objects), and it is $B_A$-linear by Lemma 3.1 of the lesson on differential graded modules. Tensoring with the K-flat module $P'$ preserves it. The complex $P'\otimes_{B_A}\operatorname{Tot}NW$ is the total complex of the double complex with columns $P'\otimes_{B_A}N\mathbb Z^{\mathrm{st}}(P_q)$, normalized in $q$, and its map to $\operatorname{Tot}$ of the columns $P_q$, which is $NP$, is given by the maps $c_{P_q}$. These are quasi-isomorphisms by Step 4, so the map of total complexes is one, by the row filtration in $q$. Composed with the shuffle map, it is $c_P$, because the shuffle map of a bisimplicial object constant in one index is the identity on normalizations. $\square$

4. Bar resolutions and the Mac Lane resolution

Proposition 4.1 (bar resolutions). Let $K\to R$ be a homomorphism of rings and $X$ an $R$-module. Let $\operatorname{Bar}(R/K,X)$ be the simplicial $R$-module with $\operatorname{Bar}_n=R^{\otimes(n+1)}\otimes X$, tensor products over $K$, with $R$ acting on the first factor, faces $\begin{aligned}&d_i(r_1\otimes\cdots\otimes r_{n+1}\otimes x)\\ &\quad =r_1\otimes\cdots\otimes r_{i+1}r_{i+2}\otimes\cdots\otimes x,\\ &d_n(r_1\otimes\cdots\otimes r_{n+1}\otimes x)\\ &\quad =r_1\otimes\cdots\otimes r_n\otimes r_{n+1}x,\end{aligned}$ for $0\leq i<n$, degeneracies inserting $1$ after $r_{i+1}$, and augmentation $\operatorname{Bar}_0=R\otimes X\to X$, $r\otimes x\mapsto rx$.

The augmented simplicial object is $K$-linearly contractible: $h(y)=1\otimes y$ satisfies $d_0h=\mathrm{id}$ and $d_{i+1}h=hd_i$, so its normalized complex, augmented to $X$, is contractible over $K$.
If $R$ and $X$ are flat over $K$, then every $\operatorname{Bar}_n$ is a flat $R$-module, and $\operatorname{Bar}(R/K,X)\to X$ is a resolution of $X$ by flat $R$-modules.

The same holds for a homomorphism of simplicial rings and a simplicial module, in each simplicial degree.

Proof. (1) The identities hold on generators: inserting $1$ in front and then multiplying it into $r_1$ gives back $y$, and the other faces commute with inserting $1$ in front, with an index shift. So $\pm h$ is a contracting homotopy of the augmented complex $\sum(-1)^id_i$, as for every augmented simplicial object with an extra degeneracy; its normalized complex is a direct summand, as a complex, and is contractible as well. (2) $\operatorname{Bar}_n=R\otimes_KY$ with $Y=R^{\otimes n}\otimes_KX$ flat over $K$. For a right $R$-module $N$, $N\otimes_R(R\otimes_KY)=N\otimes_KY$, which is exact in $N$. $\square$

The flatness of $R$ over $K$ cannot be omitted in (2) (Exercise 5.3).

Theorem 4.2 (the Mac Lane resolution). Let $M$ be an $A$-module, $R=\mathbb Z^{\mathrm{st}}(A)$ and $\operatorname{Bar}=\operatorname{Bar}(R/\mathbb Z,\mathbb Z^{\mathrm{st}}(M))$, a bisimplicial $R$-module, $(n,r)\mapsto R_r^{\otimes(n+1)}\otimes\mathbb Z^{\mathrm{st}}(M)_r$. The map $A\otimes_R\Delta\operatorname{Bar}\to M,$ given by the augmentation of the bar construction and $a\otimes x\mapsto a\,\varepsilon_M(x)$, is a quasi-isomorphism of simplicial $A$-modules. Its term of degree $n$ is $A\otimes_{\mathbb Z}\mathbb Z^{\mathrm{st}}(A)_n^{\otimes n}\otimes_{\mathbb Z}\mathbb Z^{\mathrm{st}}(M)_n,$ a direct sum of direct summands of free modules $A(A^p\times M^q)$.

