Picard group of the product of two projective lines

In the previous posting, I wanted to know how to show that the Picard group of a $\mathbb{P}^{n}$-bundle of a nice enough scheme $X$ (over a fixed field $k$) is given by $\mathrm{Pic}(X) \times \mathbb{Z}$. The easiest form of such bundle is of the form $X \times \mathbb{P}^{n}$, but this still sounds hard. Hence, let's be a bit more specific about it and take $X = \mathbb{P}^{m}$, so for now, our goal would be to show $\mathrm{Pic}(\mathbb{P}^{m} \times \mathbb{P}^{n}) \simeq \mathbb{Z}^{2}$. The simplest case would be $m = n = 1$, which is $X = \mathbb{P}^{1} \times \mathbb{P}^{1}$. This may sound pathetically simple compared to our original goal on thinking about projective bundles, but I am more curious about learning how to compute things rather than a glorious result.

Example II.6.6.1 in Hartshorne's book has the answer for this case based on the fact that the projection $\mathbb{A}^{1} \times \mathbb{P}^{1}$ onto the second component induces an isomorphism $$\mathrm{Cl}(\mathbb{A}^{1} \times \mathbb{P}^{1}) \simeq \mathrm{Cl}(\mathbb{P}^{1}).$$ I remember reading this proof a while ago, although I am sure I was super confused given my low mathematical maturity before my candidacy. (As depressing as it may sound, I am not claiming that it is high now.)

Review of Hartshorne's proof. Denote by $p_{i}$ the $i$-th projection of the product $\mathbb{P}^{1} \times \mathbb{P}^{1}$ for $i = 1, 2$. Consider the induced maps $p_{i}^{*} : \mathrm{Cl}(\mathbb{P}^{1}) \rightarrow \mathrm{Cl}(\mathbb{P}^{1} \times \mathbb{P}^{1})$ for $i = 1, 2$. (The fact that we can induce such maps seem quite particular to the forms of these schemes.) We have noted that $$\mathbb{A}^{1} \times \mathbb{P}^{1} \hookrightarrow \mathbb{P}^{1} \times \mathbb{P}^{1} \overset{p_{2}}{\longrightarrow} \mathbb{P}^{1}$$ induces the isomorphism on the class groups of the first term and the third term, so $p_{2}^{*}$ must be injective. A similar argument would show that $p_{1}^{*}$ is injective as well.

Now, note that $$(\mathbb{P}^{1} \times \mathbb{P}^{1}) \setminus (\{\infty\} \times \mathbb{P}^{1}) = \mathbb{A}^{1} \times \mathbb{P}^{1},$$ which lets us have the exact sequence $$\mathbb{Z} \rightarrow \mathrm{Cl}(\mathbb{P}^{1} \times \mathbb{P}^{1}) \rightarrow \mathrm{Cl}(\mathbb{A}^{1} \times \mathbb{P}^{1}) \rightarrow 0,$$ where the second map is induced by the inclusion and the first map is given by mapping $1$ to the class of $\{\infty\} \times \mathbb{P}^{1}$. Since $\mathbb{Z} \simeq \mathrm{Cl}(\mathbb{P}^{1})$ by mapping $1$ to any class of a hyperplane section of $\mathbb{P}^{1}$ (i.e., a $k$-point), we may exchange the first map in the exact sequence with $$\mathrm{Cl}(\mathbb{P}^{1}) \rightarrow \mathrm{Cl}(\mathbb{P}^{1} \times \mathbb{P}^{1})$$ such that the class of $\{\infty\}$ maps to the class of $\{\infty\} \times \mathbb{P}^{1}$, but this is precisely $p_{1}^{*}$, so we get the exact sequence $$0 \rightarrow \mathrm{Cl}(\mathbb{P}^{1}) \overset{p_{1}^{*}}{\longrightarrow} \mathrm{Cl}(\mathbb{P}^{1} \times \mathbb{P}^{1}) \overset{\iota^{*}}{\longrightarrow} \mathrm{Cl}(\mathbb{A}^{1} \times \mathbb{P}^{1}) \rightarrow 0,$$ where we denoted $\iota$ for the relevant inclusion. The left exactness came from our observation that $p_{1}^{*}$ is injective. As we know that the map $\iota^{*}$ sends $p_{2}^{*}(\mathrm{Cl}(\mathbb{P}^{1})) \subset \mathrm{Cl}(\mathbb{P}^{1} \times \mathbb{P}^{1})$ isomorphically to $\mathrm{Cl}(\mathbb{A}^{1} \times \mathbb{P}^{1})$, we may conclude that our exact sequence is split. Thus, we have $$\mathrm{Cl}(\mathbb{P}^{1} \times \mathbb{P}^{1}) \simeq \mathrm{Cl}(\mathbb{P}^{1}) \oplus \mathrm{Cl}(\mathbb{A}^{1} \times \mathbb{P}^{1}) \simeq \mathbb{Z}^{2}.$$ Since $$\mathrm{Pic}(\mathbb{P}^{1} \times \mathbb{P}^{1}) \simeq \mathrm{Cl}(\mathbb{P}^{1} \times \mathbb{P}^{1}),$$ as the local ring of $\mathbb{P}^{1} \times \mathbb{P}^{1}$ at any point is a UFD,  this finishes the proof. $\Box$

