The Stacks project

Proof. The question is local on $X$ hence we may assume $X$, $Y$, $S$ are affine. Say $S = \mathop{\mathrm{Spec}}(R)$, $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(B)$. Write $B = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $x$. The local ring $\mathcal{O}_{X, x} = A_\mathfrak p$ is a G-ring by More on Algebra, Proposition 15.50.10. The map $\varphi $ is a map

where $\mathfrak q \subset B$ is the prime corresponding to $y$. Let $a_1, \ldots , a_ n \in A_\mathfrak p^\wedge $ be the images of $x_1, \ldots , x_ n$ via $R[x_1, \ldots , x_ n] \to B \to B_\mathfrak q^\wedge \to A_\mathfrak p^\wedge $. Then we can apply Smoothing Ring Maps, Lemma 16.13.4 to get an étale ring map $A \to A'$ and a prime ideal $\mathfrak p' \subset A'$ and $b_1, \ldots , b_ n \in A'$ such that $\kappa (\mathfrak p) = \kappa (\mathfrak p')$, $a_ i - b_ i \in (\mathfrak p')^ N(A'_{\mathfrak p'})^\wedge $, and $f_ j(b_1, \ldots , b_ n) = 0$ for $j = 1, \ldots , n$. This determines an $R$-algebra map $B \to A'$ by sending the class of $x_ i$ to $b_ i \in A'$. This finishes the proof by taking $U = \mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(B)$ as the morphism $f$ and $u = \mathfrak p'$. $\square$

Proof. We may assume $N \geq 2$. Apply Lemma 37.39.1 to get $(U, u) \to (X, x)$ and $f : (U, u) \to (Y, y)$. We claim that $f$ is étale at $u$ which will finish the proof. In fact, we will show that the induced map $\mathcal{O}_{Y, y}^\wedge \to \mathcal{O}_{U, u}^\wedge $ is an isomorphism. Having proved this, Lemma 37.12.1 will show that $f$ is smooth at $u$ and of course $f$ is unramified at $u$ as well, so Morphisms, Lemma 29.36.5 tells us $f$ is étale at $u$. For a local ring $(R, \mathfrak m)$ we set $\text{Gr}_\mathfrak m(R) = \bigoplus _{n \geq 0} \mathfrak m^ n/\mathfrak m^{n + 1}$. To prove the claim we look at the induced diagram of graded rings

Since $N \geq 2$ this diagram is actually commutative as the displayed graded algebras are generated in degree $1$! By assumption the lower arrow is an isomorphism. By More on Algebra, Lemma 15.43.9 (for example) the map $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{U, u}^\wedge $ is an isomorphism and hence the north-west arrow in the diagram is an isomorphism. We conclude that $f$ induces an isomorphism $\text{Gr}_{\mathfrak m_ x}(\mathcal{O}_{X, x}) \to \text{Gr}_{\mathfrak m_ y}(\mathcal{O}_{U, u})$. Using induction and the short exact sequences

for both local rings we conclude (from the snake lemma) that $f$ induces isomorphisms $\mathcal{O}_{Y, y}/\mathfrak m_ y^ n \to \mathcal{O}_{U, u}/\mathfrak m_ u^ n$ for all $n$ which is what we wanted to show. $\square$

Proof. Denote $X_\sigma = X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^\wedge )$ and $x_\sigma \in X_\sigma $ the unique point lying over $x$. Observe that $\mathcal{O}_{S, s}^\wedge $ is a G-ring by More on Algebra, Proposition 15.50.6. By Lemma 37.39.2 we can choose

where both arrows are elementary étale neighbourhoods.

After replacing $S$ by an open neighbourhood of $s$, we may assume $S = \mathop{\mathrm{Spec}}(R)$ is affine. Since $\mathcal{O}_{S, s}$ is a G-ring by Smoothing Ring Maps, Theorem 16.12.1 the ring $\mathcal{O}_{S, s}^\wedge $ is a filtered colimit of smooth $R$-algebras. Thus we can write

as a directed limit of affine schemes $S_ i$ smooth over $S$. Denote $s_ i \in S_ i$ the image of the closed point of $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^\wedge )$. Observe that $\kappa (s) = \kappa (s_ i)$. Set $X_ i = X \times _ S S_ i$ and denote $x_ i \in X_ i$ the unique point mapping to $x$. Note that $\kappa (x) = \kappa (x_ i)$. Since $T$ is of finite type over $\mathbf{Z}$ by Limits, Proposition 32.6.1 we can choose an $i$ and a morphism $\sigma _ i : (S_ i, s_ i) \to (T, t)$ of pointed schemes whose composition with $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^\wedge ) \to S_ i$ is equal to $\sigma $. Set $Y_ i = Y \times _ T S_ i$ and denote $y_ i$ the image of $y_\sigma $. Note that $\kappa (y_ i) = \kappa (y_\sigma )$. By Limits, Lemma 32.10.1 we can choose an $i$ and a diagram

