# 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?

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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