Naive Line Bundle Functor isn't a Zariski Sheaf

Question

I'm reading a set of notes on algebraic stacks by Anatoly Preygel, and in the opening paragraphs it says

Meanwhile moduli problems... often have the property that the functor

$$T \mapsto \{ \text{isomorphism classes of ... over $T$} \}$$

is not only not a scheme but not even a sheaf. Indeed taking ... to be "line bundles" shows that it is not even a Zariski sheaf: The stalks of the presheaf are all trivial, but there can exist non-trivial sections.

I think I know what this means, but I'm relying on two guesses (which I'm charitably calling "intuition" in my attempt below). If someone could please verify them, and provide any other insight you think might be valuable, I would really appreciate it!

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Write $\mathsf{Aff}$ for the opposite of the category of finitely presented rings.

Let's call this functor $\mathcal{L} : \mathsf{Aff}^\text{op} \to \mathsf{Set}$, so it's a presheaf on $\mathsf{Aff}$. We want to show that it's not a sheaf for the (big) Zariski topology. We'll do that by showing that

1. each stalk is trivial
2. there are nontrivial sections

After all, if each stalk is trivial, then by uniqueness of gluing for sheaves, we should only get trivial sections. So if (1) and (2) are both satisfied, we'll show that $\mathcal{L}$ isn't a sheaf.

Let's start with (2), which is easier. A section of $\mathcal{L}$ is just $\mathcal{L}(\text{Spec} S)$ for your favorite (finitely presented) ring $S$. Of course, there are plenty of rings admitting nontrivial line bundles (said another way, there are plenty of rings $S$ with nontrivial picard group) so that nontrivial sections of $\mathcal{L}$ are abundant.

Next up is (1). According to this MO question, points of $\mathcal{E} = \mathsf{Sh}(\mathsf{Aff}, \text{Zar})$ are local rings $R$. Indeed, the inverse image part of the point (read: geometric morphism) $\mathsf{Set} \to \mathcal{E}$ will be given by $R^* : \mathcal{E} \to \mathsf{Set}$, the kan extension of $\text{Hom}(\text{Spec}(R),-) : \mathsf{Aff} \to \mathsf{Set}$ along the (yoneda) embedding $y : \mathsf{Aff} \to \mathcal{E}$. The good news is that we only need to understand $R^*$, since the stalk of $\mathcal{L}$ at $R$ is exactly $R^* \mathcal{L}$. The bad news is that I'm inexperienced with making these calculations...

Intuition says that $R$ is a colimit of its finitely presented subrings, each of which is an object of $\mathsf{Aff}$. So each finitely presented subring is like an "open set containing $R$", and we should be able to compute the stalk of $\mathcal{L}$ at $R$ as $\text{colim } \mathcal{L}(\text{Spec} R_i)$, where the $R_i$ are the finitely presented subrings of $R$... Much weaker intuition says that this should be the same thing as invertible sheaves on $R$. This gives the expected answer since we know the picard group for local rings is trivial (remember $R$ was local).

So indeed, if the above is correct, each stalk is trivial, showing (1), and we're done.

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Thanks in advance for any thoughts! ^_^

Answer 1

Your first idea is correct, that the stalk can be computed as a colimit $$\mathrm{colim}\, \mathcal{L}(R_i)$$ where the $R_i$ are the finitely presented subrings of $R$.

The reason is as follows. The category of points of a presheaf topos $\mathbf{PSh}(\mathcal{C})$ agrees with the opposite of the category of pro-objects of $\mathcal{C}$. Equivalently, it agrees with the category of ind-objects in $\mathcal{C}^\mathrm{op}$. An ind-object is written formally as $\varinjlim_i R_i$ for $(R_i)_i$ a filtered diagram in $\mathcal{C}^\mathrm{op}$. The stalk of a presheaf $F$ at a point $\varinjlim_i R_i$ is given by $\varinjlim F(R_i)$. (I write $\varinjlim$ rather than $\mathrm{colim}$ to emphasize that the colimit is filtered.)

Now if there is a Grothendieck topology $J$ on $\mathcal{C}$, then a point $\varinjlim R_i$ of $\mathbf{PSh}(\mathcal{C})$ factors through the sheaf topos $\mathbf{Sh}(\mathcal{C},J)$ if and only if the point is "$J$-local" in some sense. See for example Gabber–Kelly, "Points in algebraic geometry", Section 1. The stalk of a sheaf $F$ at a point $\varinjlim R_i$ is still given by $\varinjlim F(R_i)$.