whose base change to $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^\wedge )$ recovers $X_\sigma \leftarrow U \rightarrow Y_\sigma $. By Limits, Lemma 32.8.10 after increasing $i$ we may assume the morphisms $X_ i \leftarrow U_ i \rightarrow Y_ i$ are étale. Let $u_ i \in U_ i$ be the image of $u$. Then $u_ i \mapsto x_ i$ hence $\kappa (x) = \kappa (x_\sigma ) = \kappa (u) \supset \kappa (u_ i) \supset \kappa (x_ i) = \kappa (x)$ and we see that $\kappa (u_ i) = \kappa (x_ i)$. Hence $(X_ i, x_ i) \leftarrow (U_ i, u_ i)$ is an elementary étale neighbourhood. Since also $\kappa (y_ i) = \kappa (y_\sigma ) = \kappa (u)$ we see that also $(U_ i, u_ i) \to (Y_ i, y_ i)$ is an elementary étale neighbourhood.

At this point we have constructed a diagram

as in the statement of the lemma, except that $S_ i \to S$ is smooth. By Lemma 37.38.5 and after shrinking $S_ i$ we can assume there exists a closed subscheme $V \subset S_ i$ passing through $s_ i$ such that $V \to S$ is étale. Setting $W$ equal to the scheme theoretic inverse image of $V$ in $U_ i$ we conclude. $\square$

Proof. After replacing $X$, $S$, $T$, $Y$ by affine open subschemes we may assume the diagram in the statement of the lemma comes from applying $\mathop{\mathrm{Spec}}$ to a diagram

of Noetherian rings and finite type ring maps. In this proof every ring $E$ will be a Noetherian $R$-algebra endowed with a prime ideal $\mathfrak p_ E$ lying over $\mathfrak p_ R$ and all ring maps will be $R$-algebra maps compatible with the given primes. Moreover, if we write $E^\wedge $ we mean the completion of the localization of $E$ at $\mathfrak p_ E$. We will also use without further mention that an étale ring map $E_1 \to E_2$ such that $\kappa (\mathfrak p_{E_1}) = \kappa (\mathfrak p_{E_2})$ induces an isomorphism $E_1^\wedge = E_2^\wedge $ by More on Algebra, Lemma 15.43.9.

With this notation $\sigma $ and $\varphi $ correspond to ring maps

Here is a picture

Observe that $R^\wedge $ is a G-ring by More on Algebra, Proposition 15.50.6. Thus $B \otimes _{C, \sigma } R^\wedge $ is a G-ring by More on Algebra, Proposition 15.50.10. By Lemma 37.39.1 (translated into algebra) there exists an étale ring map $B \otimes _{C, \sigma } R^\wedge \to B'$ inducing an isomorphism $\kappa (\mathfrak p_{B \otimes _{C, \sigma } R^\wedge }) \to \kappa (\mathfrak p_{B'})$ and an $R$-algebra map $A \to B'$ such that the composition

is the same as $\varphi $ modulo $(\mathfrak p_{(B \otimes _{C, \sigma } R^\wedge )^\wedge })^ N$. Thus we may replace $\varphi $ by this composition because the only way $\varphi $ enters the conclusion is via the commutativity requirement in part (5) of the statement of the lemma. Picture:

Next, we use that $R^\wedge $ is a filtered colimit of smooth $R$-algebras (Smoothing Ring Maps, Theorem 16.12.1) because $R_{\mathfrak p_ R}$ is a G-ring by assumption. Since $C$ is of finite presentation over $R$ we get a factorization

for some $R \to R'$ smooth, see Algebra, Lemma 10.127.3. After increasing $R'$ we may assume there exists an étale $B \otimes _ C R'$-algebra $B''$ whose base change to $B \otimes _{C, \sigma } R^\wedge $ is $B'$, see Algebra, Lemma 10.143.3. Then $B'$ is the filtered colimit of these $B''$ and we conclude that after increasing $R'$ we may assume there is an $R$-algebra map $A \to B''$ such that $A \to B'' \to B'$ is the previously constructed map (same reference as above). Picture

and

This means that we may replace $C$ by $R'$, $\sigma : C \to R^\wedge $ by $R' \to R^\wedge $, and $B$ by $B''$ so that we simplify to the diagram

