## 90.5 Constructing a resolution

In the Noetherian finite type case we can construct a “small” simplicial resolution for finite type ring maps.

Lemma 90.5.1. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. Let $\mathcal{A}$ be the category of $A$-algebra maps $C \to B$. Let $n \geq 0$ and let $P_\bullet $ be a simplicial object of $\mathcal{A}$ such that

$P_\bullet \to B$ is a trivial Kan fibration of simplicial sets,

$P_ k$ is finite type over $A$ for $k \leq n$,

$P_\bullet = \text{cosk}_ n \text{sk}_ n P_\bullet $ as simplicial objects of $\mathcal{A}$.

Then $P_{n + 1}$ is a finite type $A$-algebra.

**Proof.**
Although the proof we give of this lemma is straightforward, it is a bit messy. To clarify the idea we explain what happens for low $n$ before giving the proof in general. For example, if $n = 0$, then (3) means that $P_1 = P_0 \times _ B P_0$. Since the ring map $P_0 \to B$ is surjective, this is of finite type over $A$ by More on Algebra, Lemma 15.5.1.

If $n = 1$, then (3) means that

\[ P_2 = \{ (f_0, f_1, f_2) \in P_1^3 \mid d_0f_0 = d_0f_1,\ d_1f_0 = d_0f_2,\ d_1f_1 = d_1f_2 \} \]

where the equalities take place in $P_0$. Observe that the triple

\[ (d_0f_0, d_1f_0, d_1f_1) = (d_0f_1, d_0f_2, d_1f_2) \]

is an element of the fibre product $P_0 \times _ B P_0 \times _ B P_0$ over $B$ because the maps $d_ i : P_1 \to P_0$ are morphisms over $B$. Thus we get a map

\[ \psi : P_2 \longrightarrow P_0 \times _ B P_0 \times _ B P_0 \]

The fibre of $\psi $ over an element $(g_0, g_1, g_2) \in P_0 \times _ B P_0 \times _ B P_0$ is the set of triples $(f_0, f_1, f_2)$ of $1$-simplices with $(d_0, d_1)(f_0) = (g_0, g_1)$, $(d_0, d_1)(f_1) = (g_0, g_2)$, and $(d_0, d_1)(f_2) = (g_1, g_2)$. As $P_\bullet \to B$ is a trivial Kan fibration the map $(d_0, d_1) : P_1 \to P_0 \times _ B P_0$ is surjective. Thus we see that $P_2$ fits into the cartesian diagram

\[ \xymatrix{ P_2 \ar[d] \ar[r] & P_1^3 \ar[d] \\ P_0 \times _ B P_0 \times _ B P_0 \ar[r] & (P_0 \times _ B P_0)^3 } \]

By More on Algebra, Lemma 15.5.2 we conclude. The general case is similar, but requires a bit more notation.

The case $n > 1$. By Simplicial, Lemma 14.19.14 the condition $P_\bullet = \text{cosk}_ n \text{sk}_ n P_\bullet $ implies the same thing is true in the category of simplicial $A$-algebras and hence in the category of sets (as the forgetful functor from $A$-algebras to sets commutes with limits). Thus

\[ P_{n + 1} = \mathop{\mathrm{Mor}}\nolimits (\Delta [n + 1], P_\bullet ) = \mathop{\mathrm{Mor}}\nolimits (\text{sk}_ n \Delta [n + 1], \text{sk}_ n P_\bullet ) \]

by Simplicial, Lemma 14.11.3 and Equation (14.19.0.1). We will prove by induction on $1 \leq k < m \leq n + 1$ that the ring

\[ Q_{k, m} = \mathop{\mathrm{Mor}}\nolimits (\text{sk}_ k \Delta [m], \text{sk}_ k P_\bullet ) \]

is of finite type over $A$. The case $k = 1$, $1 < m \leq n + 1$ is entirely similar to the discussion above in the case $n = 1$. Namely, there is a cartesian diagram

\[ \xymatrix{ Q_{1, m} \ar[d] \ar[r] & P_1^ N \ar[d] \\ P_0 \times _ B \ldots \times _ B P_0 \ar[r] & (P_0 \times _ B P_0)^ N } \]

where $N = {m + 1 \choose 2}$. We conclude as before.

