Remark 15.40.7. The assertion of Lemma 15.40.6 is quite strong. Namely, suppose that we have a diagram

of local homomorphisms of Noetherian complete local rings where $A \to A'$ induces an isomorphism of residue fields $k = A/\mathfrak m_ A = A'/\mathfrak m_{A'}$ and with $B \otimes _{A'} k$ formally smooth over $k$. Then we can extend this to a commutative diagram

of local homomorphisms of Noetherian complete local rings where $A \to C$ is formally smooth in the $\mathfrak m_ C$-adic topology and where $C \otimes _ A k \cong B \otimes _{A'} k$. Namely, pick $A \to C$ as in Lemma 15.40.6 lifting $B \otimes _{A'} k$ over $k$. By formal smoothness we can find the arrow $C \to B$, see Lemma 15.37.5. Denote $C \otimes _ A^\wedge A'$ the completion of $C \otimes _ A A'$ with respect to the ideal $C \otimes _ A \mathfrak m_{A'}$. Note that $C \otimes _ A^\wedge A'$ is a Noetherian complete local ring (see Algebra, Lemma 10.97.5) which is flat over $A'$ (see Algebra, Lemma 10.99.11). We have moreover

$C \otimes _ A^\wedge A' \to B$ is surjective,

if $A \to A'$ is surjective, then $C \to B$ is surjective,

if $A \to A'$ is finite, then $C \to B$ is finite, and

if $A' \to B$ is flat, then $C \otimes _ A^\wedge A' \cong B$.

Namely, by Nakayama's lemma for nilpotent ideals (see Algebra, Lemma 10.20.1) we see that $C \otimes _ A k \cong B \otimes _{A'} k$ implies that $C \otimes _ A A'/\mathfrak m_{A'}^ n \to B/\mathfrak m_{A'}^ nB$ is surjective for all $n$. This proves (1). Parts (2) and (3) follow from part (1). Part (4) follows from Algebra, Lemma 10.99.1.

## Comments (0)

There are also: