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?



Automorphisms of curves in positive characteristic
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](https://mathoverflow.net/questions/74214/examples-where-its-useful-to-know-that-a-mathematical-object-belongs-to-some-fa)


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 $R$.


