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The theory of algebraically closed field, say in characteristic zero, and of differentially closed fields (of characteristic zero) have much in common: quantifier elimination and (hence) decidability; or more precise things like ordinary Galois theory and differential Galois theory.

However, while examples of algebraically closed fields are pretty easy to give (starting with $\mathbb{C}$ and the field of Puiseux series $\varinjlim \mathbb{C}(\!(t^{1/n})\!)$ over it), differentially closed fields... are not. This paper (Spodzieja, “A geometric model of an arbitrary differentially closed field of characteristic zero”) gives an “explicit” construction of one (for some value of “explicit”), but it is highly technical and involves making choices (in a sense in which $\mathbb{C}$ and $\varinjlim \mathbb{C}(\!(t^{1/n})\!)$ do not), and leaves me nonplussed as to why the whole matter needs to be so complicated.

So, with apologies for asking a perhaps vague question: is there some reason why this must be so? Is there some argument why a differentially closed field “must” be complicated to construct?

A little more specifically, while the field of transseries (or any of the many variations thereupon) is “not too far” from being differentially closed, it isn't: is there some reason why any approach to construct a differentially closed variant of transseries must fail? (The answer may be contained in this book, but I failed to locate it.) I could ask the same thing with germs of meromorphic functions at a point.

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    $\begingroup$ The constuction in the paper is equipping the algebraic closure of some infinite rational function space $\Bbb{Q}(\Lambda_t)$ with a "directional derivative". I believe the index set $T$ can be chosen better to avoid using orderings in the construction, but it is scarcely possible to prove the closedness under differential equations without using choice. $\endgroup$
    – Zerox
    Commented Mar 3, 2022 at 14:01
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    $\begingroup$ The complexity can be seen from that, in order to construct a nontrivial differential operator, we must introduce some transcendental element $t$ out of nowhere to the usual number field. $\endgroup$
    – Zerox
    Commented Mar 3, 2022 at 14:06
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    $\begingroup$ Similar complexity caused by '$t$' occurs when completing the value-at-infinity order of real rational functions. $\endgroup$
    – Zerox
    Commented Mar 3, 2022 at 14:07
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    $\begingroup$ @Zerox That's not correct. Existence of algebraic closures is known to not be equivalent to AC. It is also known that existence of algebraic closures in general is not provable in ZF alone, though it is provable for a lot of fields one encounters in practice, like subfields of $\mathbb C$ or fields of rational functions over them. $\endgroup$
    – Wojowu
    Commented Mar 3, 2022 at 15:29
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    $\begingroup$ At least the existence of the algebraic closure of $\mathbb{C}(t)$ follows from ZF, since it is the relative algebraic closure of $\mathbb{C}(t)$ in the field of Puiseux series (which is algebraically closed by a theorem of Newton), so no axiom of choice is needed here. This is why it would be interesting to know if the existence of its differential closure follows from ZF. $\endgroup$
    – Gro-Tsen
    Commented Mar 3, 2022 at 18:21

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