# 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.

# Spectral Sequence of a Double Complex

I knew my posting would be sparser this quarter, but this was a pretty large gap. I sort of lost motivation for posting, since my class moved on from spectral sequences to group cohomology, so I wasn’t constantly reminded about things to post about. Now the class is coming back around to spectral sequences in order to compute the group cohomology, so I’m sort of in the mood again.

We’ve mostly been looking at how to get a spectral sequence out of a filtered complex, and last time I promised I’d talk about convergence. Since my posting is so sparse, and I only wanted to give an idea about it rather than a proof of all the details of the convergence, I’m going to keep it brief in order to post about something more important.

The theorem is that if we have a filtration ${\{K^p\}}$ of a complex ${K}$ and the following three conditions are satisfied

1. ${K^p=0}$ when ${p<0}$
2. ${H_{p+q}(K^p/K^{p-1})=0}$ when ${q\leq -1}$
3. ${K=\cup K^p}$

then if ${r\geq \max\{p, q+1\}+1}$, then ${E^r_{p,q}\simeq E^\infty_{p,q}}$. Where we define ${E^\infty_{p,q}:= F_{p,q}/F_{p-1, q+1}}$ where ${F_{p,q}:=im (H_{p+q}(K^p)\rightarrow H_{p+q}(K))}$.

Basically what this says is that under some “mild" conditions that were naturally satisfied in our examples, the spectral sequence converges. And not only does it converge, but it converges to an associated graded object of the homology of the original complex.

The fact that it converges you can work out without much trouble if you've followed the previous posts. It basically amounts to noticing the first two conditions force all non-zero ${E^r}$ terms to be in the “first quadrant" when ${r>1}$. Since the differential maps get longer and longer as ${r}$ increases, paying attention to some specific ${E^*_{p,q}}$, the differential pointing to this term will have to come from the fourth quadrant, and eventually the differential will point to the second quadrant. But we said that these two quadrants were filled with 0 groups. Thus our sequence we are taking homology of is ${0\rightarrow E^r_{p,q}\rightarrow 0}$ for all ${r}$ after some point. So they stabilize. Figuring out exactly what these things are that they stabilize to is far more tedious and I won't do it.

The next topic that should be addressed is how to get a spectral sequence out of a double complex. This is actually much more useful it turns out, since there are two ways to do it that will both converge to the same thing which will give us lots of information.

First, let's recall that a double complex ${C_{p,q}}$ is just sets of complexes where the ${q}$ index the rows and the ${p}$ index the columns. So for a fixed ${q}$, we get ${\cdots \leftarrow C_{p-1, q}\leftarrow C_{p, q}\leftarrow \cdots }$ a chain complex with differential, say ${d^h}$ for horizontal. Likewise we get them going down in columns for any fixed ${p}$ with differential ${d^v}$ for vertical. The differentials are compatible in the sense that ${d^h\circ d^v+d^v\circ d^h=0}$. These also have to fit together into what we call the total complex ${(Tot C)_n=\bigoplus_{p+q=n} C_{p,q}}$ where the differential is ${d^h+d^v: C_{p,q}\rightarrow C_{p-1, q}\bigoplus C_{p,q-1}}$.

Now if we consider just the single complex ${Tot C}$, and put a filtration on it, we'll have a spectral sequence as before. One natural filtration is at the ${p}$th filtered part to chop everything from the ${p}$-th column to the right off and replace it with ${0}$. Notationally, this is ${(F_pC)_n=\bigoplus_{j+q=n \ \text{and} \ j\leq p} C_{j, q}}$.

The other natural one is to chop everything from the ${p}$-th row and up off and replace it with ${0}$. We'll notate this with a prime on the left ${F_pC}$. Both of these certainly satisfy ${\cdots \subset F_pC\subset F_{p+1}C\subset \cdots}$ since from ${p}$ to ${p+1}$ everything stays exactly the same except the ${p+1}$ column of ${C_{p,q}}$ gets stuck on. We also satisfy ${\cup F_pC=Tot C}$. We do need to assume that ${C_{p,q}=0}$ when ${p<0}$ or ${q<0}$ to get all the convergence conditions above, but this isn't too bad since most naturally arising double complexes (that we'll look at) are zero before some point and we can just shift indices.

The spectral sequence we get has ${E^1_{p,q}=H_{p+q}(F_pC/F_{p-1}C)}$. Note that for the first spectral sequence the ${E^2}$ term is computed by first taking vertical homology, and then horizontal homology: ${E^2_{p,q}\simeq H_p^h(H_q^v(C))}$ and for the other filtration it is ${E^2_{p,q}\simeq H_q^v(H^h_p(C))}$.

As usual, the formal aspect of this might seem overwhelming and confusing right now, so I'll hold off on working out what the maps actually are and let this absorb a little. It should become more clear what is going on when we do some examples.