A distribution of finite abelian groups

Much of this posting is a brief regurgitation of (a corner of) a paper by Cohen and Lenstra. However, I will use some weird words that the authors do not use to feed my (and hopefully others') intuition.

Given a nonempty category $\mathcal{C}$ such that

• the automorphism group of each object is finite and
• there are finitely many isomorphism classes of $\mathcal{C},$

we define the number of points of $\mathcal{C}$ to be $$|\mathcal{C}| := \sum_{[G] \in \mathcal{C}/\simeq}\frac{1}{|\mathrm{Aut}(G)|},$$ where $\mathcal{C}/\simeq$ is the set of isomorphism classes of $\mathcal{C}.$ (This definition is not so "made-up"; see this Wikipedia page.) If $\mathscr{P}$ is a property defined on $\mathcal{C}/\simeq,$ we define $$\mathrm{Prob}_{G \in \mathcal{C}}(G \text{ satsifies } \mathscr{P}) := \frac{|\mathcal{C}_{\mathscr{P}}|}{|\mathcal{C}|},$$ where $\mathcal{C}_{\mathscr{P}}$ is the full subcategory of $\mathcal{C}$ given by the objects satisfying $\mathscr{P}.$ Note that we have $$|\mathcal{C}_{\mathscr{P}}| = \sum_{[G] \in \mathcal{C}/\simeq}\frac{\boldsymbol{1}_{\mathscr{P}}(G)}{|\mathrm{Aut}(G)|},$$ where $\boldsymbol{1}_{\mathscr{P}} : \mathcal{C}/\simeq \rightarrow \{0, 1\}$ is the indicator function for the property $\mathscr{P}.$ Replacing this indicator function to any other complex function $f : \mathcal{C}/\simeq \rightarrow \mathbb{C},$ we may define the average of $f$ in $\mathcal{C}$ as $$M_{\mathcal{C}}(f) := \frac{|\mathcal{C}|_{f}}{|\mathcal{C}|},$$ where $$|\mathcal{C}|_{f} = \sum_{[G] \in \mathcal{C}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}.$$ In particular, we have $$M_{\mathcal{C}}(\boldsymbol{1}_{\mathscr{P}}) = \mathrm{Prob}_{G \in \mathcal{C}}(G \text{ satisfies } \mathscr{P}).$$ What if the category $\mathcal{C}$ has infinitely many objects? An example we are going to focus on is the category $\textbf{Ab}^{<\infty}$ of finite abelian groups which was the focus of the paper of Cohen and Lenstra. Note that each object $\textbf{Ab}^{<\infty}$ has finite automorphism group, but the category itself has infinitely many isomorphism classes (e.g., take $\mathbb{Z}/(n)$ for infinitely many $n.$)

Limiting distributions. Now, let $\mathcal{C}$ be a category such that the automorphism group of each object is finite, but unlike before, suppose that the category has infinitely many isomorphism classes. How can we define the probability and the average? First off, we may hope that the number $$|\mathcal{C}| = \sum_{[G] \in \mathcal{C}/\simeq}\frac{1}{|\mathrm{Aut}(G)|}$$ of points on $\mathcal{C}$ is finite. In certain cases, this turns out to be true. For instance, consider the category $\textbf{Ab}_{p}^{<\infty}$ of finite abelian $p$-groups for a fixed prime $p.$ It turns out that $$|\textbf{Ab}_{p}^{<\infty}| = \sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{1}{|\mathrm{Aut}(G)|} = \prod_{i=1}^{\infty}(1 - p^{-i}),$$ and people seem to credit Hall for this computation. The first consequence of this computation is that we may give each $[G] \in \in \textbf{Ab}_{p}^{<\infty}/\simeq$ the mass $$\frac{1}{|\mathrm{Aut}(G)|}\prod_{i=1}^{\infty}(1 - p^{-i})$$ to make a discrete probability distribution on the set $\textbf{Ab}_{p}^{<\infty}/\simeq$ of isomorphism classes of fintie abelian $p$-groups. This distribution is now called Cohen-Lenstra distribution. On the other hand, note that $$|\textbf{Ab}^{<\infty}| \geq \sum_{p} |\textbf{Ab}_{p}^{<\infty}| = \infty,$$ where the sum is over all primes $p.$ Hence, we realize that even though we may easily build the notions of the probability and the average of a complex-valued function on the category $\textbf{Ab}_{p}^{<\infty},$ the same would not work for $\textbf{Ab}^{<\infty}.$

One situation that we may deal with is when we have a sequence of full subcategories $$\mathcal{C}_{1} \hookrightarrow \mathcal{C}_{2} \hookrightarrow \mathcal{C}_{3} \hookrightarrow \cdots$$ of $\mathcal{C}$ such that

