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

1. $f$ is locally of finite presentation,

2. $f$ is flat at $x$, and

3. 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. $\square$

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).