For me, the notion of a proper morphism is a map $\varphi\colon X\to Y$ such that $\varphi$ factors as $X\xrightarrow{i}Y\times\mathbb{P}^m\xrightarrow{\pi} Y$, where $\mathbb{P}^m$ is some projective space over an algebraically closed field, $i$ is a closed embedding, and $\pi$ the projection on to $Y$.

I think that a morphism $\varphi\colon X\to\{p\}$ to a point is proper iff $X$ is a projective variety. If $X$ is a projective variety, then it is closed in some $\mathbb{P}^m$. So we can embed $X\hookrightarrow\mathbb{P}^m\hookrightarrow\{p\}\times\mathbb{P}^m$, which is a closed embedding since the first map is inclusion, and the second map is just $x\mapsto (p,x)$. Composing with the projection onto $p$ gives back $\varphi$.

On the other hand, if $\varphi\colon X\to\{p\}$ is a proper morphism, then there exists a closed embedding $X\to\{p\}\times\mathbb{P}^m$, so $X$ is (isomorphic) to a closed subvariety of $\{p\}\times\mathbb{P}^m$, and hence a closed subvariety of $\mathbb{P}^m$. And so I conclude $X$ is a projective variety?

Does this look reasonable? Thanks.

  • $\begingroup$ I'm a little lost. Is the algebraically closed field fixed at the start? What's the structure on the target point? [It seems to me that it would be better to call your "proper" morphisms "projective", but that's not really a logical issue.] $\endgroup$ – Hoot Jan 15 '15 at 7:08
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    $\begingroup$ It seems to me that you've used the definition of a projective morphism instead of a proper morphism. If you're set on this definition, you need to change your question. If you actually want proper, ie finite type, separated, universally closed, then the answer to this is a no. There is a standard example of a proper non-projective toric variety- see more discussion at mathoverflow.net/questions/111504/… $\endgroup$ – KReiser Jan 15 '15 at 8:10

No: a variety such that $X\to \{p\}$ is proper is said to be complete.
A complete variety of dimension $1$ is indeeed projective but for any $d\geq 2$ there exist complete and non projective varieties of dimension $d$.
In dimension $2$ a complete smooth variety must be projective (Kodaira) , but there exists a smooth complete variety of dimension $3$ which is not projective.
This was proved by Hironaka and you can see such a variety [for some reason it is blue ...] on the cover of Shafarevich's Basic Algebraic Geometry, vol.2.
The details of Hironaka's construction are given in that volume, starting page 75.


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