• each $\mathcal{C}_{n}$ has finitely many isomorphism classes and
• $\bigcup_{n=1}^{\infty}\mathcal{C}_{n} = \mathcal{C}.$

If $|\mathcal{C}|$ is finite, then writing $\mathcal{C}_{0} := \emptyset,$ we have $$|\mathcal{C}| = \sum_{j=1}^{\infty}|\mathcal{C}_{j} \smallsetminus \mathcal{C}_{j-1}| = \lim_{n \rightarrow \infty}|\mathcal{C}_{n}|.$$ Thus, for any function $f : \mathcal{C}/\simeq \rightarrow \mathbb{C},$ we must have $$M_{\mathcal{C}}(f) = \lim_{n \rightarrow \infty}M_{\mathcal{C}_{n}}(f),$$ if $|\mathcal{C}|_{f}$ (defined above) is finite, which will necessarily yield $$|\mathcal{C}|_{f} = \lim_{n \rightarrow \infty}|\mathcal{C}_{n}|_{f}.$$ This justifies the following definition.

Definition. Let $\mathcal{C}$ be a category each of whose object has finitely many automorphisms. Suppose that there is a sequence of full subcategories $$\mathcal{C}_{1} \hookrightarrow \mathcal{C}_{2} \hookrightarrow \mathcal{C}_{3} \hookrightarrow \cdots$$ of $\mathcal{C}$ such that 

• each $\mathcal{C}_{n}$ has finitely many isomorphism classes and
• $\bigcup_{n=1}^{\infty}\mathcal{C}_{n} = \mathcal{C}.$

Given any complex-valued function on the set $\mathcal{C}/\simeq$ of isomorphism classes of $\mathcal{C},$ we define the average of $f$ with respect to the sequence to be $$M_{\mathcal{C}}(f) := \lim_{n \rightarrow \infty}M_{\mathcal{C}_{n}}(f),$$ if the limit on the right-hand side exists. Given any property $\mathscr{P}$ on $\mathcal{C}/\simeq,$ we define $$\mathrm{Prob}_{G \in \mathcal{C}}(G \text{ satisfies } \mathscr{P}) := M_{\mathcal{C}}(\boldsymbol{1}_{\mathscr{P}}),$$ if the limit in the definition of the right-hand side exists.

Remark/Conjecture. The above definition a priori depends on the given sequence of full subcategories with finitely many isomorphism classes. My guess is that any such two sequences should give the same limit (if exists, under perhaps some mild conditions), but I cannot immediately come up with a proof nor an enlightening counterexample.

Finite abelian groups. Now, we focus on the category $\mathcal{C} = \textbf{Ab}^{<\infty}$ of finite abelian groups. Our goal is to compute $$M_{\textbf{Ab}^{<\infty}}(f) = \lim_{n \rightarrow \infty}\frac{\sum_{[G] \in \textbf{Ab}^{\leq n}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}}{\sum_{[G] \in \textbf{Ab}^{\leq n}/\simeq}\frac{1}{|\mathrm{Aut}(G)|}}$$ for some "reasonable" complex-valued function $f$ defined on the set $\textbf{Ab}^{< \infty}/\simeq$ of isomorphism classes of finite abelian groups with respect to the sequence $$\textbf{Ab}^{\leq 1} \hookrightarrow  \textbf{Ab}^{\leq 2} \hookrightarrow \textbf{Ab}^{\leq 3} \hookrightarrow \cdots,$$ where $\textbf{Ab}^{\leq n}$ denotes the full subcategory of $\textbf{Ab}^{< \infty},$ consisting of finite abelian groups of size $\leq n.$

Ideas of Cohen-Lenstra. Cohen and Lenstra approached the computation of $M_{\textbf{Ab}^{<\infty}}(f)$ as follows: with some "reasonable" conditions on $f$, they showed that $$M_{\textbf{Ab}^{<\infty}}(f) = \lim_{s \rightarrow 0}\frac{\sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|}}{\sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|}}.$$ It turns out that manipulating the sum $$\sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|}$$ with complex variable $s$ is quite doable for certain $f$, so computing the above limit on the right-hand side became doable. It turns out that $$\zeta_{\textbf{Ab}^{<\infty}}(s) := \sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|} = \prod_{j=1}^{\infty}\zeta(s + j),$$ where $s \mapsto \zeta(s)$ denotes the Riemann zeta function. We know that $$\zeta(s) = 1^{-s} + 2^{-s} + 3^{-s} + \cdots$$ is well-defined for $\mathrm{Re}(s) > 1,$ and this ensures that $\zeta_{\textbf{Ab}^{<\infty}}(s)$ is well-defined for $\mathrm{Re}(s) > 0.$ Using the fact that $\zeta(s)$ has analytic continuation to $\mathbb{C}$ except at the unique pole at $s = 1$ of order $1,$ we see that $\zeta_{\textbf{Ab}^{<\infty}}(s)$ has analytic continuation to $\mathbb{C}$ except the poles at $s = 0, 1, 2, \dots,$ each of whose order is $1.$