The above proof would generalize to give a proof of what we want if we know that the projection $\mathbb{A}^{m} \times \mathbb{P}^{n}$ onto the second component induces an isomorphism $$\mathrm{Cl}(\mathbb{A}^{m} \times \mathbb{P}^{n}) \simeq \mathrm{Cl}(\mathbb{P}^{n}).$$ This is probably a valid approach, but I decided to look at a different reference, because even though this approach is mathematically clear, I still felt like I did not really get what was going on.

Vakil's book (14.2.O) had a different suggestion for the same problem, which I liked a lot.

Review of Vakil's proof. For convenience, write $X := \mathbb{P}^{1} \times \mathbb{P}^{1}$, and note that we are still working over a fixed field $k$. Let $L := \{\infty\} \times \mathbb{P}^{1}$ and $M := \mathbb{P}^{1} \times \{\infty\}$. These are codimension $1$ closed subsets of $X$, so their classes sit in $\mathrm{Cl}(X)$. Moreover, we have the exact sequence $$\mathbb{Z}^{2} \rightarrow \mathrm{Cl}(X) \rightarrow \mathrm{Cl}(X \setminus (L \cup M)) \rightarrow 0,$$ whose second map is given by the restriction, while the first map is given by $(1, 0) \mapsto [L]$ and $(0, 1) \mapsto [M]$. The exactness can be checked by constructing the relevant exact sequence with the groups of Weil divisors before taking their classes modulo principal divisors. Since $$X \setminus (L \cup M) = \mathbb{A}^{2}$$ and $\mathrm{Cl}(\mathbb{A}^{2}) = 0$, we get a (group) surjection $\mathbb{Z}^{2} \twoheadrightarrow \mathrm{Cl}(X)$. This means that $[L]$ and $[M]$ generate $\mathrm{Cl}(X)$, so we would be done if we show that these two elements are $\mathbb{Z}$-linearly independent.

At this point, I was so puzzled. The classes of divisors are quite abstract objects to me, and I did not have much handle. However, Vakil gave a quite enlightening hint: $\mathscr{O}_{X}(L)$ restricts to $\mathscr{O}$ on $L \simeq \mathbb{P}^{1}$ and $\mathscr{O}(1)$ on $M \simeq \mathbb{P}^{1}$, while $\mathscr{O}_{X}(M)$ restricts to $\mathscr{O}$ on $M$ and $\mathscr{O}(1)$ on $L$.

Ad-hoc interpretation of Vakil's hint. We know $[\mathscr{O}_{X}(L)]$ and $\mathscr{O}_{X}(M)]$ generate $\mathrm{Pic}(X)$, so we just need to show that these two are $\mathbb{Z}$-linearly independent. Hence, suppose that we have $$n_{1}[\mathscr{O}_{X}(L)] + n_{2}[\mathscr{O}_{X}(M)] = 0$$ in $\mathrm{Pic}(X)$. Vakil's hint seems to suggest that we have a group homomorphisms $\mathrm{Pic}(X) \rightarrow \mathrm{Pic}(L)$ and $\mathrm{Pic}(X) \rightarrow \mathrm{Pic}(M)$ where

• the first one sends $[\mathscr{O}_{X}(L)]$ and $[\mathscr{O}_{X}(M)]$ to $[\mathscr{O}]$ and $[\mathscr{O}(1)]$ respectively in $\mathrm{Pic}(L)$ (by seeing $L$ as a copy of $\mathbb{P}^{1}$) and 
• the second one sends $[\mathscr{O}_{X}(L)]$ and $[\mathscr{O}_{X}(M)]$ to $[\mathscr{O}(1)]$ and $[\mathscr{O}]$ respectively in $\mathrm{Pic}(M)$ (by seeing $M$ as a copy of $\mathbb{P}^{1}$).

