The Stacks project

Proof. Note that for $n \geq 2$ we have the equality of relative cotangent spaces

and similarly for $S$. Hence by Lemma 90.3.5 we see that $R_ n \to S_ n$ is surjective for all $n$. Now let $K_ n$ be the kernel of $R_ n \to S_ n$. Then the sequences

form an exact sequence of directed inverse systems. The system $(K_ n)$ is Mittag-Leffler since each $K_ n$ is Artinian. Hence by Algebra, Lemma 10.86.4 taking limits preserves exactness. So $\mathop{\mathrm{lim}}\nolimits R_ n \to \mathop{\mathrm{lim}}\nolimits S_ n$ is surjective, i.e., $f$ is surjective. $\square$

Proof. Let $R \to S_1$ and $R \to S_2$ be morphisms of $\widehat{\mathcal{C}}_\Lambda$. Consider the ring $C = S_1 \otimes _ R S_2$. This ring has a finitely generated maximal ideal $\mathfrak m = \mathfrak m_{S_1} \otimes S_2 + S_1 \otimes \mathfrak m_{S_2}$ with residue field $k$. Set $C^\wedge$ equal to the completion of $C$ with respect to $\mathfrak m$. Then $C^\wedge$ is a Noetherian ring complete with respect to the maximal ideal $\mathfrak m^\wedge = \mathfrak mC^\wedge$ whose residue field is identified with $k$, see Algebra, Lemma 10.97.5. Hence $C^\wedge$ is an object of $\widehat{\mathcal{C}}_\Lambda$. Then $S_1 \to C^\wedge$ and $S_2 \to C^\wedge$ turn $C^\wedge$ into a pushout over $R$ in $\widehat{\mathcal{C}}_\Lambda$ (details omitted). $\square$

Proof. Let $R$ and $S$ be objects of $\widehat{\mathcal{C}}_\Lambda$. Consider the ring $C = R \otimes _\Lambda S$. There is a canonical surjective map $C \to R \otimes _\Lambda S \to k \otimes _\Lambda k \to k$ where the last map is the multiplication map. The kernel of $C \to k$ is a maximal ideal $\mathfrak m$. Note that $\mathfrak m$ is generated by $\mathfrak m_ R C$, $\mathfrak m_ S C$ and finitely many elements of $C$ which map to generators of the kernel of $k \otimes _\Lambda k \to k$. Hence $\mathfrak m$ is a finitely generated ideal. Set $C^\wedge$ equal to the completion of $C$ with respect to $\mathfrak m$. Then $C^\wedge$ is a Noetherian ring complete with respect to the maximal ideal $\mathfrak m^\wedge = \mathfrak mC^\wedge$ with residue field $k$, see Algebra, Lemma 10.97.5. Hence $C^\wedge$ is an object of $\widehat{\mathcal{C}}_\Lambda$. Then $R \to C^\wedge$ and $S \to C^\wedge$ turn $C^\wedge$ into a coproduct in $\widehat{\mathcal{C}}_\Lambda$ (details omitted). $\square$

Proof. Let $x_1, \ldots , x_ n \in \mathfrak m_ S$ map to a $k$-basis for the relative cotangent space $\mathfrak m_ S/(\mathfrak m_\Lambda S + \mathfrak m_ S^2)$. Choose $y_1, \ldots , y_ m \in S$ whose images in $k$ generate $k$ over $k'$. We claim that $\dim _ k \text{Der}_\Lambda (S, k) \leq n + m$. To see this it suffices to prove that if $D(x_ i) = 0$ and $D(y_ j) = 0$, then $D = 0$. Let $a \in S$. We can find a polynomial $P = \sum \lambda _ J y^ J$ with $\lambda _ J \in \Lambda$ whose image in $k$ is the same as the image of $a$ in $k$. Then we see that $D(a - P) = D(a) - D(P) = D(a)$ by our assumption that $D(y_ j) = 0$ for all $j$. Thus we may assume $a \in \mathfrak m_ S$. Write $a = \sum a_ i x_ i$ with $a_ i \in S$. By the Leibniz rule

as we assumed $D(x_ i) = 0$. We have $\sum x_ iD(a_ i) = 0$ as multiplication by $x_ i$ is zero on $k$. $\square$

