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I've been working on the Hartshorne exercise in the title for quite a while, which goes like this: let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes, $\mathscr{L}$ a very ample invertible sheaf on $X$ relative to $Y$, and $\mathscr{M}$ a very ample invertible sheaf on $Y$ relative to $Z$. Show that $\mathscr{L} \otimes f^*\mathscr{M}$ is a very ample invertible sheaf on $X$ relative to $Z$.

After getting thoroughly stuck, I found the corresponding statement in EGA, namely Proposition 4.4.10(ii). The reason I am asking this question is that in EGA the claim is proved under some hypotheses (namely that $Z$ is quasi-compact, $f$ is of finite type, and $g$ is quasi-compact), and the conclusion is weaker: one can only say that there exists $n \geq 0$ such that $\mathscr{L} \otimes f^*(\mathscr{M}^{\otimes m})$ is very ample relative to $Z$ for all $m \geq n$. So is Hartshorne wrong, or is EGA using unnecessary hypotheses to reach a weak conclusion (I find this harder to believe), or am I misinterpreting one of the two?

Edit: there is another possibility that just occurred to me: Hartshorne remarks that EGA uses a slightly different definition of very ample, and having consulted EGA I see that this is the case. So I should extend my question to ask if this is the reason for my difficulty, and if so how does it make a difference?

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Without looking it up, if I recall correctly EGA only assumes a very ample sheaf is the pullback of $\mathcal{O}(1)$ from a projective bundle whereas Hartshorne requires it to come from $\mathbb{P}^n$. This is definitely weaker as a later Hartshorne exercise is to construct a projective bundle such $\mathcal{O}(1)$ is not very ample, so I believe they both are correct. I haven't worked it out in awhile, but I assume you can use some Segre embedding and chase the diagrams around to prove this exercise. Sorry. This comment probably isn't very useful. – Matt Nov 26 '11 at 3:29

Because $Z$ and ${\bf Z}$ look too much alike, I'm going to work instead with projective morphisms $f:X \rightarrow Y$ and $g:Y \rightarrow W$. Also, the little I have to say about the difference between EGA and Hartshorne I will put at the end of this answer (briefly, Hartshorne is not wrong and I suspect the hypotheses in EGA are necessary); from now on I'm going to stick to Hartshorne's definitions.

A natural way to think about this problem is the following: we know $X$ embeds in some projective space over $Y$, and $Y$ embeds in some projective space over $W$, and that each of these embeddings equips its domain with a particular line bundle (the restriction of the twisting sheaf for the embedding). Stability of closed embeddings under base change together with the Segre embedding $$\sigma:{\bf P}_{\bf Z}^m \times {\bf P}_{\bf Z}^n \hookrightarrow {\bf P}_{\bf Z}^{mn+m+n}$$ implies that $X$ embeds in a projective space over $W$, and we are simply left with the task of calculating the very ample bundle for this embedding in terms of the data we started with (in fact, proving only that $X$ is projective over $W$ without keeping track of the corresponding bundles is the content of Hartshorne's exercise II.4.9). The details are in the next paragraph.

First observe that the closed embedding $j:Y \hookrightarrow {\bf P}_W^n$ that we start with gives, by base extension, a closed embedding $$(j \times 1):{\bf P}^m_Y=Y \times_{{\bf Z}}{\bf P}^m_{{\bf Z}} \hookrightarrow (W \times_{\bf Z} {\bf P}^n_{\bf Z}) \times_{\bf Z} {\bf P}^m_{\bf Z}.$$ Similarly, base-changing the Segre embedding $\sigma:{\bf P}_{\bf Z}^m \times {\bf P}_{\bf Z}^n \hookrightarrow {\bf P}_{\bf Z}^{mn+m+n}$ gives a closed embedding $$(1 \times \sigma):W \times_{\bf Z} {\bf P}^n_{\bf Z} \times_{\bf Z} {\bf P}^m_{\bf Z} \hookrightarrow {\bf P}^{mn+m+n}_W$$ and composing these with the given embedding $i:X \hookrightarrow {\bf P}^m_Y$ we have obtained a closed embedding $$(1 \times \sigma) \circ (j \times 1) \circ i :X \hookrightarrow {\bf P}^{mn+m+n}_W.$$ A routine diagram chase (using the compatibilities of $i$ and $j$ with $f$ and $g$) shows that composing this embedding with the projection on $W$ gives the map $g \circ f$, proving that $g \circ f$ is a projective morphism.