If this is what Vakil meant, we can just apply the first map to our $\mathbb{Z}$-linear equation to have $$n_{1}[\mathscr{O}] + n_{2}[\mathscr{O}(1)] = 0$$ in $\mathrm{Pic}(L) \simeq \mathbb{Z}$, but since $[\mathscr{O}]$ and $[\mathscr{O}(1)]$ correspond to $0$ and $1$, respectively, in $\mathbb{Z}$, this shows that $n_{2} = 0$. Similarly, if we apply the second map to our $\mathbb{Z}$-linear equation, we get $n_{1} = 0$, which would finish our proof.

For a scheme map $\pi : X \rightarrow Y$ and a quasi-cohrent sheaf $\mathscr{F}$ on $Y$, if $\mathscr{F}$ is locally free of rank $r$, then so is its pullback $\pi^{*}\mathscr{F}$. Thus, we do have an induced map $\mathrm{Pic}(Y) \rightarrow \mathrm{Pic}(X)$. If we have an open embedding $\iota : U \hookrightarrow X$, then $\iota^{*}\mathscr{F} = \mathscr{F}|_{U}$, so if we are in a nice situation where $\mathrm{Cl}(X) \simeq \mathrm{Pic}(X)$ by $D \mapsto \mathscr{O}_{X}(D)$, the pullback map $\mathrm{Pic}(X) \rightarrow \mathrm{Pic}(U)$ would be the same as $\mathrm{Cl}(X) \rightarrow \mathrm{Cl}(U)$ given by taking the intersections of the codimension $1$ irreducible closed subsets of $X$ with $U.$ Taking such intersection is not valid for closed subsets, but the pullback map on the picard groups should be a formalism making such imagination possible.

Going back to our situation (e.g., $X = \mathbb{P}^{1} \times \mathbb{P}^{1}$), we do have the maps $\mathrm{Pic}(X) \rightarrow \mathrm{Pic}(L)$ and $\mathrm{Pic}(X) \rightarrow \mathrm{Pic}(M)$ given by the pullbacks of the inclusions, namely $\iota_{L}$ and $\iota_{H}$ of $L$ and $H$, respectively, into $X$. Thus, there will be maps $\mathrm{Cl}(X) \rightarrow \mathrm{Cl}(L)$ and $\mathrm{Cl}(X) \rightarrow \mathrm{Cl}(M),$ accordingly. If taking intersections was possible, then the map $\mathrm{Cl}(X) \rightarrow \mathrm{Cl}(L)$ would send $[L]$ to $[L]$ and $[M]$ to $[\infty]$, where $\infty := \{\infty\} \times \{\infty\}.$ Since $L$ has codimension $0$ in $L$, we "should" interpret $[L]$ as $0$ and $[\infty]$ is a generator of $\mathrm{Cl}(L) \simeq \mathbb{Z},$ so our plan should work at the level (ad-hoc) intuition.

The above paragraphs is probably the correct intuition behind the following goal: we want to show

$\iota_{L}^{*}\mathscr{O}_{X}(L) \simeq \mathscr{O}_{L}$ and $\iota_{M}^{*}\mathscr{O}_{X}(L) \simeq \mathscr{O}_{M}(1)$, while

$\iota_{L}^{*}\mathscr{O}_{X}(M) \simeq \mathscr{O}_{L}(1)$ and $\iota_{M}^{*}\mathscr{O}_{X}(M) \simeq \mathscr{O}_{M}.$

We would be done with the proof if we show them.

