As is mentioned in the introduction of this paper of Spodzieja there is a lack of 'natural' examples of differentially closed fields. The immediate naive guesses, namely the field of germs of meromorphic functions at some point, only partially work in that these structures contain differentially closed fields as substructures by Seidenberg's embedding theorem but are not differentially closed themselves.

It seems that the while the sheaf of meromorphic functions is fairly rich, it's always possible to find differential equations that are badly behaving at a given point. As someone who has been in the same room as people talking about set theoretic forcing, this made me wonder whether or not a 'direct' construction of a differentially closed field could be achieved with forcing. Essentially I want to take the field of germs of meromorphic functions at a generic point.

Does there exist a forcing extension $V[G]$ such that for some $z\in\mathbb{C}^{V[G]}\setminus \mathbb{C}^V$ the field of germs at $z$ of meromorphic functions in $V$ is differentially closed?

Of course this is equivalent to considering the ring of germs of holomorphic functions in $V$ at $z$, since no such function can have a zero or pole at $z$. A potentially more general question could be asked in terms of direct limits along directed sets of open subsets of $\mathbb{C}$, but I wanted to focus on this simpler form first.

  • $\begingroup$ How familiar are you with forcing? The point $z$ is in some sense 'new'. Here's an analogous situation: Suppose I'm considering computable meromorphic functions on simple domains (like circles of rational radius with rational centers) and I want to find some field of germs at a point that is algebraically closed. This can never work if my point is computable because for any given computable complex number $a$, the function $f(z) = z-a$ cannot have a square root in any neighborhood of $a$. $\endgroup$ Apr 30 '19 at 14:57
  • $\begingroup$ On the other hand if I choose a non-computable complex number $a$ and look at the field of germs of (computable) meromorphic functions at that point, then this field should be algebraically closed, since the zeroes and poles of a computable meromorphic function are always computable, so given any meromorphic function $f$ we can find a neighborhood of $a$ small enough that $f$ has no zeroes or poles in that neighborhood. Then there is a meromorphic square root of this function in that neighborhood. $\endgroup$ Apr 30 '19 at 15:00
  • $\begingroup$ I am very familiar with forcing. I was misreading "functions in $V$" as "functions in $V[G]$". In the sentence before that you write "the field of germs of meromorphic functions at a generic point", which implied, for me, that you mean meromorphic functions in $V[G]$. $\endgroup$
    – Jonathan
    Apr 30 '19 at 20:20
  • $\begingroup$ Ah, sorry about that. $\endgroup$ May 1 '19 at 3:14

I had a conversation about this with James Freitag at a conference recently and he pointed me to a fundamental result of Seidenberg that seems to resolves this problem in the positive.

The relevant result is from Seidenberg's paper 'Abstract Differential Algebra and the Analytic Case II.' Here it is restated slightly:

Let $U$ be a non-empty connected open subset of $\mathbb{C}$ and let $K$ be a finitely generated differential field of meromorphic functions on $U$. If $L \supseteq K$ is a finitely generated abstract differential field extension of $K$, then there exists a non-empty connected open set $W\subseteq U$, a finitely generated differential field of meromorphic functions $L^\prime$, and a differential field isomorphism $f:L\rightarrow L^\prime$ extending the restriction map $\mathrm{res}_{W,U}$ on $K$.

For anyone familiar with forcing, this result has clearly done most of the work for us. Define a forcing poset $\mathbb{P}$ whose conditions are non-empty connected open subsets of $\mathbb{C}$, with $\mathbf{1}=\mathbb{C}$, ordered by inclusion, i.e. $W \leq U$ if and only if $W \subseteq U$. (This should be equivalent to Cohen forcing building a single new real, or should I say a single new complex? It is certainly $ccc$.)

Now in the extension $V[G]$, we have that $\bigcap G$ contains a single complex number, $z$. I want to argue that the field of germs of meromorphic functions in $V$ at $z$ is differentially closed. Call this field $K$. (Note that since these are meromorphic functions it is automatically a field and not just a ring, but since no non-zero element of this collection can vanish at $z$ it's actually the same as the ring of germs of holomorphic functions at $z$.)

Let $L \supseteq K$ be any abstract differentially closed field extension of $K$ (in $V[G]$). Let $b\in L$ be any element and $\overline{a} \in K$ be any finite tuple of elements. Let $L_0$ be the differential field generated by $\overline{a}b$. Since $\mathrm{DCF}_0$ is $\omega$-stable, it doesn't gain any new types over $\varnothing$ in $V[G]$ (since there are only countably many of them), so there exists a differential field $K_0 \in V$ isomorphic to $L_0$.

Let $U\in G$ be a common domain of the elements of the tuple $\overline{a}$. Now consider the set $D\subseteq \mathbb{P}$ of opens $W \subseteq U$ such that there exists a meromorphic function $c$ on $W$ such that the field generated by $\overline{a}c$ is isomorphic to $K_0$. By Seidenberg's result this is clearly dense below $U$ (as a condition), so by genericity some $W\in D$ is also in $G$. So the germ of $c$ at $z$ is also in $K$.

This says that every quantifier free type over a finite tuple $\overline{a} \in K$ is realized in $K$, implying that it is existentially closed and therefore differentially closed. Moreover by quantifier elimination we actually get that $K$ is $\omega$-saturated. Since $\mathbb{P}$ is $ccc$, $K$ has size continuum, since it contains $\mathbb{C}^V$.

I'd be really curious to know more about the structure $K$ model theoretically, since it seems fairly natural to me. In particular, is it $\omega_1$-saturated or more? Can it consistently have a Vaughtian pair (by which I mean a definable infinite subset of size strictly less than continuum)? Does its isomorphism type depend on the choice of the generic point $z$? What if we chose some other $w\in \mathbb{C}^{V[G]} \setminus \mathbb{C}^V$? What are its non-trivial automorphisms?

  • $\begingroup$ +1, but you probably shouldn't overload "V" like this. $\endgroup$ Jun 15 '19 at 0:56
  • $\begingroup$ It's worse than that the next letter they use for fields after $L$ is $M$. $\endgroup$ Jun 15 '19 at 0:58
  • $\begingroup$ Oof. FWIW I think renaming objects in quotations is fine (as long as you indicate that you've done so). Another thing I've seen is to have models of set theory replaced with either blackboardbold, fraktur, or genuine bold; the former looks nicest IMO, but clashes with the forcing font. $\endgroup$ Jun 15 '19 at 0:59
  • 2
    $\begingroup$ @Noah: And frankly, I first thought that $\Bbb C$ was Cohen forcing... :| $\endgroup$
    – Asaf Karagila
    Jun 15 '19 at 5:47

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