Proof. The terms: $A\otimes_R(R^{\otimes(n+1)}\otimes V)=A\otimes R^{\otimes n}\otimes V$, and by Proposition 2.2 (1) and Construction 3.1 these are sums of direct summands of $A\otimes\mathbb Z(A^{k_1})\otimes\cdots\otimes\mathbb Z(M^l)=$ $A(A^{k_1+\cdots}\times M^l)$.

For each $r$, the row $\operatorname{Bar}_{\bullet,r}\to\mathbb Z^{\mathrm{st}}(M)_r$ is a homotopy equivalence by Proposition 4.1 (1), so $\Delta\operatorname{Bar}\to\mathbb Z^{\mathrm{st}}(M)$ is a quasi-isomorphism (Lemma 1.1 of the lesson on derived tensor products). With $P'\to A$ as in the proof of Theorem 3.2, it follows that $P'\otimes_{B_A}N\Delta\operatorname{Bar}\to P'\otimes_{B_A}N\mathbb Z^{\mathrm{st}}(M)$ is a quasi-isomorphism, and by Theorem 3.2 the latter maps quasi-isomorphically to $M$. It remains to compare $P'\otimes_{B_A}N\Delta\operatorname{Bar}$ with $N(A\otimes_R\Delta\operatorname{Bar})$.

For fixed $n$, the column $\operatorname{Bar}_{n,\bullet}$ is the induced module $R\otimes V_n$, with $V_n=R^{\otimes n}\otimes\mathbb Z^{\mathrm{st}}(M)$, a simplicial abelian sheaf with flat terms, and its faces and degeneracies in the index $r$ act on all factors. The shuffle map $B_A\otimes NV_n\to N(R\otimes V_n)$ is a $B_A$-linear quasi-isomorphism, by the proof of Lemma 3.2 of the lesson on differential graded modules, which does not use commutativity. Tensoring with $P'$ preserves it, and $P'\otimes_{B_A}(B_A\otimes NV_n)=$ $P'\otimes NV_n\to$ $A\otimes NV_n$ is a quasi-isomorphism because $NV_n$ is K-flat [Tag 06YQ]. Since $A$ is constant, $A\otimes NV_n=$ $N(A\otimes V_n)=$ $N(A\otimes_R\operatorname{Bar}_{n,\bullet})$. So the natural map $P'\otimes_{B_A}N\operatorname{Bar}_{n,\bullet}\to N(A\otimes_R\operatorname{Bar}_{n,\bullet})$ is a quasi-isomorphism for each $n$, compatibly with the faces in $n$. By the row filtration in $n$, the map of total complexes is a quasi-isomorphism. The shuffle maps from the total complexes to the normalized diagonals are quasi-isomorphisms (Lemma 2.6 of the lesson on multisimplicial objects), $B_A$-linear on the side of $\Delta\operatorname{Bar}$ (Lemma 3.1 of the lesson on differential graded modules), and natural. So $P'\otimes_{B_A}N\Delta\operatorname{Bar}\to N(A\otimes_R\Delta\operatorname{Bar})$ is a quasi-isomorphism, compatible with the maps to $M$. $\square$

Since the terms are sums of direct summands of free modules on products $A^p\times M^q$, Theorem 4.2 gives, functorially in $M$, a resolution of $M$ whose terms are built from the sheaves of sets underlying $A$ and $M$ alone. Computing $\operatorname{Ext}$ groups of $M$ with such a resolution expresses them through diagrams built from these products.

5. Exercises

5.1. Let $A$ and $M$ be as in Theorem 3.2. Show directly that $H_0(A\otimes^{\mathbf L}_{B_A}N\mathbb Z^{\mathrm{st}}(M))\cong M$, using $H_0\mathbb Z^{\mathrm{st}}(A)=A$ and $H_0\mathbb Z^{\mathrm{st}}(M)=M$.

