Proposition 6.6 in Hartshorne II: Bijectivity

Question

Proposition 6.6 in Hartshorne II seeks to show that $\operatorname{Cl} (X \times \mathbb A^1) \simeq \operatorname{Cl} X$ for $X$ Noetherian, integral, separated, and regular in codimension 1.

A point of codimension 1 in $X \times \mathbb A^1$ is called type I if its image $y$ under the projection to $X$ is codimension 1. We can see such an $x$ is the generic point of the fiber of $y$.

A point of codimension 1 in $X \times \mathbb A^1$ is called type II if its image is the generic point of $X$.

I understand on a technical level, thanks to other questions on this site, the first two paragraphs of the proof. However, when Hartshorne goes to show the map

$$\pi^* : \operatorname{Cl} X \to \operatorname{Cl} (X \times \mathbb A^1)$$

given by $\sum n_i Y_i \mapsto \sum n_i \pi^{-1} Y_i$, where $\pi$ is the projection to $X$, I do not understand the following statements:

1. Injectivity: If $D = \operatorname{div} f \in \operatorname{div} X$, then $\pi^*D$ involves only prime divisors of type I. Why?

2. Injectivity: The above forces $f \in K$, the function field of $X$. Why? He gives the reasoning that otherwise $f = g/h$, with $g, h \in K[t]$, which I understand, but then claims that if both are not in $K$, the divisor of $f$ will involve some type II divisor. Why is this true?

3. Surjectivity: Why does it suffice to check that prime type II divisors are linearly equivalent to sums of prime type I divisors? Do we already know we are surjective on type I divisors?

4. Surjectivity: Why does localizing a type II prime divisor at the generic point give us a prime divisor in the spectrum of $K[t]$?

5. Assuming the above statement, we see this divisor corresponds to a principal prime $P = (f)$ in $K[t]$. I see that just fine. Hartshorne then claims the divisor of $f$ is $Z$ plus possibly some purely type I divisors. Why?

I get the sense I am not thinking about divisors in general correctly, and, more specifically, I am not truly understanding the definition and implementation of type I and II divisors.

Answer 1

1. Writing $D=\sum n_iD_i$ for $D_i$ prime, we have $\pi^*D=\sum n_i\pi^*D_i$. But $\pi^*D_i=\pi^{-1}(D_i)$, and $\pi(\pi^{-1}(D_i))=D_i$, hence the image of $D$ does not contain the generic point of $X$, and all the $\pi^{-1}D_i$ are divisors of type I.

2. If $f=g/h$ in lowest terms for $g,h\in K[t]$ so that at least one is in $K[t]\setminus K$, then the valuation of $f$ at a maximal ideal containing a zero of whichever $g,h$ is in $K[t]\setminus K$ is not zero. Such a maximal ideal is a point in the fiber of $X\times\Bbb A^1$ over the generic point of $X$, so the divisor of $f$ contains a type II divisor.

3. Yes, we already know we're surjective on type I divisors: if $D\subset X\times\Bbb A^1$ is a prime divisor of type 1, then $\pi(D)\subset X$ is a prime divisor which maps to $\pi^{-1}(\pi(D))=D$.

4. Your statement is not super clear, but the point is that if $D$ is a prime divisor of type II, then it must intersect the fiber of $X\times\Bbb A^1\to X$ over the generic point of $X$ by definition of a type II divisor. As this fiber is exactly $\operatorname{Spec} K[t]$ and the divisor cannot contain the generic point of $\operatorname{Spec} K[t]$ (this is the generic point of $X\times\Bbb A^1$), we get a collection of closed points in $\operatorname{Spec} K[t]$. If we had multiple points in this, we'd have multiple irreducible components of our divisor, so there's only one point, and thus we get a prime divisor on $\operatorname{Spec} K[t]$. (Alternatively, any prime divisor on a scheme $X$ has a unique point with $\dim \mathcal{O}_{X,p}=1$ and no points with $\dim = 0$.)

5. Take $X=\operatorname{Spec} k[x]$ and $f=xt$. Then the divisor associated to $f$ on $\Bbb A^1\times\Bbb A^1=\Bbb A^2$ is the union of the axes, which is a type one divisor plus a type two divisor. There are worse examples, but this is one very accessible one; conversely, you might be able to fix this most of the time. Digging in to the stuff required to make it just a type II divisor is more work than you need to do here because we already know we're surjective on type I divisors, and so we don't have to figure it out to finish the proof.

Part 5 could make an interesting problem to think about further (when can one guarantee the divisor associated to $f$ is just a type II divisor?), but it's not necessary to finish this proof.