The Stacks project

Proof. The implication (1) $\Rightarrow $ (2) follows from the definitions. Consider a diagram

as in Definition 15.37.1 for the $\mathfrak m$-adic topology on $R$ and the $\mathfrak n$-adic topology on $S$. Pick $m > 0$ with $\mathfrak n^ m(A/J) = 0$ (possible by continuity of maps in diagram). Consider the subring $A'$ of $A$ which is the inverse image of the image of $S$ in $A/J$. Set $J' = J$ viewed as an ideal in $A'$. Then $J'$ is an ideal of square zero in $A'$ and $A'/J'$ is a quotient of $S/\mathfrak n^ m$. Hence $A'$ is local and $\mathfrak m_{A'}^{2m} = 0$. Thus we get a diagram

as in (2). If we can construct the dotted arrow in this diagram, then we obtain the dotted arrow in the original one by composing with $A' \to A$. In this way we see that (2) implies (1).

Assume $S$ Noetherian. The implication (1) $\Rightarrow $ (3) is immediate. Assume (3) and suppose a diagram as in (2) is given. Then $\mathfrak m_ A^ n J = 0$ for some $n > 0$. Considering the maps

we see that it suffices to produce the lifting if $\mathfrak m_ A J = 0$. Assume $\mathfrak m_ A J = 0$ and let $A' \subset A$ be the ring constructed above. Then $A'/J'$ is Artinian as a quotient of the Artinian local ring $S/\mathfrak n^ m$. Thus it suffices to show that given property (3) we can find the dotted arrow in diagrams as in (2) with $A/J$ Artinian and $\mathfrak m_ A J = 0$. Let $\kappa $ be the common residue field of $A$, $A/J$, and $S$. By (3), if $J_0 \subset J$ is an ideal with $\dim _\kappa (J/J_0) = 1$, then we can produce a dotted arrow $S \to A/J_0$. Taking the product we obtain

Clearly the image of this arrow is contained in the sub $R$-algebra $A'$ of elements which map into the small diagonal $A/J \subset \prod _{J_0} A/J$. Let $J' \subset A'$ be the elements mapping to zero in $A/J$. Then $J'$ is an ideal of square zero and as $\kappa $-vector space equal to

Thus the map $J \to J'$ is injective. By the theory of vector spaces we can choose a splitting $J' = J \oplus M$. It follows that

as an $R$-algebra. Hence the map $S \to A'$ can be composed with the projection $A' \to A$ to give the desired dotted arrow thereby finishing the proof of the lemma. $\square$

Proof. Let $k_0 \subset k$ be the prime field. Then $k_0$ is perfect, hence $k / k_0$ is separable, hence formally smooth by Algebra, Lemma 10.158.7. By Lemmas 15.37.2 and 15.37.7 we see that $k_0 \to A$ is formally smooth for the $\mathfrak m$-adic topology on $A$. Hence we may assume $k = \mathbf{Q}$ or $k = \mathbf{F}_ p$.

By Algebra, Lemmas 10.97.3 and 10.110.9 it suffices to prove the completion $A^\wedge $ is regular. By Lemma 15.37.4 we may replace $A$ by $A^\wedge $. Thus we may assume that $A$ is a Noetherian complete local ring. By the Cohen structure theorem (Algebra, Theorem 10.160.8) there exist a map $K \to A$. As $k$ is the prime field we see that $K \to A$ is a $k$-algebra map.

Let $x_1, \ldots , x_ n \in \mathfrak m$ be elements whose images form a basis of $\mathfrak m/\mathfrak m^2$. Set $T = K[[X_1, \ldots , X_ n]]$. Note that

and

Let $A/\mathfrak m^2 \to T/m_ T^2$ be the local $K$-algebra isomorphism given by mapping the class of $x_ i$ to the class of $X_ i$. Denote $f_1 : A \to T/\mathfrak m_ T^2$ the composition of this isomorphism with the quotient map $A \to A/\mathfrak m^2$. The assumption that $k \to A$ is formally smooth in the $\mathfrak m$-adic topology means we can lift $f_1$ to a map $f_2 : A \to T/\mathfrak {m}_ T^3$, then to a map $f_3 : A \to T/\mathfrak {m}_ T^4$, and so on, for all $n \geq 1$. Warning: the maps $f_ n$ are continuous $k$-algebra maps and may not be $K$-algebra maps. We get an induced map $f : A \to T = \mathop{\mathrm{lim}}\nolimits T/\mathfrak m_ T^ n$ of local $k$-algebras. By our choice of $f_1$, the map $f$ induces an isomorphism $\mathfrak m/\mathfrak m^2 \to \mathfrak m_ T/\mathfrak m_ T^2$ hence each $f_ n$ is surjective and we conclude $f$ is surjective as $A$ is complete. This implies $\dim (A) \geq \dim (T) = n$. Hence $A$ is regular by definition. (It also follows that $f$ is an isomorphism.) $\square$

Proof. By Algebra, Proposition 10.158.9 the extension $\kappa /k$ is formally smooth. By Lemma 15.37.2 $k \to \kappa $ is formally smooth in the sense of Definition 15.37.1. Then we get $\kappa \to A$ from Lemma 15.37.5. $\square$

Proof. Choose $\kappa \to A$ as in Lemma 15.38.3 and apply Algebra, Lemma 10.160.10. $\square$

Proof. It suffices to prove that the completion of $A$ is formally smooth over $k$, see Lemma 15.37.4. Hence we may assume that $A$ is a complete local regular $k$-algebra with residue field $K$ separable over $k$. By Lemma 15.38.4 we see that $A = K[[x_1, \ldots , x_ n]]$.

The power series ring $K[[x_1, \ldots , x_ n]]$ is formally smooth over $k$. Namely, $K$ is formally smooth over $k$ and $K[x_1, \ldots , x_ n]$ is formally smooth over $K$ as a polynomial algebra. Hence $K[x_1, \ldots , x_ n]$ is formally smooth over $k$ by Algebra, Lemma 10.138.3. It follows that $k \to K[x_1, \ldots , x_ n]$ is formally smooth for the $(x_1, \ldots , x_ n)$-adic topology by Lemma 15.37.2. Finally, it follows that $k \to K[[x_1, \ldots , x_ n]]$ is formally smooth for the $(x_1, \ldots , x_ n)$-adic topology by Lemma 15.37.4. $\square$

Proof. The implication (2) $\Rightarrow $ (1) follows from Algebra, Lemma 10.141.2. Conversely, if $A \to B$ is smooth at $\mathfrak q$, then $A \to B_ g$ is smooth for some $g \in B$, $g \not\in \mathfrak q$. Then $A \to B_ g$ is formally smooth by Algebra, Proposition 10.138.13. Hence $A_\mathfrak p \to B_\mathfrak q$ is formally smooth as localization preserves formal smoothness (for example by the criterion of Algebra, Proposition 10.138.8 and the fact that the cotangent complex behaves well with respect to localization, see Algebra, Lemmas 10.134.11 and 10.134.13). Finally, Lemma 15.37.2 implies that $A_\mathfrak p \to B_\mathfrak q$ is formally smooth in the $\mathfrak q$-adic topology. $\square$