TITLE: Coarse moduli space of relative Picard functor for affine line
QUESTION [9 upvotes]: 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.

REPLY [3 votes]: 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.