ag.algebraic geometry - Is it known that the ring of periods is not a field?

Maybe I can sketch an argument for your first question.

Let $P$ be the ring of effective "formal" periods, generated by quadruples $[X,D,omega,gamma]$ consisting of a smooth projective $Q$-variety $X$, a normal crossing divisor $D$, top-form $omega$, and singular cycle $gamma$, as discussed in Kontsevich-Zagier.

Let $omega: P rightarrow C$ be the ring homomorphism obtained by integration, whose image is the ring of periods that you mention. You ask whether the image $omega(P)$ could be a field.

If it were, then the induced map $Spec(C) rightarrow Spec(P)$ would be a $C$-point of the scheme $Spec(P)$, whose image is a closed point of the scheme. As $P$ is the inductive limit of finitely-generated subrings $P_M$, we find that the induced map $Spec(C) rightarrow Spec(P_M)$ has image a closed point of $Spec(P_M)$ for any motive $M$ which generates a sufficiently large ring of periods (or am I making a mistake about projective limits of affine varieties?).

This, incidentally, is opposite to the expectations of Grothendieck's period conjecture, which states that the image of $Spec(C) rightarrow Spec(P_M)$ should be a generic point!

If the map $Spec(C) rightarrow Spec(P_M)$ has image a closed point, then (since $Spec(P_M)$ is defined over $Q$), its image is a point defined over $bar Q$. It follows that the dimension of the $Q$-Zariski closure of this point is zero.

But this dimension (zero) is equal (as remarked by Yves Andre in his paper "Galois Theory, Motives, and Transcendental Numbers") to the transcendence degree $TrDeg_Q[Per(M)]$, where $Per(M)$ is the set of periods of the motive $M$.

So, if one chooses a motive whose periods generate a transcendental extension of $Q$ (e.g., whose periods include $2 pi i$), one should find a contradiction.

As for the second question... I'll try to say something when I'm not about to meet with students.