The Stacks project

Proof. Let $n$ be an integer bounding the degree of the fibres of $X \to S$. By Morphisms, Lemma 29.57.5 we see that any base change has degrees of fibres bounded by $n$ also. In particular, all the integers $r$ that occur in the statement of the lemma will be $\leq n$. We will prove the lemma by induction on $n$. The base case is $n = 0$ which is obvious.

We claim the set of points $s \in S$ with $\deg _{\kappa (s)}(X_ s) = n$ is an open subset $S_ n \subset S$ and that $X \times _ S S_ n \to S_ n$ is finite locally free of degree $n$. Namely, suppose that $s \in S$ is such a point. Choose an elementary étale morphism $(U, u) \to (S, s)$ and a decomposition $U \times _ S X = W \amalg V$ as in Lemma 37.41.6. Since $V \to U$ is finite, flat, and locally of finite presentation, we see that $V \to U$ is finite locally free, see Morphisms, Lemma 29.48.2. After shrinking $U$ to a smaller neighbourhood of $u$ we may assume $V \to U$ is finite locally free of some degree $d$, see Morphisms, Lemma 29.48.5. As $u \mapsto s$ and $W_ u = \emptyset $ we see that $d = n$. Since $n$ is the maximum degree of a fibre we see that $W = \emptyset $! Thus $U \times _ S X \to U$ is finite locally free of degree $n$. By Descent, Lemma 35.23.30 we conclude that $X \to S$ is finite locally free of degree $n$ over $\mathop{\mathrm{Im}}(U \to S)$ which is an open neighbourhood of $s$ (Morphisms, Lemma 29.36.13). This proves the claim.

Let $S' = S \setminus S_ n$ endowed with the reduced induced scheme structure and set $X' = X \times _ S S'$. Note that the degrees of fibres of $X' \to S'$ are universally bounded by $n - 1$. By induction we find a stratification $S' = S_0 \amalg \ldots \amalg S_{n - 1}$ adapted to the morphism $X' \to S'$. We claim that $S = \coprod _{r = 0, \ldots , n} S_ r$ works for the morphism $X \to S$. Let $g : T \to S$ be a morphism of schemes and assume that $X \times _ S T \to T$ is finite locally free of degree $r$. As remarked above this implies that $r \leq n$. If $r = n$, then it is clear that $T \to S$ factors through $S_ n$. If $r < n$, then $g(T) \subset S' = S \setminus S_ d$ (set theoretically) hence $T_{red} \to S$ factors through $S'$, see Schemes, Lemma 26.12.7. Note that $X \times _ S T_{red} \to T_{red}$ is also finite locally free of degree $r$ as a base change. By the universal property of the stratification $S' = \coprod _{r = 0, \ldots , n - 1} S_ r$ we see that $g(T) = g(T_{red})$ is contained in $S_ r$. Conversely, suppose that we have $g : T \to S$ such that $g(T) \subset S_ r$ (set theoretically). If $r = n$, then $g$ factors through $S_ n$ and it is clear that $X \times _ S T \to T$ is finite locally free of degree $n$ as a base change. If $r < n$, then $X \times _ S T \to T$ is a morphism which is separated, flat, and locally of finite presentation, such that the restriction to $T_{red}$ is finite locally free of degree $r$. Since $T_{red} \to T$ is a universal homeomorphism, we conclude that $X \times _ S T_{red} \to X \times _ S T$ is a universal homeomorphism too and hence $X \times _ S T \to T$ is universally closed (as this is true for the finite morphism $X \times _ S T_{red} \to T_{red}$). It follows that $X \times _ S T \to T$ is finite, for example by Lemma 37.44.1. Then we can use Morphisms, Lemma 29.48.2 to see that $X \times _ S T \to T$ is finite locally free. Finally, the degree is $r$ as all the fibres have degree $r$. $\square$