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In 2007 (with more work done later), J. Hamkins and B. Löwe found that the ZFC provably valid principles of forcing are the assertions of S4.2. In the introduction, they mention field extension as a sort of algebraic analog of forcing. This got me thinking (perhaps naively)...Is there a similar connection to the theory of fields?

Say we read $\Box \varphi$ as "$\varphi$ holds in every field extension," and we define the provably valid principles of field extensions to be the modal sentences $\psi(p_1, \cdots, p_n)$ such that, for any substitution $\psi(\varphi_1, \cdots, \varphi_n)$ of the letters $p_j$ for first-order formulas in the language of fields, $\psi(\varphi_1, \cdots, \varphi_n)$ holds in every field. What can we say about the provably valid principles of field extensions? Similar question: What if we only care about the valid principles of finite field extensions of $\mathbb Q$ (number fields)? What about in other classes of fields?

In either case, the property "is a field extension of" is transitive and reflexive, so the modal rules $\Box p \to p$ and $\Box p \to \Box\Box p$ both hold. Necessitation holds rather trivially, and rule $K$ holds because if $\varphi_1 \to \varphi_2$ holds in every field and $\varphi$ hold in every field, so should $\varphi_2$. Also, as pointed out in the comments, $\Diamond \Box p \to \Box \Diamond p$ is a valid principle of field extensions as well, since fields have the amalgamation property. (Of course, one could replace "field" with "number field" in this paragraph.)

Here are the links to the papers:

Here is a related discussion

TiddSchmod
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  • As far as I can see, fields do have amalgamation in the sense that for any embeddings $f_0\colon k\to K_0$ and $f_1\colon k\to K_1$, there is a field $L$ and embeddings $g_0\colon K_0\to L$ and $g_1\colon K_1\to L$ such that $g_0\circ f_0=g_1\circ f_1$. (Proof: take a sufficiently large algebraically closed field for $L$.) In terms of extensions, this means that if $K_0,K_1\supseteq k$, there are $L_0\supseteq K_0$ and $L_1\supseteq K_1$ such that $L_0\simeq_kL_1$. Since $L_0$ and $L_1$ are then elementarily equivalent, this is enough to make S4.2 valid under your interpretation. – Emil Jeřábek Nov 01 '18 at 16:35
  • Actually, the way it is defined, it is not at all clear if the set of valid formulas is closed under substitution, hence it may not be a normal modal logic (or any logic, for that matter). – Emil Jeřábek Nov 01 '18 at 17:28
  • Yes, you are right about the amalgamation property. I was thinking about number fields since these are the field extensions I am most familiar with. I will edit my question. – TiddSchmod Nov 01 '18 at 20:23
  • I am not sure I understand your second comment, @EmilJeřábek – TiddSchmod Nov 01 '18 at 20:23
  • Nevermind, I see what you are saying @EmilJeřábek. I am willing to check this, but it would be nice to know some sufficient conditions. Is there a reference you recommend? – TiddSchmod Nov 01 '18 at 20:37
  • Well, a sufficient condition would be if for every first-order sentence $\phi$ (in the language of fields), there existed a first-order sentence $\psi$ such that $\psi\leftrightarrow\Diamond\phi$ holds in all fields. Now, it follows from elementary model theory that $F\models\Diamond\phi$ iff $F\models\mathrm{Th}\forall(T_F+\phi)$, where $T_F$ is the theory of fields (that is, $F$ extends to a model of $T_F+\phi$ iff it satisfies all universal consequences thereof). Thus, $\Diamond\phi$ is definable by a first-order sentence if and only if the theory $\mathrm{Th}\forall(T_F+\phi)$ is ... – Emil Jeřábek Nov 02 '18 at 12:31
  • ... finitely axiomatizable. However, it is very likely that $\mathrm{Th}_\forall(T_F+\phi)$ is not finitely axiomatizable for general $\phi$. – Emil Jeřábek Nov 02 '18 at 12:32
  • Also, number fields do have the amalgamation property. It’s even easier than for general fields, as number fields are all contained in a single field ($\tilde{\mathbb Q}$), hence you can just take the compositum of the two fields inside $\tilde{\mathbb Q}$. – Emil Jeřábek Nov 02 '18 at 12:36
  • Oh my goodness. Yes, you are right; I am being silly. I'm also rather surprised that $\operatorname{Th}_\forall (T_F + \phi)$ is not finitely axiomatizable. – TiddSchmod Nov 03 '18 at 17:22

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