Lemma 23.6.9. Let $R \to S$ be a homomorphism of commutative rings. There exists a factorization

$R \to A \to S$

with the following properties:

1. $(A, \text{d}, \gamma )$ is as in Definition 23.6.5,

2. $A \to S$ is a quasi-isomorphism (if we endow $S$ with the zero differential),

3. $A_0 = R[x_ j: j\in J] \to S$ is any surjection of a polynomial ring onto $S$, and

4. $A$ is a graded divided power polynomial algebra over $R$.

The last condition means that $A$ is constructed out of $A_0$ by successively adjoining a set of variables $T$ in each degree $> 0$ as in Example 23.6.2 or 23.6.3. Moreover, if $R$ is Noetherian and $R\to S$ is of finite type, then $A$ can be taken to have only finitely many generators in each degree.

Proof. We write out the construction for the case that $R$ is Noetherian and $R\to S$ is of finite type. Without those assumptions, the proof is the same, except that we have to use some set (possibly infinite) of generators in each degree.

Start of the construction: Let $A(0) = R[x_1, \ldots , x_ n]$ be a (usual) polynomial ring and let $A(0) \to S$ be a surjection. As grading we take $A(0)_0 = A(0)$ and $A(0)_ d = 0$ for $d \not= 0$. Thus $\text{d} = 0$ and $\gamma _ n$, $n > 0$, is zero as well.

Choose generators $f_1, \ldots , f_ m \in R[x_1, \ldots , x_ n]$ for the kernel of the given map $A(0) = R[x_1, \ldots , x_ n] \to S$. We apply Example 23.6.2 $m$ times to get

$A(1) = A(0)\langle T_1, \ldots , T_ m\rangle$

with $\deg (T_ i) = 1$ as a graded divided power polynomial algebra. We set $\text{d}(T_ i) = f_ i$. Since $A(1)$ is a divided power polynomial algebra over $A(0)$ and since $\text{d}(f_ i) = 0$ this extends uniquely to a differential on $A(1)$ by Lemma 23.6.8.

Induction hypothesis: Assume we are given factorizations

$R \to A(0) \to A(1) \to \ldots \to A(m) \to S$

where $A(0)$ and $A(1)$ are as above and each $R \to A(m') \to S$ for $2 \leq m' \leq m$ satisfies properties (1) and (4) of the statement of the lemma and (2) replaced by the condition that $H_ i(A(m')) \to H_ i(S)$ is an isomorphism for $m' > i \geq 0$. The base case is $m = 1$.

Induction step: Assume we have $R \to A(m) \to S$ as in the induction hypothesis. Consider the group $H_ m(A(m))$. This is a module over $H_0(A(m)) = S$. In fact, it is a subquotient of $A(m)_ m$ which is a finite type module over $A(m)_0 = R[x_1, \ldots , x_ n]$. Thus we can pick finitely many elements

$e_1, \ldots , e_ t \in \mathop{\mathrm{Ker}}(\text{d} : A(m)_ m \to A(m)_{m - 1})$

which map to generators of this module. Applying Example 23.6.2 or 23.6.3 $t$ times we get

$A(m + 1) = A(m)\langle T_1, \ldots , T_ t\rangle$

with $\deg (T_ i) = m + 1$ as a graded divided power algebra. We set $\text{d}(T_ i) = e_ i$. Since $A(m+1)$ is a divided power polynomial algebra over $A(m)$ and since $\text{d}(e_ i) = 0$ this extends uniquely to a differential on $A(m + 1)$ compatible with the divided power structure. Since we've added only material in degree $m + 1$ and higher we see that $H_ i(A(m + 1)) = H_ i(A(m))$ for $i < m$. Moreover, it is clear that $H_ m(A(m + 1)) = 0$ by construction.

To finish the proof we observe that we have shown there exists a sequence of maps

$R \to A(0) \to A(1) \to \ldots \to A(m) \to A(m + 1) \to \ldots \to S$

and to finish the proof we set $A = \mathop{\mathrm{colim}}\nolimits A(m)$. $\square$

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