Writing $$\zeta_{\textbf{Ab}^{<\infty}, f}(s) := \sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|},$$ Cohen and Lenstra showed (in Corollary 5.5 of their paper) that if $\zeta_{\textbf{Ab}^{<\infty}, f}(s)$ has a unique pole at $s = 0$ (for $\mathrm{Re}(s) \geq 0$) then we may compute $$M_{\textbf{Ab}^{<\infty}}(f) = \lim_{s \rightarrow 0}\frac{\zeta_{\textbf{Ab}^{<\infty}, f}(s)}{\zeta_{\textbf{Ab}^{<\infty}}(s)},$$ just as we desired above.

More flexible definition. For any full subcategory $\mathcal{C}$ of $\textbf{Ab}^{<\infty},$ we define $$\zeta_{\mathcal{C}}(s) := \sum_{[G] \in \mathcal{C}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|},$$ just a notation we will use later for a computation.

Independence among primes. Let $T$ be any nonempty set of primes. Suppose that

• $f$ is multiplicative among primes in $T$ (i.e., $f(G) = \prod_{p \in T}f(G_{(p)})$).

This incorporates that $f(\text{trivial group}) = 1.$ Write $\textbf{Ab}_{T}^{<\infty}$ to be the category of finite abelian groups that are supported by the primes in $T,$ and $f|_{\textbf{Ab}_{T}^{<\infty}}$ the function given by $$f|_{\textbf{Ab}_{T}^{<\infty}}(G) := f(G_{T}),$$ where $$G_{T} := \bigoplus_{p \in T}G_{(p)}.$$

We have $$\zeta_{\textbf{Ab}^{<\infty}, f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}}(s) = \left( \prod_{p \notin T}\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|} \right) \left( \prod_{p \in T}\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|} \right).$$ Thus, we have $$\frac{\zeta_{\textbf{Ab}^{<\infty}, f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}}(s)}{\zeta_{\textbf{Ab}^{<\infty}}(s)} = \prod_{p \in T} \left( \frac{\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|}}{\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|}} \right).$$ When we are in a situation where we can exchange the limit of taking $s \rightarrow 0$ with the product (e.g., when $T$ is a finite set), then this implies that 

$$\begin{align*}M_{\textbf{Ab}^{<\infty}}(f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}) &= \lim_{s \rightarrow 0}\frac{\zeta_{\textbf{Ab}^{<\infty}, f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}}(s)}{\zeta_{\textbf{Ab}^{<\infty}}(s)} \\ &= \prod_{p \in T} \left( \frac{\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}}{\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{1}{|\mathrm{Aut}(G)|}} \right) \\ &= \prod_{p \in T}M_{\textbf{Ab}_{p}^{<\infty}}(f) \\ &= M_{\textbf{Ab}_{T}^{<\infty}}(f).\end{align*}$$ However, note that (assuming  that $f(\text{trivial group}) = 1$) we get the identity $$\zeta_{\textbf{Ab}^{<\infty}, f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}}(s) = \zeta_{\textbf{Ab}_{T}^{<\infty}, f}(s)\zeta_{\textbf{Ab}_{|\mathrm{Spec}(\mathbb{Z})| \smallsetminus T}^{<\infty}}(s),$$ where $|\mathrm{Spec}(\mathbb{Z})|$ denotes the set of all primes, even without assuming that $f$ is multiplicative. That is, we are only using that $f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}$ is multiplicative between the complements $T$ and $|\mathrm{Spec}(\mathbb{Z})| \smallsetminus T$. Therefore, we must have $$M_{\textbf{Ab}^{<\infty}}(f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}) = M_{\textbf{Ab}_{T}^{<\infty}}(f)$$ for more general family of $f.$ It is known (due to Hall) that $$\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{1}{|\mathrm{Aut}(G)|} = \prod_{i=1}^{\infty}(1 - p^{-i}),$$ so $$M_{\textbf{Ab}^{<\infty}}(f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}) = \frac{\sum_{[G] \in \textbf{Ab}_{T}^{<\infty}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}}{\prod_{p \in T}\prod_{i=1}^{\infty}(1 - p^{-i})}.$$ If $f$ is multiplicative over each element of $T,$ then we also have $$\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{1}{|\mathrm{Aut}(G)|} = \prod_{i=1}^{\infty}(1 - p^{-i})^{-1},$$ so $$M_{\textbf{Ab}^{<\infty}}(f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}) = \prod_{p \in T} \left( \sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}  \prod_{i=1}^{\infty}(1 - p^{-i})\right).$$

