In the Noetherian finite type case we can construct a “small” simplicial resolution for finite type ring maps.
Lemma 92.5.1. 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
where the equalities take place in $P_0$. Observe that the triple
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
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
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
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
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
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
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
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 92.5.2. 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 92.5.1 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
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 92.5.3. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. Let $\pi $, $\underline{B}$ be as in (92.4.0.1). 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 92.4.1 and Proposition 92.5.2 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 92.5.4. 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 $n \in \mathbf{Z}$.
Remark 92.5.5 (Resolutions). Let $A \to B$ be any ring map. Let us call an augmented simplicial $A$-algebra $\epsilon : P_\bullet \to B$ a resolution of $B$ over $A$ if each $P_ n$ is a polynomial algebra and $\epsilon $ is a trivial Kan fibration of simplicial sets. If $P_\bullet \to B$ is an augmentation of a simplicial $A$-algebra with each $P_ n$ a polynomial algebra surjecting onto $B$, then the following are equivalent
$\epsilon : P_\bullet \to B$ is a resolution of $B$ over $A$,
$\epsilon : P_\bullet \to B$ is a quasi-isomorphism on associated complexes,
$\epsilon : P_\bullet \to B$ induces a homotopy equivalence of simplicial sets.
To see this use Simplicial, Lemmas 14.30.8, 14.31.9, and 14.31.8. A resolution $P_\bullet $ of $B$ over $A$ gives a cosimplicial object $U_\bullet $ of $\mathcal{C}_{B/A}$ as in Cohomology on Sites, Lemma 21.39.7 and it follows that
functorially in $\mathcal{F}$, see Lemma 92.4.1. The (formal part of the) proof of Proposition 92.5.2 shows that resolutions exist. We also have seen in the first proof of Lemma 92.4.2 that the standard resolution of $B$ over $A$ is a resolution (so that this terminology doesn't lead to a conflict). However, the argument in the proof of Proposition 92.5.2 shows the existence of resolutions without appealing to the simplicial computations in Simplicial, Section 14.34. Moreover, for any choice of resolution we have a canonical isomorphism
in $D(B)$ by Lemma 92.4.3. The freedom to choose an arbitrary resolution can be quite useful.
Lemma 92.5.6. Let $A \to B$ be a ring map. Let $\pi $, $\mathcal{O}$, $\underline{B}$ be as in (92.4.0.1). For any $\mathcal{O}$-module $\mathcal{F}$ we have in $D(\textit{Ab})$.
\[ L\pi _!(\mathcal{F}) = L\pi _!(Li^*\mathcal{F}) = L\pi _!(\mathcal{F} \otimes _\mathcal {O}^\mathbf {L} \underline{B}) \]
Proof. It suffices to verify the assumptions of Cohomology on Sites, Lemma 21.39.12 hold for $\mathcal{O} \to \underline{B}$ on $\mathcal{C}_{B/A}$. We will use the results of Remark 92.5.5 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 21.39.12 and we are done with the proof. $\square$
Lemma 92.5.7. Let $A \to B$ be a ring map. Let $\pi $, $\mathcal{O}$, $\underline{B}$ be as in (92.4.0.1). We have in $D(\textit{Ab})$.
\[ L\pi _!(\mathcal{O}) = L\pi _!(\underline{B}) = B \quad \text{and}\quad L_{B/A} = L\pi _!(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}) = L\pi _!(\Omega _{\mathcal{O}/A}) \]
Proof. This is just an application of Lemma 92.5.6 (and the first equality on the right is Lemma 92.4.3). $\square$
Here is a special case of the fundamental triangle that is easy to prove.
Lemma 92.5.8. 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 92.5.5 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 92.4.7) we obtain the result. $\square$
Example 92.5.9. 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 92.5.2, see also the discussion in Remark 92.5.5. 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 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 92.5.2 is to choose a polynomial algebra $P_2$ and a surjection Recall that Thinking of $g_ i \in P_1$ as a polynomial in $x_ t$ the conditions are 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 $P_0$
\[ P_1 \xrightarrow {d_0 - d_1} P_0 \to B \to 0 \]
\[ P_2 \longrightarrow \text{cosk}_1(P_\bullet )_2 \]
\[ \text{cosk}_1(P_\bullet )_2 = \{ (g_0, g_1, g_2) \in P_1^3 \mid d_0(g_0) = d_0(g_1), d_1(g_0) = d_0(g_2), d_1(g_1) = d_1(g_2)\} \]
\[ g_0(0) = g_1(0),\quad g_0(f_ t) = g_2(0),\quad g_1(f_ t) = g_2(f_ t) \]
are elements of $P_1$ of the desired form. Let
We set $P_2 = A[u_ i, y_ t, z_ t, v_ r, w_{t, t'}]$ where $r = (r_ t) \in Rel$, with map
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 92.5.2 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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