Making the ad-hoc interpretation (hopefully) legit. A concrete way to think about the pullback of a sheaf under a scheme map is tensor product. Writing $[x_{0} : x_{1}]$ and $[y_{0} : y_{1}]$ the coordinates for the copies of $\mathbb{P}^{1}$ appearing in $X = \mathbb{P}^{1} \times \mathbb{P}^{1}$, we may consider the open cover of $L = \{\infty\} \times \mathbb{P}^{1}$ given by $$\{\infty\} \times U_{0} = \mathrm{Spec}(k[y_{1}/y_{0}])$$ and $$\{\infty\} \times U_{1} = \mathrm{Spec}(k[y_{0}/y_{1}]).$$ The first open subset of $L$ is a closed subscheme of $$U_{1} \times U_{0} = \mathrm{Spec}(k[x_{0}/x_{1}, y_{1}/y_{0}]),$$ an open subset of $X$, and the second open subset of $L$ is a closed subscheme of $$U_{1} \times U_{1} = \mathrm{Spec}(k[x_{0}/x_{1}, y_{0}/y_{1}]),$$ another open subset of $X$. We first note that $$\Gamma(\{\infty\} \times U_{0}, \iota_{L}^{*}\mathscr{O}_{X}(L)) = \Gamma(U_{1} \times U_{0}, \mathscr{O}_{X}(L)) \otimes_{\Gamma(U_{1} \times U_{0}, \mathscr{O}_{X})} \Gamma(\{\infty\} \times U_{0}, \mathscr{O}_{L}).$$ In other words, we have $$\begin{align*}\Gamma(k[y_{1}/y_{0}] \times U_{0}, \iota_{L}^{*} & \mathscr{O}_{X}(L)) \\
&= \Gamma(\mathrm{Spec}(k[x_{0}/x_{1}, y_{1}/y_{0}]), \mathscr{O}_{X}(L)) \otimes_{k[x_{0}/x_{1}, y_{1}/y_{0}]} k[y_{1}/y_{0}].\end{align*}$$ For the sake of sanity, let us temporarily denote $x = x_{0}/x_{1}$ and $y = y_{1}/y_{0}$. In this notation $L$ is defined by cutting out the prime ideal $(x)$ in $k[x, y]$, so $$\begin{align*}\Gamma(\mathrm{Spec}(k[x, y]), \mathscr{O}_{X}(L)) &= \{f(x,y)/x \in k(x,y): f(x,y) \in k[x,y]\} \\ &= k[x,y]y^{-1}.\end{align*}$$ Hence, we have

• $\Gamma(\mathrm{Spec}(k[x_{0}/x_{1}, y_{1}/y_{0}]), \mathscr{O}_{X}(L)) = k[x_{0}/x_{1}, y_{1}/y_{0}]x_{1}/y_{0}$ and
• $\Gamma(\mathrm{Spec}(k[x_{0}/x_{1}, y_{0}/y_{1}]), \mathscr{O}_{X}(L)) = k[x_{0}/x_{1}, y_{0}/y_{1}]x_{1}/x_{0}.$

Continuing this, we note that

• $k[x_{0}/x_{1}, y_{1}/y_{0}]x_{1}/x_{0} \otimes_{k[x_{0}/x_{1}, y_{1}/y_{0}]} k[y_{1}/y_{0}] \simeq k[y_{1}/y_{0}]$ and
• $k[x_{0}/x_{1}, y_{0}/y_{1}]x_{1}/x_{0} \otimes_{k[x_{0}/x_{1}, y_{0}/y_{1}]} k[y_{0}/y_{1}] \simeq k[y_{0}/y_{1}],$

where the first map is given by $$f(x_{0}/x_{1}, y_{1}/y_{0})x_{1}/x_{0} \otimes g(y_{1}/y_{0}) \mapsto f(0, y_{1}/y_{0})g(y_{0}/y_{1})$$ and the second map is given by $$f(x_{0}/x_{1}, y_{0}/y_{1})x_{1}/x_{0} \otimes g(y_{0}/y_{1}) \mapsto f(0, y_{0}/y_{1})g(y_{0}/y_{1}).$$ These two maps are compatible with the isomorphism $$k[y_{1}/y_{0}, y_{1}/y_{0}] \simeq k[y_{0}/y_{1}, y_{0}/y_{1}]$$ given by the multiplication with $1$ (i.e., the idenity) on the intersection the two open subsets of $L$. This isomorphism is the transition function for the trivial sheaf $\mathscr{O}_{L}$ of $L,$ so gluing the two observed isomorphisms, we may conclude that $$\iota_{L}^{*}\mathscr{O}_{X}(L) \simeq \mathscr{O}_{L}.$$ OK. This is not so bad. We just need to figure out affine descriptions of sheaf isomorphisms we want to establish and then we perform our commutative algebra skills!