Computing some distributions. By the last paragraph, for any fixed prime $p,$ we may compute $$\mathrm{Prob}_{G \in \textbf{Ab}^{<\infty}}(G_{(p)} = 0) = \prod_{i=1}^{\infty}(1 - p^{-i}),$$ and for any nonzero finite abelian $p$-group $H,$ we have $$\mathrm{Prob}_{G \in \textbf{Ab}^{<\infty}}(G_{(p)} \simeq H \text{ or } 0) = \left(1 + \frac{1}{|\mathrm{Aut}(G)|}\right) \prod_{i=1}^{\infty}(1 - p^{-i}).$$ This implies that $$\mathrm{Prob}_{G \in \textbf{Ab}^{<\infty}}(G_{(p)} \simeq H) = \frac{1}{|\mathrm{Aut}(G)|} \prod_{i=1}^{\infty}(1 - p^{-i}).$$ That is, Cohen and Lenstra showed that the $p$-part of a random finite abelian group follows the Cohen-Lenstra distribution.

Counting with weight

Let $G$ be a finite group acting on a nonempty finite set $X.$ The groupoid $[X/G]$ of this action is a category defined by the following data:

• objects: elements of $X$;
• morphism: $\mathrm{Hom}_{[X/G]}(x, y) = \{g \in G : gx = y\}.$

In this category, every morphism is invertible, so $[X/G]$ is indeed a groupoid. We have $$\mathrm{Aut}_{[X/G]}(x) = \mathrm{Stab}_{G}(x),$$ the stabilizer subgroup of $G$ at $x \in X.$ The set $X/G$ of orbits of this action is precisely the set of isomorpihsm classes of $[X/G].$ 

Given an isomorphism class $[x]$ of $[X/G]$ (i.e., an orbit of the given action), the orbit-stabilizer theorem translates to $$|[x]| = |G|/|\mathrm{Aut}_{[X/G]}(x)|.$$ This implies that $$\sum_{[x] \in X/G}\frac{1}{|\mathrm{Aut}_{[X/G]}(x)|} = \frac{|X|}{|G|}.$$ Now, say we have a subset $Y \subset X$ where the $G$-action restricts to. Then

$$\frac{\sum_{[y] \in Y/G}\frac{1}{|\mathrm{Aut}(y)|}}{\sum_{[x] \in X/G}\frac{1}{|\mathrm{Aut}(x)|}} = \frac{|Y|/|G|}{|X|/|G|} = \frac{|Y|}{|X|}.$$ Hence, we just obtained another expression of the probability that a random element of $X$ lies in $Y,$ when $X$ is given the uniform distribution.

Why do we bother? We could just write $|Y|/|X|$ instead of another fraction of some complicated summations. The first reason I can think of is quite philosophical. If we are just given categories $\mathcal{B} \subset \mathcal{A}$, where the latter is nonempty, with finitely many isomorphism classes that are not necessarily given as the action groupoids, how should we define the probability that a random object of $\mathcal{A}$ is in $\mathcal{B}$? One may say $|\mathrm{Ob}(\mathcal{B})|/|\mathrm{Ob}(\mathcal{A})|,$ but it is rarely the case that a category has finitely many objects. It would be more reasonable to define $$\mathrm{Prob}_{x \in \mathcal{A}}(x \in \mathcal{B}) := \frac{\sum_{[y] \in \mathcal{B}/\simeq}\frac{1}{|\mathrm{Aut}(y)|}}{\sum_{[x] \in \mathcal{A}/\simeq}\frac{1}{|\mathrm{Aut}(x)|}},$$ which is a "generalization" of the case of action groupoids. Perhaps, we should define the number of points on the category $\mathcal{A}$ as $$\| \mathcal{A} \| := \sum_{[x] \in \mathcal{A}/\simeq}\frac{1}{|\mathrm{Aut}(x)|}.$$ OK. Now I am going to speak about something that I am not quite familiar with. The sum over the isomorphism classes with the weights above also seems to be the definition of the number of points over a finite field for a Deligne-Mumford stack (e.g. van den Bogaard and Edixhoven). When we have a quotient stack, this seems to fit our discussion even more evidently, but I am quite daunted to check any formal "stacky" axioms, so let's think about an example.

