The Stacks project

Proof. Fix $n$. If $r = 0$, then we can just pick $A = R$. Assume $r > 0$. By More on Algebra, Lemma 15.94.1 (translated into the language of chain complexes) we can choose

such that the transition maps $K_{n_{i + 1}} \to K_{n_ i}$ on Koszul algebras (see above) induce the zero map on homology in degrees $> 0$. We will prove the lemma with $N = n_ r$.

We will construct $A$ exactly as in the statement and proof of Lemma 23.6.9. Thus we will have

This will immediately give us properties (1), (2), (5), and (6). To finish the proof we will construct the $R$-algebra maps $K_ N \to A \to K_ n$. To do this we will construct

1. an isomorphism $A(1) \to K_ N = K_{n_ r}$,

2. a map $A(2) \to K_{n_{r - 1}}$,

3. $\ldots$

4. a map $A(r) \to K_{n_1}$,

5. a map $A(r + 1) \to K_{n_0} = K_ n$, and

6. a map $A \to K_ n$.

In each of these steps the map constructed will be between differential graded algebras compatibly endowed with divided powers and each of the maps will be compatible with the previous one via the transition maps between the Koszul algebras and each of the maps will induce the obvious canonical map on homology in degree $0$.

Recall that $A(0) = R$. For $m = 1$, the proof of Lemma 23.6.9 chooses $A(1) = R\langle T_1, \ldots , T_ r\rangle$ with $T_ i$ of degree $1$ and with $\text{d}(T_ i) = f_ i^ N$. Namely, the $f_ i^ N$ are generators of the kernel of $A(0) \to R/(f_1^ N, \ldots , f_ r^ N)$. Thus for $A(1) \to K_ N = K_{n_ r}$ we use the map

which is an isomorphism.

For $m = 2$, the construction in the proof of Lemma 23.6.9 chooses generators $e_1, \ldots , e_ t \in \mathop{\mathrm{Ker}}(\text{d} : A(1)_1 \to A(1)_0)$. The construction proceeds by taking $A(2) = A(1)\langle T_1, \ldots , T_ t\rangle$ as a divided power polynomial algebra with $T_ i$ of degree $2$ and with $\text{d}(T_ i) = e_ i$. Since $\varphi _1(e_ i)$ is a cocycle in $K_{n_ r}$ we see that its image in $K_{n_{r - 1}}$ is a coboundary by our choice of $n_ r$ and $n_{r - 1}$ above. Hence we can construct the following commutative diagram

by sending $T_ i$ to an element in degree $2$ whose boundary is the image of $\varphi _1(e_ i)$. The map $\varphi _2$ exists and is compatible with the differential and the divided powers by the universal of the divided power polynomial algebra.

The algebra $A(m)$ and the map $\varphi _ m : A(m) \to K_{n_{r + 1 - m}}$ are constructed in exactly the same manner for $m = 2, \ldots , r$.

Given the map $A(r) \to K_{n_1}$ we see that the composition $H_ r(A(r)) \to H_ r(K_{n_1}) \to H_ r(K_{n_0}) \subset (K_{n_0})_ r$ is zero, hence we can extend this to $A(r + 1) \to K_{n_0} = K_ n$ by sending the new polynomial generators of $A(r + 1)$ to zero.

Having constructed $A(r + 1) \to K_{n_0} = K_ n$ we can simply extend to $A(r + 2), A(r + 3), \ldots$ in the only possible way by sending the new polynomial generators to zero. This finishes the proof. $\square$

Proof. For $d = 1$, we have a short exact sequence of complexes

and the result (1) follows easily from this. For $d = 2$ we view $A\langle T \rangle$ as a filtered chain complex with subcomplexes

Applying the spectral sequence of Homology, Section 12.24 (translated into chain complexes) we obtain (2). $\square$

Proof. We first prove this for $r = 1$. Set $f = f_1$. Write $K_ t = R\langle x \rangle$, $K_ n = R\langle y \rangle$, and $K_ N = R\langle z \rangle$ with $x$, $y$, $z$ of degree $1$ and $\text{d}(x) = f^ t$, $\text{d}(y) = f^ n$, and $\text{d}(z) = f^ N$. For all $N > t$ we claim there is a quasi-isomorphism

Here the left hand side denotes the divided power polynomial algebra in variables $x$ and $z$ of degree $1$ and $u$ of degree $2$ with $\text{d}(x) = f^ t$, $\text{d}(z) = f^ N$, and $\text{d}(u) = z - f^{N - t}x$. To prove the claim, we observe that the following three submodules of $H_*(R\langle x, z\rangle )$ are the same

1. the kernel of $H_*(R\langle x, z\rangle ) \to H_*(K_ t)$,

2. the image of $z - f^{N - t}x : H_*(R\langle x, z\rangle ) \to H_*(R\langle x, z\rangle )$, and

3. the kernel of $z - f^{N - t}x : H_*(R\langle x, z\rangle ) \to H_*(R\langle x, z\rangle )$.

This observation is proved by a direct computation¹ which we omit. Then we can apply Lemma 23.12.3 part (2) to see that the claim is true.

Via the homomorphism $K_ N \to B_{N, t}$ of differential graded $R$-algebras sending $z$ to $z$, we may view $B_{N, t} \to K_ t$ as a quasi-isomorphism of left differential graded $K_ N$-modules. To define the arrow in the statement of the lemma we use the homomorphism

This makes sense as long as we assume $N \geq n + t$. It is a pleasant computation to show that the (pre or post) composition with the multiplication map is the transition map.

For $r > 1$ we proceed by writing each of the Koszul algebras as a tensor product of Koszul algebras in $1$ variable and we apply the previous construction. In other words, we write

where $x_ i$ is in degree $1$ and $\text{d}(x_ i) = f_ i^ t$. In the case $r > 1$ we then use

where $x_ i, z_ i$ have degree $1$ and $u_ i$ has degree $2$ and we have $\text{d}(x_ i) = f_ i^ t$, $\text{d}(z_ i) = f_ i^ N$, and $\text{d}(u_ i) = z_ i - f_ i^{N - t}x_ i$. The tensor product map $B_{N, t} \to K_ t$ will be a quasi-isomorphism as it is a tensor product of quasi-isomorphisms between bounded above complexes of free $R$-modules. Finally, we define the map

as the tensor product of the maps constructed in the case of $r = 1$ or simply by the rules $x_ i \mapsto 1 \otimes x_ i$, $z_ i \mapsto f_ i^{N - n}y_ i \otimes 1$, and $u_ i \mapsto - f_ i^{N - n - t}y_ i \otimes x_ i$ which makes sense as long as $N \geq n + t$. We omit the details. $\square$