I'm trying to get a feel for Grothendieck topologies by considering some really simple examples (and this relates to a question in Kleiman's chapter on the Picard Scheme in FGA explained).
I know that the etale covers of $\text{Spec}(k)$ are precisely the spectra of finite separable field extensions of $k$ (or perhaps finite disjoint unions of such spectra). Thus if $k$ is algebraically closed, there are no non-trivial etale covers.
So, moving to the finer fppf topology, I'd like to know what the fppf covers of $\text{Spec}(k)$ for $k$ an algebraically closed field look like. This is how far I've got:
If $T\xrightarrow{f} \text{Spec}(k)$ is fppf, then for any open affine $U = \text{Spec}(A)\subset T$, $A$ must be a finitely presented $k$-algebra. It will automatically be flat over $k$, since $k$ is a field.
I can't see any other restriction on $T$, so it apears as if the fppf covers of $\text{Spec}(k)$ should be all $k$-schemes of finite type. But this seems like way too many covers, especially considering how there weren't any etale covers. So have I done this right or am I missing something?
Thanks