The Stacks project

Proof. Pick $x \in D$. It suffices to find an open neighbourhood $U \subset X$ of $x$ such that $f|_ U$ is flat. Hence the lemma reduces to the case that $X = \mathop{\mathrm{Spec}}(B)$ and $S = \mathop{\mathrm{Spec}}(A)$ are affine and that $D$ is given by a nonzerodivisor $h \in B$. By assumption $B$ is a finitely presented $A$-algebra and $B/hB$ is a flat $A$-algebra. We are going to use absolute Noetherian approximation.

Write $B = A[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. Assume $h$ is the image of $h' \in A[x_1, \ldots , x_ n]$. Choose a finite type $\mathbf{Z}$-subalgebra $A_0 \subset A$ such that all the coefficients of the polynomials $h', g_1, \ldots , g_ m$ are in $A_0$. Then we can set $B_0 = A_0[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$ and $h_0$ the image of $h'$ in $B_0$. Then $B = B_0 \otimes _{A_0} A$ and $B/hB = B_0/h_0B_0 \otimes _{A_0} A$. By Algebra, Lemma 10.168.1 we may, after enlarging $A_0$, assume that $B_0/h_0B_0$ is flat over $A_0$. Let $K_0 = \mathop{\mathrm{Ker}}(h_0 : B_0 \to B_0)$. As $B_0$ is of finite type over $\mathbf{Z}$ we see that $K_0$ is a finitely generated ideal. Let $A_1 \subset A$ be a finite type $\mathbf{Z}$-subalgebra containing $A_0$ and denote $B_1$, $h_1$, $K_1$ the corresponding objects over $A_1$. By More on Algebra, Lemma 15.31.3 the map $K_0 \otimes _{A_0} A_1 \to K_1$ is surjective. On the other hand, the kernel of $h : B \to B$ is zero by assumption. Hence every element of $K_0$ maps to zero in $K_1$ for sufficiently large subrings $A_1 \subset A$. Since $K_0$ is finitely generated, we conclude that $K_1 = 0$ for a suitable choice of $A_1$.

Set $f_1 : X_1 \to S_1$ equal to $\mathop{\mathrm{Spec}}$ of the ring map $A_1 \to B_1$. Set $D_1 = \mathop{\mathrm{Spec}}(B_1/h_1B_1)$. Since $B = B_1 \otimes _{A_1} A$, i.e., $X = X_1 \times _{S_1} S$, it now suffices to prove the lemma for $X_1 \to S_1$ and the relative effective Cartier divisor $D_1$, see Morphisms, Lemma 29.25.7. Hence we have reduced to the case where $A$ is a Noetherian ring. In this case we know that the ring map $A \to B$ is flat at every prime $\mathfrak q$ of $V(h)$ by Lemma 31.18.5. Combined with the fact that the flat locus is open in this case, see Algebra, Theorem 10.129.4 we win. $\square$