Example. Consider the scheme $\mathrm{Mat}_{n} := \mathbb{A}^{n^{2}} = \mathrm{Spec}(\mathbb{F}_{q}[x_{ij}]_{1 \leq i, j \leq n})$ over a fixed finite field $\mathbb{F}_{q}.$ We chose the notation because we would like to think about $n \times n$ matrices $(x_{ij})_{i,j=1}^{n}$ in the back of our head. We can then think about the group scheme $$\mathrm{GL}_{n} := \mathrm{Mat}_{n} \smallsetminus V(\det),$$ which is an open subset of $\mathrm{Mat}_{n}.$ We have a scheme-theoretic action of $\mathrm{GL}_{n}$ on $\mathrm{Mat}_{n},$ given by the conjugation, and this gives a quotient stack $[\mathrm{Mat}_{n}/\mathrm{GL}_{n}].$ The set $[\mathrm{Mat}_{n}/\mathrm{GL}_{n}](\mathbb{F}_{q})$ of $\mathbb{F}_{q}$-points of this quotient stack is precisely the groupoid of the conjugate action of the group $\mathrm{GL}_{n}(\mathbb{F}_{q})$ of invertible matrices on the set $\mathrm{Mat}_{n}(\mathbb{F}_{q})$ of $n \times n$ matrices over $\mathbb{F}_{q},$ so in notation, we have $$[\mathrm{Mat}_{n}/\mathrm{GL}_{n}](\mathbb{F}_{q}) = [\mathrm{Mat}_{n}(\mathbb{F}_{q})/\mathrm{GL}_{n}(\mathbb{F}_{q})].$$ Moreover, note that the isomorphism classes of $\mathbb{F}_{q}[t]$-modules of $\mathbb{F}_{q}$-dimension $n$ are in one-to-one correspondence with the orbits of this conjugation action. The category $\mathrm{Mod}_{\mathbb{F}_{q}[t]}^{\dim_{\mathbb{F}_{q}} = n}$ of $\mathbb{F}_{q}[t]$-modules of $\mathbb{F}_{q}$-dimension $n$ surely has infinitely many objects, but it has only finitely many isomorphism classes, each of corresponds to an element of $\mathrm{Mat}_{n}(\mathbb{F}_{q})/\mathrm{GL}_{n}(\mathbb{F}_{q}).$ If we unravel the definition of probability we defined above, for any property $\mathscr{P}$ on the objects of the category $\mathrm{Mod}_{\mathbb{F}_{q}[t]}^{\dim_{\mathbb{F}_{q}} = n}$ that is invariant under its isomorphism classes, we have: $$\mathrm{Prob}_{M \in \mathrm{Mod}_{\mathbb{F}_{q}[t]}^{\dim_{\mathbb{F}_{q}} = n}}(M \text{ satisfies } \mathscr{P}) = \mathrm{Prob}_{A \in \mathrm{Mat}_{n}(\mathbb{F}_{q})}(A \text{ satisfies } \mathscr{P}),$$ where the right-hand side is given by the uniform distribution on $\mathrm{Mat}_{n}(\mathbb{F}_{q}).$

Remark. Another reason for taking the definition for the number of points with the weight $1/|\mathrm{Aut}|$ is due to Behrend's trace formula.

Avoiding some points over an infinite field

Prime Avoidance states that if $I$ is an ideal of a ring $R$ (commutative and unital) contained in the union of finitely many prime ideals $\mathfrak{p}_{1}, \dots, \mathfrak{p}_{r}$ of $R$, then $I$ is contained in one $\mathfrak{p}_{i}$ of these prime ideals. 

Geometrically, we look at the ideal $I$ as a closed subscheme $$\mathrm{Spec}(R/I) \hookrightarrow \mathrm{Spec}(R),$$ and in this setting, Prime Avoidance says that if $\mathrm{Spec}(R/I)$ (or more precisely its image $V_{R}(I)$) contains the intersection of a finitely many irreducible closed subsets $Z_{1}, \dots, Z_{r}$ of $\mathrm{Spec}(R)$, then it must contain one $Z_{i}$ of the closed subsets.

When $R$ contains an infinite field, then this can be strengthened: $\mathfrak{p}_{i}$ do note need to be prime ideals, but just ideals. Geometrically, this means that we do not need to require $Z_{i}$ to be irreducible.

Prime Avoidance Lemma (over infinite fields). Let $k$ be an infinite field and $R$ a vector space over $k$ (e.g., $R$ can be a $k$-algebra). Let $I$ be a subspace of $R$. Given finitely many proper subspaces $\mathfrak{p}_{1}, \dots, \mathfrak{p}_{r}$ of $R$, if $I$ is contained in the union of the subspaces, then there is a single $\mathfrak{p}_{i}$ containing $I$.

Proof. We follow Hochster's notes. Replacing $\mathfrak{p}_{i}$ with $\mathfrak{p}_{i} \cap I,$ we may assume that $\mathfrak{p}_{i} \subsetneq I.$ Then we show that $I = \mathfrak{p}_{i}$ for some $i$.

Suppose for contradiction that $I$ is not equal to any of $\mathfrak{p}_{1}, \dots, \mathfrak{p}_{r}$. Under this assumption, we may choose $v_{i} \in I \setminus \mathfrak{p}_{i}$ for each $1 \leq i \leq r$. Thus, the $k$-subspace $J$ of $I$ generated by $v_{1}, \dots, v_{r}$ will not be equal to any $\mathfrak{p}_{1} \cap J, \dots, \mathfrak{p}_{r} \cap J$. 

