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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)

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Perfect fields are fields all of whose algebraic extensions are separable, so their algebraic and separable closure coincide. Finite fields are perfect; also fields of characteristic zero (including number fields) are perfect. The simplest example of a non-perfect field is $\mathbb{F}_p(X)$, the field of rational functions in one variable over the field of $p$ elements. –  Robin Chapman Jun 7 '10 at 6:34
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May I ask why the above comment was not posted as an answer instead? It is manifestly an answer. –  Ryan Reich Jun 7 '10 at 16:14

2 Answers 2

up vote 31 down vote accepted

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).

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Even though both this answer and Martin B's are great, this has the geometric insight I was after. –  David Roberts Jun 9 '10 at 5:39

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 $k$.

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Can I cheekily say I was using the <a href="ncatlab.org/nlab/show/generalized+the">generalised 'the'</a>? The algebraic closure of k is unique up to k-isomorphism. More seriously, we might say 'Galois theory can't work' for the algebraic closure, but there is still a group Gal(k^alg/k) of k-automorphisms. How does it relate to Gal(k^sep/k)? –  David Roberts Jun 7 '10 at 6:47
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@David: Algebraic closure is not unique in the sense that $k \to k^{alg}$ cannot be extended to a functor such that $k \subseteq k^{alg}$ is a natural transformation. This shows that everytime you have to choose an algebraic closure. Two algebraic closures are isomorphic, but the isomorphism is not unique and not constructive. Thus the lnab-the does not fit here. Concerning your other question: I think that the canonical homomorphism $Gal(k^{alg} / k) \to Aut(k^{sep} / k)$ is an isomorphism. @Torsten: You may post this as an answer, I'll upvote it. –  Martin Brandenburg Jun 7 '10 at 7:43
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Doc, you can see this already in the case of $\mathbb R$. You choose your complex numbers and I choose mine. We won't be able to decide algebraically whether my $i$ is your $i$ or your $-i$. –  Bugs Bunny Jun 7 '10 at 8:29
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Bugs, according to Bourbaki, the boldface font is reserved for objects which are unique up to unique isomorphism (with the exception of $\mathbf{F}_ q$), and in particular Bourbaki regards $\mathbf{C}$ as equipped with a choice of $i$. That is, the notation $\mathbf{C}$ is meant to encode more structure than just an algebraic closure of $\mathbf{R}$. I think this viewpoint on $\mathbf{C}$ is misguided, but just keep it in mind if you find yourself discussing $i$ with Serre. (He says that there is a canonical $\sqrt{-1}$ in $\mathbf{C}$, due to this convention.) –  BCnrd Jun 7 '10 at 10:54
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BCnrd, it is only normal to suggest that complex numbers come with complex structure but sometimes they don't:-)) –  Bugs Bunny Jun 7 '10 at 15:28

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