There exists a flat proper morphism f : X —> S all of whose geometric fibres are connected nodal curves such that f is not of finite presentation. An explicit example can be found in the examples chapter of the stacks project. Once you’ve understood why the example works, you easily see that you can even make an example where all the fibres are stable curves, S is connected, and the genus of the fibres jumps!

But let me go out on a limb here and make a wild guess: If you assume that there exists an integer g > 1 such that f is flat, proper, and all fibres are stable curves of genus g, then f is of finite presentation.

Why do I think this is true? I think it is true by analogy with the following results: A finite flat module need not be projective. A finite flat module over a local ring is free. Thus given a finite flat module over a scheme S then you get a well defined rank function. Then the module is finite locally free if and only if the rank function is locally constant in the Zariski toplogy (yet another characterization of finite projective modules, see Bourbaki, Commutative Algebra, Chapter II, Theorem 1).

I also think the following may be true: Given an integer d >= 0. If R —> A is a finite type, flat ring map all of whose geometric fibres are smooth and irreducible of dimension d, then R —> A is of finite presentation. (Irreducible implies nonempty. For this one I actually have a pretty good idea for how to prove it.)

Don’t do this at home kids!

What I mean by the last sentence is that, if you are doing actual moduli, you should just **assume** that X —> S is of finite presentation. In regards to this, note that if my wild guess is correct, then Definition 1.1 of Deligne-Mumford is the correct one. Thanks to Michael Thaddeus for pointing out that Deligne and Mumford only assume proper + flat + conditions on fibres.

Deligne-Mumford aside, I can’t imagine there is any good reason apart from sloppiness or typos to ever omit finite presentation hypotheses when considering moduli issues. That is, although it is very cool that such examples exist, these kind of results you mention go in the box labelled “nifty, but useless”. Do you use some kind of improvement upon 3.4.6 in Raynaud-Gruson (exploiting their notion of purity) to prove your guess about smooth maps with geometrically irreducible fibers?

Let me now go our on a limb and propose what I think is a proof of your “wild guess” about stable curves of common genus. First, some generalities. Since every finite type map to a qcqs base is closed in a finitely presented scheme, by 3.4.1 in R-G (and fpqc descent from henselization) it follows that a proper flat scheme over a *local* scheme is always finitely presented. So by the usual yoga, given a proper flat map X –> S over a general base and a point s in S, if we work Zariski-locally around s we can arrange there to be a map f:X —> X’ to a proper flat and finitely presented object over the base such that the induced map over O_{S,s} is an isomorphism. In particular, by openness of the locus of stability for proper flat *finitely presented* families of curves, if X_s is a stable curve of genus g then we can arrange the same for all fibers of X’ –> S.

Now the key point: by EGA IV_3, sec. 13.1 Chevalley’s semi-continuity results (on the source) for fiber dimension are valid with “loc. finite type” rather than “loc. finite presentation”, so in particular openness of the “locally quasi-finite locus” on the source is true for finite type maps. Hence, by properness we can shrink more on S to arrange that f is actually quasi-finite, so (by EGA IV_4, 18.12 stuff of Deligne to allow finite type in place of finite presentation) f is actually *finite*. Since f is an isomorphism over O_{S,s} and X’ is S-proper, by shrinking more around s we can arrange that f is a closed immersion.

A closed immersion between proper stable curves of the *same* genus over an alg. closed field is surely an isomorphism (it’s been a long day, I am too lazy to think right now…), so now assuming that X —> S has all fibers stable curves of the *same* genus, it follows that f is an isomorphism on fibers. Now we’re in business. For *any* s in S, consider the pullback situation over O_{S,s}. By the RG-result mentioned above, over this O_{S,s} our original X becomes finitely presented. So in fact over O_{S,s} the map f must be an isomorphism. In particular, f is flat and finite. But it has constant fiber rank, namely 1. So f is actually finitely presented, so X is finitely presented over S (actually, f is even an isomorphism). QED

Please let me know if this looks correct, or if I have made a blunder somewhere.

Wonderful! For the part where you are lazy, I think you can easily see that you morphism X —> X’ is a flat (!) closed immersion. The fact that both X and X’ have all fibres of genus g means that it is also bijective. A bijective flat closed immersion is an isomorphism.

The proof of the algebra result is similar: In schemes language you find (possibly after etale localization) a closed immersion X —> X’ as in your proof with now X, X’ affine flat over S and X/S fin type, X’/S fin presentation, both with smooth geometrically irreducible fibres of dimension d. Then it follows that X —> X’ is bijective and the argument I just sketched above can be used to finish.