Focusing on $J$ reduces our statement to the case where $I$ is finite dimensional over $k.$ Hence, we may assume that $I = k^{n}$ for some $n \geq 0$. For each $\mathfrak{p}_{i}$, since it is not the whole $k^{n},$ there is a $k$-linear map $L_{i} : k^{n} \rightarrow k$ such that $L_{i}$ is not identically zero on $I = k^{n}$ but it is on $\mathfrak{p}_{i}$. Since $L_{i}$ is an $1 \times n$ row vector $(a_{i1}, \dots, a_{in})$, we may write $L_{i}(x_{1}, \dots, x_{n}) = a_{i1}x_{1} + \cdots + a_{in}x_{n}$. Now, consider the polynomial $$f(x_{1}, \dots, x_{n}) := \prod_{i=1}^{n}L_{i}(x_{1}, \dots, x_{n}).$$ We note that $f$ is a nonzero polynomial in $k[x_{1}, \dots, x_{n}]$ because none of $L_{i}$ are zero polynomials, but each $L_{i}$ vanishes on $\mathfrak{p}_{i}$, and the union of $\mathfrak{p}_{1}, \dots, \mathfrak{p}_{n}$ is $k^{n}$. Hence, we must have $f(x_{1}, \dots, x_{n}) = 0$ for any evaluation of $x_{1}, \dots, x_{n} \in k$. Since $k$ is infinite, the only such polynomial is $f = 0$, a contradiction. $\Box$

Remark. Although we chose ring-friendly notations, the above statement is purely a statement about vector spaces over an infinite field $k.$ It says: any vector space over an infinite field cannot be covered by finitely many proper subspaces.

Avoidance in geometry. Now, suppose that we have finitely many nonzero vectors $(x_{i0}, \dots, x_{in}) \in k^{n+1}$ for $1 \leq i \leq r.$ For any fixed $i,$ the vectors $(a_{0}, \dots, a_{n}) \in k^{n+1}$ satsifying the equation $$a_{0}x_{i0} + \cdots + a_{n}x_{in} = 0$$ form a proper subsapce $V_{i}$ of $k^{n+1}.$ If $k$ is infinite, then $V_{1}, \dots, V_{r}$ cannot cover $k^{n+1},$ so there must be a vector $(a_{0}, \dots, a_{n}) \in k^{n+1}$ that does not satisfy any of the $r$ equations. This vector is necessarily nonzero because zero is in any vector space. This shows the following.

Proposition. Given any finitely many $k$-points of $\mathbb{P}^{n},$ there is a hyperplane avoiding all those points.

Q. Can we find a hyperplane in $\mathbb{P}^{n}$ (over an infinite field $k$) that avoids any given finitely many points (not just $k$-points)?

The answer is yes. 

Proof. Let $\mathfrak{p}_{1}, \dots, \mathfrak{p}_{r}$ be homogeneous prime ideals in $k[x_{0}, \dots, x_{n}]$ not containing $(x_{0}, \dots, x_{n})$, recalling that such prime ideals parametrize the points of $$\mathbb{P}^{n}_{k} = \mathrm{Proj}(k[x_{0}, \dots, x_{n}]).$$ Using Prime Avoidance, all we need to show that there is a nonzero vector $(a_{0}, \dots, a_{n}) \in k^{n+1}$ such that $a_{0}x_{0} + \cdots + a_{n}x_{n}$ is not in $$\mathfrak{p}_{1} \cup \cdots \cup \mathfrak{p}_{r}.$$ This has to be true because otherwise all of $x_{0}, \dots, x_{n}$ must be in the union so that $(x_{0}, \dots, x_{n})$ must be in the union, but this would mean that one of $\mathfrak{p}_{i}$ must be equal to $(x_{0}, \dots, x_{n}),$ a contradiction. This finishes the proof $\Box$

Q. What about a projective variety (instead of $\mathbb{P}^{n}$)?

The answer can be yes, but this is just a matter of definition.

Let $X$ be a projective $k$-scheme. This means that we can assume $$X = \text{Proj}(k[x_{0}, \dots, x_{n}]/I) \hookrightarrow \mathbb{P}^{n}_{k},$$ for a homogenous ideal $I$ in $k[x_{0}, \dots, x_{n}]$ not containing $(x_{0}, \dots, x_{n}).$ The line bundle $\mathscr{O}_{\mathbb{P}^{n}}(1)$ pulls back to $\mathscr{O}_{X}(1)$ under this inclusion. Thus, if $k$ is an infinite field, for any finitely many points $p_{1}, \dots, p_{r} \in X,$ we can find a hyperplane $H \subset \mathbb{P}^{n}$ not containing any of $p_{1}, \dots, p_{r}.$ If we say $L$ is a nonzero linear form in $k[x_{0}, \dots, x_{n}]$ (i.e., $L$ is a global section of $\mathscr{O}_{\mathbb{P}^{n}}(1)$) presenting $H,$ then $L$ does not vanish on any of the points $p_{1}, \dots, p_{r},$ so its pull back $L|_{X}$ does not either. Hence, the locus of $L|_{X}$ is the "hyperplane" of $X$ that does not pass through any of the points $p_{1}, \dots, p_{r}.$

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.