Next, we want to show $$\iota_{L}^{*}\mathscr{O}_{X}(M) \simeq \mathscr{O}_{L}(1),$$ and by symmetry, we would be able to call it a day after this computation. 

We are still pulling back to (or restricting to) $L$, so we keep using the same open cover for $L$. However, since $M = \mathbb{P}^{1} \times \{\infty\}$ does not intersect $$U_{1} \times U_{0} = \mathrm{Spec}(k[x_{0}/x_{1}, y_{1}/y_{0}]),$$ the locus of $M$ is cut out by the unit ideal in $k[x_{0}/x_{1}, y_{1}/y_{0}]$. Therefore, we have $$\begin{align*}\Gamma(\mathrm{Spec}(k[x_{0}/x_{1}, y_{1}/y_{0}]), \mathscr{O}_{X}(M)) &= \Gamma(\mathrm{Spec}(k[x_{0}/x_{1}, y_{1}/y_{0}]), \mathscr{O}_{X}) \\ &= k[x_{0}/x_{1}, y_{1}/y_{0}].\end{align*}$$ On the other hand, the locus of $M$ in $$U_{1} \times U_{1} = \mathrm{Spec}(k[x_{0}/x_{1}, y_{0}/y_{1}])$$ is cut out by the prime ideal $(y_{0}/y_{1})$, so $$\Gamma(\mathrm{Spec}(k[x_{0}/x_{1}, y_{0}/y_{1}]), \mathscr{O}_{X}(M)) = k[x_{0}/x_{1}, y_{0}/y_{1}]y_{1}/y_{0}.$$ The intersection $$U_{1} \times (U_{0} \cap U_{1}) = \mathrm{Spec}(k[x_{0}/x_{1}, y_{1}/y_{0}, y_{0}/y_{1}])$$ does not meet $M$, so $$\Gamma(\mathrm{Spec}(k[x_{0}/x_{1}, y_{0}/y_{1}, y_{1}/y_{0}]), \mathscr{O}_{X}(M)) = k[x_{0}/x_{1}, y_{1}/y_{0}, y_{0}/y_{1}].$$

We have the isomorphisms

• $k[x_{0}/x_{1}, y_{1}/y_{0}] \otimes_{k[x_{0}/x_{1}, y_{1}/y_{0}]} k[y_{1}/y_{0}] \simeq k[y_{1}/y_{0}]$ and
• $k[x_{0}/x_{1}, y_{0}/y_{1}]y_{1}/y_{0} \otimes_{k[x_{0}/x_{1}, y_{0}/y_{1}]} k[y_{0}/y_{1}] \simeq k[y_{1}/y_{0}]$,

where the first one is given by $$f(x_{0}/x_{1}, y_{1}/y_{0}) \otimes g(y_{1}/y_{0}) \mapsto f(0, y_{1}/y_{0})g(y_{1}/y_{0}),$$ while the second one is $$f(x_{0}/x_{1}, y_{0}/y_{1})y_{1}/y_{0} \otimes g(y_{0}/y_{1}) \mapsto f(0, y_{0}/y_{1})g(y_{0}/y_{1}).$$ For a long time I thought that the two maps are not compatible with the isomorphism $$k[y_{1}/y_{0}, y_{0}/y_{1}] \simeq k[y_{0}/y_{1}, y_{1}/y_{0}]$$ given by the multiplication of $y_{0}/y_{1}$, the transition map for $\mathscr{O}_{L}(1)$.

What I struggled to notice was that the restriction maps for $\mathscr{O}_{L}(M)$ are the inclusions. This magically takes care of the issue and finishes the proof. $\Box$

Concluding remark. Recovering Vakil's proof was very hectic, but I think it gave me a good grasp of the general saying that "one should always reduce to the affine case and try commutative algebra, where the meat lies". Generalizing Vakil's proof for $\mathbb{P}^{m} \times \mathbb{P}^{n}$ seems possible, but it would be annoying to even think about many affine charts. (At a glance, however, it seems quite doable.) The specific $\mathbb{P}^{4}$-bundle for the previous posting might be resolved using a similar technique as in this argument, but I don't want to think about it right now. Vakil's book (28.1.K) also describes how to deal with more general projective bundles, so perhaps I will think about this soon.