Let $R$ be a regular (commutative associative unitial) algebra over a prime field $F$ (i.e. $F=F_p$ or $F=\mathbb{Q}$); assume that it is noetherian excellent (and even of Krull dimension $1$). What extra restrictions on $R$ will ensure that it is the colimit of a filtered system of finitely generated regular $F$algebras such that the connecting morphims are smooth (i.e. flat and unramified)? What conditions can ensure the existence of an etale cover of the spectrum of $R$ by the spectra of rings satisfying this property (I would prefer a cover by a finite family, yet an infinite one could also be sufficient for my purposes)?

$\begingroup$ The $S$ should be $R$, I think. $\endgroup$ – Mahdi MajidiZolbanin Jan 17 '14 at 1:24
Imposing the condition that the transition maps are smooth seem like a very strong condition to me. Here are some examples.
If $R$ is a field, then it need not be true, for example it isn't true if $R$ is a perfect field of transcendence degree > 0 over $\mathbf{F}_p$ (look at module of differentials). If a transcendence basis of $R/\mathbf{F}_p$ also gives a basis for $\Omega_{R/\mathbf{F}_p}$, then it is true (because the field is a colimit of finite separable extensions of a purely transcendental field). I think it doesn't hold for the field $\mathbf{F}_p((t))$ (look at what happens with power series of the form $\sum t^{p^{e_i}}$ in the colimit).
If $R$ is general, then it seems to me a necessary condition is that all the residue fields have the property (for example the fraction field if $R$ is a domain). Another necessary condition is that $\Omega_{R/\mathbf{F}_p}$ is a projective $R$module. But my guess would be these conditions together aren't sufficient.

$\begingroup$ Thank you for this answer! Yet could you give some more detail? Is there any substantial difference between $t$ and $\sum t^{p^{e_i}}$ in the field $\mathbb{F}_p((t))$? $\endgroup$ – Mikhail Bondarko Jan 18 '14 at 9:52

$\begingroup$ It seems that one should rather consider two distinct elements of this field simultaneously. $\endgroup$ – Mikhail Bondarko Jan 18 '14 at 10:03