TITLE: Section of universal curve
QUESTION [7 upvotes]: Let $\mathcal{M}_{g,1}\to \mathcal{M}_g$ be the universal genus $g$ curve, let $K_g$ denote the funciton field of $\mathcal{M}_g$. Take the generic fiber $C\to \mathrm{Spec}{(K_g)}$, after some finite extension $L/K_g$, $C_L$ will admit a rational point. Is there a good choice for $L$? If there is, do we know $\mathrm{Gal}(L/K_g)$?

REPLY [7 votes]: Let me assume $g>2$, so that there is a curve $C$ over $K_g$ of genus $g$ (the generic curve) -- the genus condition assures that the generic curve has no automorphisms.  It is known but not obvious that $C$ has no $K_g$-rational points (this is due to Hain-Matsumoto in characteristic zero and Watanabe in positive characterstic, if I remember correctly).  If you choose any closed point of $C$ with residue field $L$, the base change to $L$ will have a rational point.
There is a reasonably canonical choice of $L$.  Namely, let $W_g$ be the moduli space of curves with a Weierstrass point.  Then $W_g$ is connected and $W_g\to M_g$ is not Galois, but it is generically etale with monodromy group is $S_{g(g^2-1)}$ by this paper of Eisenbud-Harris.  Thus letting $L_g$ be the function field of $W_g$, the generic curve obtains a rational point over $L_g$ (namely, the generic Weierstrass point).  $L_g/K_g$ is not Galois, but its Galois closure has Galois group $S_{g(g^2-1)}$.
Of course this works in genus $<2$ as well, as long as you consider the generic point of $\mathcal{M}_g$ to be a stacky point rather than Spec of a field.