Some projective bundle

I have stopped studying algebraic geometry a while ago because I got scared by the following exercise.

Exercise. Say we have the coordinate $[a_{00} : a_{01} : a_{02} : a_{11} : a_{12} : a_{22}]$ for $\mathbb{P}^{5}$ and $[x_{0} : x_{1} : x_{2}]$ for $\mathbb{P}^{2}$ over a fixed field $k$. Let $X$ be the closed subscheme of $\mathbb{P}^{5} \times\mathbb{P}^{2}$ cut out by the equation $$a_{00}x_{0}^{2} + a_{01}x_{0}x_{1} + a_{02}x_{0}x_{2} + a_{11}x_{1}^{2} + a_{12}x_{1}x_{2} + a_{22}x_{2}^{2} = 0.$$ Realizing $X$ as a $\mathbb{P}^{4}$-bundle over $\mathbb{P}^{2}$, show that $X$ is a smooth sixfold, and that $\mathrm{Pic}(X) \simeq \mathbb{Z}^{2}$.

As a primitive human being, I usually just take partial derivatives and check whether the non-full-rank-Jacobian locus of them intersects the locus of original equation on every affine open set of an affine open cover. However, I could not possibly think about spending my time on doing so, which is why I stopped studying. I just realized that the author was giving me an hint all along. Once we notice that $X$ is a $\mathbb{P}^{4}$ bundle over $\mathbb{P}^{2}$ then it means that $X$ can be covered by open subsets of the form $\mathbb{P}^{4} \times U$ where $U$ are some open subsets of $\mathbb{P}^{2}$. Thus, once we can check that $X$ is such object, we would get its smoothness for free!

Q. How can we see $X$ as a $\mathbb{P}^{4}$-bundle over $\mathbb{P}^{2}$?

Ad-hoc argument. If we fix any $k$-point $[x_{0} : x_{1} : x_{2}] \in \mathbb{P}^{2}(k)$, then denoting $z_{ij} = x_{i}x_{j}$, we see that any $\boldsymbol{a} = [a_{00} : a_{01} : a_{02} : a_{11} : a_{12} : a_{22}] \in \mathbb{P}^{5}(k)$ such that $$z_{00}a_{00} + z_{01}a_{01} + z_{02}a_{02} + z_{11}a_{11} + z_{12}a_{12} + z_{22}a_{22} = 0$$ will form the points $(\boldsymbol{a}, [x_{0} : x_{1} : x_{2}]) \in X(k)$. Since $z_{ij} = x_{i}x_{j} \in k$ are fixed, these $\boldsymbol{a}$ cuts out a hyperplane of $\mathbb{P}^{5}$, which is isomorphic to $\mathbb{P}^{4}$. Now, as an ad-hoc geometor, I can imagine $\boldsymbol{x} = [x_{0} : x_{1} : x_{2}]$ "continuously" moving in $\mathbb{P}^{2}$, so we would get what we want.

Hopefully legit argument. First off, we need to understand why $X$ is a closed subscheme of $\mathbb{P}^{5} \times \mathbb{P}^{2}$. We first apply the Veronese embedding to the second component $\mathbb{P}^{2} \hookrightarrow \mathbb{P}^{5}$, which restricts to $$[x_{0} : x_{1} : x_{2}] \mapsto [x_{0}^{2} : x_{0}x_{1} : x_{0}x_{2} : x_{1}^{2} : x_{1}x_{2} : x_{2}^{2}]$$ on the sets of $k$-points. We may form such $X$ in $\mathbb{P}^{5} \times \mathbb{P}^{5}$, and this sits in $\mathbb{P}^{5} \times \mathbb{P}^{2}$ as a closed subscheme. Writing $$\boldsymbol{z} = [z_{00} : z_{01} : z_{02} : z_{11} : z_{12} : z_{22}]$$ for the coordinates for the second $\mathbb{P}^{5}$, our $X$ is cut out by the equation $$a_{00}z_{00} + a_{01}z_{01} + a_{02}z_{02} + a_{11}z_{11} + a_{12}z_{12} + a_{22}z_{22} = 0$$ and another degree $2$ homogeneous equation in $z_{ij}$, which records the image of $\mathbb{P}^{2}$ under the Veronese embedding. We are implicitly using the Segre embedding $\mathbb{P}^{5} \times \mathbb{P}^{2} \hookrightarrow \mathbb{P}^{5} \times \mathbb{P}^{5}.$ Anyways, we now understand that $X$ is a legit projective scheme.

