Zac Maeder-Wolland: Some Daily AG Problems

Hartshorne Exercise II.6.1.

Let $\pi_1,\pi_2$ be the first and second projections of $X\times\PP{k}{n}$. Clearly $X\times\PP{k}{n}$ remains integral, separated, and noetherian (these are affine local questions). To prove the regularity in codimension $1$, we will use induction on $n$ to show that if $X$ is regular in codimension 1 then so is $X\times\A{k}{n}$. By proposition 6.6, if $X$ satisfies $(*)$ then so does $X\times\A{k}{1}$. This does the base case. Then $X\times \A{k}{n}\cong (X\times \A{k}{n-1})\times\A{k}{1}$ so replacing $X$ by $X\times \A{k}{n-1}$ does the trick. Since it is no loss of generality that $\mathfrak{p}\in X\times\A{k}{n}\subset X\times\PP{k}{n}$ and $\OO_{X\times\A{k}{n},\mathfrak p}=\OO_{X\times\PP{k}{n},\mathfrak p}$, we conclude that the stalk at $\mathfrak p$ is a regular local ring of dimension 1.

Now, we will construct an isomorphism,

$$\Psi:\op{Cl}X\times\Z\ra\op{Cl}(X\times\PP{k}{n}).$$

Recall that $\op{Cl}\PP{k}{n}\cong\Z$. Let $H\subset\PP{k}{n}$ be a hyperplane. Then $X\times\PP{k}{n}-X\times H\cong X\times\A{k}{n}$. By propositions 6.5 and 6.6, we have an exact sequence,

$$\Z\ra \op{Cl}(X\times\PP{k}{n})\rightarrow \op{Cl X}\ra0\quad\quad (1)$$

where the first map is $1\mapsto X\times H=\pi_2^*H$ and the second is restriction to $X\times\A{k}{n}$ followed by $\pi_1^*$. Define $\Psi$ by $(D,n)\mapsto\pi_1^*D+n\pi_2^*H$ (recall $\pi_2^*(nH)=n\pi_2^*H$). Denote $\Bbb K(X)=:K$. First note that if $f\in K^\times$ then $\pi_1^*\op{div}(f)=\op{div}(\pi_1^*f)$ where $\pi_1^*:K\ra K(X_1,…,X_n)$ pulls back rational functions. Likewise if $f\in k(X_1,…,X_n)^\times$ then $\pi_2^*\op{div}(f)=\op{div}(\pi_2^*f)$ where $\pi_2^*:k(X_1,…,X_n)\ra K(X_1,…,X_n)$. Therefore this is a valid homomorphism. Before continuing, let us review what types of codimension 1 points in $X\times\PP{k}{n}$ there are. If $\mathfrak p\in X\times\PP{k}{n}$ has codimension 1, then let $U=\spec R\subset X$ be an affine open neighborhood containing $\pi_1(\mathfrak q)$. We see that $\mathfrak q=\mathfrak p\cap R$ and since $X\ra \A{R}{n}$ is flat, $\codim \mathfrak q\le\codim\mathfrak p$ (by Going Down). If $\mathfrak q$ has codimension 1 (call this type 1) then we must have $\mathfrak q\cdot R[X_1,…,X_n]=\mathfrak p$ and therefore $\pi_1^{-1}(V(\mathfrak q))=V(\mathfrak p)$. The image of a type 1 point in $\PP{k}{n}$ must be the generic point otherwise $\codim \mathfrak p>1$. If $\codim\mathfrak q=0$ (type 2) then $\mathfrak q$ is the generic point of $X$ and every element in $\mathfrak p-(0)$ must have nonzero degree in $R[X_1,…,X_n]$. The image of a type 2 point in $\PP{k}{n}$ can have either codimension $0$ or $1$. From this discussion, it follows that any prime divisor $Y$ on $X\times\PP{k}{n}$ corresponding to a prime $\mathfrak p$ of type 1 does not meet the generic fiber $\pi_2^{-1}(\eta_X)=\PP{K}{n}$ over $X$, as otherwise $\pi_1^{-1}(V(\mathfrak q))=V(\mathfrak p)$ would not hold.
To show injectivity, suppose $\pi_1^*D+n\pi_2^*H=\op{div}f$ for some $f\in K(X_1,…,X_n) $, and note that,

$$( \pi_1^*D+n\pi_2^*H)|_{\PP{K}{n}}=n\pi_2^*H|_{\PP{K}{n}}$$

because $\pi_1^*D$ is just a linear combination of type 1 prime divisors. Now $n\pi_2^*H|_{\PP{K}{n}}$ is just the pullback of $nH$ under the composition $\pi_1^{-1}(\eta_X)\ra X\times\PP{k}{n}\ra \PP{k}{n}$. Moreover $\pi_2^*(nH)$ is not contained in the generic fiber because $\pi_2^*H\ra X$ is a surjection. Therefore, $n\pi_2^*H|{\PP{K}{n}}$ is a divisor on $\PP{K}{n}$ that is linearly equivalent to the principle divisor $\op{div}f|{\PP{K}{n}}=\op{div}(f|{\PP{K}{n}})$ (recall that $(\cdot)|{\PP{K}{n}}$ just denotes the pullback of a divisor or function. In fact, $\Bbb K(\PP{K}{n})\cong\Bbb K(X\times\PP{k}{n})$ so pulling back rational functions in this case is an isomorphism). We conclude that $n=0$. Then $\pi_1^*D$ is principle. Restricting $\pi_1^*D$ to $X\times\A{k}{n}$ and using the isomorphism $\op{Cl}X\xrightarrow{(\pi_1\circ\iota)^{*}}\op{Cl}(X\times \A{k}{n})$ where $\iota :X\times \A{k}{n}\ra X\times \PP{k}{n}$ is inclusion, we see that $D\sim 0$. This shows injectivity.
For surjectivity, if $F\in\op{Cl(X\times\PP{k}{n}})$, then its restriction to $X\times \A{k}{n}$ is the pullback under $(\pi_1\circ\iota)^*$ of some divisor $D$ on $X$, $F|{X\times\A{k}{n}}=\pi_1^*D|{X\times\A{k}{n}}$. Therefore by the exact sequence $(1)$, $F-\pi_1^*D=n\pi_2^*H$ for some $n\in \Z$. We conclude that $\Psi$ is indeed surjective.