The notion of a projective morphism in algebraic geometry is surprisingly subtle. It is not quite clear what the definition is! For example, the definition in EGA differs from that in Hartshorne. For the purposes of this question, I take the definition in EGA$^1$.

Suppose $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ are projective morphisms. I know that $g \circ f$ is projective if $Z$ is quasicompact (see for example Exercise 18.3.B in the August 2012 version of the notes here). Is it true even without $Z$ quasicompact, or is there a counterexample?

(I suspect this is in one of the standard sources, but I haven't stumbled upon it.)

$^1$ EGA II, 5.5.1-5.5.2: $X$ is called projective over $Y$ if there is a closed $Y$-immersion $X \hookrightarrow \mathbb{P}(\mathcal{E})$ for some quasi-coherent sheaf $\mathcal{E}$ on $Y$ of finite type; equivalently if $X=\mathrm{Proj}(A)$ for some graded quasi-coherent $\mathcal{O}_Y$-algebra $A$ which is generated by $A_1$, which is of finite type.

  • $\begingroup$ Why is EGA's definition equivalent to Hartshorne's if $Y$ is quasi-compact ? Don't you need $Y$ to have an ample line bundle, or at least an ample family of line bundles, for this to be true ? $\endgroup$ Aug 21, 2012 at 17:27
  • $\begingroup$ You may very well be right --- I remember many years ago checking for hypotheses under which the two definitions agreed, and I wrote this based only on a dim memory of what I concluded. I'll edit the question right now to warn the reader. $\endgroup$
    – Ravi Vakil
    Aug 21, 2012 at 17:38
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    $\begingroup$ Shouldn't Z be quasiseparated as well? $\endgroup$ Aug 27, 2012 at 15:11
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    $\begingroup$ @Ravi: I admit I didn't check the proofs. I was asking because the corresponding statements in EGA2 have assumptions such as "Z is quasicomact and separated, or has a Noetherian underlying space", which generally mean quasiseparatedness (which was not yet invented then). $\endgroup$ Aug 28, 2012 at 13:45
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    $\begingroup$ I've added the EGA-definition. I doubt that this is the correct one if the base does not enjoy any finiteness assumptions. Perhaps one should demand that $E$ is of finite presentation as a module (equivalently, $A$ is of finite presentation as a graded algebra). What happens then? $\endgroup$ Dec 17, 2012 at 9:13

2 Answers 2


This is now Tag 0C4P with an assumption that the schemes in question are qcqs as predicted in the comments by Moret-Bailly.


This seems to be Corollary 12 in this note by Daniel Murfet (2006).

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    $\begingroup$ Thanks! I think this won't answer the question, as Daniel Murfet is using Hartshorne's definition, which is that $f: X \rightarrow Y$ is projective if there exists a closed immersion $f: X \hookrightarrow \mathbb{P}^n_Y$ for some $n$ (see his Prop. 8), which needn't be true with Grothendieck's definition in EGA. (Incidentally, I am a big fan of Daniel Murfet's notes in general.) $\endgroup$
    – Ravi Vakil
    Aug 21, 2012 at 17:03
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    $\begingroup$ (This result is also in the stacks project, but again seems to only work for Hartshorne's definition. stacks.math.columbia.edu/download/morphisms.pdf) $\endgroup$ Aug 21, 2012 at 17:06

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