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Proper is supposed to generalize compactness so something like this should be true.

Let $X$ be a proper $S$-scheme and let $f:X\rightarrow Y$ be a surjective morphism of $S$-schemes. Is it true that $Y$ also a proper $S$-scheme?

I think that I can prove that $g:Y \rightarrow S$ is universally closed. In fact, given a $S$-scheme $Z$ we have that the composition $$f_Z:X\times_S Z\rightarrow Y\times_S Z$$ is surjective as this property is invariant under base change. So if $W$ is a closed subset of $Y\times_S Z$ we have that $$g_Z(W)=(g_Z\circ f_Z)(f_Z^{-1}(W))$$ is closed as $g_Z\circ f_Z=X\times_S Z\rightarrow Z$ is a base chage of $X\rightarrow S$.

Also, as is show in here, Y should be separated. (Edit: This was not true as KReiser example shows)

Now, what about finite type?

Remark: Because of this it is enough to prove locally of finite type.

Edit: Corrected the proof of universally closed.

nowhere dense
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Walter Simon
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1 Answers1

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Unfortunately this is not true for two reasons. First of all, recall that proper is equivalent to universally closed, separated, and of finite type: if $Y$ isn't separated, it can't be proper. For example, $Y$ being a projective line with a doubled point and $X$ being two copies of the projective line with $f$ being the natural identification map provides a counterexample to your statement.

Secondly, it may be the case that a surjective map is not of finite type: let $k$ be a field, $R=k[x_1,x_2,\cdots]$, $I=(x_1,\cdots)$, and consider $\operatorname{Spec} R/I^2 \to \operatorname{Spec} k$. This map is a homeomorphism after any base change (and thus it's universally closed), but is not locally of finite type.

So one really does need the target of a surjective morphism to be separated and of finite type in order to conclude that the image of a proper scheme is again proper.

KReiser
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    Oh, now I have a contradiction in my head. Using the result proved in this question I think I can deduce that the codomain is separable. But I am also convinced that your example of two copies of $\mathbb{P}^1$ gluing in one $\mathbb{P}^1$ with a double point is a counterexample. Do you have any idea of what is happening here? – Walter Simon Dec 14 '18 at 11:08
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    Now I got it. I was assuming that $X$ proper implies $X\rightarrow Y$ proper, but actually we need $Y$ to be separated for this (see here). Also now I realized that it is intuitive to ask for $Y$ to be separated because properness generalize compactness + Hausdorff so we need to avoid the possibility that the surjective map is a weird projection that can mess with the Hausdorff part. Thank you for your answer! – Walter Simon Dec 14 '18 at 13:21
  • I have just realised that there is a surjective morphism from the (obviously proper) scheme $\text{Spec }k$ to $\text{Spec }R/I^2$ so your second example is exactly what we needed. – Walter Simon Dec 16 '18 at 10:48