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.

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    $\begingroup$ 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. $\endgroup$ Commented Mar 17, 2020 at 21:50
  • $\begingroup$ 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 $T$). $\endgroup$
    – JoS
    Commented Mar 18, 2020 at 8:46

2 Answers 2


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

enter image description here

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

enter image description here

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.

  • $\begingroup$ 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 $h^M(X)$. $\endgroup$
    – 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.

  • $\begingroup$ 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). $\endgroup$
    – JoS
    Commented May 7, 2020 at 11:33

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