# Proof of "$\mathcal{M}$ is $S$-ample if and only if $f^{*}\mathcal{M}$ is $S$-ample "( Gortz's Algebraic Geometry, Proposition 13.66 ).

I am reading the Gortz's Algebraic Geometry, p. 397, proof of Proposition 13.66. and stuck at understanidng some statement.

Propostion 13.66. Let $$S$$ be a scheme, let $$X$$ and $$Y$$ be $$S$$-schemes of finite type and let $$f: X\to Y$$ be a surjective finite locally free $$S$$-morphism. (1) Let $$\mathcal{L}$$ be an $$S$$-ample line bundle on $$X$$. Then $$N_{X/Y} (\mathcal{L})$$ is an $$S$$-ample line bundle on $$Y$$. (2) Let $$\mathcal{M}$$ be a line bundle on $$Y$$. Then $$\mathcal{M}$$ is $$S$$-ample if and only if $$f^{*}\mathcal{M}$$ is $$S$$-ample.

Here, the $$N_{X/Y} (\mathcal{L})$$ is the norm of $$\mathcal{L}$$ under $$f$$, defined in his book p.330~331. We accept the statement (1) of the Proposition.

Proof. For both assertions we may assume that $$S$$ is affine. Then a line bundle is $$S$$-ample if and only if it is ample. Moreover $$X$$ and $$Y$$ are quasi-compact, and $$X$$ is quasi-separated if and only if $$Y$$ is quasi separated ( Proposition 10.25 ). ( We also accept this argument. If needed, I will upload my own proof. )

(1) proof is omitted.

(2) As $$f$$ is affine, $$f^{*}\mathcal{M}$$ is ample if $$\mathcal{M}$$ is ample by his book Proposition 13.83. Conversely, let $$f^{*}\mathcal{M}$$ be ample. We may assume that $$f$$ has constant rank $$n$$. By (12.6.3) we have $$N_{X/Y}(f^{*}\mathcal{M}) \cong \mathcal{M}^{\otimes n}$$, and $$\mathcal{M}^{\otimes n}$$ is ample by (1). Therefore $$\mathcal{M}$$ is ample by Proposition 13.50 (1).

I am trying ot understand the bold statement. Here, I think that the notion of rank of finite locally free morphism $$f$$ was not defined in his book. But degree of $$f$$ is defined as a function ( his book p.329 ):

$$Y \to \mathbb{N}_0 , y \mapsto \operatorname{dim}_{\kappa(y)}(f_{*}\mathcal{O}_X)(y)$$

, denoted by $$\operatorname{deg}f$$. This function is locally constant on $$Y$$. I think that the author seems to be using the words 'rank' and 'degree' interchangeably. For now we accept this convention.

My first strategy to attack the bold statement is as follows.

Since $$\operatorname{deg}f$$ is locally constant, there is an open cover $$Y= \bigcup_{a\in A}V_a$$ such that $$(\operatorname{deg}f)|_{V_a} : V_a \to \mathbb{N}_0$$ is constant. Covering each $$V_a$$ by affine opens, and using the quasi-compactness of $$Y$$ (as in the above proof), we may assume that each $$V_a$$ is affine and the set $$A$$ is finite. Note that since $$f$$ is affine morphism, each $$f^{-1}(V_a)$$ is affine, so it is quasi-compact.

For each $$a\in A$$, consider $$f_a:= f|_{f^{-1}(V_a)}^{V_a} : f^{-1}(V_a) \to V_a$$.

Q. My interlude question is, $$\operatorname{deg}(f_a) = (\operatorname{deg}f)|_{V_a}$$? If so, then each $$f_a$$ has constant degree (rank).

Since $$f^{*}\mathcal{M}$$ is ample, $$(f^{*}\mathcal{M})|_{f^{-1}(V_a)}=(f_a)^{*}(\mathcal{M}|_{V_a})$$ ( This equality can be shown by easy calculation. If needed, I will upload its proof )is ample by https://stacks.math.columbia.edu/tag/0B3E.