Let $1 \leq k_0 \leq n$ and assume $Q_{k, m}$ is of finite type over $A$ for all $1 \leq k \leq k_0$ and $k < m \leq n + 1$. For $k_0 + 1 < m \leq n + 1$ we claim there is a cartesian square

\[ \xymatrix{ Q_{k_0 + 1, m} \ar[d] \ar[r] & P_{k_0 + 1}^ N \ar[d] \\ Q_{k_0, m} \ar[r] & Q_{k_0, k_0 + 1}^ N } \]

where $N$ is the number of nondegenerate $(k_0 + 1)$-simplices of $\Delta [m]$. Namely, to see this is true, think of an element of $Q_{k_0 + 1, m}$ as a function $f$ from the $(k_0 + 1)$-skeleton of $\Delta [m]$ to $P_\bullet $. We can restrict $f$ to the $k_0$-skeleton which gives the left vertical map of the diagram. We can also restrict to each nondegenerate $(k_0 + 1)$-simplex which gives the top horizontal arrow. Moreover, to give such an $f$ is the same thing as giving its restriction to $k_0$-skeleton and to each nondegenerate $(k_0 + 1)$-face, provided these agree on the overlap, and this is exactly the content of the diagram. Moreover, the fact that $P_\bullet \to B$ is a trivial Kan fibration implies that the map

\[ P_{k_0} \to Q_{k_0, k_0 + 1} = \mathop{\mathrm{Mor}}\nolimits (\partial \Delta [k_0 + 1], P_\bullet ) \]

is surjective as every map $\partial \Delta [k_0 + 1] \to B$ can be extended to $\Delta [k_0 + 1] \to B$ for $k_0 \geq 1$ (small argument about constant simplicial sets omitted). Since by induction hypothesis the rings $Q_{k_0, m}$, $Q_{k_0, k_0 + 1}$ are finite type $A$-algebras, so is $Q_{k_0 + 1, m}$ by More on Algebra, Lemma 15.5.2 once more.
$\square$

Proposition 90.5.2. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. There exists a simplicial $A$-algebra $P_\bullet $ with an augmentation $\epsilon : P_\bullet \to B$ such that each $P_ n$ is a polynomial algebra of finite type over $A$ and such that $\epsilon $ is a trivial Kan fibration of simplicial sets.

**Proof.**
Let $\mathcal{A}$ be the category of $A$-algebra maps $C \to B$. In this proof our simplicial objects and skeleton and coskeleton functors will be taken in this category.

Choose a polynomial algebra $P_0$ of finite type over $A$ and a surjection $P_0 \to B$. As a first approximation we take $P_\bullet = \text{cosk}_0(P_0)$. In other words, $P_\bullet $ is the simplicial $A$-algebra with terms $P_ n = P_0 \times _ A \ldots \times _ A P_0$. (In the final paragraph of the proof this simplicial object will be denoted $P^0_\bullet $.) By Simplicial, Lemma 14.32.3 the map $P_\bullet \to B$ is a trivial Kan fibration of simplicial sets. Also, observe that $P_\bullet = \text{cosk}_0 \text{sk}_0 P_\bullet $.

Suppose for some $n \geq 0$ we have constructed $P_\bullet $ (in the final paragraph of the proof this will be $P^ n_\bullet $) such that

$P_\bullet \to B$ is a trivial Kan fibration of simplicial sets,

$P_ k$ is a finitely generated polynomial algebra for $0 \leq k \leq n$, and

$P_\bullet = \text{cosk}_ n \text{sk}_ n P_\bullet $

By Lemma 90.5.1 we can find a finitely generated polynomial algebra $Q$ over $A$ and a surjection $Q \to P_{n + 1}$. Since $P_ n$ is a polynomial algebra the $A$-algebra maps $s_ i : P_ n \to P_{n + 1}$ lift to maps $s'_ i : P_ n \to Q$. Set $d'_ j : Q \to P_ n$ equal to the composition of $Q \to P_{n + 1}$ and $d_ j : P_{n + 1} \to P_ n$. We obtain a truncated simplicial object $P'_\bullet $ of $\mathcal{A}$ by setting $P'_ k = P_ k$ for $k \leq n$ and $P'_{n + 1} = Q$ and morphisms $d'_ i = d_ i$ and $s'_ i = s_ i$ in degrees $k \leq n - 1$ and using the morphisms $d'_ j$ and $s'_ i$ in degree $n$. Extend this to a full simplicial object $P'_\bullet $ of $\mathcal{A}$ using $\text{cosk}_{n + 1}$. By functoriality of the coskeleton functors there is a morphism $P'_\bullet \to P_\bullet $ of simplicial objects extending the given morphism of $(n + 1)$-truncated simplicial objects. (This morphism will be denoted $P^{n + 1}_\bullet \to P^ n_\bullet $ in the final paragraph of the proof.)

