The Stacks project

Proof. Since $A$ is a divided power polynomial algebra, we can find a divided power polynomial algebra $A'$ over $R'$ such that $A = A' \otimes _ R R'$. Moreover, we can lift $\text{d}$ to an $R$-linear operator $\text{d}$ on $A'$ such that

1. $\text{d}(xy) = \text{d}(x)y + (-1)^{\deg (x)}x \text{d}(y)$ for $x, y \in A'$ homogeneous, and

2. $\text{d}(\gamma _ n(x)) = \text{d}(x) \gamma _{n - 1}(x)$ for $x \in A'_{even, +}$.

We omit the details (hint: proceed one variable at the time). However, it may not be the case that $\text{d}^2$ is zero on $A'$. It is clear that $\text{d}^2$ maps $A'$ into $fA' \cong A/IA$. Hence $\text{d}^2$ annihilates $fA'$ and factors as a map $A \to A/IA$. Since $\text{d}^2$ is $R$-linear we obtain our map $\theta : A/IA \to A/IA$. The verification of the properties of a derivation is immediate. $\square$

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

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

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

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

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. $\square$

Proof. Choose a Tate resolution $R \to A \to S$ as in Lemma 23.6.9. Let $\xi \in H_2(A/IA)$ and $\theta : A/IA \to A/IA$ be the element and derivation found in Lemmas 23.7.1 and 23.7.2. Observe that

in $H_0(A/IA) = S/IS$. Hence if $g$ is not nilpotent in $S/IS$, then $\xi ^ n$ is nonzero in $H_{2n}(A/IA)$ for all $n > 0$. Since $H_{2n}(A/IA) = \text{Tor}^ R_{2n}(S, R/I)$ we conclude. $\square$

Proof. Let $f \in I$ be an element mapping to a nonzero element of $I/\mathfrak m I$ which is mapped to zero in $J/\mathfrak mJ$. We can choose an ideal $I'$ with $\mathfrak mI \subset I' \subset I$ such that $I/I'$ is generated by the image of $f$. Set $R = A/I$ and $R' = A/I'$. Let $J = (a_1, \ldots , a_ m)$ for some $a_ j \in A$. Then $f = \sum b_ j a_ j$ for some $b_ j \in \mathfrak m$. Let $r_ j, f_ j \in R$ resp. $r'_ j, f'_ j \in R'$ be the image of $b_ j, a_ j$. Then we see we are in the situation of Lemma 23.7.3 (with the ideal $I$ of that lemma equal to $\mathfrak m_ R$) and the lemma is proved. $\square$

Proof. By Lemma 23.7.4 we see that $I/\mathfrak m I \to J/\mathfrak m J$ is injective. Thus we can find $s \leq r$ and a minimal system of generators $f_1, \ldots , f_ r$ of $J$ such that $f_1, \ldots , f_ s$ are in $I$ and form a minimal system of generators of $I$. The lemma follows as any minimal system of generators of $J$ is a regular sequence by More on Algebra, Lemmas 15.30.15 and 15.30.7. $\square$

Proof. If (3) holds, then $J$ is generated by a regular sequence, see for example More on Algebra, Lemmas 15.30.13 and 15.30.7. Moreover, if (3) holds, then $S/J = (S/I)/(J/I)$ has finite projective dimension over $S/IS$ because the Koszul complex will be a finite free resolution of $S/J$ over $S/IS$. Since $R/I \to S/IS$ is flat, it then follows that $S/J$ has finite tor dimension over $R/I$ by More on Algebra, Lemma 15.66.11. Thus (3) implies (1).

The implication (1) $\Rightarrow $ (2) is trivial. Assume (2). By More on Algebra, Lemma 15.77.6 we find that $S/J$ has finite tor dimension over $S/IS$. Thus we can apply Lemma 23.7.5 to conclude that $IS$ and $J/IS$ are generated by regular sequences. Let $f_1, \ldots , f_ r \in I$ be a minimal system of generators of $I$. Since $R \to S$ is flat, we see that $f_1, \ldots , f_ r$ form a minimal system of generators for $IS$ in $S$. Thus $f_1, \ldots , f_ r \in R$ is a sequence of elements whose images in $S$ form a regular sequence by More on Algebra, Lemmas 15.30.15 and 15.30.7. Thus $f_1, \ldots , f_ r$ is a regular sequence in $R$ by Algebra, Lemma 10.68.5. $\square$