@tj_ Ah, I see. Yes, I've been wondering this as well. It doesn't seem unreasonable that if $R$ is a local noetherian ring containing $A_u$ then the map $A_u[\{x_j\}_j]\to R$ will be flat iff the $x_j$ form a regular sequence in $R,$ but I haven't been able to write down a proof of that statement.

@tj_ that's true, but $k[x,x^{-1}]$ is not the same as $k(x)$ (and similarly for more variables): the latter contains things like $1/(x-1)$, but the former does not.

This might work! But, I have to think about whether the proof of the exercise will apply to $A_u\to R,$ as $A_u$ is not necessarily going to be of the form $k'[x_1,\dots, x_r]$ for some $k'$ contained in $R.$ Is this obvious?

I mean on the left hand side that that $x$ and $u$ are formal variables mapping to the specified elements of $R,$ although I would also be happy interpreting the left hand side as the $k$-subalgebra of $R$ generated by those elements. In the cases I'm interested in, the subalgebra generated by the elements will be a polynomial algebra on those elements.

Perfect, those are exactly more concise versions of the arguments I had in mind! Thank you!