Is $\mathsf{ZFC+V=L}$ consistently $\omega$-complete?

Question

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.

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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 $T^+\in L_{\omega_1^{CK}}$.

Answer 1

Claim: $T+$"$T$ is $\omega$-complete" is inconsistent. For suppose it's consistent and now work in a model $V$ of this theory. Let $T^+$ be the resulting completion of $T$ (i.e. the unique theory of the $\omega$-models of $T$ in the sense of $V$). Then note that $T^+$ is a $\Delta^1_1$ real, so $T^+\in L_{\omega_1^{\mathrm{ck}}}$. But $L_{\omega_1^{\mathrm{ck}}}\subseteq\mathrm{wfp}(M)$ whenever $M\models T$ is an $\omega$-model, and therefore every real $x\in L_{\omega_1^{\mathrm{ck}}}$ is such that $x\leq_{\mathrm{T}} T^+$ (the sub-$\mathrm{T}$ there being "Turing", as opposed to the theory $T$). (Given $x$, fix a wellorder $W$ of $\omega$ in ordertype $\alpha$ with $x\in L_\alpha$. Then (roughly) $T^+$ models "$W$ is a wellorder", and can recover $x$ from $W$. (Edited in:) Formally, fix an integer $e$ which indexes a recursive wellorder of $\omega$ in ordertype $\alpha$ with $x\in L_\alpha$. Recall $L_\alpha$ projects to $\omega$, and $x\leq_{\mathrm{T}} t^{L_\alpha}$, the first-order theory of $L_\alpha$. Fix a Turing reduction $n$ of $x$ from $t^{L_\alpha}$. Then for $m<\omega$, we have $m\in x$ iff $T^+$ contains the statement "Let $\beta$ be the ordertype of the wellorder coded by $e$, and let $y\leq_{\mathrm{T}} t^{L_\beta}$ via the $n$th Turing program; then $m\in y$".) But with $T^+\in L_{\omega_1^{\mathrm{ck}}}$, this gives a contradiction.