5.2. Let $A=\mathbb F_2$ (in the topos of sets). List the summands of $\mathbb Z^{\mathrm{st}}(A)_0$ of the ordinals $[-1]$ and $[0]$ with bases, and check by hand that $\varepsilon$ is multiplicative on them and that the map of Remark 2.3 is not compatible with the faces.

5.3. Let $p$ be a prime, $K=\mathbb Z$, $R=\mathbb Z[t]/(pt)$ and $X=\mathbb Z$ (in the topos of sets). Show that $X$ is flat over $K$ but $\operatorname{Bar}_1(R/K,X)\cong R\otimes_{\mathbb Z}R$ is not a flat $R$-module.

6. Complete solutions

5.1. The ring $B_A$ lies in nonpositive degrees, and $H^0B_A=A$. Let $Q\to N\mathbb Z^{\mathrm{st}}(M)$ be a quasi-isomorphism from a K-flat module in nonpositive degrees (Lemma 2.1 of the lesson on differential graded modules), so that $A\otimes^{\mathbf L}_{B_A}N\mathbb Z^{\mathrm{st}}(M)$ is $A\otimes_{B_A}Q$. Since $A$ and $Q$ vanish in positive degrees, the term of degree $0$ of $A\otimes_{B_A}Q$ is $A\otimes_{B^0_A}Q^0$, and $H^0$ is its quotient by the image of $A\otimes Q^{-1}$, that is, $A\otimes_{B^0_A}H^0Q$. The ring $B^0_A$ acts on $H^0Q$ through its quotient $H^0B_A=A$, so this is $A\otimes_AH^0Q=H^0Q=H_0\mathbb Z^{\mathrm{st}}(M)=M$.

5.2. The summand of $[-1]$ is $\mathbb Z(A)$ with basis $[0],[1]$, and that of $[0]$ is $I(A)$ with basis $[1]-[0]$. The products: $[0][a]=[0]$, $[1][1]=[1]$, and in $I(A)$, $([1]-[0])([1]-[0])=[1]$ $-[0]-[0]+[0]=[1]$ $-[0]$, lying in the summand of $[1]$. The map $\varepsilon$ sends $[0]\mapsto0$, $[1]\mapsto1$ and $[1]-[0]\mapsto1$, and these values multiply correctly. For Remark 2.3: $d_1x=[1]$ in the summand of $[-1]$ and $d_0x=[1]-[0]$ in the summand of $[0]$; the projection onto the first summand followed by $s$ gives $1$ and $0$.

5.3. $X=\mathbb Z$ is flat over $\mathbb Z$. As an abelian group $R=\mathbb Z\oplus\bigoplus_{k\geq1}(\mathbb Z/p)t^k$, so $R\otimes_{\mathbb Z}R=R\oplus\bigoplus_{k\geq1}R/pR$ as left $R$-modules, and a direct summand of a flat module is flat. Here $R/pR=\mathbb F_p[t]$. The annihilator of $p$ in $R$ is $tR$, so $pR\cong R/tR$, and $pR\otimes_R\mathbb F_p[t]\cong\mathbb F_p[t]/t\mathbb F_p[t]=\mathbb F_p$. Its map to $R\otimes_R\mathbb F_p[t]=\mathbb F_p[t]$, induced by the inclusion $pR\to R$, sends the generator $p\otimes1$ to $p\cdot1=0$. So tensoring the injection $pR\to R$ with $\mathbb F_p[t]$ is not injective, and $\mathbb F_p[t]$ is not flat over $R$.

Sources

The ring structure on the stable homology complex $\mathbb Z^{\mathrm{st}}(A)$, the description of $A\otimes^{\mathbf L}_{\mathbb Z}B$ through it and the comparison with modules are due to Illusie, in his work on the cotangent complex. The theorem that $A\otimes^{\mathbf L}_{\mathbb Z^{\mathrm{st}}(A)}\mathbb Z^{\mathrm{st}}(M)$ is $M$ goes back to Mac Lane's work on the homology of rings. Resolutions of modules by sums of free modules on products of copies of $A$ and $M$ were constructed by Breen and by Deligne, answering a question of Grothendieck; the bar construction used here is classical.