## 23.12 Koszul complexes and Tate resolutions

In this section we “lift” the result of More on Algebra, Lemma 15.94.1 to the category of differential graded algebras endowed with divided powers compatible with the differential graded structure (beware that in this section we represent Koszul complexes as chain complexes whereas in locus citatus we use cochain complexes).

Let $R$ be a ring. Let $I \subset R$ be an ideal generated by $f_1, \ldots , f_ r \in R$. For $n \geq 1$ we denote

$K_ n = K_{n, \bullet } = R\langle \xi _1, \ldots , \xi _ r\rangle$

the differential graded Koszul algebra with $\xi _ i$ in degree $1$ and $\text{d}(\xi _ i) = f_ i$. There exists a unique divided power structure on this (as in Definition 23.6.5), see Example 23.6.2. For $m > n$ the transition map $K_ m \to K_ n$ is the differential graded algebra map compatible with divided powers given by sending $\xi _ i$ to $f_ i^{m - n}\xi _ i$.

Lemma 23.12.1. In the situation above, if $R$ is Noetherian, then for every $n$ there exists an $N \geq n$ and maps

$K_ N \to A \to R/(f_1^ N, \ldots , f_ r^ N)\quad \text{and}\quad A \to K_ n$

with the following properties

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

2. $A \to R/(f_1^ N, \ldots , f_ r^ N)$ is a quasi-isomorphism,

3. the composition $K_ N \to A \to R/(f_1^ N, \ldots , f_ r^ N)$ is the canonical map,

4. the composition $K_ N \to A \to K_ n$ is the transition map,

5. $A_0 = R \to R/(f_1^ N, \ldots , f_ r^ N)$ is the canonical surjection,

6. $A$ is a graded divided power polynomial algebra over $R$ with finitely many generators in each degree, and

7. $A \to K_ n$ is a homomorphism of differential graded $R$-algebras compatible with divided powers which induces the canonical map $R/(f_1^ N, \ldots , f_ r^ N) \to R/(f_1^ n, \ldots , f_ r^ n)$ on homology in degree $0$.

Condition (4) means that $A$ is constructed out of $A_0$ by successively adjoining a finite set of variables $T$ in each degree $> 0$ as in Example 23.6.2 or 23.6.3.

Proof. Fix $n$. If $r = 0$, then we can just pick $A = R$. Assume $r > 0$. By More on Algebra, Lemma 15.94.1 (translated into the language of chain complexes) we can choose

$n_{r} > n_{r - 1} > \ldots > n_1 > n_0 = n$

such that the transition maps $K_{n_{i + 1}} \to K_{n_ i}$ on Koszul algebras (see above) induce the zero map on homology in degrees $> 0$. We will prove the lemma with $N = n_ r$.

We will construct $A$ exactly as in the statement and proof of Lemma 23.6.9. Thus we will have

$A = \mathop{\mathrm{colim}}\nolimits A(m),\quad \text{and}\quad A(0) \to A(1) \to A(2) \to \ldots \to R/(f_1^ N, \ldots , f_ r^ N)$

This will immediately give us properties (1), (2), (5), and (6). To finish the proof we will construct the $R$-algebra maps $K_ N \to A \to K_ n$. To do this we will construct

1. an isomorphism $A(1) \to K_ N = K_{n_ r}$,

2. a map $A(2) \to K_{n_{r - 1}}$,

3. $\ldots$

4. a map $A(r) \to K_{n_1}$,

5. a map $A(r + 1) \to K_{n_0} = K_ n$, and

6. a map $A \to K_ n$.

In each of these steps the map constructed will be between differential graded algebras compatibly endowed with divided powers and each of the maps will be compatible with the previous one via the transition maps between the Koszul algebras and each of the maps will induce the obvious canonical map on homology in degree $0$.

Recall that $A(0) = R$. For $m = 1$, the proof of Lemma 23.6.9 chooses $A(1) = R\langle T_1, \ldots , T_ r\rangle$ with $T_ i$ of degree $1$ and with $\text{d}(T_ i) = f_ i^ N$. Namely, the $f_ i^ N$ are generators of the kernel of $A(0) \to R/(f_1^ N, \ldots , f_ r^ N)$. Thus for $A(1) \to K_ N = K_{n_ r}$ we use the map

$\varphi _1 : A(1) \longrightarrow K_{n_ r},\quad T_ i \longmapsto \xi _ i$

which is an isomorphism.

