The Stacks project

Lemma 91.5.1. 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

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

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

  3. $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. $\square$


Comments (0)


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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08PW. Beware of the difference between the letter 'O' and the digit '0'.