Coarse moduli space of relative Picard functor for affine line

Consider the relative Picard functor $$\mathrm{Pic}_{\mathbb A^1/\mathrm{Spec}(\mathbb C)}$$ sending a complex scheme $$X$$ to $$\mathrm{Pic}(X \times \mathbb A^1)/\pi_X^* \mathrm{Pic}(X)$$. Since $$\mathrm{Pic}(\mathbb A^1) = \{\mathcal O_{\mathbb A^1}\}$$, the only possible coarse or fine moduli space to this is $$\mathrm{Spec}(\mathbb C)$$. But there is an example of a scheme $$X$$ such that $$\mathrm{Pic}(X \times \mathbb A^1) \neq \mathrm{Pic}(X)$$ and thus the point cannot be a fine moduli space.

My question is: Is $$\mathrm{Spec}(\mathbb C)$$ a coarse moduli space for $$\mathrm{Pic}_{\mathbb A^1/\mathrm{Spec}(\mathbb C)}$$?

In other words, does every natural transformation $$\mathrm{Pic}_{\mathbb A^1/\mathrm{Spec}(\mathbb C)} \to \mathrm{Morphisms}(-, M')$$ for a scheme $$M'$$ factor through $$\mathrm{Morphisms}(-, \mathrm{Spec}(\mathbb C))$$? This would follow if for every line bundle $$\mathcal L$$ on a product $$X \times \mathbb A^1$$ we can find an epimorphism $$X' \to X$$ such that the pullback of $$\mathcal L$$ to $$X' \times \mathbb A^1$$ is a pullback from $$X'$$ (e.g. since $$\mathrm{Pic}(X' \times \mathbb A^1) = \mathrm{Pic}(X')$$). This is true e.g. for $$X'$$ normal, but I don't see how this helps if $$X$$ is non-reduced.

• I haven’t checked the details carefully but you can define normalization more generally (stacks.math.columbia.edu/tag/035E) and the map factors through taking the reduced scheme structure on $X$, so I think taking normalization in this generality should work. Commented Mar 17, 2020 at 21:50
• I think it is true that normalization works in the required generality, but then it does not give an epimorphism I think. The normalization of S=Spec($\mathbb C[t]/t^2$) should be T=Spec($\mathbb C$), but the map $T \to S$ is not an epimorphism (since maps $S \to X$ to some other scheme are not determined by the image of the closed point $T$).
– JoS
Commented Mar 18, 2020 at 8:46

After thinking about this for some time after today's lecture, I believe the answer to your question is "yes" and I'll attempt to give a detailed proof (which might be a bit long for a post, but here we go).

Let $$M$$ be any $$\mathbb C$$-scheme and $$h^M=\operatorname{Hom}_{\mathrm{Sch}/\mathbb C}(-,M)$$ its represented functor. Let $$\eta\colon \operatorname{Pic}_{\mathbb A^1/\mathbb C}\rightarrow h^M$$ be any natural transformation; we must show that $$\eta_X\colon \operatorname{Pic}_{\mathbb A^1/\mathbb C}(X)\rightarrow h^M(X)$$ has image a single point for all $$\mathbb C$$-schemes $$X$$.

Step 0. We reduce to the case where $$X$$ is affine. Consider any affine open cover $$X=\bigcup U_i$$ and the commutative diagram

The right vertical morphism is injective because $$h^M$$ is a Zariski-sheaf. Thus it suffices to show that each $$\eta_{U_i}\colon \operatorname{Pic}_{\mathbb A^1/\mathbb C}(U_i)\rightarrow h^M(U_i)$$ has image a single point.

Step 1. We do the case where $$X$$ is reduced (and affine). For a point $$x\in X$$ let $$\kappa(x)$$ denote its residue field. We first claim that $$h^M(X)\rightarrow \prod_{x\in X} h^M(\operatorname{Spec}\kappa(x))$$ is injective again. After passing to an affine cover of $$X$$ that is mapped into an affine cover of $$M$$ and using some simple arguments as in Step 0, this boils down to the following question: Let $$A$$ be a reduced ring and $$f,g\colon B\rightarrow A$$ two ring morphisms that agree after composition with $$A\rightarrow \kappa(\mathfrak p)$$ for all $$\mathfrak p\in \operatorname{Spec} A$$. Then $$f=g$$. Indeed, the difference $$f-g$$ must have image contained in $$\bigcap_{\mathfrak p\in \operatorname{Spec} A}\mathfrak p=0$$ since $$A$$ is reduced.

Now consider the diagram

and observe that the bottom left product is a single point because both $$\operatorname{Pic}(\operatorname{Spec} \kappa(x)[t])$$ and $$\operatorname{Pic}(\operatorname{Spec} \kappa(x))$$ are trivial for all $$x\in X$$ (since line bundles over a UFD are trivial). So injectivity of the right vertical arrow does the trick.

