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

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 flat*^{1} 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

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

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

is the desired map.

**Proof.**
Set $T^1(A) = T(A)$ and $T^{n + 1}(A) = T(T^ n(A))$. Let

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.

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.

**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. $\square$

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