# Finite and Proper and Separated, Oh My!

First note that I finally got some form of the last post to work (this morning). It involved switching some less than signs to less than or equal to, which I think is less confusing for wordpress due to the backslash.

Today will be a slightly off topic post. It is something I should have done ages ago, and I expect should be quite useful for beginning algebraic geometers that stumble upon this post by some means. I’m sure most people that have tried to work through Hartshorne chapter II get to some point around section 4 and start to panic. Morphisms have lots of properties. There are locally of finite type, finite type, finite, quasi-finite, generically finite. There are closed immersions, open immersions, just plain immersions. There is proper, compact, quasi-compact. There is separated, projective, quasi-projective.

It is overwhelming. If you manage to get all the definitions down, you still have the tremendous task of figuring out which one imply which and under which hypotheses. Is it every projective morphism is proper, or every proper morphism is projective? For one of these to work, does my scheme have to be Noetherian…locally Noetherian…is it always true?

Then once you finally have all of this down, was it just because you flat out memorized them all, or have you now developed some intuition about why different morphisms are actually different from some sort of geometric standpoint. Do you have at your fingertips examples for why each of these implications don’t reverse (moreover, are these examples well motivated, or are they cheap “one point examples”)?

After a quick glance through sections 2, 3, and 4 of Hartshorne here are the statements of the form “*blank* is always *blank*”.

1. Finite type is always locally of finite type
2. Finite is always finite type
3. Finite is always quasi-finite
4. Closed immersions are of finite type
5. Quasi-compact open immersions are of finite type
6. Open and closed immersions are separated
7. Proper maps are separated
8. Closed immersions are proper
9. Projective are quasi-projective
10. Projective are proper (need Noetherian)
11. Finite are proper

First we’ll assume at least that our schemes are locally Noetherian, since this is Hartshorne. Also, quasi-finite here means finite fibers over a point (Grothendieck I believe assumes finite type in the definitions of quasi-finite which we’ll not do).

My goal of this post is to give examples as to why none of these reverse. I’ll skip ones that aren’t very illuminating. There are two main ways in which I’ll try to accomplish this: with as few examples as possible. Namely, I should be able to write down a single example to knock off a bunch of them. This way when trying to recall whether a certain statement is true, you only have to check against a few things in your memory. The other way is to try to give very geometrically motivated examples, so that it sheds some light on what in particular each of these properties says about a morphism.

I have eleven statements listed. A few things not listed that quickly pop into my mind we’ll call 3a) A quasi-finite but not finite type map and 3b) A finite type map that is not quasi-finite. Lastly, we’ll call 12a) Proper that is not quasi-finite and 12b) Quasi-finite that is not proper.

1,2, and 3 are standards.

1. Project an infinite disjoint union of affine lines onto an affine line. Geometrically this illuminates that “locally” of finite type means we can have an infinite cover, but just plain finite type needs a finite cover. Algebraically we give the map explicitly by, ${\amalg_{i} Spec(k[x_i])\rightarrow Spec(k[x])}$ induced by the natural projection map.

2. The easiest way to violate finite, is to make a non-affine map, so just include ${\mathbb{A}^2\setminus \{0\}\hookrightarrow \mathbb{A}^2}$. The pre-image of the affine set ${\mathbb{A}^2}$ is ${\mathbb{A}^2\setminus \{0\}}$ which is not affine, so the map is not finite. If we want to violate the other way a map can be non-finite, project ${\mathbb{A}^1\rightarrow \{pt\}}$ to a point by ${Spec(k[t])\rightarrow Spec(k)}$ induced by ${k\hookrightarrow k[t]}$. Since ${k[t]}$ is finitely generated as an algebra, it is of finite type, since it is not finitely generated as a module, it is not finite. Geometrically, this is reflecting the fact that the fiber over the point is “infinite”, but in that it is not too infinite since it is “finite dimensional”.

3. Use the first example of 2. The fiber over any point is a single point or empty since it is an inclusion.

3a. Here is our first “point to point” example. The idea being that you can always get a quasi-finite morphism by mapping Spec of a field to Spec of a field, since the topological space is a single point. This means we can make the field as horrendous as we like and get a quasi-finite map. Let ${L}$ be a field extension of ${k}$ of infinite transcendence degree. Then ${k\hookrightarrow L}$ induces ${Spec(L)\rightarrow Spec(k)}$, a quasi-finite map. But this isn’t even locally of finite type.

