The Stacks project

Lemma 93.14.1. Let $A' \to A$ be a surjective map of Noetherian rings with nilpotent kernel. Let $A \to B$ be a finite type flat ring map. Let $\mathfrak b \subset B$ be an ideal such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is syntomic on the complement of $V(\mathfrak b)$. Let $(B^ h, \mathfrak b^ h)$ be the henselization of the pair $(B, \mathfrak b)$. Then $B$ has a flat lift to $A'$ if and only if $B^ h$ has a flat lift to $A'$.

First proof. This proof is a cheat. Namely, if $B$ has a flat lift $B'$, then taking the henselization $(B')^ h$ we obtain a flat lift of $B^ h$ (compare with the proof of Lemma 93.8.8). Conversely, suppose that $C'$ is an $A'$-flat lift of $(B')^ h$. Then let $\mathfrak c' \subset C'$ be the inverse image of the ideal $\mathfrak b^ h$. Then the completion $(C')^\wedge $ of $C'$ with respect to $\mathfrak c'$ is a lift of $B^\wedge $ (details omitted). Hence we see that $B$ has a flat lift by Lemma 93.12.3. $\square$

Second proof. Choose an $A$-algebra surjection $P = A[x_1, \ldots , x_ n] \to B$. Let $\mathfrak p \subset P$ be the inverse image of $\mathfrak b$. Set $P' = A'[x_1, \ldots , x_ n]$ and denote $\mathfrak p' \subset P'$ the inverse image of $\mathfrak p$. (Of course $\mathfrak p$ and $\mathfrak p'$ do not designate prime ideals here.) We will denote $P^ h$ and $(P')^ h$ the respective henselizations. We will use that taking henselizations is functorial and that the henselization of a quotient is the corresponding quotient of the henselization, see More on Algebra, Lemmas 15.11.16 and 15.12.7.

Suppose $A' \to B'$ is a flat lift of $A \to B$, in other words, $A' \to B'$ is flat and there is an $A$-algebra isomorphism $B = B' \otimes _{A'} A$. Then we can choose an $A'$-algebra map $P' \to B'$ lifting the given surjection $P \to B$. By Nakayama's lemma (Algebra, Lemma 10.20.1) we find that $B'$ is a quotient of $P'$. In particular, we find that we can endow $B'$ with an $A'$-flat $P'$-module structure lifting $B$ as an $A$-flat $P$-module. Conversely, if we can lift $B$ to a $P'$-module $M'$ flat over $A'$, then $M'$ is a cyclic module $M' \cong P'/J'$ (using Nakayama again) and setting $B' = P'/J'$ we find a flat lift of $B$ as an algebra.

Set $C = B^ h$ and $\mathfrak c = \mathfrak bC$. Suppose that $A' \to C'$ is a flat lift of $A \to C$. Then $C'$ is henselian with respect to the inverse image $\mathfrak c'$ of $\mathfrak c$ (by More on Algebra, Lemma 15.11.9 and the fact that the kernel of $C' \to C$ is nilpotent). We choose an $A'$-algebra map $P' \to C'$ lifting the $A$-algebra map $P \to C$. These maps pass through henselizations to give surjections $P^ h \to C$ and $(P')^ h \to C'$ (for the second again using Nakayama's lemma). In particular, we find that we can endow $C'$ with an $A'$-flat $(P')^ h$-module structure lifting $C$ as an $A$-flat $P^ h$-module. Conversely, if we can lift $C$ to a $(P')^ h$-module $N'$ flat over $A'$, then $N'$ is a cyclic module $N' \cong (P')^ h/\tilde J$ (using Nakayama again) and setting $C' = (P')^ h/\tilde J$ we find a flat lift of $C$ as an algebra.

Observe that $P' \to (P')^ h$ is a flat ring map which induces an isomorphism $P'/\mathfrak p' = (P')^ h/\mathfrak p'(P')^ h$ (More on Algebra, Lemma 15.12.2). We conclude that our lemma is a consequence of Lemma 93.12.2 provided we can show that $B_ g$ lifts to an $A'$-flat $P'_ g$-module for $g \in \mathfrak p'$. However, the ring map $A \to B_ g$ is syntomic and hence lifts to an $A'$-flat algebra $B'$ by Smoothing Ring Maps, Proposition 16.3.2. Since $A' \to P'_ g$ is smooth, we can lift $P_ g \to B_ g$ to a surjective map $P'_ g \to B'$ as before and we get what we want. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DZB. Beware of the difference between the letter 'O' and the digit '0'.