So, by the reduction assumption, each $$\mathcal{M}|_{V_a}$$ is ample. Can we deduce from this that $$\mathcal{M}$$ is ample? An issue that makes me stuck here is, 'ampleness of sheaf is not local property' : Ampleness of sheaf on qcqs scheme is local property?

Am I following the author's intentions well? Or perhaps, does there other method to prove that the reduction assumption works well? Can we breakthrough this difficulty? Can anyone helps?

$$f_{\ast}\mathcal{O}_X$$ is a locally free $$\mathcal{O}_Y$$-module of finite rank, so the set $$Y_n$$ of $$y \in Y$$ where $$(f_{\ast}\mathcal{O}_X)_y$$ is free over $$\mathcal{O}_{Y,y}$$ of rank $$n$$ is an open subset of $$Y$$.

Hence $$Y=\bigcup_n{Y_n}$$, and it’s a disjoint union of open subsets (by quasi-compactness, it’s finite, ie $$Y_n=\emptyset$$ if $$n \geq N$$). Therefore, $$X_n=f^{-1}(Y_n)$$ is open in $$X$$, $$X=\bigcup_{n is (again) a disjoint union of open subsets, and $$f: X_n \rightarrow Y_n$$ is finite locally free of rank $$n$$.

As you point out, ampleness for a line bundle is a global property and is very much not local (I think that every line bundle is locally ample!). However, it behaves well with respect to disjoint unions of open subsets.

In other words, if $$W$$ is a scheme which is the disjoint union of the open subschemes $$U$$ and $$V$$, then a line bundle $$\mathcal{L}$$ on $$W$$ is ample iff $$\mathcal{L}_{|U}$$ and $$\mathcal{L}_{|V}$$ are ample. (We need, of course, to adapt this to the case of any finite collection of pairwise disjoint open subschemes).

This is the ingredient that you were missing to reduce to the case where $$f$$ has fixed rank.

• Thanks for pointing out ! First, can I ask some delicated(?) question? Let $X$ be a scheme and let $\mathcal{F}$ be an $\mathcal{O}_X$-module of finite type. The function $\operatorname{rk}(\mathcal{F}) : X\to \mathbb{N}_0$, $x\mapsto \operatorname{rk}_x(\mathcal{F}) := \operatorname{dim}_{\kappa(x)}\mathcal{F}(x)$ is called the rank of $\mathcal{F}$. (Here $\mathcal{F}(x):=\mathcal{F}_x/\mathfrak{m}_x\mathcal{F}_x = \mathcal{F}_x\otimes_{\mathcal{O}_{X,x}} \kappa(x)$ ). So the degree $\operatorname{deg}(f)$ in the above my question is by definition $\operatorname{rk}(f_{*}\mathcal{O}_{X})$. Jul 20 at 9:12
• On the other hand, there is also an ordinary definition of rank of (finite) locally free $\mathcal{O}_X$-module $\mathcal{F}$ ( Gortz's book p.177 ). I guess that your definition of rank of finite locally free morphism $f : X \to Y$ is $\operatorname{rk}(f_*\mathcal{O}_X)$ as this latter definition. My small question is, for the finite locally free $\mathcal{O}_Y$-module $f_*\mathcal{O}_X$, perhaps, the two definition of $\operatorname{rk}(f_*\mathcal{O}_X)$ are equivalent? Anyway, I will try to figure out it. Jul 20 at 9:12
• Comments 1 and 2: I’d say they’re the same (in the locally free setting). Proof: it’s local on $Y$, so reduce to the case where $Y$ (hence $X$) is affine and $O(X)$ is free over $O(Y)$ of finite rank, then compute explicitly. Comments 5 and 6 seem correct, but needlessly convoluted. The set of $y \in Y$ with a given rank is closed and open, what more do you need? Comment 3: it’s an application of the definition, no special insight needed. Maybe it’s done in the Stacks section about ample line bundles. Jul 20 at 11:47
• $Y$ is the disjoint union of the open subschemes $Y_n$, so they are closed. Jul 20 at 16:35
• Hint: given any sheaf $\mathcal{G}$ on $X$, $H^0(X,\mathcal{G})=H^0(U,\mathcal{G}) \oplus H^0(V,\mathcal{G})$. Jul 21 at 7:24