Proof. If $f$ is not surjective, then $\mathfrak m_ S/(\mathfrak m_ R S + \mathfrak m_ S^2)$ is nonzero by Lemma 90.4.2. Then also $Q = S/(f(R) + \mathfrak m_ R S + \mathfrak m_ S^2)$ is nonzero. Note that $Q$ is a $k = R/\mathfrak m_ R$-vector space via $f$. We turn $Q$ into an $S$-module via $S \to k$. The quotient map $D : S \to Q$ is an $R$-derivation: if $a_1, a_2 \in S$, we can write $a_1 = f(b_1) + a_1'$ and $a_2 = f(b_2) + a_2'$ for some $b_1, b_2 \in R$ and $a_1', a_2' \in \mathfrak m_ S$. Then $b_ i$ and $a_ i$ have the same image in $k$ for $i = 1, 2$ and

in $Q$ which proves the Leibniz rule. Hence $D : S \to Q$ is a $\Lambda$-derivation which is zero on composing with $R \to S$. Since $Q \not= 0$ there also exist derivations $D : S \to k$ which are zero on composing with $R \to S$, i.e., $\text{Der}_\Lambda (S, k) \to \text{Der}_\Lambda (R, k)$ is not injective. $\square$

Proof. It is clear that $\mathfrak m_{R/J}^ n \subset J_ n/J$. Thus it suffices to show that for every $n$ there exists an $N$ such that $J_ N/J \subset \mathfrak m_{R/J}^ n$. This is equivalent to $J_ N \subset \mathfrak m_ R^ n + J$. For each $n$ the ring $R/\mathfrak m_ R^ n$ is Artinian, hence there exists a $N_ n$ such that

Set $E_ n = (J_{N_ n} + \mathfrak m_ R^ n)/\mathfrak m_ R^ n$. Set $E = \mathop{\mathrm{lim}}\nolimits E_ n \subset \mathop{\mathrm{lim}}\nolimits R/\mathfrak m_ R^ n = R$. Note that $E \subset J$ as for any $f \in E$ and any $m$ we have $f \in J_ m + \mathfrak m_ R^ n$ for all $n \gg 0$, so $f \in J_ m$ by Krull's intersection theorem, see Algebra, Lemma 10.51.4. Since the transition maps $E_ n \to E_{n - 1}$ are all surjective, we see that $J$ surjects onto $E_ n$. Hence for $N = N_ n$ works. $\square$

Proof. We will use freely that the maps $S \to A_ n$ are surjective for all $n$. Note that the maps $\mathfrak m_{A_{n + 1}}/\mathfrak m_{A_{n + 1}}^2 \to \mathfrak m_{A_ n}/\mathfrak m_{A_ n}^2$ are surjective, see Lemma 90.4.2. Hence for $n$ sufficiently large the dimension $\dim _ k (\mathfrak m_{A_ n}/\mathfrak m_{A_ n}^2)$ stabilizes to an integer, say $r$. Thus we can find $x_1, \ldots , x_ r \in \mathfrak m_ S$ whose images in $A_ n$ generate $\mathfrak m_{A_ n}$. Moreover, pick $y_1, \ldots , y_ t \in S$ whose images in $k$ generate $k$ over $\Lambda$. Then we get a ring map $P = \Lambda [z_1, \ldots , z_{r + t}] \to S$, $z_ i \mapsto x_ i$ and $z_{r + j} \mapsto y_ j$ such that the composition $P \to S \to A_ n$ is surjective for all $n$. Let $\mathfrak m \subset P$ be the kernel of $P \to k$. Let $R = P^\wedge$ be the $\mathfrak m$-adic completion of $P$; this is an object of $\widehat{\mathcal{C}}_\Lambda$. Since we still have the compatible system of (surjective) maps $R \to A_ n$ we get a map $R \to S$. Set $J_ n = \mathop{\mathrm{Ker}}(R \to A_ n)$. Set $J = \bigcap J_ n$. By Lemma 90.4.7 we see that $R/J = \mathop{\mathrm{lim}}\nolimits R/J_ n = \mathop{\mathrm{lim}}\nolimits A_ n = S$ and that the ideals $J_ n/J = I_ n$ define the $\mathfrak m$-adic topology. (Note that for each $n$ we have $\mathfrak m_ R^{N_ n} \subset J_ n$ for some $N_ n$ and not necessarily $N_ n = n$, so a renumbering of the ideals $J_ n$ may be necessary before applying the lemma.) $\square$

Proof. As $R = R' \oplus I$ we have

and similarly

Hence for $n > 1$ the map $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ induces an isomorphism on cotangent spaces. Thus a left inverse $h_ n : R/\mathfrak m_ R^ n \to S/\mathfrak m_ S^ n$ is surjective by Lemma 90.4.2. Since $h_ n$ is injective as a left inverse it is an isomorphism. Thus the canonical surjections $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ are all isomorphisms and we win. $\square$