Why do some curves not admit etale coverings?

Let $C$ be a non-singular projective curve of genus $g\geq2$ over $\mathbb{C}$.

If $g\geq 3$ and $C$ is general, why does $C$ not admit etale coverings $\pi:C\rightarrow C'$ of degree $>1$? Is this due to the fact, that a general curve of genus $g\geq 3$ does not have automorphisms?

If $g=2$, why does no curve $C$ of this genus admit etale coverings $C\rightarrow C'$? (I think I figured this one out: this follows from the Riemann-Hurwitz formula.)

• I don't understand your claim. Just pick a finite unramified field extension of the function field and then the normalization. – user40276 Aug 4 '16 at 10:23
• I forgot that the curve should be projective. See for example See for example arxiv.org/abs/math/9902145 . It is stated in Corollary 2.6. – Bernie Aug 4 '16 at 11:23
• @user40276: the OP wants the given curve $C$ to be the source of the etale map. Your construction would give a covering with $C$ as the target. – Nefertiti Aug 4 '16 at 12:02
• If $g=2$, then you're right that $C\to C'$ finite etale forces $C\to C'$ to be an isomorphism. This is simply because $2= 2g-2 = d(2g'-2)$. This clearly can only hold if $d=1$ and $g'=2$ (otherwise $d(2g'-2) >2$). Now, you just need to use that finite \'etale of degree one is an isomorphism. – Ariyan Javanpeykar Aug 4 '16 at 15:55
• @Nefertiti Oh! I see, now. I haven't noticed the ${}^'$ in the $C'$. Thanks for pointing it. – user40276 Aug 5 '16 at 0:36

To answer the question about $C \to C'$, there are probably various ways to this, probably including easier ones than those I am going to suggest.
If you fix a genus $g',$ and consider etale covers of some fixed degree $d$ of $C'$, you get curves $C$ of some prescribed genus $g$ (given by the Riemann--Hurwitz formula). Now $C'$ depends on $3g' - 3$ parameters (this is the dimension of the moduli space of genus $g'$ curves), and it has only finitely many etale covers of fixed degree. So the curves $C$ which admits etale morphisms $C \to C'$ depend only on $3g'-3$ parameters. But the moduli space of genus $g$ curves is of dimension $3g -3$, so most curves do not arise as such etale covers.
Another way to think about the problem (but I don't know how to actually prove it this way) is that if we have an etale cover $C \to C'$ then we obtain a morphism of Jacobians $\mathrm{Jac}(C') \to \mathrm{Jac}(C)$, and so $\mathrm{Jac}(C)$ is not a simple abelian variety. But, heuristically, there is no reason to expect that for a general curve $C$, its Jacobian wouldn't be simple. (One problem with this approach is that most abelian varieties don't arise as Jacobians at all, and so, while I find it suggestive, it isn't more than that, as it stands.)