Lemma 15.38.1. 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 10.141.1).

**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 15.37.1 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

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

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.
$\square$

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