with $\varphi $ equal to the composition of the horizontal arrows followed by the canonical map from $B \otimes _{C, \sigma } R^\wedge $ to its completion. The final step in the proof is to apply Lemma 37.39.1 (or its proof) one more time to $\mathop{\mathrm{Spec}}(C)$ and $\mathop{\mathrm{Spec}}(R)$ over $\mathop{\mathrm{Spec}}(R)$ and the map $C \to R^\wedge $. The lemma produces a ring map $C \to D$ such that $R \to D$ is étale, such that $\kappa (\mathfrak p_ R) = \kappa (\mathfrak p_ D)$, and such that

is equal to $\sigma : C \to R^\wedge $ modulo $(\mathfrak p_{R^\wedge })^ N$. Then we can take

as our solution to the problem posed by the lemma. Namely the diagram

commutes because $C \to D \to D^\wedge = R^\wedge $ is equal to $\sigma $ modulo $(\mathfrak p_{R^\wedge })^ N$. This proves part (5) and the other properties are immediate from the construction. $\square$

Proof. We may assume $S$ and $T$ are affine. Say $S = \mathop{\mathrm{Spec}}(R)$ and $T = \mathop{\mathrm{Spec}}(C)$. Let $c_1, \ldots , c_ n \in C$ be generators of $C$ as an $R$-algebra. Let $\mathfrak p \subset R$ be the prime ideal corresponding to $s$. Say $\mathfrak p = (f_1, \ldots , f_ m)$. After replacing $R$ by a principal localization (to clear denominators in $R_\mathfrak p$) we may assume there exist $r_1, \ldots , r_ n \in R$ and $a_{i, I} \in \mathcal{O}_{S, s}^\wedge $ where $I = (i_1, \ldots , i_ m)$ with $\sum i_ j = N$ such that

in $\mathcal{O}_{S, s}^\wedge $. Then we consider

with $\mathfrak p' = \mathfrak pC' + (t_{i, I})$ and factorization of $\sigma : C \to \mathcal{O}_{S, s}^\wedge $ through $C'$ given by sending $t_{i, I}$ to $a_{i, I}$. Taking $T' = \mathop{\mathrm{Spec}}(C')$ works because any $\sigma '$ as in the statement of the lemma will send $c_ i$ to $r_ i$ modulo the maximal ideal to the power $N$. $\square$

Proof. A useful observation is that $\kappa (s) = \kappa (t)$ by the existence of $\sigma $. Observe that the statement makes sense as the fibres of $Y_\sigma $ and $Y_{\sigma '}$ over $s \in \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^\wedge )$ are both canonically isomorphic to $Y_ t$. We will think of the property “$\sigma '$ factors through $\mathcal{O}_{T, t} \to \mathcal{O}_{T', t'}$” as a constraint on $\sigma '$. If we have several such constraints, say given by $(T'_ i, t'_ i) \to (T, t)$, $i = 1, \ldots , n$ then we can combined them by considering $(T'_1 \times _ T \ldots \times _ T T'_ n, (t'_1, \ldots , t'_ n)) \to (T, t)$. We will use this without further mention in the following.

By Lemma 37.39.5 we can assume that any $\sigma '$ as in the statement of the lemma is the same as $\sigma $ modulo $\mathfrak m_ s^2$. Note that the conormal algebra of $Y_ t$ in $Y_\sigma $ is just the quasi-coherent graded $\mathcal{O}_{Y_ t}$-algebra

and similarly for $Y_{\sigma '}$. Since $\sigma $ and $\sigma '$ agree modulo $\mathfrak m_ s^2$ we see that these two algebras are the same in degrees $0$ and $1$. On the other hand, these conormal algebras are generated in degree $1$ over degree $0$. Hence if there is an isomorphism extending the isomorphism just constructed in degrees $0$ and $1$, then it is unique.

We may assume $S$ and $T$ are affine. Let $Y = Y_1 \cup \ldots \cup Y_ n$ be an affine open covering. If we can construct $(T_ i', t'_ i) \to (T, t)$ as in the lemma such that the desired isomorphism (see previous paragraph) exists for $Y_ i \to T \to S$ and $\sigma $, then these glue by uniqueness to prove the result for $Y \to T$. Thus we may assume $Y$ is affine.