Note that conditions (b) and (c) are satisfied for $P'_\bullet $ with $n$ replaced by $n + 1$. We claim the map $P'_\bullet \to P_\bullet $ satisfies assumptions (1), (2), (3), and (4) of Simplicial, Lemmas 14.32.1 with $n + 1$ instead of $n$. Conditions (1) and (2) hold by construction. By Simplicial, Lemma 14.19.14 we see that we have $P_\bullet = \text{cosk}_{n + 1}\text{sk}_{n + 1}P_\bullet $ and $P'_\bullet = \text{cosk}_{n + 1}\text{sk}_{n + 1}P'_\bullet $ not only in $\mathcal{A}$ but also in the category of $A$-algebras, whence in the category of sets (as the forgetful functor from $A$-algebras to sets commutes with all limits). This proves (3) and (4). Thus the lemma applies and $P'_\bullet \to P_\bullet $ is a trivial Kan fibration. By Simplicial, Lemma 14.30.4 we conclude that $P'_\bullet \to B$ is a trivial Kan fibration and (a) holds as well.

To finish the proof we take the inverse limit $P_\bullet = \mathop{\mathrm{lim}}\nolimits P^ n_\bullet $ of the sequence of simplicial algebras

\[ \ldots \to P^2_\bullet \to P^1_\bullet \to P^0_\bullet \]

constructed above. The map $P_\bullet \to B$ is a trivial Kan fibration by Simplicial, Lemma 14.30.5. However, the construction above stabilizes in each degree to a fixed finitely generated polynomial algebra as desired.
$\square$

Lemma 90.5.3. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. Let $\pi $, $\underline{B}$ be as in (90.4.0.1). If $\mathcal{F}$ is an $\underline{B}$-module such that $\mathcal{F}(P, \alpha )$ is a finite $B$-module for all $\alpha : P = A[x_1, \ldots , x_ n] \to B$, then the cohomology modules of $L\pi _!(\mathcal{F})$ are finite $B$-modules.

**Proof.**
By Lemma 90.4.1 and Proposition 90.5.2 we can compute $L\pi _!(\mathcal{F})$ by a complex constructed out of the values of $\mathcal{F}$ on finite type polynomial algebras.
$\square$

Lemma 90.5.4. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. Then $H^ n(L_{B/A})$ is a finite $B$-module for all $n \in \mathbf{Z}$.

**Proof.**
Apply Lemmas 90.4.3 and 90.5.3.
$\square$

Lemma 90.5.6. Let $A \to B$ be a ring map. Let $\pi $, $\mathcal{O}$, $\underline{B}$ be as in (90.4.0.1). For any $\mathcal{O}$-module $\mathcal{F}$ we have

\[ L\pi _!(\mathcal{F}) = L\pi _!(Li^*\mathcal{F}) = L\pi _!(\mathcal{F} \otimes _\mathcal {O}^\mathbf {L} \underline{B}) \]

in $D(\textit{Ab})$.

**Proof.**
It suffices to verify the assumptions of Cohomology on Sites, Lemma 21.38.12 hold for $\mathcal{O} \to \underline{B}$ on $\mathcal{C}_{B/A}$. We will use the results of Remark 90.5.5 without further mention. Choose a resolution $P_\bullet $ of $B$ over $A$ to get a suitable cosimplicial object $U_\bullet $ of $\mathcal{C}_{B/A}$. Since $P_\bullet \to B$ induces a quasi-isomorphism on associated complexes of abelian groups we see that $L\pi _!\mathcal{O} = B$. On the other hand $L\pi _!\underline{B}$ is computed by $\underline{B}(U_\bullet ) = B$. This verifies the second assumption of Cohomology on Sites, Lemma 21.38.12 and we are done with the proof.
$\square$

Lemma 90.5.7. Let $A \to B$ be a ring map. Let $\pi $, $\mathcal{O}$, $\underline{B}$ be as in (90.4.0.1). We have

\[ L\pi _!(\mathcal{O}) = L\pi _!(\underline{B}) = B \quad \text{and}\quad L_{B/A} = L\pi _!(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}) = L\pi _!(\Omega _{\mathcal{O}/A}) \]

in $D(\textit{Ab})$.

**Proof.**
This is just an application of Lemma 90.5.6 (and the first equality on the right is Lemma 90.4.3).
$\square$

Here is a special case of the fundamental triangle that is easy to prove.

Lemma 90.5.8. Let $A \to B \to C$ be ring maps. If $B$ is a polynomial algebra over $A$, then there is a distinguished triangle $L_{B/A} \otimes _ B^\mathbf {L} C \to L_{C/A} \to L_{C/B} \to L_{B/A} \otimes _ B^\mathbf {L} C[1]$ in $D(C)$.

