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?



$k[[x]]$ as a $(k[[x]])^p$ module for ugly fields
The answer to the question is no. 


Let $R = k[[x]]$, $S = k[[y]]$, and $f:R \to S$ is the absolute Frobenius. We must show that $S$ is not a free $R$-module. By assumption on $k$, we know that $S$ must have infinite rank if it were free. On the other hand, $S$ is $x$-adically complete (since $x^p = y \in S$). Hence, the claim follows from:


**Lemma**: The $R$-module $M := R^{\oplus I}$ is $x$-adically complete if and only if $I$ is finite.


*Proof*: The "if" direction is clear. For the reverse, note that direct summands of complete $R$-modules are complete. By replacing $I$ with a subset, we may assume $I = \mathbf{N}$. Now consider the sequence $m\_n := \sum\_{i=1}^n x^i \cdot e\_i \in M$, where $e\_i$ is the $i$-th basis vector. This sequence is Cauchy (clear); I claim it has no limit in $M$. If there was a limit $m = \sum\_{i=1}^N a\_i \cdot e\_i \in M$ with $a\_i \in R$, then the neighbhourhood $m \in m + x^k M$ would contain infinitely many $m\_n$ for any fixed $k$, but staring at the coefficient of $e\_{N+1}$ shows this is false as long as $k \geq N+2$.