Write $S = \mathop{\mathrm{Spec}}(R)$, $T = \mathop{\mathrm{Spec}}(C)$, and $Y = \mathop{\mathrm{Spec}}(B)$. Choose a presentation $B = C[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Denote $R^\wedge = \mathcal{O}_{S, s}^\wedge $. Let $a_{kj} \in R^\wedge [x_1, \ldots , x_ n]$ be polynomials such that

is a set of generators for the module of relations among the $\sigma (f_ j) \in R^\wedge [x_1, \ldots , x_ n]$. Thus we have an exact sequence

Let $c$ be an integer which works in the Artin-Rees lemma for both the first and the second map in this sequence and the ideal $\mathfrak m_{R^\wedge }R^\wedge [x_1, \ldots , x_ n]$ as defined in More on Algebra, Section 15.4. Write

in multiindex notation where $a_{kj, I} \in R^\wedge $, $f_{j, I} \in C$, and $\Omega $ a finite set of multiindices. Then we see that

in $R^\wedge $. Thus we take

Then $\sigma $ factors through a map $\tilde\sigma : C' \to R^\wedge $ sending $t_{kj, I}$ to $a_{jk, I}$. Thus $T' = \mathop{\mathrm{Spec}}(C')$ comes with a point $t' \in T'$ such that $\sigma $ factors through $\mathcal{O}_{T, t} \to \mathcal{O}_{T', t'}$. Let $t_{kj} = \sum t_{kj, I} x^ I$ in $C'[x_1, \ldots , x_ n]$. Then we see that we have a complex

which is exact at $C'[x_1, \ldots , x_ n]$ and whose base change by $\tilde\sigma $ gives (37.39.6.1).

By Lemma 37.39.5 we can find a further morphism $(T'', t'') \to (T', t')$ such that $\tilde\sigma $ factors through $\mathcal{O}_{T', t'} \to \mathcal{O}_{T'', t''}$ and such that if $\sigma ' : C \to R^\wedge $ factors through $\mathcal{O}_{T'', t''}$, then the induced map $\tilde\sigma ' : C' \to R^\wedge $ agrees modulo $\mathfrak m_ s^{c + 1}$ with $\tilde\sigma $. Thus if $\sigma '$ is such a map, then we obtain a complex

over $R^\wedge [x_1, \ldots , x_ n]$ by applying $\tilde\sigma '$ to the polynomials $t_{kj}$ and $f_ j$. In other words, this is the base change of the complex (37.39.6.2) by $\tilde\sigma '$. The matrices defining this complex are congruent modulo $\mathfrak m_ s^{c + 1}$ to the matrices defining the complex (37.39.6.1) because $\tilde\sigma $ and $\tilde\sigma '$ are congruent modulo $\mathfrak m_ s^{c + 1}$. Since (37.39.6.1) is exact, we can apply More on Algebra, Lemma 15.4.2 to conclude that

as desired. $\square$

Proof. We may assume $N \geq 2$ and we may replace $(T, t)$ with $(T', t')$ as in Lemma 37.39.6. Since $(V, v) \to (S, s)$ is an elementary étale neighbourhood, so is $(X \times _ S V, (x, v)) \to (X, x)$. Thus $\mathcal{O}_{X, x} \to \mathcal{O}_{X \times _ S V, (x, v)}$ induces an isomorphism on completions by More on Algebra, Lemma 15.43.9. We claim $\mathcal{O}_{X, x} \to \mathcal{O}_{W, w}$ induces an isomorphism on completions. Having proved this, Lemma 37.12.1 will show that $f$ is smooth at $w$ and of course $f$ is unramified at $u$ as well, so Morphisms, Lemma 29.36.5 tells us $f$ is étale at $w$.

First we use the commutativity in part (5) of Lemma 37.39.4 to see that for $i \leq N$ there is a commutative diagram

This implies that $f^\sharp _ w$ defines an isomorphism $\kappa (x) \to \kappa (w)$ on residue fields and an isomorphism $\mathfrak m_ x/\mathfrak m_ x^2 \to \mathfrak m_ w/\mathfrak m_ w^2$ on cotangent spaces. Hence $f^\sharp _ w$ defines a surjection $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{W, w}^\wedge $ on complete local rings.

By Lemma 37.39.6 there is an isomorphism of $\text{Gr}_{\mathfrak m_ s}(\mathcal{O}_{(Y \times _{T, \tau } V, (y, v)})$ with $\text{Gr}_{\mathfrak m_ s}(\mathcal{O}_{Y_\sigma , y_\sigma })$. This follows by taking stalks of the isomorphism of conormal sheaves at the point $y$. Since our local rings are Noetherian taking associated graded with respect to $\mathfrak m_ s$ commutes with completion because completion with respect to an ideal is an exact functor on finite modules over Noetherian rings. This produces the right vertical isomorphism in the diagram of graded rings

We do not claim the diagram commutes. By the result of the previous paragraph the left arrow is surjective. The other three arrows are isomorphisms. It follows that the left arrow is a surjective map between isomorphic Noetherian rings. Hence it is an isomorphism by Algebra, Lemma 10.31.10 (you can argue this directly using Hilbert functions as well). In particular $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{W, w}^\wedge $ must be injective as well as surjective which finishes the proof. $\square$