Lemma 23.7.1. Let $R$ be a ring. Let $(A, \text{d}, \gamma )$ be as in Definition 23.6.5. 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 23.6.5. 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 $S/IS$.

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 23.6.9. 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. $\square$

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 15.30.15 and 15.30.7.
$\square$

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 15.30.13 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)