For $m = 2$, the construction in the proof of Lemma 23.6.9 chooses generators $e_1, \ldots , e_ t \in \mathop{\mathrm{Ker}}(\text{d} : A(1)_1 \to A(1)_0)$. The construction proceeds by taking $A(2) = A(1)\langle T_1, \ldots , T_ t\rangle$ as a divided power polynomial algebra with $T_ i$ of degree $2$ and with $\text{d}(T_ i) = e_ i$. Since $\varphi _1(e_ i)$ is a cocycle in $K_{n_ r}$ we see that its image in $K_{n_{r - 1}}$ is a coboundary by our choice of $n_ r$ and $n_{r - 1}$ above. Hence we can construct the following commutative diagram

$\xymatrix{ A(1) \ar[d] \ar[r]_{\varphi _1} & K_{n_ r} \ar[d] \\ A(2) \ar[r]^{\varphi _2} & K_{n_{r - 1}} }$

by sending $T_ i$ to an element in degree $2$ whose boundary is the image of $\varphi _1(e_ i)$. The map $\varphi _2$ exists and is compatible with the differential and the divided powers by the universal of the divided power polynomial algebra.

The algebra $A(m)$ and the map $\varphi _ m : A(m) \to K_{n_{r + 1 - m}}$ are constructed in exactly the same manner for $m = 2, \ldots , r$.

Given the map $A(r) \to K_{n_1}$ we see that the composition $H_ r(A(r)) \to H_ r(K_{n_1}) \to H_ r(K_{n_0}) \subset (K_{n_0})_ r$ is zero, hence we can extend this to $A(r + 1) \to K_{n_0} = K_ n$ by sending the new polynomial generators of $A(r + 1)$ to zero.

Having constructed $A(r + 1) \to K_{n_0} = K_ n$ we can simply extend to $A(r + 2), A(r + 3), \ldots$ in the only possible way by sending the new polynomial generators to zero. This finishes the proof. $\square$

Remark 23.12.2. In the situation above, if $R$ is Noetherian, we can inductively choose a sequence of integers $1 = n_0 < n_1 < n_2 < \ldots$ such that for $i = 1, 2, 3, \ldots$ we have maps $K_{n_ i} \to A_ i \to R/(f_1^{n_ i}, \ldots , f_ r^{n_ i})$ and $A_ i \to K_{n_{i - 1}}$ as in Lemma 23.12.1. Denote $A_{i + 1} \to A_ i$ the composition $A_{i + 1} \to K_{n_ i} \to A_ i$. Then the diagram

$\xymatrix{ K_{n_1} \ar[d] & K_{n_2} \ar[d] \ar[l] & K_{n_3} \ar[d] \ar[l] & \ldots \ar[l] \\ A_1 \ar[d] & A_2 \ar[l] \ar[d] & A_3 \ar[l] \ar[d] & \ldots \ar[l] \\ K_1 & K_{n_1} \ar[l] & K_{n_2} \ar[l] & \ldots \ar[l] }$

commutes. In this way we see that the inverse systems $(K_ n)$ and $(A_ n)$ are pro-isomorphic in the category of differential graded $R$-algebras with compatible divided powers.

Lemma 23.12.3. Let $(A, \text{d}, \gamma )$, $d \geq 1$, $f \in A_{d - 1}$, and $A\langle T \rangle$ be as in Lemma 23.6.8.

1. If $d = 1$, then there is a long exact sequences

$\ldots \to H_0(A) \xrightarrow {f} H_0(A) \to H_0(A\langle T \rangle ) \to 0$
2. For $d = 2$ there is a bounded spectral sequence $(E_1)_{i, j} = H_{j - i}(A) \cdot T^{[i]}$ converging to $H_{i + j}(A\langle T \rangle )$. The differential $(d_1)_{i, j} : H_{j - i}(A) \cdot T^{[i]} \to H_{j - i + 1}(A) \cdot T^{[i - 1]}$ sends $\xi \cdot T^{[i]}$ to the class of $f \xi \cdot T^{[i - 1]}$.

3. Add more here for other degrees as needed.

Proof. For $d = 1$, we have a short exact sequence of complexes

$0 \to A \to A\langle T \rangle \to A \cdot T \to 0$

and the result (1) follows easily from this. For $d = 2$ we view $A\langle T \rangle$ as a filtered chain complex with subcomplexes

$F^ pA\langle T \rangle = \bigoplus \nolimits _{i \leq p} A \cdot T^{[i]}$

Applying the spectral sequence of Homology, Section 12.24 (translated into chain complexes) we obtain (2). $\square$

The following lemma will be needed later.