Now use the facts that (1) the pullback by the Segre embedding $\sigma$ of $\mathcal{O}(1)$ is the tensor product $\pi_1^* \mathcal{O}(1) \otimes \pi_2^* \mathcal{O}(1)$ of the twisting sheaves of the factors, and (2) pullback of a tensor product is tensor product of the pullbacks to calculate $$((1 \times \sigma) \circ (j \times 1))^* \mathcal{O}(1)=\pi^* \mathscr{M} \otimes \mathcal{O}(1), \quad \hbox{where $\pi:{\bf P}^m_Y \rightarrow Y$ is the projection,}$$ and then $$i^*(\pi^* \mathscr{M} \otimes \mathcal{O}(1))=f^* \mathscr{M} \otimes \mathscr{L}$$ as desired.

As you are probably aware, the most significant difference between's Harthshorne's definition and the EGA definition of projective morphism is that Hartshorne requires that the given map come from an embedding in some trivial projective space bundle, while EGA allows "twisting"---i.e., only requires that the morphism come from an embedding of the form $X \hookrightarrow P$ where $P$ is the projective bundle associated to some quasi-coherent $\mathcal{O}_Y$-module (the latter seems to be a more natural generality to me, but here we are in a situation where it seems to make the statement of some basic facts more complicated!). So as you suspected there is no obvious contradiction.

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I actually think something like the quasi-compactness of $Z$ is necessary (or at least, I haven't been clever enough to figure out how to do it otherwise), but not really for the reasons why Justin asked the question. But the hypotheses on $f,g$ seem unnecessary…

Stephen's answer is completely right if we assume the morphisms $f \colon X \to Y$ and $g \colon Y \to Z$ are projective, i.e., $i \colon X \hookrightarrow \mathbb{P}^m_Z$ and $j \colon Y \hookrightarrow \mathbb{P}^n_Z$ are closed immersions, but according to Hartshorne, a sheaf is very ample if it is induced by the pullback of $\mathcal{O}(1)$ through an arbitrary immersion, defined as the composition of an open immersion followed by a closed immersion.

This has the consequence that the composition of two immersions may not be an immersion since the composition of a closed immersion followed by an open immersion is not necessarily an immersion in Hartshorne's sense (see [Stacks, Tag 01QW]—note the difference in terminology). So, the composition $(1 \times \sigma) \circ (j \times 1) \circ i$ in Stephen's answer might not be an immersion!

There are a few ways to make this exercise solvable, disregarding the fact that Hartshorne's definition of very ample should probably be fixed:

a) Redefine an immersion to be what is in EGA—that is, an open immersion followed by a closed immersion, what I will call a "locally closed" immersion [EGAI, 4.1.3 and 4.2.1]. In this case, the composition of two locally closed immersions is again a locally closed immersion by [EGAI, 4.2.5], and so Stephen's argument goes through. In particular, it seems the assumptions on $f$ and $g$ are unnecessary for the statement of the problem with Hartshorne's definition of very ample.

b) Assume that $j \colon Y \hookrightarrow \mathbb{P}^n_W$ is quasi-compact. Factor $i \colon X \hookrightarrow \mathbb{P}^m_Z$ as $i = i_2 \circ i_1$, where $i_2$ is a closed immersion and $i_1$ is an open immersion. Then, $(j \times 1)$ is quasi-compact, hence $(j \times 1) \circ i_2$ is quasi-compact, and so by [Stacks, Tag 01QV], $(j \times 1) \circ i_2$ is an immersion hence $(1 \times \sigma) \circ (j \times 1) \circ i_2 \circ i_1 = (1 \times \sigma) \circ (j \times 1) \circ i$ is an immersion.

c) Assume $Z$ is locally noetherian (which jives well with Hartshorne's use of noetherian hypotheses—see p. 100). Then, $\mathbb{P}_Z^{mn+m+n}$ is locally noetherian, hence by [Stacks, Tag 01OX] a locally closed immersion into $\mathbb{P}_Z^{mn+m+n}$ is quasi-compact, so $(1 \times \sigma) \circ (j \times 1) \circ i_2$ is again quasi-compact and we are in situation b).

Now the question becomes how these assumptions relate to those in [EGAII, 4.4.10(ii)] you mentioned. Does $g$ being quasi-compact and $Z$ being quasi-compact imply $j$ is quasi-compact, so we are situation b)? I don't know…

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