Lemma 23.7.1. Let $R$ be a ring. Let $(A, \text{d}, \gamma )$ be as in Definition 23.6.4. Let $R' \to R$ be a surjection of rings whose kernel has square zero and is generated by one element $f$. If $A$ is a graded divided power polynomial algebra over $R$ with finitely many variables in each degree, then we obtain a derivation $\theta : A/IA \to A/IA$ where $I$ is the annihilator of $f$ in $R$.

## 23.7 Application to complete intersections

Let $R$ be a ring. Let $(A, \text{d}, \gamma )$ be as in Definition 23.6.4. A *derivation* of degree $2$ is an $R$-linear map $\theta : A \to A$ with the following properties

$\theta (A_ d) \subset A_{d - 2}$,

$\theta (xy) = \theta (x)y + x\theta (y)$,

$\theta $ commutes with $\text{d}$,

$\theta (\gamma _ n(x)) = \theta (x) \gamma _{n - 1}(x)$ for all $x \in A_{2d}$ all $d$.

In the following lemma we construct a derivation.

**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

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

$\text{d}(\gamma _ n(x)) = \text{d}(x) \gamma _{n - 1}(x)$ 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. $\square$

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

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

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. $\square$

The method of proof of the following lemma is apparently due to Gulliksen.

Lemma 23.7.3. Let $R' \to R$ be a surjection of Noetherian rings whose kernel has square zero and is generated by one element $f$. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Let $\sum r_ j f_ j = 0$ be a relation in $R[x_1, \ldots , x_ n]$. Assume that

each $r_ j$ has coefficients in the annihilator $I$ of $f$ in $R$,

for some lifts $r'_ j, f'_ j \in R'[x_1, \ldots , x_ n]$ we have $\sum r'_ j f'_ j = gf$ where $g$ is not nilpotent in $S$.

Then $S$ does not have finite tor dimension over $R$ (i.e., $S$ is not a perfect $R$-algebra).

**Proof.**
Choose a Tate resolution $R \to A \to S$ as in Lemma 23.6.8. 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

Hence if $g$ is not nilpotent, 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. $\square$

The following result can be found in [Rodicio].

Lemma 23.7.4. Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $I \subset J \subset A$ be proper ideals. If $A/J$ has finite tor dimension over $A/I$, then $I/\mathfrak m I \to J/\mathfrak m J$ is injective.

**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.
$\square$

Lemma 23.7.5. Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $I \subset J \subset A$ be proper ideals. Assume

$A/J$ has finite tor dimension over $A/I$, and

$J$ is generated by a regular sequence.

Then $I$ is generated by a regular sequence and $J/I$ is generated by a regular sequence.

**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 15.29.15 and 15.29.7.
$\square$

Lemma 23.7.6. Let $R \to S$ be a local ring map of Noetherian local rings. Let $I \subset R$ and $J \subset S$ be ideals with $IS \subset J$. If $R \to S$ is flat and $S/\mathfrak m_ RS$ is regular, then the following are equivalent

$J$ is generated by a regular sequence and $S/J$ has finite tor dimension as a module over $R/I$,

$J$ is generated by a regular sequence and $\text{Tor}^{R/I}_ p(S/J, R/\mathfrak m_ R)$ is nonzero for only finitely many $p$,

$I$ is generated by a regular sequence and $J/IS$ is generated by a regular sequence in $S/IS$.

**Proof.**
If (3) holds, then $J$ is generated by a regular sequence, see for example More on Algebra, Lemmas 15.29.13 and 15.29.7. 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 15.63.11. Thus (3) implies (1).

The implication (1) $\Rightarrow $ (2) is trivial. Assume (2). By More on Algebra, Lemma 15.71.9 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 15.29.15 and 15.29.7. Thus $f_1, \ldots , f_ r$ is a regular sequence in $R$ by Algebra, Lemma 10.67.5. $\square$

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