## 29.48 Finite locally free morphisms

In many papers the authors use finite flat morphisms when they really mean finite locally free morphisms. The reason is that if the base is locally Noetherian then this is the same thing. But in general it is not, see Exercises, Exercise 109.5.3.

Definition 29.48.1. Let $f : X \to S$ be a morphism of schemes. We say $f$ is finite locally free if $f$ is affine and $f_*\mathcal{O}_ X$ is a finite locally free $\mathcal{O}_ S$-module. In this case we say $f$ is has rank or degree $d$ if the sheaf $f_*\mathcal{O}_ X$ is finite locally free of degree $d$.

Note that if $f : X \to S$ is finite locally free then $S$ is the disjoint union of open and closed subschemes $S_ d$ such that $f^{-1}(S_ d) \to S_ d$ is finite locally free of degree $d$.

Lemma 29.48.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. $f$ is finite locally free,

2. $f$ is finite, flat, and locally of finite presentation.

If $S$ is locally Noetherian these are also equivalent to

1. $f$ is finite and flat.

Proof. Let $V \subset S$ be affine open. In all three cases the morphism is affine hence $f^{-1}(V)$ is affine. Thus we may write $V = \mathop{\mathrm{Spec}}(R)$ and $f^{-1}(V) = \mathop{\mathrm{Spec}}(A)$ for some $R$-algebra $A$. Assume (1). This means we can cover $S$ by affine opens $V = \mathop{\mathrm{Spec}}(R)$ such that $A$ is finite free as an $R$-module. Then $R \to A$ is of finite presentation by Algebra, Lemma 10.7.4. Thus (2) holds. Conversely, assume (2). For every affine open $V = \mathop{\mathrm{Spec}}(R)$ of $S$ the ring map $R \to A$ is finite and of finite presentation and $A$ is flat as an $R$-module. By Algebra, Lemma 10.36.23 we see that $A$ is finitely presented as an $R$-module. Thus Algebra, Lemma 10.78.2 implies $A$ is finite locally free. Thus (1) holds. The Noetherian case follows as a finite module over a Noetherian ring is a finitely presented module, see Algebra, Lemma 10.31.4. $\square$

Lemma 29.48.3. A composition of finite locally free morphisms is finite locally free.

Proof. Omitted. $\square$

Lemma 29.48.4. A base change of a finite locally free morphism is finite locally free.

Proof. Omitted. $\square$

Lemma 29.48.5. Let $f : X \to S$ be a finite locally free morphism of schemes. There exists a disjoint union decomposition $S = \coprod _{d \geq 0} S_ d$ by open and closed subschemes such that setting $X_ d = f^{-1}(S_ d)$ the restrictions $f|_{X_ d}$ are finite locally free morphisms $X_ d \to S_ d$ of degree $d$.

Proof. This is true because a finite locally free sheaf locally has a well defined rank. Details omitted. $\square$

Lemma 29.48.6. Let $f : Y \to X$ be a finite morphism with $X$ affine. There exists a diagram

$\xymatrix{ Z' \ar[rd] & Y' \ar[l]^ i \ar[d] \ar[r] & Y \ar[d] \\ & X' \ar[r] & X }$

where

1. $Y' \to Y$ and $X' \to X$ are surjective finite locally free,

2. $Y' = X' \times _ X Y$,

3. $i : Y' \to Z'$ is a closed immersion,

4. $Z' \to X'$ is finite locally free, and

5. $Z' = \bigcup _{j = 1, \ldots , m} Z'_ j$ is a (set theoretic) finite union of closed subschemes, each of which maps isomorphically to $X'$.

Proof. Write $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. See also More on Algebra, Section 15.21. Let $x_1, \ldots , x_ n \in B$ be generators of $B$ over $A$. For each $i$ we can choose a monic polynomial $P_ i(T) \in A[T]$ such that $P(x_ i) = 0$ in $B$. By Algebra, Lemma 10.136.9 (applied $n$ times) there exists a finite locally free ring extension $A \subset A'$ such that each $P_ i$ splits completely:

