This is a question related to


[Semistable curves of genus $g\geq 2$ form an Artin algebraic stack in the etale topology?](https://mathoverflow.net/questions/429754/semistable-curves-of-genus-g-geq-2-form-an-artin-algebraic-stack-in-the-etale)


Let $\mathcal C\rightarrow \mathcal M^{ss}\_g$ be the universal curve over the Artin stack of semistable curves of a fixed genus $g\geq 2$. Is the morphism $\mathcal C\rightarrow \mathcal M^{ss}\_g$ representable by a scheme ?


It is well-known that the fibres of the morphisms are schemes because algebraic space curves are actually schemes. But it does not mean that the total space $\mathcal C\times\_{\mathcal M^{ss}\_g} T$ is a scheme for any scheme $T$ over $\mathcal M^{ss}\_g$. The total space can be easily seen to be an algebraic space even for the stack of presetable curves. But is it possible that for the sub functor of semistable curves, the morphism $\mathcal C\rightarrow \mathcal M^{ss}\_g$ is actually represented by a scheme?



On the stack of semistable curves
I am posting an answer to correct my wrong comments above. The answer is based on the comment by user @johan.


For every $g\geq 0$,
denote by $\mathfrak{M}\_g$
the stack of genus-$g$ curves that are proper, geometrically connected, reduced, and at-worst-nodal. Denote the universal curve over this stack by the $1$-morphism, $$\pi\_g:\mathfrak{M}\_{g,1}\to \mathfrak{M}\_g.$$


**Theorem** [Kai Behrend, perhaps earlier sources]. These stacks are algebraic stacks (i.e., Artin stacks).


If we mark enough points on the irreducible components, the curves are projective (appropriate linear combinations of the marked points give ample divisors). Thus, the $1$-morphism $\pi\_g$ is projective after an étale base change of the target. Hence the $1$-morphism $\pi\_g$ is representable by algebraic spaces.


There is an explicit analysis by Johan de Jong and myself of the open substack of $\mathfrak{M}\_0$ parameterizing curves with at most two irreducible components, and we use this to prove many divisor class relations on the stack $\mathfrak{M}\_0$. If you restrict $\pi\_g$ over this open substack, it is projective (the dual of the relative dualizing sheaf is relatively ample). However, the next open substack is more complicated.


**Theorem** [Damiano Fulghesu]. For every $g\geq 0$, the $1$-morphism $\pi\_g$ is not representable by schemes. In fact, this already fails over the open substack of $\mathfrak{M}\_0$ parameterizing curves with at most three irreducible components.


The basic idea of the proof is Hironaka's example. Alternatively, consider the versal deformation space of a genus-$0$ curve with precisely three irreducible components. There is an extra $\mathbb{Z}/2\mathbb{Z}$-symmetry to this deformation space coming from the symmetry of the curve that interchanges the two "leaves" of the curve and maps the "middle" component back to itself. You can use that $\mathbb{Z}/2\mathbb{Z}$ symmetry to form a quotient of the family which then gives a versal family of curves whose total space is an algebraic space that is not a scheme.


Contrary to what I wrote above, we can mark points on this family. Once we mark points on the family, we can glue higher genus curves to those marked points to get similar families where the arithmetic genus is as large as we like.


It is interesting to think about what goes wrong with the suggestion in my comment. The exceptional locus for the retraction (with its reduced structure) is a Cartier divisor in $\mathfrak{M}\_g$. The image in the pullback over $\mathfrak{M}\_g$ of the universal curve does give a closed subscheme. The blowing up of this ideal sheaf is representable by projective morphisms. Fulghesu's example shows that this blowing up is not the same as $\mathfrak{M}\_{g,1}$.