In the case of the big Zariski topos, $\mathcal{C}^\mathrm{op}$ is the category of finitely presented rings. The ind-category of the category of finitely presented rings is the category of all rings, with the formal symbol $\varinjlim_i R_i$ in the ind-category corresponding to the actual colimit $\varinjlim_i R_i$ in the category of rings. A ring is $J$-local, with $J$ the Zariski topology, if and only if it is a local ring. A ring $R$ can be written as a filtered colimit of its finitely presented subrings $R_i$, and then the stalk of your sheaf $\mathcal{L}$ is given by $\varinjlim \mathcal{L}(R_i)$.

To show that the stalk is trivial, take some finitely presented subring $S$ of $R$, and let $L$ be a line bundle on $\mathrm{Spec}(S)$. Take an element $f \in S$ such that $L$ is trivial on $\mathrm{Spec}(S_f)$ and such that the map $\mathrm{Spec}(R) \to \mathrm{Spec}(S)$ factorizes through $\mathrm{Spec}(S_f)$. Then $f$ is invertible in $R$, so $S_f$ is a (finitely presented) subring of $R$. We conclude that the line bundle $L$ becomes trivial after base change to some larger finitely presented subring. This shows that the colimit $\varinjlim \mathcal{L}(R_i)$ is trivial.

Your second idea avoids the last part of the above proof by using instead the fact that the Picard group of a local ring is trivial. This also works, but it relies on more technical machinery. You can compute: $$\varinjlim \mathcal{L}(R_i) \simeq \varinjlim \mathrm{H}^1(\mathrm{Spec}(R_i), \mathbb{G}_{m,R_i}) \simeq \mathrm{H}^1(\mathrm{Spec}(R), \mathbb{G}_{m,R}) \simeq \mathrm{Pic}(R) \simeq 0.$$ The first and third isomorphism are based on the cohomological formulation of the Picard group. I didn't specify on which site the cohomology groups are computed (big or small, Zariski or étale) because in this case it doesn't matter. The second isomorphism uses SGA4, Exposé VII, Théorème 5.7 and the fact that $\varinjlim \phi_i^*\mathbb{G}_{m,R_i} \simeq \mathbb{G}_{m,R}$, for $\phi_i : \mathrm{Spec}(R) \to \mathrm{Spec}(R_i)$ the natural map. See also the answer here.

Answer 2

$\def\Pic{\text{Pic}}$You are writing things in a very sophisticated way. I'm going to respond to the question at the top but not try to follow all of the thoughts in the middle. For a scheme $X$, I'll write $\Pic(X)$ for the abelian group of isomorphisms classes of line bundles on $X$.

(1) The functor $U \mapsto \Pic(U)$ is not a sheaf.

The "locality" condition in the definition of a sheaf says that, if $X = U \cup V$ is an open cover, and $A$ and $B$ are line bundles on $X$ with $A|_U \cong B|_U$ and $A|_V \cong B|_V$, then $A \cong B$. This is the opposite of how line bundles work!

Take $X = \mathbb{P}^1$ and let $U$ and $V$ be the standard cover of $\mathbb{P}^1$ by two $\mathbb{A}^1$'s. Every line bundle becomes trivial on $U$ and on $V$, but $\mathbb{P}^1$ has nontrivial line bundles.

If you for want $X$ to be affine as well (since your original question talked about $\text{Aff}$) take $X = \text{Spec} \mathbb{Z}[\sqrt{-5}]$, $U = \text{Spec} \mathbb{Z}[\sqrt{-5}, 2^{-1}]$ and $V = \text{Spec} \mathbb{Z}[\sqrt{-5}, 3^{-1}]$. Again, $\mathbb{Z}[\sqrt{-5}]$ is not a UFD but $\mathbb{Z}[\sqrt{-5}, 2^{-1}]$ and $\mathbb{Z}[\sqrt{-5}, 3^{-1}]$ are, so there are nontrivial line bundles on $X$ which become trivial on each of $U$ and $V$.

(2) The stronger statement that the stalks of $U \mapsto \Pic(U)$ are all trivial. I find the language of stalks on the big Zariski site awkward, so let's just talk about open sets on a fixed scheme $X$. Then $U \mapsto \Pic(U)$ is a presheaf on $X$.

Let $p$ be any point of $X$. Then the stalk of this presheaf is $\lim_{\to} \Pic(V)$ where the limit is over all open neighborhoods $V$ of $p$. Let $U$ be any open neighborhood of $p$, and let $L$ be any line bundle on $U$. Then there is some neighborhood $V$ of $p$ such that $L|_V$ is trivial (that's what a line bundle means!). So $[L]$ maps to $1$ in $\Pic(V)$. So every class becomes trivial in the stalk.