Lemma 23.12.4. In the situation above, for all $n \geq t \geq 1$ there exists an $N > n$ and a map

$K_ t \longrightarrow K_ n \otimes _ R K_ t$

in the derived category of left differential graded $K_ N$-modules whose composition with the multiplication map is the transition map (in either direction).

Proof. We first prove this for $r = 1$. Set $f = f_1$. Write $K_ t = R\langle x \rangle$, $K_ n = R\langle y \rangle$, and $K_ N = R\langle z \rangle$ with $x$, $y$, $z$ of degree $1$ and $\text{d}(x) = f^ t$, $\text{d}(y) = f^ n$, and $\text{d}(z) = f^ N$. For all $N > t$ we claim there is a quasi-isomorphism

$B_{N, t} = R\langle x, z, u \rangle \longrightarrow K_ t,\quad x \mapsto x,\quad z \mapsto f^{N - t}x,\quad u \mapsto 0$

Here the left hand side denotes the divided power polynomial algebra in variables $x$ and $z$ of degree $1$ and $u$ of degree $2$ with $\text{d}(x) = f^ t$, $\text{d}(z) = f^ N$, and $\text{d}(u) = z - f^{N - t}x$. To prove the claim, we observe that the following three submodules of $H_*(R\langle x, z\rangle )$ are the same

1. the kernel of $H_*(R\langle x, z\rangle ) \to H_*(K_ t)$,

2. the image of $z - f^{N - t}x : H_*(R\langle x, z\rangle ) \to H_*(R\langle x, z\rangle )$, and

3. the kernel of $z - f^{N - t}x : H_*(R\langle x, z\rangle ) \to H_*(R\langle x, z\rangle )$.

This observation is proved by a direct computation1 which we omit. Then we can apply Lemma 23.12.3 part (2) to see that the claim is true.

Via the homomorphism $K_ N \to B_{N, t}$ of differential graded $R$-algebras sending $z$ to $z$, we may view $B_{N, t} \to K_ t$ as a quasi-isomorphism of left differential graded $K_ N$-modules. To define the arrow in the statement of the lemma we use the homomorphism

$B_{N, t} = R\langle x, z, u \rangle \to K_ n \otimes _ R K_ t,\quad x \mapsto 1 \otimes x,\quad z \mapsto f^{N - n}y \otimes 1,\quad u \mapsto - f^{N - n - t}y \otimes x$

This makes sense as long as we assume $N \geq n + t$. It is a pleasant computation to show that the (pre or post) composition with the multiplication map is the transition map.

For $r > 1$ we proceed by writing each of the Koszul algebras as a tensor product of Koszul algebras in $1$ variable and we apply the previous construction. In other words, we write

$K_ t = R\langle x_1, \ldots , x_ r\rangle = R\langle x_1\rangle \otimes _ R \ldots \otimes _ R R\langle x_ r\rangle$

where $x_ i$ is in degree $1$ and $\text{d}(x_ i) = f_ i^ t$. In the case $r > 1$ we then use

$B_{N, t} = R\langle x_1, z_1, u_1 \rangle \otimes _ R \ldots \otimes _ R R\langle x_ r, z_ r, u_ r \rangle$

where $x_ i, z_ i$ have degree $1$ and $u_ i$ has degree $2$ and we have $\text{d}(x_ i) = f_ i^ t$, $\text{d}(z_ i) = f_ i^ N$, and $\text{d}(u_ i) = z_ i - f_ i^{N - t}x_ i$. The tensor product map $B_{N, t} \to K_ t$ will be a quasi-isomorphism as it is a tensor product of quasi-isomorphisms between bounded above complexes of free $R$-modules. Finally, we define the map

$B_{N, t} \to K_ n \otimes _ R K_ t = R\langle y_1, \ldots , y_ r\rangle \otimes _ R R\langle x_1, \ldots , x_ r\rangle$

as the tensor product of the maps constructed in the case of $r = 1$ or simply by the rules $x_ i \mapsto 1 \otimes x_ i$, $z_ i \mapsto f_ i^{N - n}y_ i \otimes 1$, and $u_ i \mapsto - f_ i^{N - n - t}y_ i \otimes x_ i$ which makes sense as long as $N \geq n + t$. We omit the details. $\square$

 Hint: setting $z' = z - f^{N - t}x$ we see that $R\langle x, z\rangle = R\langle x, z'\rangle$ with $\text{d}(z') = 0$ and moreover the map $R\langle x, z'\rangle \to K_ t$ is the map killing $z'$.

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