$P_ i(T) = \prod \nolimits _{k = 1, \ldots , d_ i} (T - \alpha _{ik})$

for certain $\alpha _{ik} \in A'$. Set

$C = A'[T_1, \ldots , T_ n]/(P_1(T_1), \ldots , P_ n(T_ n))$

and $B' = A' \otimes _ A B$. The map $C \to B'$, $T_ i \mapsto 1 \otimes x_ i$ is an $A'$-algebra surjection. Setting $X' = \mathop{\mathrm{Spec}}(A')$, $Y' = \mathop{\mathrm{Spec}}(B')$ and $Z' = \mathop{\mathrm{Spec}}(C)$ we see that (1) – (4) hold. Part (5) holds because set theoretically $\mathop{\mathrm{Spec}}(C)$ is the union of the closed subschemes cut out by the ideals

$(T_1 - \alpha _{1k_1}, T_2 - \alpha _{2k_2}, \ldots , T_ n - \alpha _{nk_ n})$

for any $1 \leq k_ i \leq d_ i$. $\square$

The following lemma is stated in the correct generality in Lemma 29.55.4 below.

Lemma 29.48.7. Let $f : Y \to X$ be a finite morphism of schemes. Let $T \subset Y$ be a closed nowhere dense subset of $Y$. Then $f(T) \subset X$ is a closed nowhere dense subset of $X$.

Proof. By Lemma 29.44.11 we know that $f(T) \subset X$ is closed. Let $X = \bigcup X_ i$ be an affine covering. Since $T$ is nowhere dense in $Y$, we see that also $T \cap f^{-1}(X_ i)$ is nowhere dense in $f^{-1}(X_ i)$. Hence if we can prove the theorem in the affine case, then we see that $f(T) \cap X_ i$ is nowhere dense. This then implies that $T$ is nowhere dense in $X$ by Topology, Lemma 5.21.4.

Assume $X$ is affine. Choose a diagram

$\xymatrix{ Z' \ar[rd] & Y' \ar[l]^ i \ar[d]^{f'} \ar[r]_ a & Y \ar[d]^ f \\ & X' \ar[r]^ b & X }$

as in Lemma 29.48.6. The morphisms $a$, $b$ are open since they are finite locally free (Lemmas 29.48.2 and 29.25.10). Hence $T' = a^{-1}(T)$ is nowhere dense, see Topology, Lemma 5.21.6. The morphism $b$ is surjective and open. Hence, if we can prove $f'(T') = b^{-1}(f(T))$ is nowhere dense, then $f(T)$ is nowhere dense, see Topology, Lemma 5.21.6. As $i$ is a closed immersion, by Topology, Lemma 5.21.5 we see that $i(T') \subset Z'$ is closed and nowhere dense. Thus we have reduced the problem to the case discussed in the following paragraph.

Assume that $Y = \bigcup _{i = 1, \ldots , n} Y_ i$ is a finite union of closed subsets, each mapping isomorphically to $X$. Consider $T_ i = Y_ i \cap T$. If each of the $T_ i$ is nowhere dense in $Y_ i$, then each $f(T_ i)$ is nowhere dense in $X$ as $Y_ i \to X$ is an isomorphism. Hence $f(T) = f(T_ i)$ is a finite union of nowhere dense closed subsets of $X$ and we win, see Topology, Lemma 5.21.2. Suppose not, say $T_1$ contains a nonempty open $V \subset Y_1$. We are going to show this leads to a contradiction. Consider $Y_2 \cap V \subset V$. This is either a proper closed subset, or equal to $V$. In the first case we replace $V$ by $V \setminus V \cap Y_2$, so $V \subset T_1$ is open in $Y_1$ and does not meet $Y_2$. In the second case we have $V \subset Y_1 \cap Y_2$ is open in both $Y_1$ and $Y_2$. Repeat sequentially with $i = 3, \ldots , n$. The result is a disjoint union decomposition

$\{ 1, \ldots , n\} = I_1 \amalg I_2, \quad 1 \in I_1$

and an open $V$ of $Y_1$ contained in $T_1$ such that $V \subset Y_ i$ for $i \in I_1$ and $V \cap Y_ i = \emptyset$ for $i \in I_2$. Set $U = f(V)$. This is an open of $X$ since $f|_{Y_1} : Y_1 \to X$ is an isomorphism. Then

$f^{-1}(U) = V\ \amalg \ \bigcup \nolimits _{i \in I_2} (Y_ i \cap f^{-1}(U))$

As $\bigcup _{i \in I_2} Y_ i$ is closed, this implies that $V \subset f^{-1}(U)$ is open, hence $V \subset Y$ is open. This contradicts the assumption that $T$ is nowhere dense in $Y$, as desired. $\square$

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