The Stacks project

Proof. By Lemma 29.25.13 we see that $q$ is flat. Hence it suffices to show that the fibres of $Y \to S$ are local complete intersections, see Lemma 29.30.11. Let $s \in S$. Consider the morphism $X_ s \to Y_ s$. This is a base change of the morphism $X \to Y$ and hence surjective, and syntomic (Lemma 29.30.4). For the same reason $X_ s$ is syntomic over $\kappa (s)$. Moreover, $Y_ s$ is locally of finite type over $\kappa (s)$ (Lemma 29.15.4). In this way we reduce to the case where $S$ is the spectrum of a field.

Assume $S = \mathop{\mathrm{Spec}}(k)$. Let $y \in Y$. Choose an affine open $\mathop{\mathrm{Spec}}(A) \subset Y$ neighbourhood of $y$. Let $\mathop{\mathrm{Spec}}(B) \subset X$ be an affine open such that $f(\mathop{\mathrm{Spec}}(B)) \subset \mathop{\mathrm{Spec}}(A)$, containing a point $x \in X$ such that $f(x) = y$. Choose a surjection $k[x_1, \ldots , x_ n] \to A$ with kernel $I$. Choose a surjection $A[y_1, \ldots , y_ m] \to B$, which gives rise in turn to a surjection $k[x_ i, y_ j] \to B$ with kernel $J$. Let $\mathfrak q \subset k[x_ i, y_ j]$ be the prime corresponding to $y \in \mathop{\mathrm{Spec}}(B)$ and let $\mathfrak p \subset k[x_ i]$ the prime corresponding to $x \in \mathop{\mathrm{Spec}}(A)$. Since $x$ maps to $y$ we have $\mathfrak p = \mathfrak q \cap k[x_ i]$. Consider the following commutative diagram of local rings:

We claim that the hypotheses of Algebra, Lemma 10.135.12 are satisfied. Conditions (1) and (2) are trivial. Condition (4) follows as $X \to Y$ is flat. Condition (3) follows as the rings $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{X_ y, x} = \mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ are complete intersection rings by our assumptions that $f$ and $p$ are syntomic, see Lemma 29.30.10. The output of Algebra, Lemma 10.135.12 is exactly that $\mathcal{O}_{Y, y}$ is a complete intersection ring! Hence by Lemma 29.30.10 again we see that $Y$ is syntomic over $k$ at $y$ as desired. $\square$