In the case of a local homomorphism of local rings one can limit the diagrams for which the lifting property has to be checked. Please compare with Algebra, Lemma 10.139.2.

Lemma 15.37.1. Let $(R, \mathfrak m) \to (S, \mathfrak n)$ be a local homomorphism of local rings. The following are equivalent

$R \to S$ is formally smooth in the $\mathfrak n$-adic topology,

for every solid commutative diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/J \\ R \ar[r] \ar[u] & A \ar[u] } \]

of local homomorphisms of local rings where $J \subset A$ is an ideal of square zero, $\mathfrak m_ A^ n = 0$ for some $n > 0$, and $S \to A/J$ induces an isomorphism on residue fields, a dotted arrow exists which makes the diagram commute.

If $S$ is Noetherian these conditions are also equivalent to

same as in (2) but only for diagrams where in addition $A \to A/J$ is a small extension (Algebra, Definition 10.139.1).

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

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/J \\ R \ar[r] \ar[u] & A \ar[u] } \]

as in Definition 15.36.1 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

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A'/J' \\ R \ar[r] \ar[u] & A' \ar[u] } \]

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

\[ A \to A/\mathfrak m_ A^{n - 1}J \to \ldots \to A/\mathfrak mJ \to A/J \]

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

\[ S \longrightarrow \prod \nolimits _{J_0 \text{ as above}} A/J_0 \]

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

\[ J' = \prod \nolimits _{J_0 \text{ as above}} J/J_0 \]

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

\[ A' = A \oplus M \]

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.
$\square$

The following lemma will be improved on in Section 15.39.

Lemma 15.37.2. Let $k$ be a field and let $(A, \mathfrak m, K)$ be a Noetherian local $k$-algebra. If $k \to A$ is formally smooth for the $\mathfrak m$-adic topology, then $A$ is a regular local ring.

**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 10.152.7. By Lemmas 15.36.2 and 15.36.7 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 10.96.3 and 10.109.9 it suffices to prove the completion $A^\wedge $ is regular. By Lemma 15.36.4 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 10.154.8) 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

\[ A/\mathfrak m^2 \cong K[x_1, \ldots , x_ n]/(x_ ix_ j) \]

and

\[ T/\mathfrak m_ T^2 \cong K[X_1, \ldots , X_ n]/(X_ iX_ j). \]

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.)
$\square$

Lemma 15.37.3. Let $k$ be a field. Let $(A, \mathfrak m, \kappa )$ be a complete local $k$-algebra. If $\kappa /k$ is separable, then there exists a $k$-algebra map $\kappa \to A$ such that $\kappa \to A \to \kappa $ is $\text{id}_\kappa $.

**Proof.**
By Algebra, Proposition 10.152.9 the extension $\kappa /k$ is formally smooth. By Lemma 15.36.2 $k \to \kappa $ is formally smooth in the sense of Definition 15.36.1. Then we get $\kappa \to A$ from Lemma 15.36.5.
$\square$

Lemma 15.37.4. Let $k$ be a field. Let $(A, \mathfrak m, \kappa )$ be a complete local $k$-algebra. If $\kappa /k$ is separable and $A$ regular, then there exists an isomorphism of $A \cong \kappa [[t_1, \ldots , t_ d]]$ as $k$-algebras.

**Proof.**
Choose $\kappa \to A$ as in Lemma 15.37.3 and apply Algebra, Lemma 10.154.10.
$\square$

The following result will be improved on in Section 15.39

Lemma 15.37.5. Let $k$ be a field. Let $(A, \mathfrak m, K)$ be a regular local $k$-algebra such that $K/k$ is separable. Then $k \to A$ is formally smooth in the $\mathfrak m$-adic topology.

**Proof.**
It suffices to prove that the completion of $A$ is formally smooth over $k$, see Lemma 15.36.4. Hence we may assume that $A$ is a complete local regular $k$-algebra with residue field $K$ separable over $k$. By Lemma 15.37.4 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 10.136.3. 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 15.36.2. 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 15.36.4.
$\square$

Lemma 15.37.6. Let $A \to B$ be a finite type ring map with $A$ Noetherian. Let $\mathfrak q \subset B$ be a prime ideal lying over $\mathfrak p \subset A$. The following are equivalent

$A \to B$ is smooth at $\mathfrak q$, and

$A_\mathfrak p \to B_\mathfrak q$ is formally smooth in the $\mathfrak q$-adic topology.

**Proof.**
The implication (2) $\Rightarrow $ (1) follows from Algebra, Lemma 10.139.2. 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 10.136.13. Hence $A_\mathfrak p \to B_\mathfrak q$ is formally smooth as localization preserves formal smoothness (for example by the criterion of Algebra, Proposition 10.136.8 and the fact that the cotangent complex behaves well with respect to localization, see Algebra, Lemmas 10.132.11 and 10.132.13). Finally, Lemma 15.36.2 implies that $A_\mathfrak p \to B_\mathfrak q$ is formally smooth in the $\mathfrak q$-adic topology.
$\square$

## Comments (0)