Lemma 61.8.1. Given a ring $A$ there exists a faithfully flat ind-étale $A$-algebra $C$ such that every faithfully flat étale ring map $C \to B$ has a retraction.
61.8 Constructing ind-étale algebras
Let $A$ be a ring. Recall that any étale ring map $A \to B$ is isomorphic to a standard smooth ring map of relative dimension $0$. Such a ring map is of the form
\[ A \longrightarrow A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n) \]
where the determinant of the $n \times n$-matrix with entries $\partial f_ i/\partial x_ j$ is invertible in the quotient ring. See Algebra, Lemma 10.143.2.
Let $S(A)$ be the set of all faithfully flat1 standard smooth $A$-algebras of relative dimension $0$. Let $I(A)$ be the partially ordered (by inclusion) set of finite subsets $E$ of $S(A)$. Note that $I(A)$ is a directed partially ordered set. For $E = \{ A \to B_1, \ldots , A \to B_ n\} $ set
\[ B_ E = B_1 \otimes _ A \ldots \otimes _ A B_ n \]
Observe that $B_ E$ is a faithfully flat étale $A$-algebra. For $E \subset E'$, there is a canonical transition map $B_ E \to B_{E'}$ of étale $A$-algebras. Namely, say $E = \{ A \to B_1, \ldots , A \to B_ n\} $ and $E' = \{ A \to B_1, \ldots , A \to B_{n + m}\} $ then $B_ E \to B_{E'}$ sends $b_1 \otimes \ldots \otimes b_ n$ to the element $b_1 \otimes \ldots \otimes b_ n \otimes 1 \otimes \ldots \otimes 1$ of $B_{E'}$. This construction defines a system of faithfully flat étale $A$-algebras over $I(A)$ and we set
\[ T(A) = \mathop{\mathrm{colim}}\nolimits _{E \in I(A)} B_ E \]
Observe that $T(A)$ is a faithfully flat ind-étale $A$-algebra (Algebra, Lemma 10.39.20). By construction given any faithfully flat étale $A$-algebra $B$ there is a (non-unique) $A$-algebra map $B \to T(A)$. Namely, pick some $(A \to B_0) \in S(A)$ and an isomorphism $B \cong B_0$. Then the canonical coprojection
\[ B \to B_0 \to T(A) = \mathop{\mathrm{colim}}\nolimits _{E \in I(A)} B_ E \]
is the desired map.
Proof. Set $T^1(A) = T(A)$ and $T^{n + 1}(A) = T(T^ n(A))$. Let
\[ C = \mathop{\mathrm{colim}}\nolimits T^ n(A) \]
This algebra is faithfully flat over each $T^ n(A)$ and in particular over $A$, see Algebra, Lemma 10.39.20. Moreover, $C$ is ind-étale over $A$ by Lemma 61.7.4. If $C \to B$ is étale, then there exists an $n$ and an étale ring map $T^ n(A) \to B'$ such that $B = C \otimes _{T^ n(A)} B'$, see Algebra, Lemma 10.143.3. If $C \to B$ is faithfully flat, then $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(T^ n(A))$ is surjective, hence $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(T^ n(A))$ is surjective. In other words, $T^ n(A) \to B'$ is faithfully flat. By our construction, there is a $T^ n(A)$-algebra map $B' \to T^{n + 1}(A)$. This induces a $C$-algebra map $B \to C$ which finishes the proof. $\square$
Remark 61.8.2. Let $A$ be a ring. Let $\kappa $ be an infinite cardinal bigger or equal than the cardinality of $A$. Then the cardinality of $T(A)$ is at most $\kappa $. Namely, each $B_ E$ has cardinality at most $\kappa $ and the index set $I(A)$ has cardinality at most $\kappa $ as well. Thus the result follows as $\kappa \otimes \kappa = \kappa $, see Sets, Section 3.6. It follows that the ring constructed in the proof of Lemma 61.8.1 has cardinality at most $\kappa $ as well.
