*This was previously asked and bountied on MSE:*

For brevity, let $T$ be $\mathsf{ZFC+V=L}$.

Say that an extension of $\mathsf{ZFC}$ is *$\omega$-complete* iff it has exactly one $\omega$-model up to elementary equivalence. While the $\omega$-incompleteness of $T$ is easily provable in theories only slightly stronger than $T$ itself, I don't immediately see how to do it in $T$ alone. My question is:

Is the theory $S:=T+$ "$T$ is $\omega$-complete" consistent?

Here are a couple observations:

If we replace "$\omega$-model" by "well-founded model," the answer is obviously

**yes**under standard assumptions. Let $\alpha$ be the second-smallest ordinal such that $L_\alpha\models\mathsf{ZFC}$. Then $L_\alpha$ also satisfies "$\mathsf{ZFC+V=L}$ has exactly one well-founded model." Unfortunately, we have no analogous hierarchy of $\omega$-models, so this is a non-starter here.As to the specific choice of theory in question, the point is that (something like) $\mathsf{V=L}$ is needed to block an easy proof of a negative answer via forcing. For example, reasoning in $\mathsf{ZFC}$, if $\mathsf{ZFC}$ had an $\omega$-model $\mathcal{M}$ it would have a countable one $\hat{\mathcal{M}}$, and we could force over $\hat{\mathcal{M}}$ to get a non-elementarily-equivalent $\omega$-model $\hat{\mathcal{N}}$.

*(Forcing over ill-founded countable models is no harder really than forcing over well-founded ones.)*The key point here is that forcing preserves $\mathsf{ZFC}$. This breaks down of course for $\mathsf{V=L}$ and so this argument is irrelevant here. Given the paucity of techniques we currently have for building models of $\mathsf{ZFC}$ in the first place, this seems to be a real issue.

Ultimately I suspect that the answer is negative, but the above two points between them rule out all the lines of attack I've been able to think of so far.

EDIT: In light of Farmer S's answer below, let me explicitly mention a rule of thumb which I forgot: **when thinking about properties which are not too far from first-order definable, always consider the hyperarithmetic hierarchy!**

For example, for every $\mathcal{L}_{\omega_1,\omega}$-sentence $\varphi$, if $\varphi$ has a model then it has a model $M$ which is countable in $L$, and moreover the $L$-least (real coding a) model of $\varphi$ is hyperarithmetic relative to (any real coding) $\varphi$. The property "Is an $\omega$-model of $T$" is expressible as a computable $\mathcal{L}_{\omega_1,\omega}$-sentence, and this drives Farmer S's point that $T^+\in L_{\omega_1^{CK}}$.

is alreadya statement in the object language: every model $M$ of $\mathsf{ZFC}$ has a notion of "$\omega$-model of $\mathsf{ZFC}$," namely "model of $\mathsf{ZFC}$ whose $\omega$ is isomorphic to my $\omega$." There's no linguistic or otherwise "meta" issue here. (In particular, note that if $M\models\mathsf{ZFC}$ is not an $\omega$-model itself then the $\omega$-models of $\mathsf{ZFC}$ in the sense of $M$ - assuming $M$ thinks there are any at all - willnotbe genuine $\omega$-models.) So the $S$ of the OPisa genuine first-order theory. $\endgroup$ – Noah Schweber Mar 8 at 20:556more comments