To show that $X$ is a $\mathbb{P}^{4}$ bundle over $\mathbb{P}^{2}$, we need to find a locally free sheaf $\mathcal{E}$ of rank $5$ on $\mathbb{P}^{2}$ such that $$X \simeq \textbf{Proj}(\mathrm{Sym}^{\bullet}(\mathcal{E})).$$ This seems very scary, but let's see what I can do. The standard affine open cover for $\mathbb{P}^{2}$ is given by the three affine open subsets $$U_{i} := \mathrm{Spec}(k[x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}]) \simeq \mathbb{A}^{2}$$ for $i = 0, 1, 2$. Over each $U_{i}$, we should get an open subset $X_{i}$ of $X$ such that $$X_{i} \simeq \mathrm{Proj}(\mathrm{Sym}^{\bullet}(\mathcal{E}(U_{i}))).$$ Okay. This is less scary because we ran away from the relative stuff. We would be done, if we could magically make this happen so that $X_{i} \simeq \mathbb{P}^{4}$. The defining equation for $X$ is $$a_{00}x_{0}^{2} + a_{01}x_{0}x_{1} + a_{02}x_{0}x_{2} + a_{11}x_{1}^{2} + a_{12}x_{1}x_{2} + a_{22}x_{2}^{2} = 0,$$ so for $X_{i}$, we would need $$a_{00}\left(\frac{x_{0}}{x_{i}}\right)^{2} + a_{01}\left(\frac{x_{0}}{x_{i}}\right)\left(\frac{x_{1}}{x_{i}}\right) + a_{02}\left(\frac{x_{0}}{x_{i}}\right)\left(\frac{x_{2}}{x_{i}}\right) + a_{11}\left(\frac{x_{1}}{x_{i}}\right)^{2} \\ + a_{12}\left(\frac{x_{1}}{x_{i}}\right)\left(\frac{x_{2}}{x_{i}}\right) + a_{22}\left(\frac{x_{2}}{x_{i}}\right)^{2} = 0.$$ We may look at $x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i} \in k[x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}]$ as if they are scalars. With this viewpoint, our $X_{i}$ can be defined by taking $\mathrm{Proj}$ of $k[x_{0}/x_{1}, x_{1}/x_{i}, x_{2}/x_{i}, \boldsymbol{a}]$ modulo the equation above, which we may denote as $$f(x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}, \boldsymbol{a}) = 0.$$ Continuing our reverse-engineering, we want $$\mathrm{Sym}^{\bullet}(\mathcal{E}(U_{i})) = k\left[\frac{x_{0}}{x_{i}}, \frac{x_{1}}{x_{i}}, \frac{x_{2}}{x_{i}}, \boldsymbol{a}\right]/\left(f\left(\frac{x_{0}}{x_{i}}, \frac{x_{1}}{x_{i}}, \frac{x_{2}}{x_{i}}, \boldsymbol{a}\right)\right).$$ Hence, it would be nice if we can find a rank $5$ free module $\mathcal{E}(U_{i})$ over $k[x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}]$, whose symmetric powers give the above ring. This seems like a cheating, but I think we can just take degree $1$ part (in $\boldsymbol{a} = (a_{ij})$) of the above ring, namely $$\mathcal{E}(U_{i}) := \frac{ R_{i}a_{00} \oplus R_{i}a_{01} \oplus R_{i}a_{02} \oplus R_{i}a_{11} \oplus R_{i}a_{12} \oplus R_{i}a_{22} }{ R_{i} ( x_{0/i}^{2}a_{00} + x_{0/i}x_{1/i}a_{01} + x_{0/i}x_{2/i}a_{02} + x_{1/i}^{2}a_{11} + x_{1/i}x_{2/i}a_{12} + x_{2/i}^{2}a_{22})},$$ where $x_{j/i} := x_{j}/x_{i}$ and $R_{i} := k[x_{0/i}, x_{1/i}, x_{2/i}]$ and call it a day. Gluing over different $U_{i}$'s should work out to construct $\mathcal{E}$.

Q. What about the Picard group?

Let's do this later. Exercise II.7.9 of Hartshorne's book says that for a very general scheme $Y$, which includes $\mathbb{P}^{2}$, for any locally free sheaf $\mathcal{E}$ of $Y$ with finite rank $\geq 2$, we have $$\mathrm{Pic}(\textbf{Proj}(\mathrm{Sym}^{\bullet}(\mathcal{E}))) \simeq \mathrm{Pic}(Y) \times \mathbb{Z}.$$ Since $\mathrm{Pic}(\mathbb{P}^{n}) \simeq \mathbb{Z}$ for any $n \geq 1$ (over any field), taking $Y = \mathbb{P}^{2}$, this general statement will finish the exercise we started.