# Hartshorne exercise II.5.12(b)

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?

-
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.