Remark 88.5.2 (0GAP): Lifting maps

Remark 88.5.2 (Lifting maps). Let $A$ be a Noetherian ring and $I \subset A$ be an ideal. Let $B$ be an object of (88.2.0.2). Let $C$ be an $I$-adically complete $A$-algebra. Let $\psi _ n : B \to C/I^ nC$ be an $A$-algebra homomorphism. The obstruction to lifting $\psi _ n$ to an $A$-algebra homomorphism into $C/I^{2n}C$ is an element

\[ o(\psi _ n) \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^ nC/I^{2n}C) \]

as we will explain. Namely, choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Choose a lift $\psi : A[x_1, \ldots , x_ r]^\wedge \to C$ of $\psi _ n$. Since $\psi (J) \subset I^ nC$ we get $\psi (J^2) \subset I^{2n}C$ and hence we get a $B$-linear homomorphism

\[ o(\psi ) : J/J^2 \longrightarrow I^ nC/I^{2n}C, \quad g \longmapsto \psi (g) \]

which of course extends to a $C$-linear map $J/J^2 \otimes _ B C \to I^ nC/I^{2n}C$. Since $\mathop{N\! L}\nolimits _{B/A}^\wedge = (J/J^2 \to \bigoplus B \text{d}x_ i)$ we get $o(\psi _ n)$ as the image of $o(\psi )$ by the identification

\begin{align*} & \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^ nC/I^{2n}C) \\ & = \mathop{\mathrm{Coker}}\left(\mathop{\mathrm{Hom}}\nolimits _ B(\bigoplus B\text{d}x_ i, I^ nC/I^{2n}C) \to \mathop{\mathrm{Hom}}\nolimits _ B(J/J^2, I^ nC/I^{2n}C)\right) \end{align*}

See More on Algebra, Lemma 15.84.4 part (1) for the equality.

Suppose that $o(\psi _ n)$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^{n'}C/I^{2n'}C)$ for some integer $n'$ with $n > n' > n/2$. We claim that this means we can find an $A$-algebra homomorphism $\psi '_{2n'} : B \to C/I^{2n'}C$ which agrees with $\psi _ n$ as maps into $C/I^{n'}C$. The extreme case $n' = n$ explains why we previously said $o(\psi _ n)$ is the obstruction to lifting $\psi _ n$ to $C/I^{2n}C$. Proof of the claim: the hypothesis that $o(\psi _ n)$ maps to zero tells us we can find a $B$-module map

\[ h : \bigoplus B\text{d}x_ i \longrightarrow I^{n'}C/I^{2n'}C \]

such that $o(\psi )$ and $h \circ \text{d}$ agree as maps into $I^{n'}C/I^{2n'}C$. Say $h(\text{d}x_ i) = \delta _ i \bmod I^{2n'}C$ for some $\delta _ i \in I^{n'}C$. Then we look at the map

\[ \psi ' : A[x_1, \ldots , x_ r]^\wedge \to C,\quad x_ i \longmapsto \psi (x_ i) - \delta _ i \]

A computation with power series shows that $\psi '(J) \subset I^{2n'}C$. Namely, for $g \in J$ we get

\[ \psi '(g) \equiv \psi (g) - \sum \psi (\partial g/\partial x_ i)\delta _ i \equiv o(\psi )(g) - (h \circ \text{d})(g) \equiv 0 \bmod I^{2n'}C \]

See Remark 88.5.1 for the first equality. Hence $\psi '$ induces an $A$-algebra homomorphism $\psi '_{2n'} : B \to C/I^{2n'}C$ as desired.

The Stacks project

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