Lemma 37.22.1. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point with image $s \in S$. Let $h \in \mathfrak m_ x \subset \mathcal{O}_{X, x}$. Assume

$f$ is locally of finite presentation,

$f$ is flat at $x$, and

the image $\overline{h}$ of $h$ in $\mathcal{O}_{X_ s, x} = \mathcal{O}_{X, x}/\mathfrak m_ s\mathcal{O}_{X, x}$ is a nonzerodivisor.

Then there exists an affine open neighbourhood $U \subset X$ of $x$ such that $h$ comes from $h \in \Gamma (U, \mathcal{O}_ U)$ and such that $D = V(h)$ is an effective Cartier divisor in $U$ with $x \in D$ and $D \to S$ flat and locally of finite presentation.

**Proof.**
We are going to prove this by reducing to the Noetherian case. By openness of flatness (see Theorem 37.15.1) we may assume, after replacing $X$ by an open neighbourhood of $x$, that $X \to S$ is flat. We may also assume that $X$ and $S$ are affine. After possible shrinking $X$ a bit we may assume that there exists an $h \in \Gamma (X, \mathcal{O}_ X)$ which maps to our given $h$.

We may write $S = \mathop{\mathrm{Spec}}(A)$ and we may write $A = \mathop{\mathrm{colim}}\nolimits _ i A_ i$ as a directed colimit of finite type $\mathbf{Z}$ algebras. Then by Algebra, Lemma 10.168.1 or Limits, Lemmas 32.10.1, 32.8.2, and 32.10.1 we can find a cartesian diagram

\[ \xymatrix{ X \ar[r] \ar[d]_ f & X_0 \ar[d]^{f_0} \\ S \ar[r] & S_0 } \]

with $f_0$ flat and of finite presentation, $X_0$ affine, and $S_0$ affine and Noetherian. Let $x_0 \in X_0$, resp. $s_0 \in S_0$ be the image of $x$, resp. $s$. We may also assume there exists an element $h_0 \in \Gamma (X_0, \mathcal{O}_{X_0})$ which restricts to $h$ on $X$. (If you used the algebra reference above then this is clear; if you used the references to the chapter on limits then this follows from Limits, Lemma 32.10.1 by thinking of $h$ as a morphism $X \to \mathbf{A}^1_ S$.) Note that $\mathcal{O}_{X_ s, x}$ is a localization of $\mathcal{O}_{(X_0)_{s_0}, x_0} \otimes _{\kappa (s_0)} \kappa (s)$, so that $\mathcal{O}_{(X_0)_{s_0}, x_0} \to \mathcal{O}_{X_ s, x}$ is a flat local ring map, in particular faithfully flat. Hence the image $\overline{h}_0 \in \mathcal{O}_{(X_0)_{s_0}, x_0}$ is contained in $\mathfrak m_{(X_0)_{s_0}, x_0}$ and is a nonzerodivisor. We claim that after replacing $X_0$ by a principal open neighbourhood of $x_0$ the element $h_0$ is a nonzerodivisor in $B_0 = \Gamma (X_0, \mathcal{O}_{X_0})$ such that $B_0/h_0B_0$ is flat over $A_0 = \Gamma (S_0, \mathcal{O}_{S_0})$. If so then

\[ 0 \to B_0 \xrightarrow {h_0} B_0 \to B_0/h_0B_0 \to 0 \]

is a short exact sequence of flat $A_0$-modules. Hence this remains exact on tensoring with $A$ (by Algebra, Lemma 10.39.12) and the lemma follows.

It remains to prove the claim above. The corresponding algebra statement is the following (we drop the subscript ${}_0$ here): Let $A \to B$ be a flat, finite type ring map of Noetherian rings. Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak p \subset A$. Assume $h \in \mathfrak q$ maps to a nonzerodivisor in $B_{\mathfrak q}/\mathfrak p B_{\mathfrak q}$. Goal: show that after possible replacing $B$ by $B_ g$ for some $g \in B$, $g \not\in \mathfrak q$ the element $h$ becomes a nonzerodivisor and $B/hB$ becomes flat over $A$. By Algebra, Lemma 10.99.2 we see that $h$ is a nonzerodivisor in $B_{\mathfrak q}$ and that $B_{\mathfrak q}/hB_{\mathfrak q}$ is flat over $A$. By openness of flatness, see Algebra, Theorem 10.129.4 or Theorem 37.15.1 we see that $B/hB$ is flat over $A$ after replacing $B$ by $B_ g$ for some $g \in B$, $g \not\in \mathfrak q$. Finally, let $I = \{ b \in B \mid hb = 0\} $ be the annihilator of $h$. Then $IB_{\mathfrak q} = 0$ as $h$ is a nonzerodivisor in $B_{\mathfrak q}$. Also $I$ is finitely generated as $B$ is Noetherian. Hence there exists a $g \in B$, $g \not\in \mathfrak q$ such that $IB_ g = 0$. After replacing $B$ by $B_ g$ we see that $h$ is a nonzerodivisor.
$\square$

## Comments (0)