Lemma 23.7.2. Assumption and notation as in Lemma 23.7.1. Suppose $S = H_0(A)$ is isomorphic to $R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ for some $n$, $m$, and $f_ j \in R[x_1, \ldots , x_ n]$. Moreover, suppose given a relation

$\sum r_ j f_ j = 0$

with $r_ j \in R[x_1, \ldots , x_ n]$. Choose $r'_ j, f'_ j \in R'[x_1, \ldots , x_ n]$ lifting $r_ j, f_ j$. Write $\sum r'_ j f'_ j = gf$ for some $g \in R/I[x_1, \ldots , x_ n]$. If $H_1(A) = 0$ and all the coefficients of each $r_ j$ are in $I$, then there exists an element $\xi \in H_2(A/IA)$ such that $\theta (\xi ) = g$ in $S/IS$.

Proof. Let $A(0) \subset A(1) \subset A(2) \subset \ldots$ be the filtration of $A$ such that $A(m)$ is gotten from $A(m - 1)$ by adjoining divided power variables of degree $m$. Then $A(0)$ is a polynomial algebra over $R$ equipped with an $R$-algebra surjection $A(0) \to S$. Thus we can choose a map

$\varphi : R[x_1, \ldots , x_ n] \to A(0)$

lifting the augmentations to $S$. Next, $A(1) = A(0)\langle T_1, \ldots , T_ t \rangle$ for some divided power variables $T_ i$ of degree $1$. Since $H_0(A) = S$ we can pick $\xi _ j \in \sum A(0)T_ i$ with $\text{d}(\xi _ j) = \varphi (f_ j)$. Then

$\text{d}\left(\sum \varphi (r_ j) \xi _ j\right) = \sum \varphi (r_ j) \varphi (f_ j) = \sum \varphi (r_ jf_ j) = 0$

Since $H_1(A) = 0$ we can pick $\xi \in A_2$ with $\text{d}(\xi ) = \sum \varphi (r_ j) \xi _ j$. If the coefficients of $r_ j$ are in $I$, then the same is true for $\varphi (r_ j)$. In this case $\text{d}(\xi )$ dies in $A_1/IA_1$ and hence $\xi$ defines a class in $H_2(A/IA)$.

The construction of $\theta$ in the proof of Lemma 23.7.1 proceeds by successively lifting $A(i)$ to $A'(i)$ and lifting the differential $\text{d}$. We lift $\varphi$ to $\varphi ' : R'[x_1, \ldots , x_ n] \to A'(0)$. Next, we have $A'(1) = A'(0)\langle T_1, \ldots , T_ t\rangle$. Moreover, we can lift $\xi _ j$ to $\xi '_ j \in \sum A'(0)T_ i$. Then $\text{d}(\xi '_ j) = \varphi '(f'_ j) + f a_ j$ for some $a_ j \in A'(0)$. Consider a lift $\xi ' \in A'_2$ of $\xi$. Then we know that

$\text{d}(\xi ') = \sum \varphi '(r'_ j)\xi '_ j + \sum fb_ iT_ i$

for some $b_ i \in A(0)$. Applying $\text{d}$ again we find

$\theta (\xi ) = \sum \varphi '(r'_ j)\varphi '(f'_ j) + \sum f \varphi '(r'_ j) a_ j + \sum fb_ i \text{d}(T_ i)$

The first term gives us what we want. The second term is zero because the coefficients of $r_ j$ are in $I$ and hence are annihilated by $f$. The third term maps to zero in $H_0$ because $\text{d}(T_ i)$ maps to zero. $\square$

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