**Proof.**
We will use the observations of Remark 90.5.5 without further mention. Choose a resolution $\epsilon : P_\bullet \to C$ of $C$ over $B$ (for example the standard resolution). Since $B$ is a polynomial algebra over $A$ we see that $P_\bullet $ is also a resolution of $C$ over $A$. Hence $L_{C/A}$ is computed by $\Omega _{P_\bullet /A} \otimes _{P_\bullet , \epsilon } C$ and $L_{C/B}$ is computed by $\Omega _{P_\bullet /B} \otimes _{P_\bullet , \epsilon } C$. Since for each $n$ we have the short exact sequence $0 \to \Omega _{B/A} \otimes _ B P_ n \to \Omega _{P_ n/A} \to \Omega _{P_ n/B}$ (Algebra, Lemma 10.138.9) and since $L_{B/A} = \Omega _{B/A}[0]$ (Lemma 90.4.7) we obtain the result.
$\square$

Example 90.5.9. Let $A \to B$ be a ring map. In this example we will construct an “explicit” resolution $P_\bullet $ of $B$ over $A$ of length $2$. To do this we follow the procedure of the proof of Proposition 90.5.2, see also the discussion in Remark 90.5.5.

We choose a surjection $P_0 = A[u_ i] \to B$ where $u_ i$ is a set of variables. Choose generators $f_ t \in P_0$, $t \in T$ of the ideal $\mathop{\mathrm{Ker}}(P_0 \to B)$. We choose $P_1 = A[u_ i, x_ t]$ with face maps $d_0$ and $d_1$ the unique $A$-algebra maps with $d_ j(u_ i) = u_ i$ and $d_0(x_ t) = 0$ and $d_1(x_ t) = f_ t$. The map $s_0 : P_0 \to P_1$ is the unique $A$-algebra map with $s_0(u_ i) = u_ i$. It is clear that

\[ P_1 \xrightarrow {d_0 - d_1} P_0 \to B \to 0 \]

is exact, in particular the map $(d_0, d_1) : P_1 \to P_0 \times _ B P_0$ is surjective. Thus, if $P_\bullet $ denotes the $1$-truncated simplicial $A$-algebra given by $P_0$, $P_1$, $d_0$, $d_1$, and $s_0$, then the augmentation $\text{cosk}_1(P_\bullet ) \to B$ is a trivial Kan fibration. The next step of the procedure in the proof of Proposition 90.5.2 is to choose a polynomial algebra $P_2$ and a surjection

\[ P_2 \longrightarrow \text{cosk}_1(P_\bullet )_2 \]

Recall that

\[ \text{cosk}_1(P_\bullet )_2 = \{ (g_0, g_1, g_2) \in P_1^3 \mid d_0(g_0) = d_0(g_1), d_1(g_0) = d_0(g_2), d_1(g_1) = d_1(g_2)\} \]

Thinking of $g_ i \in P_1$ as a polynomial in $x_ t$ the conditions are

\[ g_0(0) = g_1(0),\quad g_0(f_ t) = g_2(0),\quad g_1(f_ t) = g_2(f_ t) \]

Thus $\text{cosk}_1(P_\bullet )_2$ contains the elements $y_ t = (x_ t, x_ t, f_ t)$ and $z_ t = (0, x_ t, x_ t)$. Every element $G$ in $\text{cosk}_1(P_\bullet )_2$ is of the form $G = H + (0, 0, g)$ where $H$ is in the image of $A[u_ i, y_ t, z_ t] \to \text{cosk}_1(P_\bullet )_2$. Here $g \in P_1$ is a polynomial with vanishing constant term such that $g(f_ t) = 0$ in $P_0$. Observe that

$g = x_ t x_{t'} - f_ t x_{t'}$ and

$g = \sum r_ t x_ t$ with $r_ t \in P_0$ if $\sum r_ t f_ t = 0$ in $P_0$

are elements of $P_1$ of the desired form. Let

\[ Rel = \mathop{\mathrm{Ker}}(\bigoplus \nolimits _{t \in T} P_0 \longrightarrow P_0),\quad (r_ t) \longmapsto \sum r_ tf_ t \]

We set $P_2 = A[u_ i, y_ t, z_ t, v_ r, w_{t, t'}]$ where $r = (r_ t) \in Rel$, with map

\[ P_2 \longrightarrow \text{cosk}_1(P_\bullet )_2 \]

given by $y_ t \mapsto (x_ t, x_ t, f_ t)$, $z_ t \mapsto (0, x_ t, x_ t)$, $v_ r \mapsto (0, 0, \sum r_ t x_ t)$, and $w_{t, t'} \mapsto (0, 0, x_ t x_{t'} - f_ t x_{t'})$. A calculation (omitted) shows that this map is surjective. Our choice of the map displayed above determines the maps $d_0, d_1, d_2 : P_2 \to P_1$. Finally, the procedure in the proof of Proposition 90.5.2 tells us to choose the maps $s_0, s_1 : P_1 \to P_2$ lifting the two maps $P_1 \to \text{cosk}_1(P_\bullet )_2$. It is clear that we can take $s_ i$ to be the unique $A$-algebra maps determined by $s_0(x_ t) = y_ t$ and $s_1(x_ t) = z_ t$.

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