Hi Will, indeed I knew about the casus irreducibilis which might one of the key reason (swing factor) for the acceptation by certain mathematicians of complex numbers for solving cubic polynomials. Since as you explained, when an irreducible real cubic polynomial has 3 real roots (so Galois group $\simeq C_3$) then none of the roots can be written in terms of radicals of positive quantities. This means that the cancellation of the imaginary part (which can only takes place in the complex world) is an unavoidable phenomenon when you shoehorn yourself to only allow the use of radicals.

So @Ari, do you know what kind of obstruction when $g\geq 4$ prevents an abelian variety to be the Jacobian of a curve?

Ari, is $\mathcal{A}_g$ the moduli space of principally polarized abelian varieties? Even if this Torelli locus is dense (Zariski or analytically) how does it say anything about the Jacobian origin of a given principally polarized CM abelian variety?

Thanks Francois for the simple but instructive examples. So are you suggesting that in general the motivic galois group should act on the ring generated by the periods obtained from the comparison isomorphism?

Thanks Keerthi, so the key point is that the cycle class map commutes with $\omega$.

Karl, so I've tried to show that the Jacobian's criterion implies flatness and no ramification but it does not seem as easy as what I first thought...

@Karl, I'm confused about one point, if $A=F_p[x]/x^p$ and $m=(\bar{x})$ then $Fr_p(m)=m'=0$ and therefore is not a maximal ideal...

I guess that the equivalence of these 2 definitions is probably not too difficult to prove, what do you think?

Dear @Karl, thanks a lot for your proof, it seems to work. Note though that in your proof you are using another definition of etale namely that the map is flat an unramified. You use the flatness (at least) in order to reduce to the case where $B$ is a free $A$ module (a flat finitely presented modules over a local ring is free) and you use the fact that it is unramified to guarantee that $B/mB$ is a separable field extension of $A/mA$. But in my set up, my definition of etale was different since I used the invertibility of the jacobian.

Yes of course, @Karl