3b. The second example of 2 works since the fiber is infinite.

4. The first example of 2 works since the image is open, not closed.

5, 6, and 8 are about immersions which is a different flavor of property than what I’m generally concerned with in this post, so I’ll skip them (maybe I’ll do a full post just on immersions and types of immersions at a later date).

7. Now we get to some of the more subtle morphism properties. We need to violate the “universally closed” property somehow using a separated morphism. The second example of 2 works since it is just the “structure map” of ${\mathbb{A}^1}$ as a ${k}$-scheme, and structure maps of affines are always separated. It is not proper, however, because ${\mathbb{A}^1\times_{Spec(k)}\mathbb{A}^1\rightarrow \mathbb{A}^1}$ is not a closed map. Take a hyperbola in ${\mathbb{A}^2}$. It is a closed set (namely, $Spec\left(\frac{k[x,y]}{(xy-1)}\right)$), but the projection onto ${\mathbb{A}^1}$ is ${\mathbb{A}^1\setminus \{0\}}$ which is open.

9. The second example of 2 works yet again, since including ${\mathbb{A}^1\hookrightarrow \mathbb{P}^1}$ is an open immersion and it respects the structure map, so we’ve now factored ${\mathbb{A}^1\rightarrow Spec(k)}$ by an open immersion followed by a projective morphism ${\mathbb{A}^1\hookrightarrow \mathbb{P}^1\rightarrow Spec(k)}$ and hence it is quasi-projective. All projective maps (of Noetherian schemes) are proper, so since by 7 this isn’t proper it isn’t projective.

10. Let ${\mathbb{A}^1\rightarrow\mathbb{A}^1}$ be the identity map. Then it is finite and hence proper. But this cannot be factored through projective space as a closed immersion (I honestly don’t have an argument for that). See comments.

11. Let ${\mathbb{P}^n_\mathbb{Z}\rightarrow Spec(\mathbb{Z})}$ be the natural structure map. Then it is projective (of Noetherian schemes) and hence proper, but it is not affine so it is not finite.

12a. Use 3a since proper maps must be of finite type.

12b. Use 11. The justification here is a rather deep theorem that says a proper and quasi-finite map is finite. So if this were quasi-finite, then it would be finite which we said was not the case.

We have now completed what I listed. Maybe I’ll do a continuation of this next time, since I realized I skipped quite a bit with the immersions and things like finite type but not quasi-projective and separated but not quasi-projective. Also, I became increasingly less geometric and clear, relying on theorems to figure out whether the example satisfied the right criteria. It would be nice to figure out geometrically what is going on with examples like 11 and why there are infinite fibers there. I hope this helps some people.

### Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

### 6 thoughts on “Finite and Proper and Separated, Oh My!”

1. I already realized that in trying to make things simplest with the fewest examples, I’ve made things more complicated.

For 12b, just use $\mathbb{P}^n_k\to Spec(k)$. The preimage of the point is all of projective space, so it is clearly not quasi-finite, but it is projective almost by definition and hence proper.

2. Interesting-some incarnation of Gelbaum-Olmsted ought to make a “Counterexamples in Algebraic Geometry.”

For 10, can’t you use $\mathbb{A}^n_k \to \mathbb{P}^0_k \times \mathbb{A}^n_k \to \mathbb{A}^n_k$? Not sure if this is cheating. Hartshorne does give an example in an exercise in Chapter III.5 of a proper nonprojective scheme.

3. Unless someone does it first, this is my grand money-making scheme for the future. Keep collecting examples and write a “Counterexamples in Algebraic Geometry” book. It is quite surprising no one has done it yet.

As for 10, you’re right. I knew that was too easy. This one is tough. I was trying to do it without any advanced tools, but to show something is not projective is hard without anything to go on.

4. Funny, I’ve also been working on a “counterexamples in AG” type of database to perhaps one day be a book (and MO has been a wonderful source of them) as for the proper but nonprojective, that’s tricky. The only examples I have off the top of my head are nonprojective, complete varieties, like the Hironaka Threefold and a few Toric things, but I don’t have an elementary example either (and if you’re going to work over a field, I believe that you need to use something like this, because proper implies fiberwise complete, etc, standard argument here.)

5. It is false that spectrum of product of infinite many copies of K[X] is coproduct of corresponding affine lanes because that coproduct is not quasicompact.

6. Alright. The counterexample is still to project an infinite disjoint union of affine lines to a single affine line. Just ignore the algebra thing I wrote.