Suppose that $k$ is a field of characteristic $p$ such that $k$ is not a finite $k^p$-module. For example, $k = \mathbb{F}_p(x_1, x_2, x_3, ...)$.

Is it true that $k[[x]]$ is a free $(k[[x]])^p$-module? We know that it is flat by a theorem of Kunz. However, it is certainly not finite, so flat is not the same as free.

The naive thing to try (in terms of a basis) is to do the following. Choose {$\lambda_i$} a basis for $k$ over $k^p$. Consider the set {$\lambda_i x^j$} for $0 \leq j \leq p-1$. If $k$ is a finite $k^p$-vector space, this set is easily seen to be a basis for $k[[x]]$ over $(k[[x]])^p$.

However, because of our (ugly) field, we can consider the power series:

$$\lambda_0 + \lambda_1 x^1 + \lambda_2 x^2 + ... + \lambda_n x^n + ...$$

where say $\lambda_0, \lambda_1, ...$ runs over some countably infinite subset of the {$\lambda_i$}. It is easy to see that this cannot be written as a finite $(k[[x]])^p$-linear combination of subset of the $\lambda_i x^j$ (where again $0 \leq j \leq p-1$).

But of course, maybe there's some clever way to choose a basis that actually does work?