Composition of finite flat morphisms cancellation

Outline/some thoughts: Suppose we have morphisms of schemes $$X \overset{f}{\rightarrow} Y \overset{g}{\rightarrow} S$$ where $$f$$ is finite, the composition $$g \circ f$$ is finite, and $$S$$ is locally Noetherian. Does it follow that $$g$$ is finite? Is it true that $$g$$ is at least affine? It's true that the composition of finite morphisms is finite, and if instead we assume $$g$$ and the composition are both finite then $$f$$ is finite. Assuming $$g$$ is affine we can reduce the problem to showing that if $$A \overset{\alpha}{\rightarrow} B \overset{\beta}{\rightarrow} C$$ are morphisms of rings with $$A$$ Noetherian making $$C$$ into a f.g. (equiv coherent) $$A$$-module as well as a f.g. $$B$$-module, then $$B$$ is a f.g. $$A$$-module. Since $$A$$ is Noetherian $$C$$ is a Noetherian $$A$$-module, so $$\beta(B)$$ is a finite $$A$$-module, but I can't figure out why $$\ker(\beta)$$ is a f.g. $$A$$-module. Actually, it is not necessarily a finite $$A$$-module, consider for example the maps $$k \to k[x] \to k$$ fixing $$k$$ and taking $$x \mapsto 0$$.

In the above situation, suppose that also $$f$$ is faithfully flat and $$g \circ f$$ is flat. Then we can conclude that $$g$$ is flat. With these added assumptions, can we show that $$g$$ is finite? Note that the above example is ruled out as $$k$$ is not flat over $$k[x]$$.

Here's an example showing that the request for $$g$$ to be finite or affine is not true. Let $$X=S=\Bbb A^1$$, and $$Y=\Bbb A^1\coprod\Bbb P^1$$. Let $$f:X\to Y$$ be given by the identity morphism on $$\Bbb A^1$$ and let $$g:Y\to S$$ be given by the identity morphism on $$\Bbb A^1$$ and sending $$\Bbb P^1$$ to $$0$$. Then $$f$$ is finite, $$g\circ f$$ is finite, but $$g$$ is not finite (indeed, $$g$$ is not even affine). This shows that some strange stuff can be going on with $$g$$ which isn't captured by knowing everything about $$f$$ and $$g\circ f$$.
Once you require that $$f$$ is faithfully flat, things get better, mostly due to surjectivity. We use the equivalent definition of a finite morphism as a morphism which is proper and quasi-finite (EGA IV, Part 4, Corollaire 18.12.4).
Proof that $$g$$ is quasi-finite when $$f$$ is surjective and $$g\circ f$$ is finite: let $$s\in S$$ be a point. We wish to show that $$Y_s$$ has finitely many points for every $$s\in S$$. Suppose that there is at least one $$s$$ where this fails. Then as $$f$$ is surjective, we have the fiber $$X_s$$ of $$g\circ f$$ over $$s$$ must have at least $$|Y_s|$$ elements, but this is impossible since $$g\circ f$$ is finite and thus quasi-finite.
Next, by StacksProject 03GN, if $$g$$ is separated and of finite type while $$f$$ is surjective and $$g\circ f$$ is proper, then $$g$$ is also proper.
So if one has $$f,g\circ f$$ finite, one can guarantee $$g$$ finite if $$f$$ is surjective and $$g$$ is separated of finite type.
• Your example clears up the first question well. The argument when $f$ is surjective is great, but as you pointed out we need $g$ to be separated finite type to conclude that $g$ is finite. All I have is that $S$ is locally Noetherian. Tate (Finite Flat Group Schemes, Prop 3.1) seems to suggest this is clearly true without any extra conditions. Do you have any thoughts on this? – ggg Oct 28 '18 at 17:39
• It's certainly not true that $g$ may be finite without being separated or finite type (as finite implies proper which implies separated + finite type). I would go back and take a read through any "we assume blah to always mean an (adjective) blah" type sentences that are in the text. Often authors will make some such statement at the beginning of their text or the beginning of the chapter. – KReiser Oct 29 '18 at 3:54