if the determinant of the matrix is nowhere zero, then for fixed $a$ and varying $b \in \mathbb{C}$ the integral $F(b) = \int\limits_a^b A^{-1}(z)dz$ along some fixed choice of path from $a$ to $b$ is well-defined, and should be path-independent. Can we show this is a logarithm?

@LaurentBerger I have not, but I will. Thank you. I have also gotten a lot of insight from your own writings on period rings

Perfectoid rings are typically not Noetherian, although this is not strictly about “infinitely many indeterminates”

No, by the Auslander-Buchsbaum theorem the map $\mathbb{C}[x]_{(x)} \to \mathbb{C}[x, y]_{(x,y)}$ is a counterexample

The same counterexample I gave for your previous question ($\mathbb{Z}_p \to \mathbb{F}_p$) seems to show the answer is no

I downvoted since it seems the argument is incomplete.

(I have not checked details)

See here: stacks.math.columbia.edu/tag/01JO . It seems like some statement for tensor products of two rings should be possible to the effect: take the images of Spec$R_1$ and Spec $R_2$ in Spec $\mathbb{Z}$, intersect them, and take the preimages $U_1, U_2$ of this intersection in Spec of each ring. The dimension of the tensor product should be something like the sum of the dimensions of $U_1$ and $U_2$.

For curves it’s fairly simple. You can resolve all singulairites by taking the normalization and reduce to the smooth case. Then because any birational map of smooth projective curves extends to an isomorphism, you’re classifying projective curves up to isomorphism. All spelled out here: math.stackexchange.com/questions/190127/… . Saying much more requires more sophisticated tools. Riemann-Hurwitz is kind of incidental to this