Call a sentence $\varphi$ in the language of arithmetic $Q$-like iff $\mathbb{N}\models\varphi$ and $\{\varphi\}$ is essentially incomplete (= no computably axiomatizable theory containing $\varphi$ is complete and consistent). The standard example is the conjunction of the finitely many axioms of Robinson's $Q$ (hence the name), but this is of course not unique – and indeed the partial order $\mathfrak{Q}$ of (equivalence classes of) $Q$-like sentences under entailment is not linear. On the positive side, $\mathfrak{Q}$ is clearly a distributive lattice.
My question is:
What exactly is $\mathfrak{Q}$, up to isomorphism?
There's an obvious candidate, based on the idea that everything that can happen does in this sort of situation: the (countable) random distributive lattice (that is, the Fraisse limit of the set of finite distributive lattices – see e.g. here). However, I'm having trouble proving this. Even showing that $\mathfrak{Q}$ has no least element isn't trivial, as far as I can see.
(As a quick remark, note that essentially undecidable theories need not come from elements of $\mathfrak{Q}$: Robinson's $R$ is essentially undecidable but each of its finitely axiomatizable subtheories has a computable completion, since each such theory is satisfied in (an appropriate analogue of) $\mathbb{N}_{\le k}\cup\mathbb{R}_{>k}$ for some integer $k$ which is decidable since it's interpretable in a decidable structure. So this is a bit more finicky than understanding the simpler class of essentially undecidable theories.)
Best Answer
Fedor Pakhomov showed at MO that $\mathfrak{Q}$ is in fact the countable random distributive lattice.
(I've made this CW to avoid reputation gain, and will delete this answer if Fedor posts an answer of his own here.)