Separable and algebraic closures? I have no intuition for field theory, so here goes. I know what the algebraic and separable closures of a field are, but I have no feeling of how different (or same!) they could be.
So, what are the differences between them (if any) for a perfect field? A finite field? A number field?
Are there geometric parallels? (say be passing to schemes, or any other analogy)
 A: Geometrically there is a very big difference between separable and algebraic
closures (in the only case where there is a difference at all, i.e., in positive
characteristic $p$). Technically, this comes from the fact that an algebraically
closed field $k$ has no non-trivial derivations $D$; for every $f\in k$ there is
a $g\in k$ such that $g^p=f$ and then $D(f)=D(g^p)=pg^{p-1}D(g)=0$. This means
that an algebraically closed field contains no differential-geometric
information. On the other hand, if $K\subseteq L$ is a separable extension, then
every derivation of $K$ extends uniquely to a derivation of $L$ so when taking a
separable closure of a field a lot of differential-geometric information remains.
Hence I tend to think of a point of a variety for a separably closed field as a
very thick point (particularly if it is a separable closure of a generic point)
while a point over an algebraically closed field is just an ordinary (very thin)
point.
Of course you lose infinitesimal information by just passing to the
perfection of a field (which is most conveniently defined as the direct
limit over the system of $p$'th power maps). Sometimes that is however exactly
what you want. That idea first appeared (I think) in Serre's theory of
pro-algebraic groups where he went one step further and took the perfection of
group schemes (for any scheme in positive characteristic the perfection is the limit,
inverse this time, of the system of Frobenius morphisms) or equivalently
restricted their representable functors to perfect schemes. This essentially
killed off all infinitesimal group schemes and made the theory much closer to
the characteristic zero theory (though interesting differences remained mainly
in the fact that there are more smooth unipotent group schemes such as the Witt
vector schemes). Another, interesting example is for  Milne's flat cohomology
duality theory which needs to invert Frobenius by passing to perfect schemes in
order to have higher $\mathrm{Ext}$-groups vanish (see SLN 868).
A: First of all, there is not the algebraic/separable closure. Choices have to be made. However, if an algebraic closure $k^{\mathrm{alg}}$ of $k$ is fixed, inside it there is a unique separable closure $k^{\mathrm{sep}}$ of $k$, namely the subfield consisting of the separable elements over $k$.
Ignoring the failure of uniqueness, you can regard $k^{\mathrm{alg}}$ as the biggest algebraic extension of $k$, whereas $k^{\mathrm{sep}}$ is the biggest galois extension of $k$. The latter is because $k^{\mathrm{sep}}$ is easily seen to be normal. In particular, you can apply Galois theory and relate the group theory of the absolute Galois group $\mathrm{Gal}(k^{\mathrm{sep}}/k)$ with the field theory of Galois extensions of $k$. The algebraic closure is too big to make Galois theory work. 
Obviously $k$ is perfect if and only if $k^{\mathrm{alg}} = k^{\mathrm{sep}}$. Finite fields and fields of characteristic $0$ (in particular number fields) are perfect. But what is the difference in the other cases? Let $p = \mathrm{char}(k) > 0$. Then $k^{\mathrm{alg}} / k^{\mathrm{sep}}$ is purely inseparable, i.e. for every $a \in k^{\mathrm{alg}}$ there is some $n \geq 1$ such that $a^{p^n} \in k^{\mathrm{sep}}$. In other words, this field extension is given by adjoining all $p^n$-th roots. A consequence of this is that the restriction map $\mathrm{Aut}_k(\overline{k}) \to \mathrm{Gal}(k^{\mathrm{sep}}/k)$ is an isomorphism.
Actually one can show that the canonical map $k^{\mathrm{sep}} \otimes_k k^{\mathrm{perf}} \to k^{\mathrm{alg}}$ is an isomorphism, where $k^{\mathrm{perf}}=\cup_{n \geq 0} k^{1/p^n}$ is the perfect hull of $k$.