Step 2. We do the case where $$X$$ is noetherian (and affine). To this end we claim $$\operatorname{Pic}(X)\cong \operatorname{Pic}(X^{\mathrm{red}})$$ for every affine noetherian scheme $$X$$, which immediately reduces everything to Step 1 [Edit: Turns out it doesn't, but fortunately Nuno has found a beautiful fix.] (this also uses $$(X\times \mathbb A^1)^{\mathrm{red}}\cong X^{\mathrm{red}}\times\mathbb A^1$$ which follows from a simple inspection). To prove the claim, let $$\mathcal J$$ be the coherent sheaf on $$X$$ cutting out its reduction $$X^{\mathrm{red}}$$. Since $$X$$ is noetherian, $$\mathcal J^n=0$$ for some $$n$$. Doing induction on $$n$$ we may assume $$\mathcal J^2=0$$. Now consider the short exact sequence $$1\longrightarrow (1+\mathcal J)\longrightarrow \mathcal O_X^\times \longrightarrow \mathcal O_{X^{\mathrm{red}}}^\times\longrightarrow 1$$ of multiplicative sheaves on $$X$$ (or rather its underlying topological space, which is the same as of $$X^{\mathrm{red}}$$). Using $$\mathcal J^2=0$$, it's straightforward to check that $$1+\mathcal J$$ is isomorphic to $$\mathcal J$$ (as an additive sheaf of abelian groups on $$X$$). Since $$X$$ is affine, $$H^1(X,\mathcal J)=0=H^2(X,\mathcal J)$$, so the long exact cohomology sequence provides the desired isomorphism $$\operatorname{Pic}(X)\cong H^1(X,\mathcal O_X^\times)\cong H^1(X^{\mathrm{red}}, \mathcal O_{X^{\mathrm{red}}}^\times)\cong \operatorname{Pic}(X^{\mathrm{red}})$$.

Step 3. We consider general affine $$\mathbb C$$-schemes $$X$$. Write $$X=\lim X_\alpha$$ as a cofiltered limit of noetherian affine $$\mathbb C$$-schemes $$X_\alpha$$ with affine transition maps. Using [Stacks project, Tag 01ZR & Tag 0B8W], one obtains $$\operatorname{Pic}(X)\cong\operatorname{colim}\operatorname{Pic}(X_\alpha)$$. The same holds for $$X\times \mathbb A^1\cong \lim(X_\alpha\times\mathbb A^1)$$, so actually $$\operatorname{Pic}_{\mathbb A^1/\mathbb C}(X)\cong\operatorname{colim}\operatorname{Pic}_{\mathbb A^1/\mathbb C}(X_\alpha)$$. Now every $$\eta_{X_\alpha}\colon \operatorname{Pic}_{\mathbb A^1/\mathbb C}(X_\alpha)\rightarrow h^M(X_\alpha)\rightarrow h^M(X)$$ has image a single point by Step 2, hence the same must be true for $$\eta_X$$ by the fact that the colimit in question is filtered. This finishes the proof.

I believe the argument in Step 3 can also be used (with some care) to show $$\operatorname{Pic}(X)\cong \operatorname{Pic}(X^{\mathrm{red}})$$ for arbitrary affine schemes, which would give an alternative to Step 3.

• Great, thanks for writing this up! I definitely agree with Steps 0 and 1. The statement that Pic of a (noetherian) affine scheme is isomorphic to Pic of the reduced scheme seems true to me (this is also mentioned in mathoverflow.net/questions/301520/… . However I agree with Nuno that this seems to not quite reduce everything to Step 1: problem is that the functoriality gives a map $h^M(X) \to h^M(X^{red})$. Knowing that $Pic_{\mathbb{A}^1/\mathbb{C}} \to h^M(X^{red})$ has image a point is not enough to show the same for the map to $h^M(X)$.
– JoS
Commented May 7, 2020 at 11:24

I don't think Step 2 in Florians answer reduces everything to Step 1, so I present an alternative to Step 1 and 2. I show that in the noetherian case for every line bundle $$\mathcal L$$ on $$X \times \mathbb A^1$$ we can find an epimorphism $$X' \to X$$ such that the pullback of $$\mathcal L$$ to $$X' \times \mathbb A^1$$ is a pullback from $$X'$$.

Step A: We reduce the noetherian affine case to the special case that X is the spectrum of a noetherian local ring. Suppose that $$X=Spec(A)$$ and let us consider $$\sqcup_{\mathfrak m \in Max A} Spec(A_{\mathfrak m}) \to Spec(A)$$. This is an epimorphism because it gives an injection on global sections.

Step B: Now assume $$A$$ is a noetherian local ring with maximal ideal $$\mathfrak m$$ then $$\sqcup_{n \in \mathbb N} Spec(A/\mathfrak m^n) \to Spec(A)$$ is an epimorphism by the Krull intersection theorem. As a topological space $$Spec(A/\mathfrak m^n)$$ is a point. Hence $$Pic(Spec(A/\mathfrak m^n)) = 0$$. We need to show $$Pic(Spec(A/\mathfrak m^n) \times \mathbb A^1) = 0$$. For $$n=1$$ this is clear. For higher $$n$$ it follows from Florian's step 2.

• Great, thanks for being so careful and spotting this point, and of course for providing such an elegant resolution! I hope it's ok that I still mark Florian's as a solution (there is no good way to share this between different posts I think).
– JoS
Commented May 7, 2020 at 11:33