Remark 61.8.3. The construction $A \mapsto T(A)$ is functorial in the following sense: If $A \to A'$ is a ring map, then we can construct a commutative diagram Namely, given $(A \to A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n))$ in $S(A)$ we can use the ring map $\varphi : A \to A'$ to obtain a corresponding element $(A' \to A'[x_1, \ldots , x_ n]/(f^\varphi _1, \ldots , f^\varphi _ n))$ of $S(A')$ where $f^\varphi $ means the polynomial obtained by applying $\varphi $ to the coefficients of the polynomial $f$. Moreover, there is a commutative diagram which is a in the category of rings. For $E \subset S(A)$ finite, set $E' = \varphi (E)$ and define $B_ E \to B_{E'}$ in the obvious manner. Taking the colimit gives the desired map $T(A) \to T(A')$, see Categories, Lemma 4.14.8.
\[ \xymatrix{ A \ar[r] \ar[d] & T(A) \ar[d] \\ A' \ar[r] & T(A') } \]
\[ \xymatrix{ A \ar[r] \ar[d] & A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n) \ar[d] \\ A' \ar[r] & A'[x_1, \ldots , x_ n]/(f^\varphi _1, \ldots , f^\varphi _ n) } \]
Lemma 61.8.4. Let $A$ be a ring such that every faithfully flat étale ring map $A \to B$ has a retraction. Then the same is true for every quotient ring $A/I$.
Proof. Let $A/I \to \overline{B}$ be faithfully flat étale. By Algebra, Lemma 10.143.10 we can write $\overline{B} = B/IB$ for some étale ring map $A \to B'$. The image $U$ of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is open and contains $V(I)$. Hence the complement $Z = \mathop{\mathrm{Spec}}(A) \setminus U$ is quasi-compact and disjoint from $V(I)$. Hence $Z \subset D(f_1) \cup \ldots \cup D(f_ r)$ for some $r \geq 0$ and $f_ i \in I$. Then $A \to B' = B \times \prod A_{f_ i}$ is faithfully flat étale and $\overline{B} = B'/IB'$. Hence the retraction $B' \to A$ to $A \to B'$, induces a retraction to $A/I \to \overline{B}$. $\square$
Lemma 61.8.5. Let $A$ be a ring such that every faithfully flat étale ring map $A \to B$ has a retraction. Then every local ring of $A$ at a maximal ideal is strictly henselian.
Proof. Let $\mathfrak m$ be a maximal ideal of $A$. Let $A \to B$ be an étale ring map and let $\mathfrak q \subset B$ be a prime lying over $\mathfrak m$. By the description of the strict henselization $A_\mathfrak m^{sh}$ in Algebra, Lemma 10.155.11 it suffices to show that $A_\mathfrak m = B_\mathfrak q$. Note that there are finitely many primes $\mathfrak q = \mathfrak q_1, \mathfrak q_2, \ldots , \mathfrak q_ n$ lying over $\mathfrak m$ and there are no specializations between them as an étale ring map is quasi-finite, see Algebra, Lemma 10.143.6. Thus $\mathfrak q_ i$ is a maximal ideal and we can find $g \in \mathfrak q_2 \cap \ldots \cap \mathfrak q_ n$, $g \not\in \mathfrak q$ (Algebra, Lemma 10.15.2). After replacing $B$ by $B_ g$ we see that $\mathfrak q$ is the only prime of $B$ lying over $\mathfrak m$. The image $U \subset \mathop{\mathrm{Spec}}(A)$ of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is open (Algebra, Proposition 10.41.8). Thus the complement $\mathop{\mathrm{Spec}}(A) \setminus U$ is closed and we can find $f \in A$, $f \not\in \mathfrak p$ such that $\mathop{\mathrm{Spec}}(A) = U \cup D(f)$. The ring map $A \to B \times A_ f$ is faithfully flat and étale, hence has a retraction $\sigma : B \times A_ f \to A$ by assumption on $A$. Observe that $\sigma $ is étale, hence flat as a map between étale $A$-algebras (Algebra, Lemma 10.143.8). Since $\mathfrak q$ is the only prime of $B \times A_ f$ lying over $A$ we find that $A_\mathfrak p \to B_\mathfrak q$ has a retraction which is also flat. Thus $A_\mathfrak p \to B_\mathfrak q \to A_\mathfrak p$ are flat local ring maps whose composition is the identity. Since a flat local homomorphism of local rings is injective we conclude these maps are isomorphisms as desired. $\square$
Lemma 61.8.6. Let $A$ be a ring such that every faithfully flat étale ring map $A \to B$ has a retraction. Let $Z \subset \mathop{\mathrm{Spec}}(A)$ be a closed subscheme. Let $A \to A_ Z^\sim $ be as constructed in Lemma 61.5.1. Then every faithfully flat étale ring map $A_ Z^\sim \to C$ has a retraction.
