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It is well known that over an algebraically closed field of characteristic zero a general curve (for an open subset of $M_g$) of genus $g\geq 3$ is automorphism-free.

Is this result still true over a non algebraically closed field and over a field of positive characteristic?

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    $\begingroup$ Yes, that is still true. The group scheme of automorphisms of curves of genus $g \geq 2$ is a priori bounded, since the representation on the vector space of global sections of the third power of the dualizing sheaf is faithful. Thus, you may base change to an uncountable field. Now consider a general deformation of a hyperelliptic curve branched over a general configuration of $2g+2$ points in $\mathbb{P}^1$. The only nontrivial automorphism of the hyperelliptic curve is the involution. For $g\geq 3$, under a general first-order deformation, the involution does not extend. $\endgroup$
    – Jason Starr
    Commented Oct 20, 2015 at 13:52

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Jason Starr already answered your question in the comments above.

Let me just point that you can also use a standard deformation argument to prove that the general curve of genus at least three has no non-trivial automorphisms, see e.g. Dan Petersen's answer to Examples where it's useful to know that a mathematical object belongs to some family of objects

This argument can also be used to prove that the general complete intersection of general type in some projective space has no non-trivial automorphisms.

Explicit examples of some varieties with no non-trivial automorphisms are given in Poonen's articles 

Varieties without extra automorphisms I: curves  
Varieties without extra automorphisms II: hyperelliptic curves  
Varieties without extra automorphisms III: hypersurfaces

They are available on his website http://www-math.mit.edu/~poonen/

Addendum: As Jason Starr points out in the comment below, using singular specializations to deduce facts about smooth fibres requires a bit of care. Let me explain where one should be careful in applying the argument given by Dan Petersen.

Fix an integer $g\geq 3$ and an algebraically closed field $k$. Let $X$ be a stable curve of genus $g$ over $k$ with no non-trivial automorphisms. (You construct this by pinching $g-2$ well-chosen points on some genus $2$ curve, as explained by Dan Petersen.) Now, let $R = k[[t]]$ and let $\mathcal X\to \mathrm{Spec} R$ be a stable curve such that

  1. The special fibre $\mathcal X_k$ is isomorphic to $X$ over $k$, and
  2. the generic fibre $\mathcal X_K$ is a smooth genus $g$ curve.

The existence of such a stable curve requires a bit of thought. A quick argument is to say that stable curves form a boundary of $\mathcal M_g$ (as mentioned by Dan Petersen in his answer). You can also try to give a more explicit construction (of a smooth curve $\mathcal X_K$ which specializes to $X$.) On the other hand, the easiest approach might be to just prove that any singular curve can be smoothened. (I can't remember whether that's a difficult fact to prove at the moment.)

Now, let $G :=\mathrm{Aut}_R(\mathcal X)$ be the scheme of automorphisms of $\mathcal X$. Since $\omega_{\mathcal X/R}$ is relative ample, the group scheme $G$ is affine and finite type over $R$. Since stable curves are ``unique'' (by the theory of minimal regular models of curves, say) the group scheme $G$ is proper. (Note: this is the subtle point that Jason Starr is alluding to below. Uniqueness of limits of all stable limits (in the preceding sense) is enough, as it implies by the valuative criterion for properness, applied to the diagonal, that the (affine) diagonal of $\overline{\mathcal M_g}$ is finite.) So $G\to \mathrm{Spec} R$ is a finite group scheme, whose special fibre is the trivial group scheme. You can now conclude that $G$ is itself the trivial group scheme over $R$.

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    $\begingroup$ There is something subtle in Dan Petersen's argument, and I also should have explained this point in my comment. For curves that embed in Abelian varieties, for a smooth specialization, the group scheme of automorphisms is proper over the base by Weil extension. For singular specializations, this can easily fail, e.g., start with a hyperelliptic curve and then let the smooth specialization be obtained by attaching a $\mathbb{P}^1$ at a non-Weierstrass point. So there is some work necessary if one approaches this by singular specializations ... $\endgroup$
    – Jason Starr
    Commented Oct 21, 2015 at 9:56
  • $\begingroup$ @JasonStarr Thank you for your comment. I tried explaining the subtle point you are alluding to. Did I understand it correctly? $\endgroup$
    – Ariyan Javanpeykar
    Commented Oct 21, 2015 at 10:16
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    $\begingroup$ I am not quite sure that the uniqueness of stable limits is sufficient. For a family of genus $1$ curves with specified degree $3$ ample invertible sheaf, the group of automorphisms is the $3$-torsion in the (corresponding) elliptic curve, and this has order $9$. As the elliptic curve specializes to a nodal plane cubic, the $3$-torsion quasi-finite, flat group scheme is non-proper and specializes to a group of order $3$ (assuming the characteristic is not $3$). I do not believe that the lost automorphisms have to do with non-uniqueness of specializations. $\endgroup$
    – Jason Starr
    Commented Oct 21, 2015 at 10:29
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    $\begingroup$ @JasonStarr Nice comment. First, uniqueness of stable limits is certainly sufficient, as it implies that the diagonal (hence inertia stack) of $\overline{\mathcal M_g}$ is proper and thus finite (by either using its affine or quasi-finite). I guess your example shows that the stack of stable cubic curves (which is not the stack of stable genus one curves with a section) is not separated. Abstractly: the stack $\overline C$ of stable cubic curves is a finite type smooth DM stack over $\mathbb Z[1/3]$. The locus of smooth cubic curves is separated. $\endgroup$
    – Ariyan Javanpeykar
    Commented Oct 21, 2015 at 14:36
  • $\begingroup$ To be clear, for me: the stack of smooth (resp. stable) cubic curves is the quotient of the "Hilbert scheme" of smooth (resp. stable) cubic curves by $PGL_3$. Of course, the term Hilbert scheme is a bit too fancy in this context. $\endgroup$
    – Ariyan Javanpeykar
    Commented Oct 21, 2015 at 14:37

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