Are the algebraic closure (of a field) and the real closure (of a totally ordered field) closure operators (when restricted to appropriate sets of fields, so that they are maps on a set instead of a class)?

From what I see, yes, but I just need a confirmation from someone knowledgeable of the subject. I'm in doubt because I haven't found one citation of the real closure being a closure operator.

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    $\begingroup$ There is no such thing as the algebraic closure (or the real closure) of the field $F$. But certainly if we are working within a single master field, algebraic closure and real closure are closure operators. $\endgroup$ – André Nicolas Apr 6 '12 at 20:36

The algebraic closure and real closure are closures in the sense that $\overline{\overline{K}} = \overline{K}$ and $(K^r)^r = K^r$ (where '=' might not literally be equality, but is definitely a canonical, natural isomorphism), and there are canonical, natural inclusions, e.g. $K \mapsto \overline{K}$. And given an inclusion $K \subseteq L$, you can choose algebraic closures with $\overline{K} \subseteq \overline{L}$. You'd have to jump through some hoops to turn it into a closure of the form you describe.

On the partially ordered set of subfields of a field $K$, you can define a closure operator $F \mapsto \overline{F} \cap K$. Similarly for a formally real field and its set of formally real subfields.

EDIT: the unbounded version is best stated in terms of category theory. Let $\bf Fld$ be the category of fields and homomorphisms. There is a functor $C : {\bf Fld} \to {\bf Fld}$ and a natural transformation $i : {\bf 1} \to C$ such that

  • $C(F)$ is the algebraic closure of $F$
  • $i_{C(F)} : C(F) \to C(C(F))$ is an isomorphism

A similar version should work for real closure too, on the category of formally real fields, or maybe ordered fields.

EDIT2: as the comments show, I got the above part wrong. You can define $C$ as a graph homomorphism, but it won't be a functor. You could switch to a suitable subcategory of $\bf Fld$ to make it a functor, but I'm not sure how informative that would be.

  • $\begingroup$ If I have an algebraically closed field, resp. real closed field $K$, and if $\mathcal{F}:=\{\text{subfields of }K\}$, then the algebraic closure, resp. real closure, viewed as maps $\mathcal{F}\to\mathcal{F}$, are closure operators with respect to $\subseteq$? Is that a better formulation? $\endgroup$ – Leon Apr 6 '12 at 20:45
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    $\begingroup$ @Leon: Yes, more or less. If you're being picky about details (and from your question it seems you might be), the map $\mathcal{F} \to \mathcal{F}$ applied to $F$ is better said as "the <foo> closure of $F$ in $K$". But since $K$ is <foo>-closed, the "<foo> closure of $F$ in $K$" does turn out to be a "<foo> closure of $F$". Of course, in an appropriate context, you can leave off the "... in $K$" part because it can be inferred. $\endgroup$ – Hurkyl Apr 6 '12 at 22:21
  • $\begingroup$ Re: Edit: How do you make algebraic closure functorial? That is, how do you define $C$ on field homomorphisms? $\endgroup$ – t.b. Apr 7 '12 at 9:27
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    $\begingroup$ The statement in the EDIT is wrong. It is possible to find a functor Fld -> Fld which maps a field to its algebraic closure (transfinite recursion on trans. degree ...), but there is no such functor equipped with a natural transformation from the identity functor, i.e. compatible inclusions from the field to its chosen algebraic closure. I don't think that category theory is the correct language for algebraic closures, they are not unique in the usual unique-sense. You should better talk about amalgamated structures or alike. $\endgroup$ – Martin Brandenburg Apr 7 '12 at 9:37
  • $\begingroup$ Thank you Hurkyl, as well as @MartinBrandenburg and t.b. $\endgroup$ – Leon Apr 8 '12 at 17:32

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