Proof. There exists an étale ring map $A \to B'$ such that $C = B' \otimes _ A A_ Z^\sim $ as $A_ Z^\sim $-algebras. The image $U' \subset \mathop{\mathrm{Spec}}(A)$ of $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$ is open and contains $V(I)$, hence we can find $f \in I$ such that $\mathop{\mathrm{Spec}}(A) = U' \cup D(f)$. Then $A \to B' \times A_ f$ is étale and faithfully flat. By assumption there is a retraction $B' \times A_ f \to A$. Localizing we obtain the desired retraction $C \to A_ Z^\sim $. $\square$
Lemma 61.8.7. Let $A \to B$ be a ring map inducing algebraic extensions on residue fields. There exists a commutative diagram with the following properties:
$A \to C$ is faithfully flat and ind-étale,
$B \to D$ is faithfully flat and ind-étale,
$\mathop{\mathrm{Spec}}(C)$ is w-local,
$\mathop{\mathrm{Spec}}(D)$ is w-local,
$\mathop{\mathrm{Spec}}(D) \to \mathop{\mathrm{Spec}}(C)$ is w-local,
the set of closed points of $\mathop{\mathrm{Spec}}(D)$ is the inverse image of the set of closed points of $\mathop{\mathrm{Spec}}(C)$,
the set of closed points of $\mathop{\mathrm{Spec}}(C)$ surjects onto $\mathop{\mathrm{Spec}}(A)$,
the set of closed points of $\mathop{\mathrm{Spec}}(D)$ surjects onto $\mathop{\mathrm{Spec}}(B)$,
for $\mathfrak m \subset C$ maximal the local ring $C_\mathfrak m$ is strictly henselian.
\[ \xymatrix{ B \ar[r] & D \\ A \ar[r] \ar[u] & C \ar[u] } \]
Proof. There is a faithfully flat, ind-Zariski ring map $A \to A'$ such that $\mathop{\mathrm{Spec}}(A')$ is w-local and such that the set of closed points of $\mathop{\mathrm{Spec}}(A')$ maps onto $\mathop{\mathrm{Spec}}(A)$, see Lemma 61.5.3. Let $I \subset A'$ be the ideal such that $V(I)$ is the set of closed points of $\mathop{\mathrm{Spec}}(A')$. Choose $A' \to C'$ as in Lemma 61.8.1. Note that the local rings $C'_{\mathfrak m'}$ at maximal ideals $\mathfrak m' \subset C'$ are strictly henselian by Lemma 61.8.5. We apply Lemma 61.5.8 to $A' \to C'$ and $I \subset A'$ to get $C' \to C$ with $C'/IC' \cong C/IC$. Note that since $A' \to C'$ is faithfully flat, $\mathop{\mathrm{Spec}}(C'/IC')$ surjects onto the set of closed points of $A'$ and in particular onto $\mathop{\mathrm{Spec}}(A)$. Moreover, as $V(IC) \subset \mathop{\mathrm{Spec}}(C)$ is the set of closed points of $C$ and $C' \to C$ is ind-Zariski (and identifies local rings) we obtain properties (1), (3), (7), and (9).
Denote $J \subset C$ the ideal such that $V(J)$ is the set of closed points of $\mathop{\mathrm{Spec}}(C)$. Set $D' = B \otimes _ A C$. The ring map $C \to D'$ induces algebraic residue field extensions. Keep in mind that since $V(J) \to \mathop{\mathrm{Spec}}(A)$ is surjective the map $T = V(JD) \to \mathop{\mathrm{Spec}}(B)$ is surjective too. Apply Lemma 61.5.8 to $C \to D'$ and $J \subset C$ to get $D' \to D$ with $D'/JD' \cong D/JD$. All of the remaining properties given in the lemma are immediate from the results of Lemma 61.5.8. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (1)
Comment #8849 by Wataru on